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Contemporary
Engineering
Economics
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Contemporary
Engineering
Economics
Fourth Edition
Chan S. Park
Depar tment of Industr ial
and Systems Engineer ing
Aubur n Univer sity
Upper Saddle River, NJ 07458
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Library of Congress Cataloging-in-Publication Data on File
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© 2007 by Pearson Education, Inc.
Pearson Prentice Hall
Upper Saddle River, New Jersey 07458
All rights reserved. No part of this book may be reproduced, in any form or by any means
without permission in writing from the publisher.
The author and publisher of this book have used their best efforts in preparing this book. These
efforts include the development, research, and testing of the theories and programs to determine
their effectiveness. The author and publisher make no warranty of any kind, expressed or
implied, with regard to these programs or the documentation contained in this book. The author
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Printed in the United States of America
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ISBN 0-13-187628-7
Pearson Education Ltd., London
Pearson Education Australia Pte. Ltd., Sydney
Pearson Education Singapore, Pte. Ltd.
Pearson Education North Asia Ltd., Hong Kong
Pearson Education Canada, Inc., Toronto
Pearson Educación de Mexico, S.A. de C.V.
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Pearson Education Malaysia, Ptd. Ltd.
Pearson Education, Inc., Upper Saddle River, New Jersey
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To my wife, Kim (Inkyung); and my children, Michael and Edward
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CONTENTS
Preface
xix
PART 1 B ASICS OF FINANCIAL DECISIONS
Chapter 1
Engineering Economic Decisions
1
2
1.1
Role of Engineers in Business
1.1.1 Types of Business Organization
1.1.2 Engineering Economic Decisions
1.1.3 Personal Economic Decisions
4
5
6
6
1.2
What Makes the Engineering Economic Decision Difficult?
7
1.3
Economic Decisions versus Design Decisions
8
1.4
Large-Scale Engineering Projects
1.4.1 How a Typical Project Idea Evolves
1.4.2 Impact of Engineering Projects on Financial Statements
1.4.3 A Look Back in 2005: Did Toyota Make the Right Decision?
9
9
12
13
1.5
Common Types of Strategic Engineering Economic Decisions
13
1.6
Fundamental Principles of Engineering Economics
15
Summary
17
Chapter 2
Understanding Financial Statements
18
2.1
Accounting: The Basis of Decision Making
21
2.2
Financial Status for Businesses
2.2.1 The Balance Sheet
2.2.2 The Income Statement
2.2.3 The Cash Flow Statement
22
24
27
30
2.3
Using Ratios to Make Business Decisions
2.3.1 Debt Management Analysis
2.3.2 Liquidity Analysis
2.3.3 Asset Management Analysis
2.3.4 Profitability Analysis
2.3.5 Market Value Analysis
2.3.6 Limitations of Financial Ratios in Business Decisions
33
34
37
38
39
41
42
Summary
43
Problems
44
Short Case Studies
50
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Chapter 3
Interest Rate and Economic Equivalence
52
3.1
Interest: The Cost of Money
3.1.1 The Time Value of Money
3.1.2 Elements of Transactions Involving Interest
3.1.3 Methods of Calculating Interest
3.1.4 Simple Interest versus Compound Interest
54
55
56
59
62
3.2
Economic Equivalence
3.2.1 Definition and Simple Calculations
3.2.2 Equivalence Calculations: General Principles
3.2.3 Looking Ahead
63
63
66
71
3.3
Development of Interest Formulas
3.3.1 The Five Types of Cash Flows
3.3.2 Single-Cash-Flow Formulas
3.3.3 Uneven Payment Series
3.3.4 Equal Payment Series
3.3.5 Linear Gradient Series
3.3.6 Geometric Gradient Series
71
72
73
80
84
96
102
3.4
Unconventional Equivalence Calculations
3.4.1 Composite Cash Flows
3.4.2 Determining an Interest Rate to Establish Economic Equivalence
107
107
114
Summary
119
Problems
119
Short Case Studies
129
Chapter 4
Understanding Money and Its Management
134
4.1
Nominal and Effective Interest Rates
4.1.1 Nominal Interest Rates
4.1.2 Effective Annual Interest Rates
4.1.3 Effective Interest Rates per Payment Period
4.1.4 Continuous Compounding
136
136
137
140
141
4.2
Equivalence Calculations with Effective Interest Rates
4.2.1 When Payment Period Is Equal to Compounding Period
4.2.2 Compounding Occurs at a Different Rate than that at
Which Payments Are Made
143
144
Equivalence Calculations with Continuous Payments
4.3.1 Single-Payment Transactions
4.3.2 Continuous-Funds Flow
152
152
152
4.3
145
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4.4
Changing Interest Rates
4.4.1 Single Sums of Money
4.4.2 Series of Cash Flows
156
156
158
4.5
Debt Management
4.5.1 Commercial Loans
4.5.2 Loan versus Lease Financing
4.5.3 Home Mortgage
159
159
167
171
4.6
Investing in Financial Assets
4.6.1 Investment Basics
4.6.2 How to Determine Your Expected Return
4.6.3 Investing in Bonds
175
175
176
179
Summary
187
Problems
188
Short Case Studies
199
PART 2 EVALUATION OF BUSINESS
AND ENGINEERING ASSETS
Chapter 5
5.1
Present-Worth Analysis
203
204
Describing Project Cash Flows
5.1.1 Loan versus Project Cash Flows
5.1.2 Independent versus Mutually Exclusive
Investment Projects
207
207
5.2
Initial Project Screening Method
5.2.1 Payback Period: The Time It Takes to Pay Back
5.2.2 Benefits and Flaws of Payback Screening
5.2.3 Discounted Payback Period
5.2.4 Where Do We Go from Here?
210
210
213
214
215
5.3
Discounted Cash Flow Analysis
5.3.1 Net-Present-Worth Criterion
5.3.2 Meaning of Net Present Worth
5.3.3 Basis for Selecting the MARR
215
216
220
222
5.4
Variations of Present-Worth Analysis
5.4.1 Future-Worth Analysis
5.4.2 Capitalized Equivalent Method
223
223
227
5.5
Comparing Mutually Exclusive Alternatives
5.5.1 Meaning of Mutually Exclusive and “Do Nothing”
5.5.2 Analysis Period
232
232
235
209
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5.5.3 Analysis Period Equals Project Lives
5.5.4 Analysis Period Differs from Project Lives
5.5.5 Analysis Period Is Not Specified
236
238
246
Summary
249
Problems
249
Short Case Studies
265
Chapter 6
Annual Equivalent-Worth Analysis
268
6.1
Annual Equivalent-Worth Criterion
6.1.1 Fundamental Decision Rule
6.1.2 Annual-Worth Calculation with Repeating Cash Flow Cycles
6.1.3 Comparing Mutually Exclusive Alternatives
270
270
273
275
6.2
Capital Costs versus Operating Costs
277
6.3
Applying Annual-Worth Analysis
6.3.1 Benefits of AE Analysis
6.3.2 Unit Profit or Cost Calculation
6.3.3 Make-or-Buy Decision
6.3.4 Pricing the Use of an Asset
280
281
281
283
286
6.4
Life-Cycle Cost Analysis
287
6.5
Design Economics
294
Summary
303
Problems
304
Short Case Studies
318
Chapter 7
Rate-of-Return Analysis
322
7.1
Rate of Return
7.1.1 Return on Investment
7.1.2 Return on Invested Capital
324
324
326
7.2
Methods for Finding the Rate of Return
7.2.1 Simple versus Nonsimple Investments
7.2.2 Predicting Multiple i*’s
7.2.3 Computational Methods
327
327
329
331
7.3
Internal-Rate-of-Return Criterion
7.3.1 Relationship to PW Analysis
7.3.2 Net-Investment Test: Pure versus Mixed Investments
7.3.3 Decision Rule for Pure Investments
7.3.4 Decision Rule for Mixed Investments
338
338
339
341
344
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7.4
Mutually Exclusive Alternatives
7.4.1 Flaws in Project Ranking by IRR
7.4.2 Incremental Investment Analysis
7.4.3 Handling Unequal Service Lives
352
352
353
360
Summary
363
Problems
364
Short Case Studies
381
PART 3 ANALY SIS OF PROJECT C ASH FLOWS
Chapter 8
Cost Concepts Relevant to Decision Making
385
386
8.1
General Cost Terms
8.1.1 Manufacturing Costs
8.1.2 Nonmanufacturing Costs
388
388
390
8.2
Classifying Costs for Financial Statements
8.2.1 Period Costs
8.2.2 Product Costs
390
391
391
8.3
Cost Classification for Predicting Cost Behavior
8.3.1 Volume Index
8.3.2 Cost Behaviors
394
394
395
8.4
Future Costs for Business Decisions
8.4.1 Differential Cost and Revenue
8.4.2 Opportunity Cost
8.4.3 Sunk Costs
8.4.4 Marginal Cost
400
400
404
406
406
8.5
Estimating Profit from Production
8.5.1 Calculation of Operating Income
8.5.2 Sales Budget for a Manufacturing Business
8.5.3 Preparing the Production Budget
8.5.4 Preparing the Cost-of-Goods-Sold Budget
8.5.5 Preparing the Nonmanufacturing Cost Budget
8.5.6 Putting It All Together: The Budgeted Income Statement
8.5.7 Looking Ahead
411
412
412
413
415
416
418
419
Summary
420
Problems
421
Short Case Study
427
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Chapter 9
Depreciation and Corporate Taxes
428
9.1
Asset Depreciation
9.1.1 Economic Depreciation
9.1.2 Accounting Depreciation
431
432
432
9.2
Factors Inherent in Asset Depreciation
9.2.1 Depreciable Property
9.2.2 Cost Basis
9.2.3 Useful Life and Salvage Value
9.2.4 Depreciation Methods: Book and Tax Depreciation
433
433
434
435
436
9.3
Book Depreciation Methods
9.3.1 Straight-Line Method
9.3.2 Accelerated Methods
9.3.3 Units-of-Production Method
437
437
439
445
9.4
Tax Depreciation Methods
9.4.1 MACRS Depreciation
9.4.2 MACRS Depreciation Rules
446
446
447
9.5
Depletion
9.5.1 Cost Depletion
9.5.2 Percentage Depletion
453
453
454
9.6
Repairs or Improvements Made to Depreciable Assets
9.6.1 Revision of Book Depreciation
9.6.2 Revision of Tax Depreciation
456
456
457
9.7
Corporate Taxes
9.7.1 Income Taxes on Operating Income
459
459
9.8
Tax Treatment of Gains or Losses on Depreciable Assets
9.8.1 Disposal of a MACRS Property
9.8.2 Calculations of Gains and Losses on MACRS Property
462
462
464
9.9
Income Tax Rate to Be Used in Economic Analysis
9.9.1 Incremental Income Tax Rate
9.9.2 Consideration of State Income Taxes
467
467
470
9.10
The Need for Cash Flow in Engineering Economic Analysis
9.10.1 Net Income versus Net Cash Flow
9.10.2 Treatment of Noncash Expenses
472
472
473
Summary
476
Problems
478
Short Case Studies
487
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Chapter 10
Developing Project Cash Flows
490
10.1
Cost–Benefit Estimation for Engineering Projects
10.1.1 Simple Projects
10.1.2 Complex Projects
492
493
493
10.2
Incremental Cash Flows
10.2.1 Elements of Cash Outflows
10.2.2 Elements of Cash Inflows
10.2.3 Classification of Cash Flow Elements
494
494
495
497
10.3
Developing Cash Flow Statements
10.3.1 When Projects Require Only Operating and Investing Activities
10.3.2 When Projects Require Working-Capital Investments
10.3.3 When Projects Are Financed with Borrowed Funds
10.3.4 When Projects Result in Negative Taxable Income
10.3.5 When Projects Require Multiple Assets
498
498
502
507
509
513
10.4
Generalized Cash Flow Approach
10.4.1 Setting up Net Cash Flow Equations
10.4.2 Presenting Cash Flows in Compact Tabular Formats
10.4.3 Lease-or-Buy Decision
516
517
518
520
Summary
524
Problems
525
Short Case Studies
537
PART 4 HANDLING RISK AND UNCERTAINTY
Chapter 11
541
Inflation and Its Impact on Project Cash Flows
542
11.1
Meaning and Measure of Inflation
11.1.1 Measuring Inflation
11.1.2 Actual versus Constant Dollars
544
544
550
11.2
Equivalence Calculations under Inflation
11.2.1 Market and Inflation-Free Interest Rates
11.2.2 Constant-Dollar Analysis
11.2.3 Actual-Dollar Analysis
11.2.4 Mixed-Dollar Analysis
553
553
553
554
558
11.3
Effects of Inflation on Project Cash Flows
11.3.1 Multiple Inflation Rates
11.3.2 Effects of Borrowed Funds under Inflation
558
562
563
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11.4
Rate-of-Return Analysis under Inflation
11.4.1 Effects of Inflation on Return on Investment
11.4.2 Effects of Inflation on Working Capital
566
566
569
Summary
572
Problems
574
Short Case Studies
582
Chapter 12
Project Risk and Uncertainty
584
12.1
Origins of Project Risk
586
12.2
Methods of Describing Project Risk
12.2.1 Sensitivity Analysis
12.2.2 Break-Even Analysis
12.2.3 Scenario Analysis
587
587
591
594
12.3
Probability Concepts for Investment Decisions
12.3.1 Assessment of Probabilities
12.3.2 Summary of Probabilistic Information
12.3.3 Joint and Conditional Probabilities
12.3.4 Covariance and Coefficient of Correlation
596
596
601
603
605
12.4
Probability Distribution of NPW
12.4.1 Procedure for Developing an NPW Distribution
12.4.2 Aggregating Risk over Time
12.4.3 Decision Rules for Comparing Mutually Exclusive
Risky Alternatives
605
605
611
12.5
Risk Simulation
12.5.1 Computer Simulation
12.5.2 Model Building
12.5.3 Monte Carlo Sampling
12.5.4 Simulation Output Analysis
12.5.5 Risk Simulation with @RISK
618
619
620
623
628
630
12.6
Decision Trees and Sequential Investment Decisions
12.6.1 Structuring a Decision-Tree Diagram
12.6.2 Worth of Obtaining Additional Information
12.6.3 Decision Making after Having Imperfect
Information
633
634
639
Summary
647
Problems
648
Short Case Studies
658
616
642
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Chapter 13
Real-Options Analysis
664
13.1
Risk Management: Financial Options
13.1.1 Buy Call Options when You Expect the Price to Go Up
13.1.2 Buy Put Options when You Expect the Price to Go Down
666
669
669
13.2
Option Strategies
13.2.1 Buying Calls to Reduce Capital That Is at Risk
13.2.2 Protective Puts as a Hedge
670
670
673
13.3
Option Pricing
13.3.1 Replicating-Portfolio Approach with a Call Option
13.3.2 Risk-Free Financing Approach
13.3.3 Risk-Neutral Probability Approach
13.3.4 Put-Option Valuation
13.3.5 Two-Period Binomial Lattice Option Valuation
13.3.6 Multiperiod Binomial Lattice Model
13.3.7 Black–Scholes Option Model
675
675
677
679
680
681
683
684
13.4
Real-Options Analysis
13.4.1 A Conceptual Framework for Real Options
in Engineering Economics
13.4.2 Types of Real-Option Models
686
13.5
13.6
687
688
Estimating Volatility at the Project Level
13.5.1 Estimating a Project’s Volatility through a
Simple Deferral Option
13.5.2 Use the Existing Model of a Financial Option to Estimate σ2
697
699
Compound Options
703
Summary
708
Problems
709
Short Case Studies
713
697
PART 5 SPECIAL TOPICS IN ENGINEERING
ECONOMICS 715
Chapter 14
14.1
Replacement Decisions
716
Replacement Analysis Fundamentals
14.1.1 Basic Concepts and Terminology
14.1.2 Opportunity Cost Approach to Comparing
Defender and Challenger
718
718
721
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14.2
Economic Service Life
723
14.3
Replacement Analysis when the Required Service Is Long
14.3.1 Required Assumptions and Decision Frameworks
14.3.2 Replacement Strategies under the Infinite Planning Horizon
14.3.3 Replacement Strategies under the Finite Planning Horizon
14.3.4 Consideration of Technological Change
728
729
730
735
738
14.4
Replacement Analysis with Tax Considerations
739
Summary
755
Problems
756
Short Case Studies
768
Chapter 15
Capital-Budgeting Decisions
776
15.1
Methods of Financing
15.1.1 Equity Financing
15.1.2 Debt Financing
15.1.3 Capital Structure
778
779
780
782
15.2
Cost of Capital
15.2.1 Cost of Equity
15.2.2 Cost of Debt
15.2.3 Calculating the Cost of Capital
787
787
792
794
15.3
Choice of Minimum Attractive Rate of Return
15.3.1 Choice of MARR when Project Financing Is Known
15.3.2 Choice of MARR when Project Financing Is Unknown
15.3.3 Choice of MARR under Capital Rationing
795
795
797
799
15.4
Capital Budgeting
15.4.1 Evaluation of Multiple Investment Alternatives
15.4.2 Formulation of Mutually Exclusive Alternatives
15.4.3 Capital-Budgeting Decisions with Limited Budgets
803
803
803
805
Summary
809
Problems
810
Short Case Studies
816
Chapter 16
16.1
Economic Analysis in the Service Sector
What Is the Service Sector?
16.1.1 Characteristics of the Service Sector
16.1.2 How to Price Service
822
824
825
825
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16.2
Economic Analysis in Health-Care Service
16.2.1 Economic Evaluation Tools
16.2.2 Cost-Effectiveness Analysis
16.2.3 How to Use a CEA
826
827
828
829
16.3 Economic Analysis in the Public Sector
16.3.1 What Is Benefit–Cost Analysis?
16.3.2 Framework of Benefit–Cost Analysis
16.3.3 Valuation of Benefits and Costs
16.3.4 Quantifying Benefits and Costs
16.3.5 Difficulties Inherent in Public-Project Analysis
832
833
833
834
836
840
16.4 Benefit–Cost Ratios
16.4.1 Definition of Benefit–Cost Ratio
16.4.2 Relationship between B/C Ratio and NPW
16.4.3 Comparing Mutually Exclusive Alternatives:
Incremental Analysis
840
840
843
16.5 Analysis of Public Projects Based on Cost-Effectiveness
16.5.1 Cost-Effectiveness Studies in the Public Sector
16.5.2 A Cost-Effectiveness Case Study
846
847
847
843
Summary
856
Problems
857
Short Case Studies
862
Appendix A
Index
899
Interest Factors for Discrete Compounding
869
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PREFACE
What is “Contemporary” About Engineering Economics?
Decisions made during the engineering design phase of product development determine
the majority of the costs associated with the manufacturing of that product (some say that
this value may be as high as 85%). As design and manufacturing processes become more
complex, engineers are making decisions that involve money more than ever before. With
more than 80% of the total GDP (Gross Domestic Product) in the United States provided
by the service sector, engineers work on various economic decision problems in the service sector as well. Thus, the competent and successful engineer in the twenty-first century
must have an improved understanding of the principles of science, engineering, and economics, coupled with relevant design experience. Increasingly, in the new world economy,
successful businesses will rely on engineers with such expertise.
Economic and design issues are inextricably linked in the product/service life cycle.
Therefore, one of my strongest motivations for writing this text was to bring the realities
of economics and engineering design into the classroom and to help students integrate
these issues when contemplating many engineering decisions. Of course my underlying
motivation for writing this book was not simply to address contemporary needs, but to
address as well the ageless goal of all educators: to help students to learn. Thus, thoroughness, clarity, and accuracy of presentation of essential engineering economics were
my aim at every stage in the development of the text.
Changes in the Fourth Edition
Much of the content has been streamlined to provide materials in depth and to reflect the
challenges in contemporary engineering economics. Some of the highlighted changes are
as follows:
• Chapter 13 “Real Options Analysis” is new and provides a new perspective on how
engineers should manage risk in their strategic economic decision problems.
Traditionally, risk is avoided in project analysis, which is a passive way of handling
the matter. The goal of the real options approach is to provide a contemporary tool
that will assist engineers so that they can actively manage the risk involved in longterm projects.
• Chapter 12 has been significantly revised to provide more probabilistic materials for
the analytical treatment of risk and uncertainty. Risk simulation has been introduced
by way of using @RISK.
• Three chapters have been merged with various materials from other chapters.
Chapter 3 on cost concepts and behaviors has been moved to Part III and now
appears as Chapter 8 “Cost Concepts Relevant to Decision Making”; it is now part
of project cash flow analysis. Chapter 6 on principles of investing is now part of
Chapter 4 “Understanding Money and Its Management.” Materials from various
chapters have been merged into a single chapter and now appear as Chapter 9
“Depreciation and Corporate Income Taxes”.
• The chapter on the economic analysis in the public sector has been expanded and now
appears as Chapter 16 “Economic Analysis in the Service Sector”; this revised chapter now provides economic analysis unique to service sectors beyond the government
xix
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xx PREFACE
•
•
•
•
sector. Increasingly, engineers seek their career in the service sector, such as healthcare, financial institutions, transportation, and logistics. In this chapter, we present
some unique features that must be considered when evaluating investment projects in
the service sector.
All the end-of-chapter problems are revised to reflect the materials changes in the
main text.
All the chapter opening vignettes—a trademark of Contemporary Engineering
Economics—have been completely replaced with more current and thought-provoking
case studies.
Self-study problems and FE practice questions are available as interactive quizzes
with instant feedback as part of the book’s new OneKey CourseCompass site.
OneKey is an online resource for instructors and students; more detailed information
about OneKey can be found in the OneKey section of this Preface. OneKey can be
accessed via www.prenhall.com/onekey.
Various Excel spreadsheet modeling techniques are introduced throughout the chapters and the original Excel files are provided online at the OneKey site .
Overview of the Text
Although it contains little advanced math and few truly difficult concepts, the introductory engineering economics course is often a curiously challenging one for the
sophomores, juniors, and seniors who take it. There are several likely explanations for
this difficulty.
1. The course is the student’s first analytical consideration of money (a resource with
which he or she may have had little direct contact beyond paying for tuition, housing, food, and textbooks).
2. The emphasis on theory may obscure for the student the fact that the course aims,
among other things, to develop a very practical set of analytical tools for measuring project worth. This is unfortunate since, at one time or another, virtually every
engineer—not to mention every individual—is responsible for the wise allocation
of limited financial resources.
3. The mixture of industrial, civil, mechanical, electrical, and manufacturing engineering, and other undergraduates who take the course often fail to “see themselves” in the skills the course and text are intended to foster. This is perhaps less
true for industrial engineering students, whom many texts take as their primary
audience, but other disciplines are often motivationally shortchanged by a text’s
lack of applications that appeal directly to them.
Goal of the Text
This text aims not only to provide sound and comprehensive coverage of the concepts of
engineering economics but also to address the difficulties of students outlined above, all
of which have their basis in inattentiveness to the practical concerns of engineering economics. More specifically, this text has the following chief goals:
1. To build a thorough understanding of the theoretical and conceptual basis upon
which the practice of financial project analysis is built.
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Preface
2. To satisfy the very practical needs of the engineer toward making informed financial decisions when acting as a team member or project manager for an engineering
project.
3. To incorporate all critical decision-making tools—including the most contemporary, computer-oriented ones that engineers bring to the task of making informed
financial decisions.
4. To appeal to the full range of engineering disciplines for which this course is often
required: industrial, civil, mechanical, electrical, computer, aerospace, chemical,
and manufacturing engineering, as well as engineering technology.
Prerequisites
The text is intended for undergraduate engineering students at the sophomore level or
above. The only mathematical background required is elementary calculus. For Chapters
12 and 13, a first course in probability or statistics is helpful but not necessary, since the
treatment of basic topics there is essentially self-contained.
Taking Advantage of the Internet
The integration of computer use is another important feature of Contemporary Engineering
Economics. Students have greater access to and familiarity with the various spreadsheet
tools, and instructors have greater inclination either to treat these topics explicitly in the
course or to encourage students to experiment independently.
A remaining concern is that the use of computers will undermine true understanding of course concepts. This text does not promote the use of trivial spreadsheet applications as a replacement for genuine understanding of and skill in applying traditional
solution methods. Rather, it focuses on the computer’s productivity-enhancing benefits
for complex project cash flow development and analysis. Specifically, Contemporary
Engineering Economics includes a robust introduction to computer automation in the
form of Computer Notes, which are included in the optional OneKey course
(www.prenhall.com/onekey).
Additionally, spreadsheets are introduced via Microsoft Excel examples. For spreadsheet coverage, the emphasis is on demonstrating a chapter concept that embodies some
complexity that can be much more efficiently resolved on a computer than by traditional
longhand solutions.
OneKey
Available as a special package, OneKey is Prentice Hall’s exclusive new resource for
instructors and students. Instructors have access online to all available course supplements
and can create and assign tests, quizzes, or graded homework assignments. Students have
access to interactive exercises, quizzes, and more. The following resources are available
when an instructor adopts the text plus OneKey package:
• Interactive self-study quizzes organized by chapter with instant feedback, plus interactive FE Exam practice questions
• Computer notes with Excel files of selected example problems from the text.
• Case Studies: A collection of actual cases, two personal-finance and six industrybased, is now available. The investment projects detailed in the cases relate to a
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xxii PREFACE
variety of engineering disciplines. Each case is based on multiple text concepts, thus
encouraging students to synthesize their understanding in the context of complex,
real-world investments. Each case begins with a list of engineering economic concepts utilized in the case and concludes with discussion questions to test students’
conceptual understanding.
• Analysis Tools: A collection of various financial calculators is available. Cash Flow
Analyzer is an integrated online Java program that is menu driven for convenience
and flexibility; it provides (1) a flexible and easy-to-use cash flow editor for data
input and modifications, and (2) an extensive array of computational modules and
user-selected graphic outputs.
• Instructor Resources: Instructors Solutions Manual, PowerPoint Lecture Notes,
Case Studies and more.
Please contact your Prentice Hall representative for details and ordering information
for OneKey packages. Detailed instructions about how to access and use this content can
be found at the site, which can be accessed at: www.prenhall.com/onekey.
The Financial Times
We are please to announce a special partnership with The Financial Times. For a small
additional charge, Prentice Hall offers students a 15-week subscription to The Financial
Times. Upon adoption of a special package containing the book and the subscription
booklet, professors will receive a free one-year subscription. Please contact your Prentice
Hall representative for details and ordering information.
Acknowledgments
This book reflects the efforts of a great many individuals over a number of years. In particular, I would like to recognize the following individuals, whose reviews and comments
on prior editions have contributed to this edition. Once again, I would like to thank each
of them:
Kamran Abedini, California Polytechnic–Pomona
James Alloway, Syracuse University
Mehar Arora, U. Wisconsin–Stout
Joel Arthur, California State University–Chico
Robert Baker, University of Arizona
Robert Barrett, Cooper Union and Pratt Institute
Tom Barta, Iowa State University
Charles Bartholomew, Widener University
Richard Bernhard, North Carolina State University
Bopaya Bidanda, University of Pittsburgh
James Buck, University of Iowa
Philip Cady, The Pennsylvania State University
Tom Carmichal, Southern College of Technology
Jeya Chandra, The Pennsylvania State University
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Preface
Max C. Deibert, Montana State University
Stuart E. Dreyfus, University of California, Berkeley
Philip A. Farrington, University of Alabama at Huntsville
W. J. Foley, RPI
Jane Fraser, University of Southern Colorado
Terry L Friesz, Penn State University
Anil K. Goyal, RPI
Bruce Hartsough, University of California, Davis
Carl Hass, University of Texas, Austin
John Held, Kansas State University
T. Allen Henry, University of Alabama
R.C. Hodgson, University of Notre Dame
Scott Iverson, University of Washington
Peter Jackson, Cornell University
Philip Johnson, University of Minnesota
Harold Josephs, Lawrence Tech
Henry Kallsen, University of Alabama
W. J. Kennedy, Clemson University
Oh Keytack, University of Toledo
Wayne Knabach, South Dakota State University
Stephen Kreta, California Maritime Academy
John Krogman, University of Wisconsin–Platteville
Dennis Kroll, Bradley University
Michael Kyte, University of Idaho
Gene Lee, University of Central Florida
William Lesso, University of Texas–Austin
Martin Lipinski, Memphis State University
Robert Lundquist, Ohio State University
Richard Lyles, Michigan State University
Gerald T. Mackulak, Arizona State University
Abu S. Masud, The Wichita State University
Sue McNeil, Carnegie-Mellon University
James Milligan, University of Idaho
Richard Minesinger, University of Massachusetts, Lowell
Gary Moynihan, The University of Alabama
James S. Noble, University of Missouri, Columbia
Michael L. Nobs, Washington University, St. Louis
Wayne Parker, Mississippi State University
Elizabeth Pate-Cornell, Stanford University
Cecil Peterson, GMI
George Prueitt, U.S. Naval Postgraduate School
J.K. Rao, California State University-Long Beach
Susan Richards, GMI
Bruce A. Reichert, Kansas State University
Mark Roberts, Michigan Tech
John Roth, Vanderbilt University
Paul L. Schillings, Montana State University
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xxiv PREFACE
Bill Shaner, Colorado State University
Fred Sheets, California Polytechnic, Pomona
Dean Shup, University of Cincinnati
Milton Smith, Texas Tech
David C. Slaughter, University of California, Davis
Charles Stavridge, FAMU/FSU
Junius Storry, South Dakota State University
Frank E. Stratton, San Diego State University
George Stukhart, Texas A&M University
Donna Summers, University of Dayton
Joe Tanchoco, Purdue University
Deborah Thurston, University of Illinois at Urbana-Champaign
Lt. Col. James Treharne, U.S. Army
L. Jackson Turaville, Tennessee Technological University
Thomas Ward, University of Louisville
Theo De Winter, Boston University
Yoo Yang, Cal Poly State University
Special Acknowledgement
Personally, I wish to thank the following individuals for their additional inputs to the
fourth edition: Michael L. Nobs, Washington University, St. Louis, Terry L. Friesz, Penn
State University, Gene Lee, University of Central Florida, Gerald T. Mackulak, Arizona
State University, and Phillip A. Farrington, University of Alabama, Huntsville. Major
Hyun Jin Han who helped me in developing the Instructor Manual; Holly Stark, my editor at Prentice Hall, who assumed responsibility for the overall project; Scott Disanno,
my production editor at Prentice Hall, who oversaw the entire book production. I also
acknowledge that many of the financial terminologies found in the marginal notes are
based on the online glossary defined by Investopedia and Investorwords.com. Finally, I
would like to thank Dr. Alice E. Smith, Chair of Industrial & Systems Engineering at
Auburn University, who provided me with the resources.
CHAN S. PARK
AUBURN, ALABAMA
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P A R T
1
Basics of Financial
Decisions
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ONE
CHAPTER
Engineering Economic
Decisions
Google Cofounder Sergey Brin Comes to Class at
Berkeley1 Sergey Brin, cofounder of Google, showed up for
class as a guest speaker at Berkeley on October 3, 2005.
Casual and relaxed, Brin talked about how Google came to be,
answered students’ questions, and showed that someone worth
$11 billion (give or take a billion) still can be comfortable in an old
pair of blue jeans. Indistinguishable in dress, age, and demeanor
from many of the students in the class, Brin covered a lot of
ground in his remarks, but ultimately it was his unspoken message
that was most powerful: To those with focus and passion, all things
are possible. In his remarks to the class, Brin stressed simplicity.
Simple ideas sometimes can change the world, he said. Likewise,
Google started out with the simplest of ideas, with a global audience
in mind. In the mid-1990s, Brin and Larry Page were Stanford students
pursuing doctorates in computer science. Brin recalled that at that
time there were some five major Internet search engines, the importance of searching was being de-emphasized, and the owners of these
major search sites were focusing on creating portals with increased
content offerings.“We believed we could build a better search.We
had a simple idea—that not all pages are created equal. Some are
more important,” related Brin. Eventually, they developed a unique
approach to solving one of computing’s biggest challenges: retrieving
relevant information from a massive set of data.
According to Google lore,1 by January of 1996 Larry and Sergey had
begun collaboration on a search engine called BackRub, named for its
unique ability to analyze the “back links” pointing to a given website.
1
2
UC Berkeley News, Oct. 4, 2005, UC Regents and Google Corporate Information: http://www.
google.com/corporate/history.html.
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How Google Works
Query
Google user
1. The Web server sends the
query to the index
servers—it tells which
pages contain the words
that match the query.
2. The query travels to the
Doc servers (which
retrieve the stored
documents) and snippets
Google Web servers
are generated to describe
each search result.
3. The search results are
returned to the user in a
fraction of a second.
Doc servers
Index servers
Larry, who had always enjoyed tinkering with machinery and had gained some
“notoriety” for building a working printer out of Lego® bricks, took on the
task of creating a new kind of server environment that used low-end PCs instead of big expensive machines. Afflicted by the perennial shortage of cash
common to graduate students everywhere, the pair took to haunting the department’s loading docks in hopes of tracking down newly arrived computers
that they could borrow for their network. A year later, their unique approach
to link analysis was earning BackRub a growing reputation among those who
had seen it. Buzz about the new search technology began to build as word
spread around campus. Eventually, in 1998 they decided to create a company
named “Google” by raising $25 million from venture capital firms Kleiner
Perkins Caufield & Byers and Sequoia Capital. Since taking their Internet
search engine public in August 2004, the dynamic duo behind Google has seen
their combined fortune soar to $22 billion. At a recent $400, Google trades at
90 times trailing earnings, after starting out at $85. The success has vaulted
both Larry and Sergey into Forbes magazine’s list of the 400 wealthiest Americans. The net worth of the pair is estimated at $11 billion each.
3
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4 CHAPTER 1 Engineering Economic Decisions
A Little Google History
• 1995
• Developed in dorm room of Larry Page and Sergey Brin, graduate students at
Stanford University
• Nicknamed BackRub
• 1998
• Raised $25 million to set up Google, Inc.
• Ran 100,000 queries a day out of a garage in Menlo Park
• 2005
• Over 4,000 employees worldwide
• Over 8 billion pages indexed
The story of how the Google founders got motivated to invent a search engine and eventually transformed their invention to a multibillion-dollar business is a typical one.
Companies such as Dell, Microsoft, and Yahoo all produce computer-related products
and have market values of several billion dollars. These companies were all started by
highly motivated young college students just like Brin. One thing that is also common to
all these successful businesses is that they have capable and imaginative engineers who
constantly generate good ideas for capital investment, execute them well, and obtain good
results. You might wonder about what kind of role these engineers play in making such
business decisions. In other words, what specific tasks are assigned to these engineers, and
what tools and techniques are available to them for making such capital investment decisions? We answer these questions and explore related issues throughout this book.
CHAPTER LEARNING OBJECTIVES
After completing this chapter, you should understand the following concepts:
The role of engineers in business.
Types of business organization.
The nature and types of engineering economic decisions.
What makes the engineering economic decisions difficult.
How a typical engineering project idea evolves in business.
Fundamental principles of engineering economics.
1.1 Role of Engineers in Business
ahoo, Apple Computer, Microsoft Corporation, and Sun Microsystems produce
computer products and have a market value of several billion dollars each. These
companies were all started by young college students with technical backgrounds.
When they went into the computer business, these students initially organized
their companies as proprietorships. As the businesses grew, they became partnerships and
were eventually converted to corporations. This chapter begins by introducing the three primary forms of business organization and briefly discusses the role of engineers in business.
Y
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Section 1.1 Role of Engineers in Business 5
1.1.1 Types of Business Organization
As an engineer, you should understand the nature of the business organization with which
you are associated. This section will present some basic information about the type of organization you should choose should you decide to go into business for yourself.
The three legal forms of business, each having certain advantages and disadvantages,
are proprietorships, partnerships, and corporations.
Proprietorships
A proprietorship is a business owned by one individual. This person is responsible for
the firm’s policies, owns all its assets, and is personally liable for its debts. A proprietorship
has two major advantages. First, it can be formed easily and inexpensively. No legal and
organizational requirements are associated with setting up a proprietorship, and organizational costs are therefore virtually nil. Second, the earnings of a proprietorship are taxed
at the owner’s personal tax rate, which may be lower than the rate at which corporate income is taxed. Apart from personal liability considerations, the major disadvantage of a
proprietorship is that it cannot issue stocks and bonds, making it difficult to raise capital for
any business expansion.
Partnerships
A partnership is similar to a proprietorship, except that it has more than one owner.
Most partnerships are established by a written contract between the partners. The contract normally specifies salaries, contributions to capital, and the distribution of profits
and losses. A partnership has many advantages, among which are its low cost and ease of
formation. Because more than one person makes contributions, a partnership typically
has a larger amount of capital available for business use. Since the personal assets of all
the partners stand behind the business, a partnership can borrow money more easily from
a bank. Each partner pays only personal income tax on his or her share of a partnership’s
taxable income.
On the negative side, under partnership law each partner is liable for a business’s
debts. This means that the partners must risk all their personal assets—even those not invested in the business. And while each partner is responsible for his or her portion of the
debts in the event of bankruptcy, if any partners cannot meet their pro rata claims, the remaining partners must take over the unresolved claims. Finally, a partnership has a limited
life, insofar as it must be dissolved and reorganized if one of the partners quits.
Corporations
A corporation is a legal entity created under provincial or federal law. It is separate from
its owners and managers. This separation gives the corporation four major advantages:
(1) It can raise capital from a large number of investors by issuing stocks and bonds;
(2) it permits easy transfer of ownership interest by trading shares of stock; (3) it allows
limited liability—personal liability is limited to the amount of the individual’s investment
in the business; and (4) it is taxed differently than proprietorships and partnerships, and
under certain conditions, the tax laws favor corporations. On the negative side, it is expensive to establish a corporation. Furthermore, a corporation is subject to numerous
governmental requirements and regulations.
As a firm grows, it may need to change its legal form because the form of a business
affects the extent to which it has control of its own operations and its ability to acquire
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6 CHAPTER 1 Engineering Economic Decisions
funds. The legal form of an organization also affects the risk borne by its owners in case
of bankruptcy and the manner in which the firm is taxed. Apple Computer, for example,
started out as a two-man garage operation. As the business grew, the owners felt constricted by this form of organization: It was difficult to raise capital for business expansion; they felt that the risk of bankruptcy was too high to bear; and as their business
income grew, their tax burden grew as well. Eventually, they found it necessary to convert
the partnership into a corporation.
In the United States, the overwhelming majority of business firms are proprietorships,
followed by corporations and partnerships. However, in terms of total business volume
(dollars of sales), the quantity of business transacted by proprietorships and partnerships
is several times less than that of corporations. Since most business is conducted by corporations, this text will generally address economic decisions encountered in that form of
ownership.
1.1.2 Engineering Economic Decisions
What role do engineers play within a firm? What specific tasks are assigned to the engineering staff, and what tools and techniques are available to it to improve a firm’s profits?
Engineers are called upon to participate in a variety of decisions, ranging from manufacturing, through marketing, to financing decisions. We will restrict our focus, however, to
various economic decisions related to engineering projects. We refer to these decisions as
engineering economic decisions.
In manufacturing, engineering is involved in every detail of a product’s production,
from conceptual design to shipping. In fact, engineering decisions account for the majority (some say 85%) of product costs. Engineers must consider the effective use of capital
assets such as buildings and machinery. One of the engineer’s primary tasks is to plan for
the acquisition of equipment (capital expenditure) that will enable the firm to design
and produce products economically.
With the purchase of any fixed asset—equipment, for instance—we need to estimate
the profits (more precisely, cash flows) that the asset will generate during its period of
service. In other words, we have to make capital expenditure decisions based on predictions about the future. Suppose, for example, you are considering the purchase of a deburring machine to meet the anticipated demand for hubs and sleeves used in the production
of gear couplings. You expect the machine to last 10 years. This decision thus involves an
implicit 10-year sales forecast for the gear couplings, which means that a long waiting period will be required before you will know whether the purchase was justified.
An inaccurate estimate of the need for assets can have serious consequences. If you
invest too much in assets, you incur unnecessarily heavy expenses. Spending too little on
fixed assets, however, is also harmful, for then the firm’s equipment may be too obsolete
to produce products competitively, and without an adequate capacity, you may lose a portion of your market share to rival firms. Regaining lost customers involves heavy marketing expenses and may even require price reductions or product improvements, both of
which are costly.
1.1.3 Personal Economic Decisions
In the same way that an engineer can play a role in the effective utilization of corporate financial assets, each of us is responsible for managing our personal financial affairs. After
we have paid for nondiscretionary or essential needs, such as housing, food, clothing, and
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Section 1.2 What Makes the Engineering Economic Decision Difficult? 7
transportation, any remaining money is available for discretionary expenditures on items
such as entertainment, travel, and investment. For money we choose to invest, we want to
maximize the economic benefit at some acceptable risk. The investment choices are unlimited and include savings accounts, guaranteed investment certificates, stocks, bonds,
mutual funds, registered retirement savings plans, rental properties, land, business ownership, and more.
How do you choose? The analysis of one’s personal investment opportunities utilizes
the same techniques that are used for engineering economic decisions. Again, the challenge
is predicting the performance of an investment into the future. Choosing wisely can be
very rewarding, while choosing poorly can be disastrous. Some investors in the energy
stock Enron who sold prior to the fraud investigation became millionaires. Others, who
did not sell, lost everything.
A wise investment strategy is a strategy that manages risk by diversifying investments. With such an approach, you have a number of different investments ranging from
very low to very high risk and are in a number of business sectors. Since you do not have
all your money in one place, the risk of losing everything is significantly reduced. (We
discuss some of these important issues in Chapters 12 and 13.)
1.2 What Makes the Engineering Economic
Decision Difficult?
The economic decisions that engineers make in business differ very little from the financial decisions made by individuals, except for the scale of the concern. Suppose, for example, that a firm is using a lathe that was purchased 12 years ago to produce pump
shafts. As the production engineer in charge of this product, you expect demand to continue into the foreseeable future. However, the lathe has begun to show its age: It has broken frequently during the last 2 years and has finally stopped operating altogether. Now
you have to decide whether to replace or repair it. If you expect a more efficient lathe to
be available in the next 1 or 2 years, you might repair the old lathe instead of replacing it.
The major issue is whether you should make the considerable investment in a new lathe
now or later. As an added complication, if demand for your product begins to decline, you
may have to conduct an economic analysis to determine whether declining profits from
the project offset the cost of a new lathe.
Let us consider a real-world engineering decision problem on a much larger scale, as
taken from an article from The Washington Post.2 Ever since Hurricane Katrina hit the
city of New Orleans in August 2005, the U.S. federal government has been under pressure to show strong support for rebuilding levees in order to encourage homeowners and
businesses to return to neighborhoods that were flooded when the city’s levees crumbled
under Katrina’s storm surge. Many evacuees have expressed reluctance to rebuild without
assurances that New Orleans will be made safe from future hurricanes, including Category
5 storms, the most severe. Some U.S. Army Corps of Engineers officers estimated that it
would cost more than $1.6 billion to restore the levee system merely to its design
strength—that is, to withstand a Category 3 storm. New design features would include
floodgates on several key canals, as well as stone-and-concrete fortification of some of
2
Joby Warrick and Peter Baker, “Bush Pledges $1.5 Billion for New Orleans—Proposal Would Double Aid
From U.S. for Flood Protection,” The Washington Post, Dec. 16, 2005, p. A03.
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8 CHAPTER 1 Engineering Economic Decisions
Existing levees
Lake
Pontchartrain
Floodgates to
be installed
Lakefront
Airport
London
Orleans Ave.
Canal Canal
17th St.
Canal
Metairie
New Orleans
French
Quarter
Superdome
Convention Ctr.
Industrial Canal
Lower
Ninth
Mis
siss Ward
ipp
i R Chalmette
ive
r
Lake
Borgne
M
iss
iss
ip
pi
G
ul
fO
ut
le
t
0
5
Miles
Figure 1.1 The White House pledged $1.5 billion to armor existing
New Orleans levees with concrete and stone, build floodgates on three
canals, and upgrade the city’s pumping system.
the city’s earthen levees, as illustrated in Figure 1.1. Donald E. Powell, the administration’s coordinator of post-Katrina recovery, insisted that the improvements would make
the levee system much stronger than it had been in the past. But he declined to say
whether the administration would support further upgrades of the system to Category 5
protection, which would require substantial reengineering of existing levees at a cost that
could, by many estimates, exceed $30 billion.
Obviously, this level of engineering decision is far more complex and more significant than a business decision about when to introduce a new product. Projects of
this nature involve large sums of money over long periods of time, and it is difficult to
justify the cost of the system purely on the basis of economic reasoning, since we do
not know when another Katrina-strength storm will be on the horizon. Even if we
decide to rebuild the levee systems, should we build just enough to withstand a Category 3 storm, or should we build to withstand a Category 5 storm? Any engineering
economic decision pertaining to this type of extreme event will be extremely difficult
to make.
In this book, we will consider many types of investments—personal investments as
well as business investments. The focus, however, will be on evaluating engineering projects on the basis of their economic desirability and on dealing with investment situations
that a typical firm faces.
1.3 Economic Decisions versus Design Decisions
Economic decisions differ in a fundamental way from the types of decisions typically
encountered in engineering design. In a design situation, the engineer utilizes known
physical properties, the principles of chemistry and physics, engineering design correlations, and engineering judgment to arrive at a workable and optimal design. If the
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Section 1.4 Large-Scale Engineering Projects 9
judgment is sound, the calculations are done correctly, and we ignore technological
advances, the design is time invariant. In other words, if the engineering design to meet
a particular need is done today, next year, or in five years’ time, the final design would not
change significantly.
In considering economic decisions, the measurement of investment attractiveness,
which is the subject of this book, is relatively straightforward. However, the information
required in such evaluations always involves predicting or forecasting product sales,
product selling prices, and various costs over some future time frame—5 years, 10 years,
25 years, etc.
All such forecasts have two things in common. First, they are never completely accurate compared with the actual values realized at future times. Second, a prediction or
forecast made today is likely to be different from one made at some point in the future. It
is this ever-changing view of the future that can make it necessary to revisit and even
change previous economic decisions. Thus, unlike engineering design, the conclusions
reached through economic evaluation are not necessarily time invariant. Economic decisions have to be based on the best information available at the time of the decision and a
thorough understanding of the uncertainties in the forecasted data.
1.4 Large-Scale Engineering Projects
In the development of any product, a company’s engineers are called upon to translate an
idea into reality. A firm’s growth and development depend largely upon a constant flow of
ideas for new products, and for the firm to remain competitive, it has to make existing
products better or produce them at a lower cost. In the next section, we present an example
of how a design engineer’s idea eventually turned into an innovative automotive product.
1.4.1 How a Typical Project Idea Evolves
The Toyota Motor Corporation introduced the world’s first mass-produced car powered
by a combination of gasoline and electricity. Known as the Prius, this vehicle is the first
of a new generation of Toyota cars whose engines cut air pollution dramatically and boost
fuel efficiency to significant levels. Toyota, in short, wants to launch and dominate a new
“green” era for automobiles—and is spending $1.5 billion to do it. Developed for the
Japanese market initially, the Prius uses both a gasoline engine and an electric motor as
its motive power source. The Prius emits less pollution than ordinary cars, and it gets
more mileage, which means less output of carbon dioxide. Moreover, the Prius gives a
highly responsive performance and smooth acceleration. The following information from
BusinessWeek3 illustrates how a typical strategic business decision is made by the engineering staff of a larger company. Additional information has been provided by Toyota
Motor Corporation.
Why Go for a Greener Car?
Toyota first started to develop the Prius in 1995. Four engineers were assigned to figure
out what types of cars people would drive in the 21st century. After a year of research, a
3
Emily Thornton (Tokyo), Keith Naughton (Detroit), and David Woodruff, “Japan’s Hybrid Cars—Toyota and
rivals are betting on pollution fighters—Will they succeed?” BusinessWeek, Dec. 4, 1997.
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10 CHAPTER 1 Engineering Economic Decisions
chief engineer concluded that Toyota would have to sell cars better suited to a world with
scarce natural resources. He considered electric motors. But an electric car can travel only
215 km before it must be recharged. Another option was fuel-cell cars that run on hydrogen. But he suspected that mass production of this type of car might not be possible for another 15 years. So the engineer finally settled on a hybrid system powered by an electric
motor, a nickel–metal hydride battery, and a gasoline engine. From Toyota’s perspective, it
is a big bet, as oil and gasoline prices are bumping along at record lows at that time. Many
green cars remain expensive and require trade-offs in terms of performance. No carmaker
has ever succeeded in selling consumers en masse something they have never wanted to
buy: cleaner air. Even in Japan, where a liter of regular gas can cost as much as $1, carmakers have trouble pushing higher fuel economy and lower carbon emissions. Toyota has
several reasons for going green. In the next century, as millions of new car owners in
China, India, and elsewhere take to the road, Toyota predicts that gasoline prices will rise
worldwide. At the same time, Japan’s carmakers expect pollution and global warming to
become such threats that governments will enact tough measures to clean the air.
What Is So Unique in Design?
It took Toyota engineers two years to develop the current power system in the Prius. The
car comes with a dual engine powered by both an electric motor and a newly developed
1.5-liter gasoline engine. When the engine is in use, a special “power split device” sends
some of the power to the driveshaft to move the car’s wheels. The device also sends some
of the power to a generator, which in turn creates electricity, to either drive the motor or
recharge the battery. Thanks to this variable transmission, the Prius can switch back and
forth between motor and engine, or employ both, without creating any jerking motion.
The battery and electric motor propel the car at slow speeds. At normal speeds, the electric motor and gasoline engine work together to power the wheels. At higher speeds, the
battery kicks in extra power if the driver must pass another car or zoom up a hill.
When the car decelerates and brakes, the motor converts the vehicle’s kinetic energy
into electricity and sends it through an inverter to be stored in the battery. (See Figure 1.2.)
Battery
Generator
Engine
Power Split
Device
Inverter
Motor
Reduction
Gears
Figure 1.2 Arrangement of components of
the Toyota Hybrid System II.
(Source: Evaluation of 2004 Toyota Prius Hybrid
Electric Drive System, Oak Ridge National Laboratory, ONR/TM2004/247, U.S. Department of Energy.)
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Section 1.4 Large-Scale Engineering Projects 11
• When engine efficiency is low, such as during start-up and with midrange speeds,
motive power is provided by the motor alone, using energy stored in the battery.
• Under normal driving conditions, overall efficiency is optimized by controlling the
power allocation so that some of the engine power is used for turning the generator to
supply electricity for the motor while the remaining power is used for turning the
wheels.
• During periods of acceleration, when extra power is needed, the generator supplements the electricity being drawn from the battery, so the motor is supplied with the
required level of electrical energy.
• While decelerating and braking, the motor acts as a generator that is driven by the
wheels, thus allowing the recovery of kinetic energy. The recovered kinetic energy is
converted to electrical energy that is stored in the battery.
• When necessary, the generator recharges the battery to maintain sufficient reserves.
• When the vehicle is not moving and when the engine moves outside of certain speed
and load conditions, the engine stops automatically.
So the car’s own movement, as well as the power from the gasoline engine, provides the
electricity needed. The energy created and stored by deceleration boosts the car’s efficiency. So does the automatic shutdown of the engine when the car stops at a light. At
higher speeds and during acceleration, the companion electric motor works harder, allowing the gas engine to run at peak efficiency. Toyota claims that, in tests, its hybrid car
has boosted fuel economy by 100% and engine efficiency by 80%. The Prius emits about
half as much carbon dioxide, and about one-tenth as much carbon monoxide, hydrocarbons, and nitrous oxide, as conventional cars.
Is It Safe to Drive on a Rainy Day?
Yet, major hurdles remain to be overcome in creating a mass market for green vehicles.
Car buyers are not anxious enough about global warming to justify a “sea-level change”
in automakers’ marketing. Many of Japan’s innovations run the risk of becoming impressive technologies meant for the masses, but bought only by the elite. The unfamiliarity of
green technology can also frighten consumers. The Japanese government sponsors festivals at which people can test drive alternative-fuel vehicles. But some turned down that
chance on a rainy weekend in May because they feared that riding in an electric car might
electrocute them. An engineer points out, “It took 20 years for the automatic transmission
to become popular in Japan.” It certainly would take that long for many of these technologies to catch on.
How Much Would It Cost?
The biggest question remaining about the mass production of the vehicle concerns its
production cost. Costs will need to come down for Toyota’s hybrid to be competitive
around the world, where gasoline prices are lower than in Japan. Economies of scale will
help as production volumes increase, but further advances in engineering also will be essential. With its current design, Prius’s monthly operating cost would be roughly twice
that of a conventional automobile. Still, Toyota believes that it will sell more than 12,000
Prius models to Japanese drivers during the first year. To do that, it is charging just
$17,000 a car. The company insists that it will not lose any money on the Prius, but rivals
and analysts estimate that, at current production levels, the cost of building a Prius could
be as much as $32,000. If so, Toyota’s low-price strategy will generate annual losses of
more than $100 million on the new compact.
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12 CHAPTER 1 Engineering Economic Decisions
Will There Be Enough Demand?
Why buy a $17,000 Prius when a $13,000 Corolla is more affordable? It does not help Toyota that it is launching the Prius in the middle of a sagging domestic car market. Nobody
really knows how big the final market for hybrids will be. Toyota forecasts that hybrids will
account for a third of the world’s auto market as early as 2005, but Japan’s Ministry of
International Trade and Industry expects 2.4 million alternative-fuel vehicles, including hybrids, to roam Japan’s backstreets by 2010. Nonetheless, the Prius has set a new standard
for Toyota’s competitors. There seems to be no turning back for the Japanese. The government may soon tell carmakers to slash carbon dioxide emissions by 20% by 2010. And it
wants them to cut nitrous oxide, hydrocarbon, and carbon monoxide emissions by 80%.
The government may also soon give tax breaks to consumers who buy green cars.
Prospects for the Prius started looking good: Total sales for Prius reached over 7,700
units as of June 1998. With this encouraging sales trend, Toyota finally announced that
the Prius would be introduced in the North American and European markets by the year
2000. The total sales volume will be approximately 20,000 units per year in the North
American and European market combined. As with the 2000 North American and European introduction, Toyota is planning to use the next two years to develop a hybrid vehicle optimized to the usage conditions of each market.
What Is the Business Risk in Marketing the Prius?
Engineers at Toyota Motors have stated that California would be the primary market for the
Prius outside Japan, but they added that an annual demand of 50,000 cars would be necessary to justify production. Despite Toyota management’s decision to introduce the hybrid
electric car into the U.S. market, the Toyota engineers were still uncertain whether there
would be enough demand. Furthermore, competitors, including U.S. automakers, just do
not see how Toyota can achieve the economies of scale needed to produce green cars at a
profit. The primary advantage of the design, however, is that the Prius can cut auto pollution
in half. This is a feature that could be very appealing at a time when government air-quality
standards are becoming more rigorous and consumer interest in the environment is strong.
However, in the case of the Prius, if a significant reduction in production cost never materializes, demand may remain insufficient to justify the investment in the green car.
1.4.2 Impact of Engineering Projects on Financial Statements
Engineers must understand the business environment in which a company’s major business decisions are made. It is important for an engineering project to generate profits, but
it also must strengthen the firm’s overall financial position. How do we measure Toyota’s
success in the Prius project? Will enough Prius models be sold, for example, to keep the
green-engineering business as Toyota’s major source of profits? While the Prius project
will provide comfortable, reliable, low-cost driving for the company’s customers, the bottom line is its financial performance over the long run.
Regardless of a business’s form, each company has to produce basic financial statements
at the end of each operating cycle (typically a year). These financial statements provide the
basis for future investment analysis. In practice, we seldom make investment decisions
solely on the basis of an estimate of a project’s profitability, because we must also consider
the overall impact of the investment on the financial strength and position of the company.
Suppose that you were the president of the Toyota Corporation. Suppose further that
you even hold some shares in the company, making you one of the company’s many
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Section 1.5 Common Types of Strategic Engineering Economic Decisions 13
owners. What objectives would you set for the company? While all firms are in business
in hopes of making a profit, what determines the market value of a company are not profits per se, but cash flow. It is, after all, available cash that determines the future investments and growth of the firm. Therefore, one of your objectives should be to increase the
company’s value to its owners (including yourself) as much as possible. To some extent,
the market price of your company’s stock represents the value of your company. Many
factors affect your company’s market value: present and expected future earnings, the
timing and duration of those earnings, and the risks associated with them. Certainly, any
successful investment decision will increase a company’s market value. Stock price can
be a good indicator of your company’s financial health and may also reflect the market’s
attitude about how well your company is managed for the benefit of its owners.
1.4.3 A Look Back in 2005: Did Toyota Make the Right Decision?
Clearly, there were many doubts and uncertainties about the market for hybrids in 1998.
Even Toyota engineers were not sure that there would be enough demand in the U.S.
market to justify the production of the vehicle. Seven years after the Prius was introduced, it turns out that Toyota’s decision to go ahead was the right decision. The continued success of the vehicle led to the launching of a second-generation Prius at the New
York Motor Show in 2003. This car delivered higher power and better fuel economy than
its predecessor. Indeed, the new Prius proved that Toyota has achieved its goal: to create
an eco-car with high-level environmental performance, but with the conventional draw of
a modern car. These features, combined with its efficient handling and attractive design,
are making the Prius a popular choice of individuals and companies alike. In fact, Toyota
has already announced that it will double Prius production for the U.S. market, from
50,000 to 100,000 units annually, but even that may fall short of demand if oil prices continue to increase in the future.
Toyota made its investors happy, as the public liked the new hybrid car, resulting in
an increased demand for the product. This, in turn, caused stock prices, and hence shareholder wealth, to increase. In fact, the new, heavily promoted, green car turned out to be
a market leader in its class and contributed to sending Toyota’s stock to an all-time high
in late 2005.4 Toyota’s market value continued to increase well into 2006. Any successful
investment decision on Prius’s scale will tend to increase a firm’s stock prices in the
marketplace and promote long-term success. Thus, in making a large-scale engineering
project decision, we must consider its possible effect on the firm’s market value. (In
Chapter 2, we discuss the financial statements in detail and show how to use them in our
investment decision making.)
1.5 Common Types of Strategic Engineering
Economic Decisions
The story of how the Toyota Corporation successfully introduced a new product and became the market leader in the hybrid electric car market is typical: Someone had a good
idea, executed it well, and obtained good results. Project ideas such as the Prius can originate
from many different levels in an organization. Since some ideas will be good, while others
4
Toyota Motor Corporation, Annual Report, 2005.
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14 CHAPTER 1 Engineering Economic Decisions
will not, we need to establish procedures for screening projects. Many large companies
have a specialized project analysis division that actively searches for new ideas, projects,
and ventures. Once project ideas are identified, they are typically classified as (1) equipment or process selection, (2) equipment replacement, (3) new product or product expansion, (4) cost reduction, or (5) improvement in service or quality. This classification
scheme allows management to address key questions: Can the existing plant, for example, be used to attain the new production levels? Does the firm have the knowledge and
skill to undertake the new investment? Does the new proposal warrant the recruitment of
new technical personnel? The answers to these questions help firms screen out proposals
that are not feasible, given a company’s resources.
The Prius project represents a fairly complex engineering decision that required the
approval of top executives and the board of directors. Virtually all big businesses face investment decisions of this magnitude at some time. In general, the larger the investment,
the more detailed is the analysis required to support the expenditure. For example, expenditures aimed at increasing the output of existing products or at manufacturing a new
product would invariably require a very detailed economic justification. Final decisions
on new products, as well as marketing decisions, are generally made at a high level within the company. By contrast, a decision to repair damaged equipment can be made at a
lower level. The five classifications of project ideas are as follows:
• Equipment or Process Selection. This class of engineering decision problems involves selecting the best course of action out of several that meet a project’s requirements. For example, which of several proposed items of equipment shall we
purchase for a given purpose? The choice often hinges on which item is expected to
generate the largest savings (or the largest return on the investment). For example,
the choice of material will dictate the manufacturing process for the body panels in
the automobile. Many factors will affect the ultimate choice of the material, and engineers should consider all major cost elements, such as the cost of machinery and
equipment, tooling, labor, and material. Other factors may include press and assembly, production and engineered scrap, the number of dies and tools, and the cycle
times for various processes.
• Equipment Replacement. This category of investment decisions involves considering the expenditure necessary to replace worn-out or obsolete equipment. For example, a company may purchase 10 large presses, expecting them to produce stamped
metal parts for 10 years. After 5 years, however, it may become necessary to produce
the parts in plastic, which would require retiring the presses early and purchasing
plastic molding machines. Similarly, a company may find that, for competitive reasons, larger and more accurate parts are required, making the purchased machines
become obsolete earlier than expected.
• New Product or Product Expansion. Investments in this category increase company revenues if output is increased. One common type of expansion decision includes decisions about expenditures aimed at increasing the output of existing
production or distribution facilities. In these situations, we are basically asking,
“Shall we build or otherwise acquire a new facility?” The expected future cash inflows in this investment category are the profits from the goods and services produced
in the new facility. A second type of expenditure decision includes considering expenditures necessary to produce a new product or to expand into a new geographic
area. These projects normally require large sums of money over long periods.
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Section 1.6 Fundamental Principles of Engineering Economics 15
• Cost Reduction. A cost-reduction project is a project that attempts to lower a firm’s
operating costs. Typically, we need to consider whether a company should buy
equipment to perform an operation currently done manually or spend money now in
order to save more money later. The expected future cash inflows on this investment
are savings resulting from lower operating costs.
• Improvement in Service or Quality. Most of the examples in the previous sections
were related to economic decisions in the manufacturing sector. The decision techniques we develop in this book are also applicable to various economic decisions
related to improving services or quality of product.
1.6 Fundamental Principles of Engineering Economics
This book is focused on the principles and procedures engineers use to make sound
economic decisions. To the first-time student of engineering economics, anything related to money matters may seem quite strange when compared to other engineering
subjects. However, the decision logic involved in solving problems in this domain is
quite similar to that employed in any other engineering subject. There are fundamental
principles to follow in engineering economics that unite the concepts and techniques
presented in this text, thereby allowing us to focus on the logic underlying the practice
of engineering economics.
• Principle 1: A nearby penny is worth a distant dollar. A fundamental concept
in engineering economics is that money has a time value associated with it. Because we can earn interest on money received today, it is better to receive money
earlier than later. This concept will be the basic foundation for all engineering
project evaluation.
Time Value of Money
If you receive $100 now, you can invest it and have more money available six months
from now
$100
$100
Today
Earning opportunity
Six months later
• Principle 2: All that counts are the differences among alternatives. An economic decision should be based on the differences among the alternatives considered.
All that is common is irrelevant to the decision. Certainly, any economic decision is
no better than the alternatives being considered. Thus, an economic decision should
be based on the objective of making the best use of limited resources. Whenever a
choice is made, something is given up. The opportunity cost of a choice is the value
of the best alternative given up.
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16 CHAPTER 1 Engineering Economic Decisions
Comparing Buy versus Lease
Whatever you decide, you need to spend the same amount of money on
fuel and maintenance
Option
Buy
Lease
Monthly
Maintenance
Cash
Outlay at
Signing
Monthly
Payment
Salvage
Value at
End of
Year 3
$960
$550
$6,500
$350
$9,000
$960
$550
$2,400
$550
0
Monthly
Fuel
Cost
5
Irrelevant items in decision making
Marginal cost:
The cost
associated
with one
additional unit
of production,
also called the
incremental
cost.
• Principle 3: Marginal revenue must exceed marginal cost. Effective decision
making requires comparing the additional costs of alternatives with the additional
benefits. Each decision alternative must be justified on its own economic merits
before being compared with other alternatives. Any increased economic activity
must be justified on the basis of the fundamental economic principle that marginal
revenue must exceed marginal cost. Here, marginal revenue means the additional revenue made possible by increasing the activity by one unit (or small unit). Marginal
cost has an analogous definition. Productive resources—the natural resources, human
resources, and capital goods available to make goods and services—are limited.
Therefore, people cannot have all the goods and services they want; as a result, they
must choose some things and give up others.
Marginal Analysis
To justify your action, marginal revenue must exceed marginal cost
$2
Cost of Goods Sold
Marginal
cost
1 unit
$4
Gross Revenue
1 unit
Marginal
revenue
• Principle 4: Additional risk is not taken without the expected additional return.
For delaying consumption, investors demand a minimum return that must be greater
than the anticipated rate of inflation or any perceived risk. If they didn’t receive
enough to compensate for anticipated inflation and the perceived investment risk, investors would purchase whatever goods they desired ahead of time or invest in assets
that would provide a sufficient return to compensate for any loss from inflation or
potential risk.
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Summary 17
Risk and Return Trade -Off
Expected returns from bonds and stocks are normally higher than the
expected return from a savings account
Investment
Class
Savings account
(cash)
Potential
Risk
Low/None
Bond (debt)
Moderate
Stock (equity)
High
Expected
Return
1.5%
4.8%
11.5%
The preceding four principles are as much statements of common sense as they are theoretical precepts. These principles provide the logic behind what is to follow. We build
on them and attempt to draw out their implications for decision making. As we continue,
keep in mind that, while the topics being treated may change from chapter to chapter,
the logic driving our treatment of them is constant and rooted in the four fundamental
principles.
SUMMARY
This chapter has given us an overview of a variety of engineering economic problems
that commonly are found in the business world. We examined the role, and the increasing importance, of engineers in the firm, as evidenced by the development at Toyota of the Prius, a hybrid vehicle. Commonly, engineers are called upon to participate
in a variety of strategic business decisions ranging from product design to marketing.
The term engineering economic decision refers to any investment decision related to
an engineering project. The facet of an economic decision that is of most interest from
an engineer’s point of view is the evaluation of costs and benefits associated with making a capital investment.
The five main types of engineering economic decisions are (1) equipment or process
selection, (2) equipment replacement, (3) new product or product expansion, (4) cost
reduction, and (5) improvement in service or quality.
The factors of time and uncertainty are the defining aspects of any investment project.
The four fundamental principles that must be applied in all engineering economic de-
cisions are (1) the time value of money, (2) differential (incremental) cost and revenue,
(3) marginal cost and revenue, and (4) the trade-off between risk and reward.
Risk-return
tradeoff:
Invested money
can render
higher profits
only if it is
subject to the
possibility of
being lost.
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TWO
CHAPTER
Understanding Financial
Statements
Dell Manages Profitability, Not Inventory1 In 1994, Dell was
a struggling second-tier PC maker. Like other PC makers, Dell ordered
its components in advance and carried a large amount of component
inventory. If its forecasts were wrong, Dell had major write-downs.
Then Dell began to implement a new business model. Its operations
had always featured a build-to-order process with direct sales to customers, but Dell took a series of ingenious steps to eliminate its inventories.The results were spectacular.
Over a four-year period, Dell’s revenues grew from $2 billion to
$16 billion, a 50 % annual growth rate. Earnings per share increased by
62 % per year. Dell’s stock price increased by more than 17,000 % in a
little over eight years. In 1998, Dell’s return on invested capital was
217 %, and the company had $1.8 billion in cash. (The rapid growth
1983–
Michael Dell starts business of preformatting IBM PC HDs on
weekends
1985–
1986–
$6 million sales, upgrading IBM compatibles for local businesses
$70 million sales; focus on assembling own line of PCs
1990–
$500 million sales with an extensive line of products
1996–
Dell goes online; $1 million per day in online sales; $5.3B in annual
sales
Dell online sales at $3 million per day; 50% growth rate for third
consecutive year, $7.8B in total annual sales.
1997–
2005–
1
$49.2B in sales
Jonathan Byrnes, “Dell Manages Profitability, Not Inventory,” Harvard Business School, Working
Knowledge, June 2, 2003.
18
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continued, and the company’s sales revenues finally reached $50 billion in
2005.)
Profitability management—coordinating a company’s day-to-day activities
through careful forethought and attentive oversight—was at the core of Dell’s
transformation in this critical period. Dell created a tightly aligned business
model that enabled it to dispense with the need for its component inventories. Not only was capital not needed, but the change generated enormous
amounts of cash that Dell used to fuel its growth. How did Dell do it?
Account selection. Dell purposely selected customers with relatively
predictable purchasing patterns and low service costs. The company developed a core competence in targeting customers and kept a massive database for this purpose.
The remainder of Dell’s business involved individual consumers. To obtain stable demand in this segment, Dell used higher-end products and
those with the latest technology to target second-time buyers who had
regular upgrade purchase patterns, required little technical support, and
paid by credit card.
Demand management. Dell’s core philosophy of actively managing
demand in real time, or “selling what you have,” rather than making what
you want to sell, was a critical driver of Dell’s successful profitability management. Without this critical element, Dell’s business model simply would
not have been effective.
Product life-cycle management. Because Dell’s customers were
largely high-end repeat buyers who rapidly adopted new technology, Dell’s
marketing could focus on managing product life-cycle transitions.
Supplier management. Although Dell’s manufacturing system featured a combination of build-product-to-order and buy-component-to-plan
processes, the company worked closely with its suppliers to introduce more
flexibility into its system.
Forecasting. Dell’s forecast accuracy was about 70 to 75 %, due to its
careful account selection. Demand management, in turn, closed the forecast
gap. When in doubt, Dell managers overforecast on high-end products because it was easier to sell up and high-end products had a longer shelf life.
19
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20 CHAPTER 2 Understanding Financial Statements
Net worth is
the amount
by which a
company’s or
individual’s
assets exceed
the company’s
or individual’s
liabilities.
Liquidity management. Direct sales were explicitly targeted toward high-end customers who paid with a credit card. These sales had a 4-day cash conversion cycle, while
Dell took 45 days to pay its vendors. This approach generated a huge amount of liquidity
that helped finance Dell’s rapid growth and limited its external financing needs. Dell’s
cash engine was a key underlying factor that enabled it to earn such extraordinarily high
returns.
If you want to explore investing in Dell stock, what information would you go by?
You would certainly prefer that Dell have a record of accomplishment of profitable operations, earning a profit (net income) year after year. The company would need a steady
stream of cash coming in and a manageable level of debt. How would you determine
whether the company met these criteria? Investors commonly use the financial statements
contained in the annual report as a starting point in forming expectations about future
levels of earnings and about the firm’s riskiness.
Before making any financial decision, it is good to understand an elementary aspect of your financial situation—one that you’ll also need for retirement planning, estate planning, and, more generally, to get an answer to the question, “How am I
doing?” It is called your net worth. If you decided to invest $10,000 in Dell stocks,
how would that affect your net worth? You may need this information for your own
financial planning, but it is routinely required whenever you have to borrow a large
sum of money from a financial institution. For example, when you are buying a home,
you need to apply for a mortgage. Invariably, the bank will ask you to submit your
net-worth statement as a part of loan processing. Your net-worth statement is a snapshot of where you stand financially at a given point in time. The bank will determine
how creditworthy you are by examining your net worth. In a similar way, a corporation prepares the same kind of information for its financial planning or to report its
financial health to stockholders or investors. The reporting document is known as the
financial statements. We will first review the basics of figuring out the personal net
worth and then illustrate how any investment decision will affect this net-worth statement. Understanding the relationship between net worth and investing decisions will
enhance one’s overall understanding of how a company manages its assets in business
operations.
CHAPTER LEARNING OBJECTIVES
After completing this chapter, you should understand the following concepts:
The role of accounting in economic decisions.
Four types of financial statements prepared for investors and
regulators.
How to read the balance sheet statement.
How to use the income statement to manage a business.
The sources and uses of cash in business operation.
How to conduct the ratio analysis and what the numbers really mean.
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Section 2.1 Accounting: The Basis of Decision Making 21
2.1 Accounting:The Basis of Decision Making
e need financial information when we are making business decisions.
Virtually all businesses and most individuals keep accounting records to
aid in making decisions. As illustrated in Figure 2.1, accounting is the
information system that measures business activities, processes the resulting information into reports, and communicates the results to decision makers. For
this reason, we call accounting “the language of business.” The better you understand
this language, the better you can manage your financial well-being, and the better your
financial decisions will be.
Personal financial planning, education expenses, loans, car payments, income taxes,
and investments are all based on the information system we call accounting. The uses of
accounting information are many and varied:
• Individual people use accounting information in their day-to-day affairs to manage
bank accounts, to evaluate job prospects, to make investments, and to decide whether
to rent an apartment or buy a house.
• Business managers use accounting information to set goals for their organizations,
to evaluate progress toward those goals, and to take corrective actions if necessary.
Decisions based on accounting information may include which building or equipment to purchase, how much merchandise to keep on hand as inventory, and how
much cash to borrow.
• Investors and creditors provide the money a business needs to begin operations.
To decide whether to help start a new venture, potential investors evaluate what
income they can expect on their investment. Such an evaluation involves analyzing
the financial statements of the business. Before making a loan, banks determine
the borrower’s ability to meet scheduled payments. This kind of evaluation includes
a projection of future operations and revenue, based on accounting information.
W
Accounting data for future d
ecis
io
ns
Post
audit
People make decisions
Company prepares reports
to show the results of
their operations
Business transactions occur
Figure 2.1
The accounting system, which illustrates the flow of information.
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22 CHAPTER 2 Understanding Financial Statements
An essential product of an accounting information system is a series of financial
statements that allows people to make informed decisions. For personal use, the networth statement is a snapshot of where you stand financially at a given point in time. You
do that by adding your assets—such as cash, investments, and pension plans—in one column and your liabilities—or debts—in the other. Then subtract your liabilities from your
assets to find your net worth. In other words, your net worth is what you would be left
with if you sold everything and paid off all you owe. For business use, financial statements are the documents that report financial information about a business entity to decision makers. They tell us how a business is performing and where it stands financially. Our
purpose is not to present the bookkeeping aspects of accounting, but to acquaint you with
financial statements and to give you the basic information you need to make sound engineering economic decisions through the remainder of the book.
2.2 Financial Status for Businesses
Just like figuring out your personal wealth, all businesses must prepare their financial status. Of the various reports corporations issue to their stockholders, the annual report is by
far the most important, containing basic financial statements as well as management’s
opinion of the past year’s operations and the firm’s future prospects. What would managers
and investors want to know about a company at the end of the fiscal year? Following are
four basic questions that managers or investors are likely to ask:
• What is the company’s financial position at the end of the fiscal period?
• How well did the company operate during the fiscal period?
• On what did the company decide to use its profits?
• How much cash did the company generate and spend during the fiscal period?
As illustrated in Figure 2.2, the answer to each of these questions is provided by one of
the following financial statements: the balance sheet statement, the income statement, the
statement of retained earnings, and the cash flow statement. The fiscal year (or operating
Beginning of fiscal period (January 1, 2006)
How much profit did the
company make during the
fiscal period?
Income Statement
What did the company
decide to use their
profit for?
Statement of
Retained Earnings
How much cash did the
company generate and
spend during the period?
Statement of
Cash Flows
What is the company’s
financial position at
the end of fiscal period?
Balance Sheet
End of fiscal period (December 31, 2006)
Figure 2.2
statements.
Information reported on the financial
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Section 2.2 Financial Status for Businesses 23
cycle) can be any 12-month term, but is usually January 1 through December 31 of a
calendar year.
As mentioned in Section 1.1.2, one of the primary responsibilities of engineers in
business is to plan for the acquisition of equipment (capital expenditure) that will enable
the firm to design and produce products economically. This type of planning will require
an estimation of the savings and costs associated with the acquisition of equipment and
the degree of risk associated with execution of the project. Such an estimation will affect
the business’ bottom line (profitability), which will eventually affect the firm’s stock
price in the marketplace. Therefore, engineers should understand the various financial
statements in order to communicate with upper management regarding the status of a
project’s profitability. The situation is summarized in Figure 2.3.
For illustration purposes, we use data taken from Dell Corporation, manufacturer of a
wide range of computer systems, including desktops, notebooks, and workstations, to discuss the basic financial statements. In 1984, Michael Dell began his computer business at
the University of Texas in Austin, often hiding his IBM PC in his roommate’s bathtub
when his family came to visit. His dorm-room business officially became Dell Computer
Corporation on May 3, 1984. Since 2001, Dell has become the number-one and fastest
growing among all major computer system companies worldwide, with 55,200 employees
around the globe. Dell’s pioneering “direct model” is a simple concept involving selling
personal computer systems directly to customers. It offers (1) in-person relationships with
corporate and institutional customers; (2) telephone and Internet purchasing (the latter averaging $50 million a day in 2001); (3) built-to-order computer systems; (4) phone and online technical support; and (5) next-day on-site product service.
The company’s revenue in 2005 totaled $49.205 billion. During fiscal 2005, Dell
maintained its position as the world’s number-one supplier of personal computer systems, with a performance that continued to outpace the industry. Over the same period,
Dell’s global market share of personal computer sales reached 17.8%. In the company’s
2005 annual report, management painted an even more optimistic picture for the future,
External constraints
Strategic policy
decisions
by management
Operating decisions
Environmental
regulations
Investment decisions
Antitrust laws
Financing decisions
Product and workplace
safety rules
Role of engineers
Evaluation of capital
expenditure related to projects
Selection of production
methods used
Assessment of engineering safety
and environmental impact
Selection of types of products or
services produced
Accounting information
The balance sheet statement
The income statement
The cash flow statement
Expected financial performance
Firm’s market value
Expected profitability
Stock price
Timing of cash flows
Degree of financial risk
Figure 2.3
Summary of major factors affecting stock prices.
Bottom line
is slang for net
income or
accounting
profit.
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24 CHAPTER 2 Understanding Financial Statements
stating that Dell will continue to invest in information systems, research, development,
and engineering activities to support its growth and to provide for new competitive products.
Of course, there is no assurance that Dell’s revenue will continue to grow at the annual
rate of 50% in the future.
What can individual investors make of all this? Actually, we can make quite a bit. As
you will see, investors use the information contained in an annual report to form expectations about future earnings and dividends. Therefore, the annual report is certainly of
great interest to investors.
2.2.1 The Balance Sheet
What is the company’s financial position at the end of the reporting period? We find the
answer to this question in the company’s balance sheet statement. A company’s balance sheet, sometimes called its statement of financial position, reports three main categories of items: assets, liabilities, and stockholders’ equity. Assets are arranged in order
of liquidity. The most liquid assets appear at the top of the page, the least liquid assets at
the bottom of the page. (See Figure 2.4.) Because cash is the most liquid of all assets, it
is always listed first. Current assets are so critical that they are separately broken out and
totaled. They are what will hold the business afloat for the next year.
Liabilities are arranged in order of payment, the most pressing at the top of the list, the
least pressing at the bottom. Like current assets, current liabilities are so critical that they
are separately broken out and totaled. They are what will be paid out during the next year.
A company’s financial statements are based on the most fundamental tool of accounting: the accounting equation. The accounting equation shows the relationship among
assets, liabilities, and owners’ equity:
Assets = Liabilities + Owners’ Equity
Every business transaction, no matter how simple or complex, can be expressed in terms
of its effect on the accounting equation. Regardless of whether a business grows or contracts, the equality between the assets and the claims against the assets is always maintained. In other words, any change in the amount of total assets is necessarily accompanied
by an equal change on the other side of the equation—that is, by an increase or decrease in
either the liabilities or the owners’ equity.
As shown in Table 2.1, the first half of Dell’s year-end 2005 and 2004 balance sheets
lists the firm’s assets, while the remainder shows the liabilities and equity, or claims
against those assets.
Assets
Current Assets
Liabilities
Current Liabilities
Long-Term Liabilities
Long-Term Assets
Equity
1. Owner Contributions
2. Retained Earnings
Figure 2.4 Using the four quadrants of the
balance sheet.
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Section 2.2 Financial Status for Businesses 25
TABLE 2.1
Consolidated Statements of Financial Position (in millions) for Dell, Inc.
January 28, January 30,
2005
2004
Assets
Current assets:
Cash and cash equivalents
$
4,747
$
4,317
Short-term investments
5,060
835
Accounts receivable, net
4,414
3,635
459
327
Inventories
Other
2,217
1,519
16,897
10,633
Property, plant, and equipment, net
1,691
1,517
Investments
4,319
6,770
Total current assets
Other noncurrent assets
Total assets
308
391
$ 23,215
$ 19,311
$
$
Liabilities and Stockholders’ Equity
Current liabilities:
Accounts payable
Accrued and other
Total current liabilities
Long-term debt
Other noncurrent liabilities
Total liabilities
Commitments and contingent liabilities (Note 8)
8,895
7,316
5,241
3,580
14,136
10,896
505
505
2,089
1,630
16,730
13,031
—
—
—
—
8,195
6,823
(10,758)
(6,539)
9,174
6,131
Stockholders’ equity:
Preferred stock and capital in excess of $.01 par value; shares
issued and outstanding: none
Common stock and capital in excess of $.01 par value; shares
authorized: 7,000; shares issued: 2,769 and 2,721, respectively
Treasury stock, at cost; 284 and 165 shares, respectively
Retained earnings
Other comprehensive loss
(82)
(83)
Other
(44)
(52)
Total stockholders’ equity
Total liabilities and stockholders’ equity
Source: Annual Report, Dell Corporation, 2005.
6,485
6,280
$ 23,215
$ 19,311
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26 CHAPTER 2 Understanding Financial Statements
Current assets
account
represents the
value of all
assets that are
reasonably
expected to be
converted into
cash within one
year.
Assets
The dollar amount shown in the assets portion of the balance sheet represents how much
the company owns at the time it issues the report. We list the asset items in the order of
their “liquidity,” or the length of time it takes to convert them to cash.
• Current assets can be converted to cash or its equivalent in less than one year. Current assets generally include three major accounts:
1. The first is cash. A firm typically has a cash account at a bank to provide for the
funds needed to conduct day-to-day business. Although assets are always stated
in terms of dollars, only cash represents actual money. Cash-equivalent items are
also listed and include marketable securities and short-term investments.
2. The second account is accounts receivable—money that is owed the firm, but that
has not yet been received. For example, when Dell receives an order from a retail
store, the company will send an invoice along with the shipment to the retailer.
Then the unpaid bill immediately falls into the accounts receivable category.
When the bill is paid, it will be deducted from the accounts receivable account
and placed into the cash category. A typical firm will have a 30- to 45-day accounts receivable, depending on the frequency of its bills and the payment terms
for customers.
3. The third account is inventories, which show the dollar amount that Dell has invested in raw materials, work in process, and finished goods available for sale.
• Fixed assets are relatively permanent and take time to convert into cash. Fixed assets
reflect the amount of money Dell paid for its plant and equipment when it acquired
those assets. The most common fixed asset is the physical investment in the business,
such as land, buildings,2 factory machinery, office equipment, and automobiles. With
the exception of land, most fixed assets have a limited useful life. For example, buildings and equipment are used up over a period of years. Each year, a portion of the usefulness of these assets expires, and a portion of their total cost should be recognized as
a depreciation expense. The term depreciation denotes the accounting process for this
gradual conversion of fixed assets into expenses. Property, plant and equipment, net
thus represents the current book value of these assets after deducting depreciation
expenses.
• Finally, other assets include investments made in other companies and intangible
assets such as goodwill, copyrights, franchises, and so forth. Goodwill appears on the
balance sheet only when an operating business is purchased in its entirety. Goodwill
indicates any additional amount paid for the business above the fair market value of
the business. (Here, the fair market value is defined as the price that a buyer is willing to pay when the business is offered for sale.)
Liabilities and Stockholders’ Equity (Owners’ Net Worth)
The claims against assets are of two types: liabilities and stockholders’ equity. The liabilities of a company indicate where the company obtained the funds to acquire its assets
and to operate the business. Liability is money the company owes. Stockholders’ equity is
that portion of the assets of a company which is provided by the investors (owners).
Therefore, stockholders’ equity is the liability of a company to its owners.
2
Land and buildings are commonly called real assets to distinguish them from equipment and machinery.
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Section 2.2 Financial Status for Businesses 27
• Current liabilities are those debts which must be paid in the near future (normally,
within one year). The major current liabilities include accounts and notes payable
within a year. Also included are accrued expenses (wages, salaries, interest, rent,
taxes, etc., owed, but not yet due for payment), and advance payments and deposits
from customers.
• Other liabilities include long-term liabilities, such as bonds, mortgages, and longterm notes, that are due and payable more than one year in the future.
• Stockholders’ equity represents the amount that is available to the owners after all
other debts have been paid. Generally, stockholders’ equity consists of preferred and
common stock, treasury stock, capital surplus, and retained earnings. Preferred stock
is a hybrid between common stock and debt. In case the company goes bankrupt, it
must pay its preferred stockholders after its debtors, but before its common stockholders. Preferred dividend is fixed, so preferred stockholders do not benefit if the
company’s earnings grow. In fact, many firms do not use preferred stock. The common
stockholders’ equity, or net worth, is a residual:
Assets - Liabilities - Preferred stock = Common stockholders’ equity
$11,471 - $6,163 - $0 = $5,308.
• Common stock is the aggregate par value of the company’s stock issued. Companies
rarely issue stocks at a discount (i.e., at an amount below the stated par). Normally,
corporations set the par value low enough so that, in practice, stock is usually sold at a
premium.
• Paid-in capital (capital surplus) is the amount of money received from the sale of
stock that is over and above the par value of the stock. Outstanding stock is the number of shares issued that actually are held by the public. If the corporation buys back
part of its own issued stock, that stock is listed as treasury stock on the balance sheet.
• Retained earnings represent the cumulative net income of the firm since its inception, less the total dividends that have been paid to stockholders. In other words, retained
earnings indicate the amount of assets that have been financed by plowing profits back
into the business. Therefore, retained earnings belong to the stockholders.
2.2.2 The Income Statement
The second financial report is the income statement, which indicates whether the company is making or losing money during a stated period, usually a year. Most businesses
prepare quarterly and monthly income statements as well. The company’s accounting period refers to the period covered by an income statement.
Basic Income Statement Equation
Revenue
–
Expenses
Net Income (Loss)
For Dell, the accounting period begins on February 1 and ends on January 31 of the
following year. Table 2.2 gives the 2005 and 2004 income statements for Dell.
Current
liabilities are
bills that are due
to creditors and
suppliers within
a short period of
time.
Treasury stock
is not taken into
consideration
when calculating
earnings per
share or
dividends.
Retained
earnings refers
to earnings not
paid out as
dividends but
instead are
reinvested in the
core business or
used to pay off
debt.
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28 CHAPTER 2 Understanding Financial Statements
TABLE 2.2
Consolidated Statements of Income (in millions, except per share
amounts) Dell, Inc.
Fiscal Year Ended
January 28,
2005
Net revenue
January 30,
2004
January 31,
2003
$ 49,205
$ 41,444
$ 35,404
Cost of revenue
40,190
33,892
29,055
Gross margin
9,015
7,552
6,349
4,298
3,544
3,050
463
464
455
Total operating expenses
4,761
4,008
3,505
Operating income
4,254
3,544
2,844
Operating expenses:
Selling, general, and administrative
Research, development, and engineering
Investment and other income, net
Income before income taxes
Income tax provision
Net income
191
180
183
4,445
3,724
3,027
1,402
1,079
905
$
3,043
$
2,645
$
2,122
Basic
$
1.21
$
1.03
$
0.82
Diluted
$
1.18
$
1.01
$
0.80
Earnings per common share:
Weighted average shares outstanding:
Basic
2,509
2,565
2,584
Diluted
2,568
2,619
2,644
Source: Annual Report, Dell Corporation, 2005.
Gross margin
represents the
amount of
money the
company
generated over
the cost of
producing its
goods or
services.
Reporting Format
Typical items that are itemized in the income statement are as follows:
• Revenue is the income from goods sold and services rendered during a given accounting period.
• Net revenue represents gross sales, less any sales return and allowances.
• Shown on the next several lines are the expenses and costs of doing business, as deductions from revenue. The largest expense for a typical manufacturing firm is the
expense it incurs in making a product (such as labor, materials, and overhead), called
the cost of revenue (or cost of goods sold).
• Net revenue less the cost of revenue gives the gross margin.
• Next, we subtract any other operating expenses from the operating income. These other
operating expenses are expenses associated with paying interest, leasing machinery or
equipment, selling, and administration. This results in the operating income.
• Finally, we determine the net income (or net profit) by subtracting the income taxes
from the taxable income. Net income is also commonly known as accounting income.
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Section 2.2 Financial Status for Businesses 29
Earnings per Share
Another important piece of financial information provided in the income statement is the
earnings per share (EPS).3 In simple situations, we compute the EPS by dividing the
available earnings to common stockholders by the number of shares of common stock outstanding. Stockholders and potential investors want to know what their share of profits is,
not just the total dollar amount. The presentation of profits on a per share basis allows the
stockholders to relate earnings to what they paid for a share of stock. Naturally, companies
want to report a higher EPS to their investors as a means of summarizing how well they
managed their businesses for the benefits of the owners. Interestingly, Dell earned $1.21
per share in 2005, up from $1.03 in 2004, but it paid no dividends.
Retained Earnings
As a supplement to the income statement, many corporations also report their retained
earnings during the accounting period. When a corporation makes some profits, it has to
decide what to do with those profits. The corporation may decide to pay out some of the
profits as dividends to its stockholders. Alternatively, it may retain the remaining profits
in the business in order to finance expansion or support other business activities.
When the corporation declares dividends, preferred stock has priority over common
stock. Preferred stock pays a stated dividend, much like the interest payment on bonds.
The dividend is not a legal liability until the board of directors has declared it. However,
many corporations view the dividend payments to preferred stockholders as a liability.
Therefore, “available for common stockholders” reflects the net earnings of the corporation, less the preferred stock dividends. When preferred and common stock dividends
are subtracted from net income, the remainder is retained earnings (profits) for the year.
As mentioned previously, these retained earnings are reinvested into the business.
EXAMPLE 2.1 Understanding Dell’s Balance Sheet
and Income Statement
With revenue of $49,205 million for fiscal year 2005, Dell is the world’s leading direct computer systems company. Tables 2.2 and 2.3 show how Dell generated its net
income during the fiscal year.
The Balance Sheet. Dell’s $23,215 million of total assets shown in Table 2.1 were
necessary to support its sales of $49,205 million.
• Dell obtained the bulk of the funds it used to buy assets
1. By buying on credit from its suppliers (accounts payable).
2. By borrowing from financial institutions (notes payable and long-term bonds).
3. By issuing common stock to investors.
4. By plowing earnings into the business, as reflected in the retained earnings
account.
3
In reporting EPS, the firm is required to distinguish between “basic EPS” and “diluted EPS.” Basic EPS is
the net income of the firm, divided by the number of shares of common stock outstanding. By contrast, the diluted EPS includes all common stock equivalents (convertible bonds, preferred stock, warrants, and rights),
along with common stock. Therefore, diluted EPS will usually be less than basic EPS.
EPS is generally
considered to be
the single
most important
variable in
determining a
share’s price.
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30 CHAPTER 2 Understanding Financial Statements
• The net increase in fixed assets was $174 million ( $1,691 - $1,517; Table 2.1).
However, this amount is after a deduction for the year’s depreciation expenses.
We should add depreciation expense back to show the increase in gross fixed assets. From the company’s cash flow statement in Table 2.3, we see that the 2005
depreciation expense is $334 million; thus, the acquisition of fixed assets equals
$508 million.
• Dell had a total long-term debt of $505 million (Table 2.1), consisting of several
bonds issued in previous years. The interest Dell paid on these long-term debts was
about $16 million.
• Dell had 2,509 million shares of common stock outstanding. Investors actually
provided the company with a total capital of $8,195 million (Table 2.1). However,
Dell has retained the current as well as previous earnings of $9,174 million since it
was incorporated. Dell also held $10,758 million worth of treasury stock, which
was financed through the current as well as previous retained earnings. The combined net stockholders’ equity was $6,485 million, and these earnings belong to
Dell’s common stockholders (Table 2.1).
• On the average, stockholders have a total investment of $2.58 per share ($6,485
million/2,509 million shares) in the company. The $2.58 figure is known as the
stock’s book value. In the fall of 2005, the stock traded in the general range from
$32 to $40 per share. Note that this market price is quite different from the stock’s
book value. Many factors affect the market price, the most important one being
how investors expect the company to perform in the future. Certainly, the company’s direct made-to-order business practices have had a major influence on the
market value of its stock.
The Income Statement. Dell’s net revenue was $49,205 million in 2005, compared
with $41,444 million in 2004, a gain of 18.73% (Table 2.2). Profits from operations
(operating income) rose 20.03% to $4,254 million, and net income was up 15.05% to
$3,043 million.
• Dell issued no preferred stock, so there is no required cash dividend. Therefore, the
entire net income of $3,043 million belongs to the common stockholders.
• Earnings per common share climbed at a faster pace than in 2004, to $1.21, an increase of 17.48% (Table 2.2). Dell could retain this income fully for reinvestment
in the firm, or it could pay it out as dividends to the common stockholders. Instead
of either of these alternatives, Dell repurchased and retired 56 million common
stocks for $1,012 million. We can see that Dell had earnings available to common
stockholders of $3,043 million. As shown in Table 2.1, the beginning balance of
the retained earnings was $6,131 million. Therefore, the total retained earnings
grew to $9,174 million.
2.2.3 The Cash Flow Statement
The income statement explained in the previous section indicates only whether the
company was making or losing money during the reporting period. Therefore, the emphasis was on determining the net income (profits) of the firm for supporting its operating
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Section 2.2 Financial Status for Businesses 31
activities. However, the income statement ignores two other important business activities
for the period: financing and investing activities. Therefore, we need another financial
statement—the cash flow statement, which details how the company generated the
cash it received and how the company used that cash during the reporting period.
Sources and Uses of Cash
The difference between the sources (inflows) and uses (outflows) of cash represents the
net cash flow during the reporting period. This is a very important piece of information,
because investors determine the value of an asset (or, indeed, of a whole firm) by the cash
flows it generates. Certainly, a firm’s net income is important, but cash flows are even
more important, particularly because the company needs cash to pay dividends and to
purchase the assets required to continue its operations. As mentioned in the previous section, the goal of the firm should be to maximize the price of its stock. Since the value of
any asset depends on the cash flows produced by the asset, managers want to maximize
the cash flows available to investors over the long run. Therefore, we should make investment decisions on the basis of cash flows rather than profits. For such investment decisions, it is necessary to convert profits (as determined in the income statement) to cash
flows. Table 2.3 is Dell’s statement of cash flows, as it would appear in the company’s
annual report.
Reporting Format
In preparing the cash flow statement such as that in Table 2.3, many companies identify
the sources and uses of cash according to the types of business activities. There are three
types of activities:
• Operating activities. We start with the net change in operating cash flows from the
income statement. Here, operating cash flows represent those cash flows related to
production and the sales of goods or services. All noncash expenses are simply added
back to net income (or after-tax profits). For example, an expense such as depreciation is only an accounting expense (a bookkeeping entry). Although we may charge
depreciation against current income as an expense, it does not involve an actual cash
outflow. The actual cash flow may have occurred when the asset was purchased.
(Any adjustments in working capital4 will also be listed here.)
• Investing activities. Once we determine the operating cash flows, we consider any
cash flow transactions related to investment activities, which include purchasing new
fixed assets (cash outflow), reselling old equipment (cash inflow), and buying and
selling financial assets.
• Financing activities. Finally, we detail cash transactions related to financing
any capital used in business. For example, the company could borrow or sell
more stock, resulting in cash inflows. Paying off existing debt will result in cash
outflows.
By summarizing cash inflows and outflows from three activities for a given accounting period, we obtain the net change in the cash flow position of the company.
4
The difference between the increase in current assets and the spontaneous increase in current liabilities is the
net change in net working capital. If this change is positive, then additional financing, over and above the
cost of the fixed assets, is needed to fund the increase in current assets. This will further reduce the cash flow
from the operating activities.
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32 CHAPTER 2 Understanding Financial Statements
TABLE 2.3
Consolidated Statements of Cash Flows (in millions) Dell, Inc.
Fiscal Year Ended
January 28, January 30, January 31,
2005
2004
2003
Cash flows from operating activities:
Net income
$ 3,043
$
2,645
$ 2,122
Adjustments to reconcile net income to net cash
provided by operating activities:
Depreciation and amortization
334
263
211
Tax benefits of employee stock plans
249
181
260
Effects of exchange rate changes on monetary
assets and liabilities denominated in foreign
currencies
(602)
(677)
(537)
78
113
60
1,755
872
1,210
Other
Changes in:
Operating working capital
Noncurrent assets and liabilities
Net cash provided by operating activities
453
273
212
5,310
3,670
3,538
(12,261)
(12,099)
(8,736)
10,469
10,078
7,660
Cash flows from investing activities:
Investments:
Purchases
Maturities and sales
Capital expenditures
(525)
(329)
(305)
Purchase of assets held in master lease facilities
—
(636)
—
Cash assumed in consolidation of Dell Financial
Services, L.P.
—
172
—
Net cash used in investing activities
(2,317)
(2,814)
(1,381)
(4,219)
(2,000)
(2,290)
617
265
Cash flows from financing activities:
Repurchase of common stock
Issuance of common stock under employee plans
and other
1,091
Net cash used in financing activities
(3,128)
Effect of exchange rate changes on cash and cash
equivalents
Net increase in cash and cash equivalents
Cash and cash equivalents at beginning of period
Cash and cash equivalents at end of period
Source: Annual Report, Dell Corporation, 2005.
565
(1,383)
612
(2,025)
459
430
85
591
4,317
4,232
3,641
$ 4,747
$ 4,317
$ 4,232
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Section 2.3 Using Ratios to Make Business Decisions 33
EXAMPLE 2.2 Understanding Dell’s Cash Flow Statement
As shown in Table 2.3, Dell’s cash flow from operations amounted to $5,310 million.
Note that this is significantly more than the $3,043 million earned during the reporting period. Where did the extra money come from?
• The main reason for the difference lies in the accrual-basis accounting principle
used by the Dell Corporation. In accrual-basis accounting, an accountant recognizes the impact of a business event as it occurs. When the business performs a
service, makes a sale, or incurs an expense, the accountant enters the transaction
into the books, regardless of whether cash has or has not been received or paid. For
example, an increase in accounts receivable of $4,414 million - $3,635 million =
$779 million during 2005 represents the amount of total sales on credit (Table 2.1).
Since the $779 million figure was included in the total sales in determining the net
income, we need to subtract it to determine the company’s true cash position. After
adjustments, the net cash provided from operating activities is $5,310 million.
• As regards investment activities, there was an investment outflow of $525 million in
new plant and equipment. Dell sold $10,469 million worth of stocks and bonds during the period, and reinvested $12,261 million in various financial securities. The net
cash flow provided from these investing activities amounted to - $2,317 million,
which means an outflow of money.
• Financing activities produced a net outflow of $4,219 million, including the repurchase of the company’s own stocks. (Repurchasing its own stock is equivalent to
investing the firm’s idle cash from operation in the stock market. Dell could have
bought another company’s stock, such as IBM or Microsoft stock, with the money,
but Dell liked its own stock better than any other stocks on the market.) Dell also
raised $1,091 million by issuing shares of common stock. The net cash used in financing activities amounted to $3,128 million (outflow).
• Finally, there was the effect of exchange rate changes on cash for foreign sales.
This amounted to a net increase of $565 million. Together, the three types of activities generated a total cash flow of $430 million. With the initial cash balance of
$4,317 million, the ending cash balance thus increased to $4,747 million. This
same amount denotes the change in Dell’s cash position, as shown in the cash accounts in the balance sheet.
2.3 Using Ratios to Make Business Decisions
As we have seen in Dell’s financial statements, the purpose of accounting is to provide information for decision making. Accounting information tells what happened at a particular point in time. In that sense, financial statements are essentially historical documents.
However, most users of financial statements are concerned about what will happen in the
future. For example,
• Stockholders are concerned with future earnings and dividends.
• Creditors are concerned with the company’s ability to repay its debts.
• Managers are concerned with the company’s ability to finance future expansion.
• Engineers are concerned with planning actions that will influence the future course
of business events.
Accrual-basis
accounting
measures the
performance
and position of
a company by
recognizing
economic events
regardless of
when cash
transactions
occur.
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34 CHAPTER 2 Understanding Financial Statements
Although financial statements are historical documents, they can still provide valuable information bearing on all of these concerns. An important part of financial analysis
is the calculation and interpretation of various financial ratios. In this section, we consider some of the ratios that analysts typically use in attempting to predict the future
course of events in business organizations. We may group these ratios into five categories
(debt management, liquidity, asset management, profitability, and market trend) as outlined in Figure 2.5. In all financial ratio calculations, we will use the 2005 financial statements for Dell Computer Corporation, as summarized in Table 2.4.
2.3.1 Debt Management Analysis
All businesses need assets to operate. To acquire assets, the firm must raise capital.
When the firm finances its long-term needs externally, it may obtain funds from the
capital markets. Capital comes in two forms: debt and equity. Debt capital is capital
borrowed from financial institutions. Equity capital is capital obtained from the owners
of the company.
The basic methods of financing a company’s debt are through bank loans and the sale
of bonds. For example, suppose a firm needs $10,000 to purchase a computer. In this situation, the firm would borrow money from a bank and repay the loan, together with the
interest specified, in a few years. This kind of financing is known as short-term debt
Debt Management
Market Trend
Ratios that show how a
firm uses debt financing
and its ability to meet
debt repayment
obligations.
A set of ratios that
relate the firm’s stock
price to its earnings and
book value per share.
Debt ratio
Financial
Ratios
Profitability
Liquidity
Ratios that show the
relationship of a firm’s
cash and other assets
to its current liabilities.
Current ratio
Quick ratio
Asset Management
A set of ratios which
measure how effectively
a firm is managing its
assets.
Inventory
turnover ratio
Day’s sales
outstanding ratio
Total assets
turnover ratio
Figure 2.5
P/E ratio
Market/book ratio
Times-interestearned ratio
A set of ratios which show
the combined effects of
liquidity, asset management, and debt on
operating results.
Profit margin on
sales
Return on total
assets
Return on common
equity
Types of ratios used in evaluating a firm’s financial health.
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Section 2.3 Using Ratios to Make Business Decisions 35
TABLE 2.4
Summary of Dell’s Key Financial Statements
Balance Sheet
January 28, January 30,
2005
2004
Cash and cash equivalent
4,747
4,317
Accounts receivables, net
4,414
3,635
459
327
Total current assets
16,897
10,633
Total assets
23,215
19,311
Total current liabilities
14,136
10,896
Inventories
Long-term debt
505
505
Total liabilities
16,730
13,031
Common stock
8,195
6,823
Retained earnings
9,174
6,131
Total stockholders’ equity
6,485
6,280
49,205
41,444
Gross income (margin)
9,015
7,522
Operating income (margin)
4,254
3,544
Net income (margin)
3,043
2,645
Beginning retained earnings
6,131
3,486
Net income
3,043
2,645
Ending retained earnings
9,174
6,131
Income Statement
Net revenue
Statements of Retained Earnings
Statement of Cash Flows
Net cash from operating activities
5,310
3,670
Net cash used in investing activities
(2,317)
(2,814)
Net cash used in financing activities
(3,128)
(1,383)
565
612
Beginning cash position
4,317
4,232
Ending cash position
4,747
4,317
Effect of exchange rate changes
financing. Now suppose that the firm needs $100 million for a construction project. Normally, it would be very expensive (or require a substantial mortgage) to borrow the
money directly from a bank. In this situation, the company would go public to borrow
money on a long-term basis. When investors lend capital to a company and the company
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36 CHAPTER 2 Understanding Financial Statements
consents to repay the loan at an agreed-upon interest rate, the investor is the creditor of
the corporation. The document that records the nature of the arrangement between the
issuing company and the investor is called a bond. Raising capital by issuing a bond is
called long-term debt financing.
Similarly, there are different types of equity capital. For example, the equity of a
proprietorship represents the money provided by the owner. For a corporation, equity
capital comes in two forms: preferred stock and common stock. Investors provide capital
to a corporation and the company agrees to endow the investor with fractional ownership in the corporation. Preferred stock pays a stated dividend, much like the interest
payment on bonds. However, the dividend is not a legal liability until the company declares it. Preferred stockholders have preference over common stockholders as regards
the receipt of dividends if the company has to liquidate its assets. We can examine the
extent to which a company uses debt financing (or financial leverage) in the operation of
its business if we
• Check the balance sheet to determine the extent to which borrowed funds have been
used to finance assets, and
• Review the income statement to see the extent to which fixed charges (interests) are
covered by operating profits.
Two essential indicators of a business’s ability to pay its long-term liabilities are the
debt ratio and the times-interest-earned ratio.
Debt ratio: A
ratio that indicates what
proportion of
debt a company
has relative to
its assets.
Debt Ratio
The relationship between total liabilities and total assets, generally called the debt ratio,
tells us the proportion of the company’s assets that it has financed with debt:
Debt ratio =
=
Total debt
Total assets
$16,730
= 72.07%.
$23,215
Total debt includes both current liabilities and long-term debt. If the debt ratio is unity,
then the company has used debt to finance all of its assets. As of January 28, 2005,
Dell’s debt ratio was 72.07%; this means that its creditors have supplied close to 72% of
the firm’s total financing. Certainly, most creditors prefer low debt ratios, because the
lower the ratio, the greater is the cushion against creditors’ losses in case of liquidation.
If a company seeking financing already has large liabilities, then additional debt payments may be too much for the business to handle. For such a highly leveraged company,
creditors generally charge higher interest rates on new borrowing to help protect
themselves.
Times-Interest-Earned Ratio
The most common measure of the ability of a company’s operations to provide protection to the long-term creditor is the times-interest-earned ratio. We find this ratio by
dividing earnings before interest and income taxes (EBIT) by the yearly interest charges
that must be met. Dell issued $500 million worth of senior notes and long-term bonds
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Section 2.3 Using Ratios to Make Business Decisions 37
with a combined interest rate of 2.259%. This results in $11.29 million in interest expenses5 in 2005:
Times-interest-earned ratio =
=
EBIT
Interest expense
$4,445 + $11.29
= 394.72 times.
$11.29
The times-interest-earned ratio measures the extent to which operating income can decline before the firm is unable to meet its annual interest costs. Failure to meet this obligation can bring legal action by the firm’s creditors, possibly resulting in the company’s
bankruptcy. Note that we use the earnings before interest and income taxes, rather than
net income, in the numerator. Because Dell must pay interest with pretax dollars, Dell’s
ability to pay current interest is not affected by income taxes. Only those earnings remaining after all interest charges are subject to income taxes. For Dell, the times-interestearned ratio for 2005 would be 395 times. This ratio is exceptionally high compared with
the rest of the industry’s 65.5 times during the same operating period.
2.3.2 Liquidity Analysis
If you were one of the many suppliers to Dell, your primary concern would be whether Dell
will be able to pay off its debts as they come due over the next year or so. Short-term creditors want to be repaid on time. Therefore, they focus on Dell’s cash flows and on its working capital, as these are the company’s primary sources of cash in the near future. The
excess of current assets over current liabilities is known as working capital, a figure that
indicates the extent to which current assets can be converted to cash to meet current obligations. Therefore, we view a firm’s net working capital as a measure of its liquidity position.
In general, the larger the working capital, the better able the business is to pay its debt.
Current Ratio
We calculate the current ratio by dividing current assets by current liabilities:
Current ratio =
=
Current assets
Current liabilities
$16,897
= 1.1953 times.
$14,136
If a company is getting into financial difficulty, it begins paying its bills (accounts
payable) more slowly, borrowing from its bank, and so on. If current liabilities are rising
faster than current assets, the current ratio will fall, and that could spell trouble. What is
an acceptable current ratio? The answer depends on the nature of the industry. The general rule of thumb calls for a current ratio of 2 to 1. This rule, of course, is subject to many
exceptions, depending heavily on the composition of the assets involved.
5
Unless the interest expenses are itemized in the income statement, you will find them in the firm’s annual report.
The current
ratio measures
a company’s
ability to pay its
short-term
obligations.
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38 CHAPTER 2 Understanding Financial Statements
Quick (Acid-Test) Ratio
The quick ratio tells us whether a company could pay all of its current liabilities if they
came due immediately. We calculate the quick ratio by deducting inventories from current assets and then dividing the remainder by current liabilities:
Quick ratio =
=
Current assets - Inventories
Current liabilities
$16,897 - $459
= 1.1628 times.
$14,136
The quick ratio measures how well a company can meet its obligations without having to
liquidate or depend too heavily on its inventory. Inventories are typically the least liquid
of a firm’s current assets; hence, they are the assets on which losses are most likely to
occur in case of liquidation. Although Dell’s current ratio may appear to be below the average for its industry, 1.4, its liquidity position is relatively strong, as it has carried very
little inventory in its current assets (only $459 million out of $16,897 million of current
assets, or 2.7%). We often compare against industry average figures and should note at
this point that an industry average is not an absolute number that all firms should strive to
maintain. In fact, some very well managed firms will be above the average, while other
good firms will be below it. However, if we find that a firm’s ratios are quite different
from the average for its industry, we should examine the reason for the difference.
2.3.3 Asset Management Analysis
The ability to sell inventory and collect accounts receivables is fundamental to business
success. Therefore, the third group of ratios measures how effectively the firm is managing
its assets. We will review three ratios related to a firm’s asset management: (1) the inventory turnover ratio, (2) the day’s sales outstanding ratio, and (3) the total asset turnover
ratio. The purpose of these ratios is to answer this question: Does the total amount of each
type of asset, as reported on the balance sheet, seem reasonable in view of current and projected sales levels? The acquisition of any asset requires the use of funds. On the one hand,
if a firm has too many assets, its cost of capital will be too high; hence, its profits will be
depressed. On the other hand, if assets are too low, the firm is likely lose profitable sales.
Inventory
turnover: A
ratio that shows
how many times
the inventory of
a firm is sold and
replaced over a
specific period.
Inventory Turnover
The inventory turnover ratio measures how many times the company sold and replaced its
inventory over a specific period—for example, during the year. We compute the ratio by
dividing sales by the average level of inventories on hand. We compute the average inventory figure by taking the average of the beginning and ending inventory figures.
Since Dell has a beginning inventory figure of $327 million and an ending inventory
figure of $459 million, its average inventory for the year would be $393 million, or
1$327 + $4592/2. Then we compute Dell’s inventory turnover for 2005 as follows:
Inventory turnover ratio =
=
Sales
Average inventory balance
$49,205
= 125.20 times.
$393
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Section 2.3 Using Ratios to Make Business Decisions 39
As a rough approximation, Dell was able to sell and restock its inventory 125.20 times
per year. Dell’s turnover of 125.20 times is much faster than its competitor HPQ (Hewlett
Packard), 9.5 times. This suggests that HPQ is holding excessive stocks of inventory; excess stocks are, of course, unproductive, and they represent an investment with a low or
zero rate of return.
Day’s Sales Outstanding (Accounts Receivable Turnover)
The day’s sales outstanding (DSO) is a rough measure of how many times a company’s
accounts receivable have been turned into cash during the year. We determine this ratio,
also called the average collection period, by dividing accounts receivable by average
sales per day. In other words, the DSO indicates the average length of time the firm must
wait after making a sale before receiving cash. For Dell,
DSO =
=
Receivables
Receivables
=
Average sales per day
Annual sales/365
$4,414
$4,414
=
$49,205/365
$134.81
Average
collection
period is often
used to help
determine if a
company is
trying to
disguise weak
sales.
= 32.74 days.
Thus, on average, it takes Dell 32.74 days to collect on a credit sale. During the same period, HPQ’s average collection period was 43–52 days. Whether the average of 32.74
days taken to collect an account is good or bad depends on the credit terms Dell is offering its customers. If the credit terms are 30 days, we can say that Dell’s customers, on the
average, are not paying their bills on time. In order to improve their working-capital position, most customers tend to withhold payment for as long as the credit terms will allow
and may even go over a few days. The long collection period may signal either that customers are in financial trouble or that the company manages its credit poorly.
Total Assets Turnover
The total assets turnover ratio measures how effectively the firm uses its total assets in
generating its revenues. It is the ratio of sales to all the firm’s assets:
Total assets turnover ratio =
=
Sales
Total assets
$49,205
= 2.12 times.
$23,215
Dell’s ratio of 2.12 times, compared with HPQ’s 1.1, is almost 93% faster, indicating that
Dell is using its total assets about 93% more intensively than HPQ is. In fact, Dell’s total
investment in plant and equipment is about one-fourth of HPQ’s. If we view Dell’s ratio
as the industry average, we can say that HPQ has too much investment in inventory, plant,
and equipment compared to the size of sale.
2.3.4 Profitability Analysis
One of the most important goals for any business is to earn a profit. The ratios examined thus
far provide useful clues about the effectiveness of a firm’s operations, but the profitability
Asset turnover
is a measure of
how well assets
are being used
to produce
revenue.
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40 CHAPTER 2 Understanding Financial Statements
ratios show the combined effects of liquidity, asset management, and debt on operating
results. Therefore, ratios that measure profitability play a large role in decision making.
The profit
margin
measures how
much out of
every dollar of
sales a company
actually keeps
in earnings.
Profit Margin on Sales
We calculate the profit margin on sales by dividing net income by sales. This ratio indicates the profit per dollar of sales:
Profit margin on sales =
=
Net income available to common stockholders
Sales
$3,043
= 6.18%.
$49,205
Thus, Dell’s profit margin is equivalent to 6.18 cents for each sales dollar generated.
Dell’s profit margin is greater than HPQ’s profit margin of 3.6%, indicating that, although
HPQ’s sales are about 76% more than Dell’s revenue during the same operating period,
HPQ’s operation is less efficient than Dell’s. HPQ’s low profit margin is also a result of
its heavy use of debt and its carrying a very high volume of inventory. Recall that net income is income after taxes. Therefore, if two firms have identical operations in the sense
that their sales, operating costs, and earnings before income tax are the same, but if one
company uses more debt than the other, it will have higher interest charges. Those interest charges will pull net income down, and since sales are constant, the result will be a
relatively low profit margin.
Return on Total Assets
The return on total assets—or simply, return on assets (ROA)—measures a company’s
success in using its assets to earn a profit. The ratio of net income to total assets measures
the return on total assets after interest and taxes:
Return on total assets =
=
The return on
equity reveals
how much profit
a company
generates with
the money its
shareholders
have invested
in it.
Net income + interest expense11 - tax rate2
Average total assets
$3,043 + $11.2911 - 0.3152
= 14.35%.
1$23,215 + $19,3112/2
Adding interest expenses back to net income results in an adjusted earnings figure that
shows what earnings would have been if the assets had been acquired solely by selling
shares of stock. (Note that Dell’s effective tax rate was 31.5% in 2005.) With this adjustment, we may be able to compare the return on total assets for companies with differing
amounts of debt. Again, Dell’s 14.35% return on assets is well above the 4.1% for HPQ.
This high return results from (1) the company’s high basic earning power and (2) its low
use of debt, both of which cause its net income to be relatively high.
Return on Common Equity
Another popular measure of profitability is rate of return on common equity. This ratio
shows the relationship between net income and common stockholders’ investment in the
company—that is, how much income is earned for every $1 invested by the common
stockholders. To compute the return on common equity, we first subtract preferred dividends from net income, yielding the net income available to common stockholders. We
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Section 2.3 Using Ratios to Make Business Decisions 41
then divide this net income available to common stockholders by the average common
stockholders’ equity during the year. We compute average common equity by using the
beginning and ending balances. At the beginning of fiscal-year 2005, Dell’s common equity balance was $6,280 million; at the end of fiscal-year 2005, the balance was $6,485
million. The average balance is then simply $6,382.50 million, and we have
Return on common equity =
Net income available to common stockholders
Average common equity
=
$3,043
1$6,485 + $6,2802/2
=
$3,043
= 47.68%.
$6,382.50
The rate of return on common equity for Dell was 47.68% during 2005. Over the same
period, HPQ’s return on common equity amounted to 8.2%, a poor performance relative
to the computer industry (12.6% in 2005) in general.
To learn more about what management can do to increase the return on common equity, or ROE, we may rewrite the ROE in terms of the following three components:
ROE =
=
Net income
Stockholders’ equity
Net income
Sales
Assets
*
*
.
Sales
Assets
Stockholders’ equity
The three principal components can be described as the profit margin, asset turnover, and
financial leverage, respectively, so that
ROE = 1Profit margin2 * 1Asset turnover2 * 1Financial leverage2
= 16.18%2 * 12.122 * a
23,215
b
6,382.5
= 47.68%.
This expression tells us that management has only three key ratios for controlling a company’s ROE: (1) the earnings from sales (the profit margin); (2) the revenue generated
from each dollar of assets employed (asset turnover); and (3) the amount of equity used
to finance the assets in the operation of the business (financial leverage).
2.3.5 Market Value Analysis
When you purchase a company’s stock, what are your primary factors in valuing the
stock? In general, investors purchase stock to earn a return on their investment. This return consists of two parts: (1) gains (or losses) from selling the stock at a price that differs
from the investors’ purchase price and (2) dividends—the periodic distributions of profits
to stockholders. The market value ratios, such as the price-to-earnings ratio and the market-to-book ratio, relate the firm’s stock price to its earnings and book value per share, respectively. These ratios give management an indication of what investors think of the
Financial
leverage: The
degree to which
an investor or
business is
utilizing
borrowed
money.
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42 CHAPTER 2 Understanding Financial Statements
company’s past performance and future prospects. If the firm’s asset and debt management is sound and its profit is rising, then its market value ratios will be high, and its
stock price will probably be as high as can be expected.
Price-to-Earnings Ratio
The price-to-earnings (P/E) ratio shows how much investors are willing to pay per dollar
of reported profits. Dell’s stock sold for $41.50 in early February of 2005, so with an EPS
of $1.21, its P/E ratio was 34.29:
The P/E ratio
shows how
much investors
are willing to pay
per dollar of
earnings.
P/E ratio =
=
Price per share
Earnings per share
$41.5
= 34.29.
$1.21
That is, the stock was selling for about 34.29 times its current earnings per share. In general, P/E ratios are higher for firms with high growth prospects, other things held constant,
but they are lower for firms with lower expected earnings. Dell’s expected annual increase
in operating earnings is 30% over the next 3 to 5 years. Since Dell’s ratio is greater than
25%, the average for other computer industry firms, this suggests that investors value
Dell’s stock more highly than most as having excellent growth prospects. However, all
stocks with high P/E ratios carry high risk whenever the expected growths fail to materialize.
Any slight earnings disappointment tends to punish the market price significantly.
Book Value per Share
Another ratio frequently used in assessing the well-being of the common stockholders is
the book value per share, which measures the amount that would be distributed to holders
of each share of common stock if all assets were sold at their balance-sheet carrying
amounts and if all creditors were paid off. We compute the book value per share for Dell’s
common stock as follows:
Book value per share =
=
Total stockholders’ equity-preferred stock
Shares outstanding
$6,485 - $0
= $2.58.
2,509
If we compare this book value with the current market price of $41.50, then we may say
that the stock appears to be overpriced. Once again, though, market prices reflect expectations about future earnings and dividends, whereas book value largely reflects the results
of events that occurred in the past. Therefore, the market value of a stock tends to exceed
its book value. Table 2.5 summarizes the financial ratios for Dell Computer Corporation in
comparison to its direct competitor Hewlett Packard (HPQ) and the industry average.
2.3.6 Limitations of Financial Ratios in Business Decisions
Business decisions are made in a world of uncertainty. As useful as ratios are, they have
limitations. We can draw an analogy between their use in decision making and a physician’s use of a thermometer. A reading of 102°F indicates that something is wrong with the
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Summary 43
TABLE 2.5
Comparisons of Dell Computer Corporation’s Key Financial Ratios with
Those of Hewlett Packard (HPQ) and the Industry Average (2005)
Category
Financial Ratios
Dell
HPQ
Industry
Debt
Management
Debt ratio
Time-interest earned
72.07%
394.72
9%
18.37
29%
65.5
Liquidity
Current ratio
1.1953
1.4
1.4
1.16
125.20
0.90
9.5
1.0
11.1
Asset
Management
Quick ratio
Inventory turnover
Day’s sales
outstanding
32.74
52.43
34
Total asset turnover
2.12
1.1
1.0
Profit margin
6.18%
3.6%
5.0%
Return on total asset
14.35%
4.1%
4.9%
Return on common equity
47.68%
8.2%
12.6%
34.29
27.5
25.3
2.58
13.05
9.26
Profitability
Market Trend
P/E ratio
Book value-to-share
ratio
patient, but the temperature alone does not indicate what the problem is or how to cure it. In
other words, ratio analysis is useful, but analysts should be aware of ever-changing market
conditions and make adjustments as necessary. It is also difficult to generalize about
whether a particular ratio is “good” or “bad.” For example, a high current ratio may indicate a strong liquidity position, which is good, but holding too much cash in a bank account (which will increase the current ratio) may not be the best utilization of funds.
Ratio analysis based on any one year may not represent the true business condition. It is
important to analyze trends in various financial ratios, as well as their absolute levels, for
trends give clues as to whether the financial situation is likely to improve or deteriorate.
To do a trend analysis, one simply plots a ratio over time. As a typical engineering student, your judgment in interpreting a set of financial ratios is understandably weak at this
point, but it will improve as you encounter many facets of business decisions in the real
world. Again, accounting is a language of business, and as you speak it more often, it can
provide useful insights into a firm’s operations.
SUMMARY
The primary purposes of this chapter were (1) to describe the basic financial statements,
(2) to present some background information on cash flows and corporate profitability,
and (3) to discuss techniques used by investors and managers to analyze financial statements. Following are some concepts we covered:
Before making any major financial decisions, it is important to understand their
impact on your net worth. Your net-worth statement is a snapshot of where you stand
financially at a given point in time.
Trend analysis
is based on the
idea that what
has happened in
the past gives
traders an idea
of what will
happen in the
future.
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44 CHAPTER 2 Understanding Financial Statements
The three basic financial statements contained in the annual report are the balance
sheet, the income statement, and the statement of cash flows. Investors use the information provided in these statements to form expectations about future levels of earnings and dividends and about the firm’s risk-taking behavior.
A firm’s balance sheet shows a snapshot of a firm’s financial position at a particular
point in time through three categories: (1) assets the firm owns, (2) liabilities the firm
owes, and (3) owners’ equity, or assets less liabilities.
A firm’s income statement reports the results of operations over a period of time and
shows earnings per share as its “bottom line.” The main items are (1) revenues and
gains, (2) expenses and losses, and (3) net income or net loss (revenue less expenses).
A firm’s statement of cash flows reports the impact of operating, investing, and fi-
nancing activities on cash flows over an accounting period.
The purpose of calculating a set of financial ratios is twofold: (1) to examine the rela-
tive strengths and weaknesses of a company compared with those of other companies
in the same industry and (2) to learn whether the company’s position has been improving or deteriorating over time.
Liquidity ratios show the relationship of a firm’s current assets to its current liabilities
and thus its ability to meet maturing debts. Two commonly used liquidity ratios are the
current ratio and the quick (acid-test) ratio.
Asset management ratios measure how effectively a firm is managing its assets. Some of
the major ratios are inventory turnover, fixed assets turnover, and total assets turnover.
Debt management ratios reveal (1) the extent to which a firm is financed with debt and
(2) the firm’s likelihood of defaulting on its debt obligations. In this category are the
debt ratio and the times-interest-earned ratio.
Profitability ratios show the combined effects of liquidity, asset management, and debt
management policies on operating results. Profitability ratios include the profit margin
on sales, the return on total assets, and the return on common equity.
Market value ratios relate the firm’s stock price to its earnings and book value per share,
and they give management an indication of what investors think of the company’s past
performance and future prospects. Market value ratios include the price-to-earnings
ratio and the book value per share.
Trend analysis, in which one plots a ratio over time, is important, because it reveals
whether the firm’s ratios are improving or deteriorating over time.
PROBLEMS
Financial Statements
2.1 Consider the balance-sheet entries for War Eagle Corporation in Table P2.1.
(a) Compute the firm’s
Current assets: $_____
Current liabilities: $_____
Working capital: $_____
Shareholders’ equity: $_____
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Problems 45
TABLE P2.1
Balance Sheet Statement as of December 31, 2000
Assets:
Cash
$ 150,000
Marketable securities
200,000
Accounts receivables
150,000
Inventories
50,000
Prepaid taxes and insurance
30,000
Manufacturing plant
at cost
Less accumulated
depreciation
Net fixed assets
Goodwill
$ 600,000
100,000
500,000
20,000
Liabilities and shareholders’ equity:
Notes payable
Accounts payable
Income taxes payable
50,000
100,000
80,000
Long-term mortgage bonds
400,000
Preferred stock, 6%, $100 par value
(1,000 shares)
100,000
Common stock, $15 par value
(10,000 shares)
150,000
Capital surplus
150,000
Retained earnings
70,000
(b) If the firm had a net income of $500,000 after taxes, what is the earnings per
share?
(c) When the firm issued its common stock, what was the market price of the
stock per share?
2.2 A chemical processing firm is planning on adding a duplicate polyethylene plant at
another location. The financial information for the first project year is shown in
Table P2.2.
(a) Compute the working-capital requirement during the project period.
(b) What is the taxable income during the project period?
(c) What is the net income during the project period?
(d) Compute the net cash flow from the project during the first year.
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46 CHAPTER 2 Understanding Financial Statements
TABLE P2.2
Financial Information for First Project Year
Sales
$1,500,000
Manufacturing costs
Direct materials
$ 150,000
Direct labor
200,000
Overhead
100,000
Depreciation
200,000
Operating expenses
150,000
Equipment purchase
400,000
Borrowing to finance equipment
200,000
Increase in inventories
100,000
Decrease in accounts receivable
20,000
Increase in wages payable
30,000
Decrease in notes payable
40,000
Income taxes
272,000
Interest payment on financing
20,000
Financial Ratio Analysis
2.3 Table P2.3 shows financial statements for Nano Networks, Inc. The closing stock
price for Nano Network was $56.67 (split adjusted) on December 31, 2005. On
the basis of the financial data presented, compute the various financial ratios and
make an informed analysis of Nano’s financial health.
TABLE P2.3
Balance Sheet for Nano Networks, Inc.
Dec. 2005
U.S. $ (000)
(Year)
Dec. 2004
U.S. $ (000)
(Year)
Balance Sheet
Summary
Cash
158,043
20,098
Securities
285,116
0
Receivables
24,582
8,056
Allowances
632
0
0
0
377,833
28,834
20,588
10,569
Inventory
Current assets
Property and equipment, net
(Continued)
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Problems 47
Dec. 2005
U.S. $ (000)
(Year)
Depreciation
Dec. 2004
U.S. $ (000)
(Year)
8,172
2,867
513,378
36,671
55,663
14,402
Bonds
0
0
Preferred mandatory
0
0
Preferred stock
0
0
Common stock
2
1
Other stockholders’ equity
457,713
17,064
Total liabilities and equity
513,378
36,671
102,606
3,807
Cost of sales
45,272
4,416
Other expenses
71,954
31,661
Loss provision
0
0
Interest income
8,011
1,301
-6,609
-69
2,425
2
Income continuing
-9,034
-30,971
Net income
9,034
30,971
$0.1
$0.80
- $0.10
- $0.80
- $0.05
- $0.40
(split adjusted)
(split adjusted)
Total assets
Current liabilities
Income Statement
Summary
Total revenues
Income pretax
Income tax
EPS primary
EPS diluted
(a) Debt ratio
(b) Times-interest-earned ratio
(c) Current ratio
(d) Quick (acid-test) ratio
(e) Inventory turnover ratio
(f) Day’s sales outstanding
(g) Total assets turnover
(h) Profit margin on sales
(i) Return on total assets
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48 CHAPTER 2 Understanding Financial Statements
(j) Return on common equity
(k) Price-to-earnings ratio
(l) Book value per share
2.4 The balance sheet that follows summarizes the financial conditions for Flex, Inc.,
an electronic outsourcing contractor, for fiscal-year 2005. Unlike Nano Network
Corporation in Problem 2.3, Flex has reported a profit for several years running.
Compute the various financial ratios and interpret the firm’s financial health during fiscal-year 2005.
Aug. 2005
U.S. $ (000)
(12 mos.)
Aug. 2004
U.S. $ (000)
(Year)
Balance Sheet
Summary
Cash
Securities
Receivables
Allowances
Inventory
Current assets
Property and equipment, net
Depreciation
Total assets
Current liabilities
Bonds
Preferred mandatory
Preferred stock
Common stock
Other stockholders’ equity
Total liabilities and equity
1,325,637
362,769
1,123,901
5,580
1,080,083
3,994,084
1,186,885
533,311
4,834,696
1,113,186
922,653
0
0
271
2,792,820
4,834,696
225,228
83,576
674,193
-3,999
788,519
1,887,558
859,831
-411,792
2,410,568
840,834
385,519
0
0
117
1,181,209
2,410,568
Income Statement
Summary
Total revenues
Cost of sales
Other expenses
Loss provision
Interest expense
8,391,409
7,614,589
335,808
2,143
36,479
5,288,294
4,749,988
237,063
2,254
24,759
(Continued)
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Problems 49
Aug. 2005
U.S. $ (000)
(12 mos.)
Income pretax
Income tax
Income continuing
Discontinued
Extraordinary
Changes
Net income
EPS primary
EPS diluted
432,342
138,407
293,935
0
0
0
293,935
$1.19
$1.13
Aug. 2004
U.S. $ (000)
(Year)
298,983
100,159
198,159
0
0
0
198,159
$1.72
$1.65
(a) Debt ratio
(b) Times-interest-earned ratio
(c) Current ratio
(d) Quick (acid-test) ratio
(e) Inventory turnover ratio
(f) Day’s sales outstanding
(g) Total assets turnover
(h) Profit margin on sales
(i) Return on total assets
(j) Return on common equity
(k) Price-to-earnings ratio
(l) Book value per share
2.5 J. C. Olson & Co. had earnings per share of $8 in year 2006, and it paid a $4 dividend. Book value per share at year’s end was $80. During the same period, the
total retained earnings increased by $24 million. Olson has no preferred stock, and
no new common stock was issued during the year. If Olson’s year-end debt (which
equals its total liabilities) was $240 million, what was the company’s year-end
debt-to-asset ratio?
2.6 If Company A uses more debt than Company B and both companies have identical operations in terms of sales, operating costs, etc., which of the following statements is true?
(a) Company B will definitely have a higher current ratio.
(b) Company B has a higher profit margin on sales than Company A.
(c) Both companies have identical profit margins on sales.
(d) Company B’s return on total assets would be higher.
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50 CHAPTER 2 Understanding Financial Statements
2.7 You are looking to buy stock in a high-growth company. Which of the following
ratios best indicates the company’s growth potential?
(a) Debt ratio
(b) Price-to-earnings ratio
(c) Profit margin
(d) Total asset turnover
2.8 Which of the following statements is incorrect?
(a) The quickest way to determine whether the firm has too much debt is to calculate the debt-to-equity ratio.
(b) The best rule of thumb for determining the firm’s liquidity is to calculate the
current ratio.
(c) From an investor’s point of view, the rate of return on common equity is a
good indicator of whether or a firm is generating an acceptable return to the
investor.
(d) The operating margin is determined by expressing net income as a percentage
of total sales.
2.9 Consider the following financial data for Northgate Corporation:
• Cash and marketable securities, $100
• Total fixed assets, $280
• Annual sales, $1,200
• Net income, $358
• Inventory, $180
• Current liabilities, $134
• Current ratio, 3.2
• Average correction period, 45 days
• Average common equity, $500
On the basis of these financial data, determine the firm’s return on (common) equity.
(a) 141.60%
(b) 71.6%
(c) 76.0%
(d) 30%
Short Case Studies
ST2.1 Consider the two companies Cisco Systems and Lucent Technologies, which compete with each other in the network equipment sector. Lucent enjoys strong relationships among Baby Bells in the telephone equipment area and Cisco has a
dominant role in the network router and switching equipment area. Get these companies’ annual reports from their websites, and answer the following questions
(Note: To download their annual reports, visit http://www.cisco.com and http://
www.lucent.com) and look for “Investors’ Relations”:
(a) On the basis of the most recent financial statements, comment on each company’s financial performance in the following areas:
• Asset management
• Liquidity
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Short Case Studies 51
• Debt management
• Profitability
• Market trend
(b) Check the current stock prices for both companies. The stock ticker symbol is
CSCO for Cisco and LU for Lucent. Based on your analysis in Problem 2.6(a),
which company would you bet your money on and why? (Lucent and Alcatel
were engaged in discussions about a potential merger of equals in late March
of 2006.)
ST2.2 Compare XM Satellite Radio, Inc., and Sirius Satellite Radio, Inc., using a thorough financial ratios analysis.
(a) For each company, compute all the ratios listed in Figure 2.5 (i.e., debt management, liquidity, asset management, market trend, and profitability) for the
year 2005.
(b) Compare and contrast the two companies, using the ratios you calculated from
part (a).
(c) Carefully read and summarize the “risk management” or “hedging” practices
described in the financial statements of each company.
(d) If you were a mutual-fund manager and could invest in only one of these companies, which one would you select and why? Be sure to justify your answer
by using your results from parts (a), (b), and (c).
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THREE
CHAPTER
Interest Rate and
Economic Equivalence
No Lump Sum for Lottery-Winner Grandma, 941 A judge
denied a 94-year-old woman’s attempt to force the Massachusetts
Lottery Commission to pay her entire $5.6 million winnings up front
on the grounds that she otherwise won’t live long enough to collect it
all. The ruling means that the commission can pay Louise Outing, a retired waitress, in installments over 20 years. After an initial gross payment
of $283,770, Outing would be paid 19 annual gross checks of $280,000.
That’s about $197,000 after taxes. Lottery Executive Director Joseph
Sullivan said all players are held to the same rules, which are printed on
the back of Megabucks tickets. Lottery winners are allowed to “assign”
their winnings to a state-approved financial company that makes the full
payment—but only in return for a percentage of the total winnings.
Outing, who won a Megabucks drawing in September, has seven
grandchildren, nine great-grandchildren, and six great-great-grandchildren.
“I’d like to get it and do what I want with it,” she said.“I’m not going to
live 20 years. I’ll be 95 in March.”
1
52
“No Lump Sum for Lottery-Winner Grandma, 94,” The Associated Press, December 30, 2004.
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The next time you play a lottery, look at the top section of the play slip. You
will see two boxes: “Cash Value” and “Annual Payments.” You need to mark
one of the boxes before you fill out the rest of the slip. If you don’t, and you
win the jackpot, you will automatically receive the jackpot as annual payments. That is what happened to Ms. Louise Outing. If you mark the “Cash
Value box” and you win, you will receive the present cash value of the announced jackpot in one lump sum. This amount will be less than the announced jackpot. With the announced jackpot of $5.6 million, Ms. Outing
could receive about 52.008%, or $2.912 million, in one lump sum (less withholding tax). This example is based on average market costs as of January
2005 of 20 annual payments funded by the U.S.Treasury Zero Coupon
Bonds (or a 7.2% coupon rate). With this option, you can look forward to a
large cash payment up front.
First, most people familiar with investments would tell Ms. Outing
that receiving a lump amount of $2.912 million today is likely to prove a
far better deal than receiving $280,000 a year for 20 years, even if the
grandma lives long enough to collect the entire annual payments.
After losing the court appeal, Ms. Outing was able to find a buyer for her
lottery in a lump-sum amount of $2.467 million. To arrive at that price,
the buyer calculated the return he wanted to earn—at that time about
9.5% interest, compounded annually—and applied that rate in reverse to
the $5.6 million he stood to collect over 20 years. The buyer says the
deals he strikes with winners applies a basic tenet of all financial transactions, the time value of money: A dollar in hand today is worth more
than one that will be paid to you in the future.
In engineering economics, the principles discussed in this chapter are
regarded as the underpinning for nearly all project investment analysis.
This is because we always need to account for the effect of interest
operating on sums of cash over time. Interest formulas allow us to place
different cash flows received at different times in the same time frame
and to compare them. As will become apparent, almost our entire
study of engineering economics is built on the principles introduced in
this chapter.
53
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54 CHAPTER 3 Interest Rate and Economic Equivalence
CHAPTER LEARNING OBJECTIVES
After completing this chapter, you should understand the following concepts:
The time value of money.
The difference between simple interest and the compound interest.
The meaning of economic equivalence and why we need it in economic
analysis.
How to compare two different money series by means of the concept
of economic equivalence.
The interest operation and the types of interest formulas used to facilitate
the calculation of economic equivalence.
3.1 Interest:The Cost of Money
ost of us are familiar in a general way with the concept of interest. We know
that money left in a savings account earns interest, so that the balance over
time is greater than the sum of the deposits. We also know that borrowing to
buy a car means repaying an amount over time, that that amount includes interest, and that it is therefore greater than the amount borrowed. What may be unfamiliar to
us is the idea that, in the financial world, money itself is a commodity and, like other
goods that are bought and sold, money costs money.
The cost of money is established and measured by a market interest rate, a percentage that is periodically applied and added to an amount (or varying amounts) of
money over a specified length of time. When money is borrowed, the interest paid is
the charge to the borrower for the use of the lender’s property; when money is lent or
invested, the interest earned is the lender’s gain from providing a good to another
(Figure 3.1). Interest, then, may be defined as the cost of having money available for
use. In this section, we examine how interest operates in a free-market economy and we
establish a basis for understanding the more complex interest relationships that follow
later on in the chapter.
M
Market interest
rate: Interest
rate quoted
by financial
institutions.
Charge or Cost to Borrower
g
Figure 3.1 The meaning of interest rate to the lender (bank) and
to the borrower.
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Section 3.1 Interest: The Cost of Money
Account Value
Cost of Refrigerator
Case 1:
Inflation
exceeds
earning power
N = 0 $100
N = 1 $106
1earning rate = 6%2
N = 0 $100
N = 1 $108
1inflation rate = 8%2
Case 2:
Earning power
exceeds
inflation
N = 0 $100
N = 1 $106
1earning rate = 6%2
N = 0 $100
N = 1 $104
1inflation rate = 4%2
Figure 3.2
55
Gains achieved or losses incurred by delaying consumption.
3.1.1 The Time Value of Money
The “time value of money” seems like a sophisticated concept, yet it is a concept that
you grapple with every day. Should you buy something today or save your money and
buy it later? Here is a simple example of how your buying behavior can have varying
results: Pretend you have $100, and you want to buy a $100 refrigerator for your dorm
room. If you buy it now, you are broke. Suppose that you can invest money at 6% interest, but the price of the refrigerator increases only at an annual rate of 4% due to inflation.
In a year you can still buy the refrigerator, and you will have $2 left over. Well, if the
price of the refrigerator increases at an annual rate of 8% instead, you will not have
enough money (you will be $2 short) to buy the refrigerator a year from now. In that
case, you probably are better off buying the refrigerator now. The situation is summarized in Figure 3.2.
Clearly, the rate at which you earn interest should be higher than the inflation rate to
make any economic sense of the delayed purchase. In other words, in an inflationary
economy, your purchasing power will continue to decrease as you further delay the purchase
of the refrigerator. In order to make up this future loss in purchasing power, your earning
interest rate should be sufficiently larger than the anticipated inflation rate. After all,
time, like money, is a finite resource. There are only 24 hours in a day, so time has to be
budgeted, too. What this example illustrates is that we must connect the “earning power”
and the “purchasing power” to the concept of time.
When we deal with large amounts of money, long periods of time, or high interest
rates, the change in the value of a sum of money over time becomes extremely significant. For example, at a current annual interest rate of 10%, $1 million will earn $100,000
in interest in a year; thus, to wait a year to receive $1 million clearly involves a significant
sacrifice. When deciding among alternative proposals, we must take into account the operation of interest and the time value of money in order to make valid comparisons of
different amounts at various times.
The way interest operates reflects the fact that money has a time value. This is why
amounts of interest depend on lengths of time; interest rates, for example, are typically
given in terms of a percentage per year. We may define the principle of the time value of
money as follows: The economic value of a sum depends on when it is received. Because
money has both earning as well as purchasing power over time, as shown in Figure 3.3
(it can be put to work, earning more money for its owner), a dollar received today has a
greater value than a dollar received at some future time.
The time value
of money: The
idea that a dollar
today is worth
more than a
dollar in the
future because
the dollar received today can
earn interest.
Purchasing
power: The
value of a
currency
expressed in
terms of the
amount of goods
or services that
one unit of
money can buy.
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56 CHAPTER 3 Interest Rate and Economic Equivalence
Time
Loss
o
f Pur
chas
ing P
owe
r
e
Incr
ase
ar
in E
er
Pow
ning
Figure 3.3 The time value of money. This is a two-edged sword whereby
earning grows, but purchasing power decreases, as time goes by.
Actual dollars:
The cash flow
measured
in terms of
the dollars at
the time of the
transaction.
When lending or borrowing interest rates are quoted by financial institutions on the marketplace, those interest rates reflect the desired amounts to be earned, as well as any protection from loss in the future purchasing power of money because of inflation. (If we want to
know the true desired earnings in isolation from inflation, we can determine the real interest
rate. We consider this issue in Chapter 11. The earning power of money and its loss of value
because of inflation are calculated by different analytical techniques.) In the meantime, we
will assume that, unless otherwise mentioned, the interest rate used in this book reflects the
market interest rate, which takes into account the earning power, as well as the effect of inflation perceived in the marketplace. We will also assume that all cash flow transactions are
given in terms of actual dollars, with the effect of inflation, if any, reflected in the amount.
3.1.2 Elements of Transactions Involving Interest
Many types of transactions (e.g., borrowing or investing money or purchasing machinery on
credit) involve interest, but certain elements are common to all of these types of transactions:
• An initial amount of money in transactions involving debt or investments is called
the principal.
• The interest rate measures the cost or price of money and is expressed as a percentage per period of time.
• A period of time, called the interest period, determines how frequently interest is
calculated. (Note that even though the length of time of an interest period can vary,
interest rates are frequently quoted in terms of an annual percentage rate. We will
discuss this potentially confusing aspect of interest in Chapter 4.)
• A specified length of time marks the duration of the transaction and thereby establishes
a certain number of interest periods.
• A plan for receipts or disbursements yields a particular cash flow pattern over a
specified length of time. (For example, we might have a series of equal monthly
payments that repay a loan.)
• A future amount of money results from the cumulative effects of the interest rate
over a number of interest periods.
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Section 3.1 Interest:The Cost of Money
57
For the purposes of calculation, these elements are represented by the following variables:
An
i
N
P
A discrete payment or receipt occurring at the end of some interest period.
The interest rate per interest period.
The total number of interest periods.
A sum of money at a time chosen as time zero for purposes of analysis; sometimes
referred to as the present value or present worth.
F = A future sum of money at the end of the analysis period. This sum may be specified as FN.
A = An end-of-period payment or receipt in a uniform series that continues for N periods. This is a special situation where A 1 = A 2 = Á = A N.
Vn = An equivalent sum of money at the end of a specified period n that considers the
effect of the time value of money. Note that V0 = P and VN = F.
=
=
=
=
Because frequent use of these symbols will be made in this text, it is important that you become familiar with them. Note, for example, the distinction between A, A n, and A N. The
symbol A n refers to a specific payment or receipt, at the end of period n, in any series of payments. A N is the final payment in such a series, because N refers to the total number of interest periods. A refers to any series of cash flows in which all payments or receipts are equal.
Example of an Interest Transaction
As an example of how the elements we have just defined are used in a particular situation,
let us suppose that an electronics manufacturing company buys a machine for $25,000
and borrows $20,000 from a bank at a 9% annual interest rate. In addition, the company
pays a $200 loan origination fee when the loan commences. The bank offers two repayment
plans, one with equal payments made at the end of every year for the next five years, the
other with a single payment made after the loan period of five years. These two payment
plans are summarized in Table 3.1.
• In Plan 1, the principal amount P is $20,000, and the interest rate i is 9%. The interest
period is one year, and the duration of the transaction is five years, which means there
are five interest periods 1N = 52. It bears repeating that whereas one year is a common
interest period, interest is frequently calculated at other intervals: monthly, quarterly, or
TABLE 3.1
Repayment Plans for Example Given in Text (for N 5 years
and i 9%)
Payments
End of Year
Receipts
Plan 1
Year 0
$20,000.00
$ 200.00
Plan 2
$
200.00
Year 1
5,141.85
0
Year 2
5,141.85
0
Year 3
5,141.85
0
Year 4
5,141.85
0
Year 5
5,141.85
30,772.48
P = $20,000, A = $5,141.85, F = $30,772.48
Note: You actually borrow $19,800 with the origination fee of $200, but you pay back on the basis of $20,000.
Present value:
The amount that
a future sum
of money is
worth today,
given a specified
rate of return.
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58 CHAPTER 3 Interest Rate and Economic Equivalence
semiannually, for instance. For this reason, we used the term period rather than year
when we defined the preceding list of variables. The receipts and disbursements
planned over the duration of this transaction yield a cash flow pattern of five equal payments A of $5,141.85 each, paid at year’s end during years 1 through 5. (You’ll have to
accept these amounts on faith for now—the next section presents the formula used to
arrive at the amount of these equal payments, given the other elements of the problem.)
• Plan 2 has most of the elements of Plan 1, except that instead of five equal repayments,
we have a grace period followed by a single future repayment F of $30,772.78.
Cash Flow Diagrams
Problems involving the time value of money can be conveniently represented in graphic
form with a cash flow diagram (Figure 3.4). Cash flow diagrams represent time by a
horizontal line marked off with the number of interest periods specified. The cash flows
over time are represented by arrows at relevant periods: Upward arrows denote positive
flows (receipts), downward arrows negative flows (disbursements). Note, too, that the
arrows actually represent net cash flows: Two or more receipts or disbursements made
at the same time are summed and shown as a single arrow. For example, $20,000 received during the same period as a $200 payment would be recorded as an upward arrow
of $19,800. Also, the lengths of the arrows can suggest the relative values of particular
cash flows.
Cash flow diagrams function in a manner similar to free-body diagrams or circuit
diagrams, which most engineers frequently use: Cash flow diagrams give a convenient
summary of all the important elements of a problem, as well as a reference point to determine whether the statement of the problem has been converted into its appropriate
parameters. The text frequently uses this graphic tool, and you are strongly encouraged
to develop the habit of using well-labeled cash flow diagrams as a means to identify and
summarize pertinent information in a cash flow problem. Similarly, a table such as Table 3.1
can help you organize information in another summary format.
$19,800
$20,000
Cash flow at n 0
is a net cash flow
after summing
$20,000 and taking
away $200
0
i 9%
$200
1
2
3
4
5
$5,141.85
$5,141.85
$5,141.85
$5,141.85
$5,141.85
0
Figure 3.4 A cash flow diagram for Plan 1 of the loan
repayment example summarized in Table 3.1.
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Section 3.1 Interest:The Cost of Money
59
Interest Period
$10
$8
$5
$12
0
Beginning of
Interest period
0
$10
1
End of interest
period
$45
1
Figure 3.5 Any cash flows occurring during the interest period are summed to a single amount and placed
at the end of the interest period.
End-of-Period Convention
In practice, cash flows can occur at the beginning or in the middle of an interest period—
or indeed, at practically any point in time. One of the simplifying assumptions we make
in engineering economic analysis is the end-of-period convention, which is the practice
of placing all cash flow transactions at the end of an interest period. (See Figure 3.5.) This
assumption relieves us of the responsibility of dealing with the effects of interest within
an interest period, which would greatly complicate our calculations.
It is important to be aware of the fact that, like many of the simplifying assumptions and estimates we make in modeling engineering economic problems, the end-ofperiod convention inevitably leads to some discrepancies between our model and
real-world results.
Suppose, for example, that $100,000 is deposited during the first month of the year in
an account with an interest period of one year and an interest rate of 10% per year. In
such a case, the difference of 1 month would cause an interest income loss of $10,000.
This is because, under the end-of-period convention, the $100,000 deposit made during
the interest period is viewed as if the deposit were made at the end of the year, as opposed
to 11 months earlier. This example gives you a sense of why financial institutions choose
interest periods that are less than one year, even though they usually quote their rate as an
annual percentage.
Armed with an understanding of the basic elements involved in interest problems, we
can now begin to look at the details of calculating interest.
3.1.3 Methods of Calculating Interest
Money can be lent and repaid in many ways, and, equally, money can earn interest in
many different ways. Usually, however, at the end of each interest period, the interest
earned on the principal amount is calculated according to a specified interest rate. The
two computational schemes for calculating this earned interest are said to yield either
simple interest or compound interest. Engineering economic analysis uses the compound-interest scheme almost exclusively.
End-of-period
convention:
Unless otherwise
mentioned,
all cash flow
transactions
occur at the
end of an
interest period.
Simple interest:
The interest rate
is applied only
to the original
principal amount
in computing
the amount of
interest.
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60 CHAPTER 3 Interest Rate and Economic Equivalence
Simple Interest
Simple interest is interest earned on only the principal amount during each interest period.
In other words, with simple interest, the interest earned during each interest period does
not earn additional interest in the remaining periods, even though you do not withdraw it.
In general, for a deposit of P dollars at a simple interest rate of i for N periods, the
total earned interest would be
I = 1iP2N.
(3.1)
The total amount available at the end of N periods thus would be
F = P + I = P11 + iN2.
(3.2)
Simple interest is commonly used with add-on loans or bonds. (See Chapter 4.)
Compound:
The ability of
an asset to
generate earnings
that are then
reinvested and
generate their
own earnings.
Compound Interest
Under a compound-interest scheme, the interest earned in each period is calculated on the
basis of the total amount at the end of the previous period. This total amount includes the
original principal plus the accumulated interest that has been left in the account. In this
case, you are, in effect, increasing the deposit amount by the amount of interest earned. In
general, if you deposited (invested) P dollars at interest rate i, you would have
P + iP = P11 + i2 dollars at the end of one period. If the entire amount (principal and
interest) is reinvested at the same rate i for another period, at the end of the second period
you would have
P11 + i2 + i[P11 + i2] = P11 + i211 + i2
= P11 + i22.
Continuing, we see that the balance after the third period is
P11 + i22 + i[P11 + i22] = P11 + i23.
This interest-earning process repeats, and after N periods the total accumulated value
(balance) F will grow to
F = P11 + i2N.
(3.3)
EXAMPLE 3.1 Compound Interest
Suppose you deposit $1,000 in a bank savings account that pays interest at a rate of
10% compounded annually. Assume that you don’t withdraw the interest earned at
the end of each period (one year), but let it accumulate. How much would you have
at the end of year 3?
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Section 3.1 Interest:The Cost of Money
SOLUTION
Given: P = $1,000, N = 3 years, and i = 10% per year.
Find: F.
Applying Eq. (3.3) to our three-year, 10% case, we obtain
F = $1,00011 + 0.1023 = $1,331.
The total interest earned is $331, which is $31 more than was accumulated under the
simple-interest method (Figure 3.6). We can keep track of the interest accruing
process more precisely as follows:
Period
Amount at Beginning
of Interest Period
Interest Earned
for Period
Amount at End
of Interest Period
1
$1,000
$1,000(0.10)
$1,100
2
1,100
1,100(0.10)
1,210
3
1,210
1,210(0.10)
1,331
COMMENTS: At the end of the first year, you would have $1,000, plus $100 in interest, or
a total of $1,100. In effect, at the beginning of the second year, you would be depositing
$1,100, rather than $1,000. Thus, at the end of the second year, the interest earned would
be 0.101$1,1002 = $110, and the balance would be $1,100 + $110 = $1,210. This
is the amount you would be depositing at the beginning of the third year, and the interest earned for that period would be 0.101$1,2102 = $121. With a beginning principal amount of $1,210 plus the $121 interest, the total balance would be $1,331 at
the end of year 3.
$1,100
0
1
$1,210
$1,000
2
$1,331
$1,100
3
$1,210
Figure 3.6 The process of computing the balance when $1,000
at 10% is deposited for three years (Example 3.1).
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62 CHAPTER 3 Interest Rate and Economic Equivalence
3.1.4 Simple Interest versus Compound Interest
From Eq. (3.3), the total interest earned over N periods is
I = F - P = P[11 + i2N - 1].
(3.4)
Compared with the simple-interest scheme, the additional interest earned with compound
interest is
¢I = P[11 + i2N - 1] - 1iP2N
= P[11 + i2N - 11 + iN2].
(3.5)
(3.6)
As either i or N becomes large, the difference in interest earnings also becomes large, so the
effect of compounding is further pronounced. Note that, when N = 1, compound interest
is the same as simple interest.
Using Example 3.1, we can illustrate the difference between compound interest and
the simple interest. Under the simple-interest scheme, you earn interest only on the principal
amount at the end of each interest period. Under the compounding scheme, you earn interest
on the principal, as well as interest on interest.
Figure 3.7 illustrates the fact that compound interest is a sum of simple interests
earned on the original principal, as well as periodic simple interests earned on a series of
simple interests.
$1,331
F $1,000(1 0.10)3
$1,331
$1,000
$100
$100
$100
Simple interest earned
on $1,000 at n 0
$10
$10
Simple interest earned on $100
at n 1
$10
Simple interest earned
on $100 at n 2
$1
n0
Figure 3.7
n1
n2
Simple interest earned
on $10 at n 2
n3
The relationship between simple interest and compound interest.
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Section 3.2 Economic Equivalence
EXAMPLE 3.2 Comparing Simple with Compound Interest
In 1626, Peter Minuit of the Dutch West India Company paid $24 to purchase
Manhattan Island in New York from the Indians. In retrospect, if Minuit had invested the $24 in a savings account that earned 8% interest, how much would it be
worth in 2007?
SOLUTION
Given: P = $24, i = 8% per year, and N = 381 years.
Find: F, based on (a) 8% simple interest and (b) 8% compound interest.
(a) With 8% simple interest,
F = $24[1 + 10.08213812] = $755.52.
(b) With 8% compound interest,
F = $2411 + 0.082381 = $130,215,319,909,015.
COMMENTS: The significance of compound interest is obvious in this example. Many
of us can hardly comprehend the magnitude of $130 trillion. In 2007, the total population in the United States was estimated to be around 300 million. If the money were
distributed equally among the population, each individual would receive $434,051.
Certainly, there is no way of knowing exactly how much Manhattan Island is worth
today, but most real-estate experts would agree that the value of the island is nowhere
near $130 trillion. (Note that the U.S. national debt as of December 31, 2007, was estimated to be $9.19 trillion.)
3.2 Economic Equivalence
The observation that money has a time value leads us to an important question: If receiving
$100 today is not the same thing as receiving $100 at any future point, how do we measure
and compare various cash flows? How do we know, for example, whether we should prefer
to have $20,000 today and $50,000 ten years from now, or $8,000 each year for the next ten
years? In this section, we describe the basic analytical techniques for making these comparisons. Then, in Section 3.3, we will use these techniques to develop a series of formulas that
can greatly simplify our calculations.
3.2.1 Definition and Simple Calculations
The central question in deciding among alternative cash flows involves comparing their
economic worth. This would be a simple matter if, in the comparison, we did not need to
consider the time value of money: We could simply add the individual payments within a
cash flow, treating receipts as positive cash flows and payments (disbursements) as negative cash flows. The fact that money has a time value, however, makes our calculations
more complicated. We need to know more than just the size of a payment in order to
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64 CHAPTER 3 Interest Rate and Economic Equivalence
Economic
equivalence:
The process of
comparing two
different cash
amounts at
different points
in time.
determine its economic effect completely. In fact, as we will see in this section, we
need to know several things:
• The magnitude of the payment.
• The direction of the payment: Is it a receipt or a disbursement?
• The timing of the payment: When is it made?
• The interest rate in operation during the period under consideration.
It follows that, to assess the economic impact of a series of payments, we must consider
the impact of each payment individually.
Calculations for determining the economic effects of one or more cash flows are
based on the concept of economic equivalence. Economic equivalence exists between
cash flows that have the same economic effect and could therefore be traded for one another
in the financial marketplace, which we assume to exist.
Economic equivalence refers to the fact that a cash flow—whether a single payment or
a series of payments—can be converted to an equivalent cash flow at any point in time. For
example, we could find the equivalent future value F of a present amount P at interest rate i
at period n; or we could determine the equivalent present value P of N equal payments A.
The preceding strict concept of equivalence, which limits us to converting a cash
flow into another equivalent cash flow, may be extended to include the comparison of alternatives. For example, we could compare the value of two proposals by finding the
equivalent value of each at any common point in time. If financial proposals that appear
to be quite different turn out to have the same monetary value, then we can be
economically indifferent to choosing between them: In terms of economic effect, one
would be an even exchange for the other, so no reason exists to prefer one over the other
in terms of their economic value.
A way to see the concepts of equivalence and economic indifference at work in the
real world is to note the variety of payment plans offered by lending institutions for consumer loans. Table 3.2 extends the example we developed earlier to include three different
repayment plans for a loan of $20,000 for five years at 9% interest. You will notice, perhaps
to your surprise, that the three plans require significantly different repayment patterns and
TABLE 3.2
Typical Repayment Plans for a Bank Loan of $20,000 (for N = 5 years
and i = 9%)
Repayments
Plan 1
Plan 2
Plan 3
Year 1
$ 5,141.85
0
$ 1,800.00
Year 2
5,141.85
0
1,800.00
Year 3
5,141.85
0
1,800.00
Year 4
5,141.85
0
1,800.00
Year 5
5,141.85
$30,772.48
21,800.00
Total of payments
$25,709.25
$30,772.48
$29,000.00
Total interest paid
$ 5,709.25
$10,772.48
$ 9,000.00
Plan 1: Equal annual installments; Plan 2: End-of-loan-period repayment of principal and interest; Plan 3:
Annual repayment of interest and end-of-loan repayment of principal
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Section 3.2 Economic Equivalence
different total amounts of repayment. However, because money has a time value, these
plans are equivalent, and economically, the bank is indifferent to a consumer’s choice of
plan. We will now discuss how such equivalence relationships are established.
Equivalence Calculations: A Simple Example
Equivalence calculations can be viewed as an application of the compound-interest relationships we developed in Section 3.1. Suppose, for example, that we invest $1,000 at
12% annual interest for five years. The formula developed for calculating compound interest, F = P11 + i2N (Eq. 3.3), expresses the equivalence between some present
amount P and a future amount F, for a given interest rate i and a number of interest periods N. Therefore, at the end of the investment period, our sums grow to
$1,00011 + 0.1225 = $1,762.34.
Thus, we can say that at 12% interest, $1,000 received now is equivalent to $1,762.34 received in five years and that we could trade $1,000 now for the promise of receiving
$1,762.34 in five years. Example 3.3 further demonstrates the application of this basic
technique.
EXAMPLE 3.3 Equivalence
Suppose you are offered the alternative of receiving either $3,000 at the end of five
years or P dollars today. There is no question that the $3,000 will be paid in full (no
risk). Because you have no current need for the money, you would deposit the P dollars
in an account that pays 8% interest. What value of P would make you indifferent to
your choice between P dollars today and the promise of $3,000 at the end of five years?
STRATEGY: Our job is to determine the present amount that is economically equivalent
to $3,000 in five years, given the investment potential of 8% per year. Note that the
statement of the problem assumes that you would exercise the option of using the
earning power of your money by depositing it. The “indifference” ascribed to you
refers to economic indifference; that is, in a marketplace where 8% is the applicable
interest rate, you could trade one cash flow for the other.
SOLUTION
Given: F = $3,000, N = 5 years, and i = 8% per year.
Find: P.
Equation: Eq. (3.3), F = P11 + i2N.
Rearranging terms to solve for P gives
P =
Substituting yields
P =
F
.
11 + i2N
$3,000
= $2,042.
11 + 0.0825
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66 CHAPTER 3 Interest Rate and Economic Equivalence
$ 3, 0 0
0 (1
0.08
)
5
$2,042
F
P
0
$2,205
$2,382
1
2
$2,572
$2,778
$3,000
3
4
5
Years
Figure 3.8 Various dollar amounts that will be economically equivalent to $3,000
in five years, given an interest rate of 8% (Example 3.3).
We summarize the problem graphically in Figure 3.8.
COMMENTS: In this example, it is clear that if P is anything less than $2,042, you
would prefer the promise of $3,000 in five years to P dollars today; if P is greater
than $2,042, you would prefer P. As you may have already guessed, at a lower interest
rate, P must be higher to be equivalent to the future amount. For example, at
i = 4%, P = $2,466.
3.2.2 Equivalence Calculations: General Principles
In spite of their numerical simplicity, the examples we have developed reflect several
important general principles, which we will now explore.
Common base
period: To
establish an
economic equivalence between
two cash flow
amounts, a
common base
period must be
selected.
Principle 1: Equivalence Calculations Made to Compare
Alternatives Require a Common Time Basis
Just as we must convert fractions to common denominators to add them together, we must
also convert cash flows to a common basis to compare their value. One aspect of this basis
is the choice of a single point in time at which to make our calculations. In Example 3.3, if
we had been given the magnitude of each cash flow and had been asked to determine
whether they were equivalent, we could have chosen any reference point and used the
compound interest formula to find the value of each cash flow at that point. As you can
readily see, the choice of n = 0 or n = 5 would make our problem simpler because we
need to make only one set of calculations: At 8% interest, either convert $2,042 at time 0
to its equivalent value at time 5, or convert $3,000 at time 5 to its equivalent value at time 0.
(To see how to choose a different reference point, take a look at Example 3.4.)
When selecting a point in time at which to compare the value of alternative cash
flows, we commonly use either the present time, which yields what is called the present
worth of the cash flows, or some point in the future, which yields their future worth.
The choice of the point in time often depends on the circumstances surrounding a particular
decision, or it may be chosen for convenience. For instance, if the present worth is known
for the first two of three alternatives, all three may be compared simply by calculating the
present worth of the third.
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Section 3.2 Economic Equivalence
EXAMPLE 3.4 Equivalent Cash Flows Are Equivalent
at Any Common Point in Time
In Example 3.3, we determined that, given an interest rate of 8% per year, receiving
$2,042 today is equivalent to receiving $3,000 in five years. Are these cash flows also
equivalent at the end of year 3?
STRATEGY: This problem is summarized in Figure 3.9. The solution consists of solving
two equivalence problems: (1) What is the future value of $2,042 after three years at 8%
interest (part (a) of the solution)? (2) Given the sum of $3,000 after five years and an interest rate of 8%, what is the equivalent sum after 3 years (part (b) of the solution)?
SOLUTION
Given:
(a) P = $2,042; i = 8% per year; N = 3 years.
(b) F = $3,000; i = 8% per year; N = 5 - 3 = 2 years.
Find: (1) V3 for part (a); (2) V3 for part (b). (3) Are these two values equivalent?
Equation:
(a) F = P11 + i2N.
(b) P = F11 + i2-N.
V3
$2,042(1 0.08)3
$2,572
$2,042
0
1
2
3
4
5
Years
(a)
$3,000(1 0.08)2
$3,000
$2,572
0
1
2
3
4
Years
(b)
Base period
Figure 3.9 Selection of a base period for an equivalence
calculation (Example 3.4).
5
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68 CHAPTER 3 Interest Rate and Economic Equivalence
Notation: The usual terminology of F and P is confusing in this example, since
the cash flow at n = 3 is considered a future sum in part (a) of the solution and a
past cash flow in part (b) of the solution. To simplify matters, we are free to arbitrarily designate a reference point n = 3 and understand that it need not to be now
or the present. Therefore, we assign the equivalent cash flow at n = 3 to a single
variable, V3.
1. The equivalent worth of $2,042 after three years is
V3 = 2,04211 + 0.0823
= $2,572.
2. The equivalent worth of the sum $3,000 two years earlier is
V3 = F11 + i2-N
= $3,00011 + 0.082-2
= $2,572.
(Note that N = 2 because that is the number of periods during which discounting
is calculated in order to arrive back at year 3.)
3. While our solution doesn’t strictly prove that the two cash flows are equivalent at
any time, they will be equivalent at any time as long as we use an interest rate of 8%.
Principle 2: Equivalence Depends on Interest Rate
The equivalence between two cash flows is a function of the magnitude and timing of individual cash flows and the interest rate or rates that operate on those flows. This principle
is easy to grasp in relation to our simple example: $1,000 received now is equivalent to
$1,762.34 received five years from now only at a 12% interest rate. Any change in the
interest rate will destroy the equivalence between these two sums, as we will demonstrate
in Example 3.5.
EXAMPLE 3.5 Changing the Interest Rate Destroys
Equivalence
In Example 3.3, we determined that, given an interest rate of 8% per year, receiving
$2,042 today is equivalent to receiving $3,000 in five years. Are these cash flows
equivalent at an interest rate of 10%?
SOLUTION
Given: P = $2,042, i = 10% per year, and N = 5 years.
Find: F: Is it equal to $3,000?
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Section 3.2 Economic Equivalence
We first determine the base period under which an equivalence value is computed.
Since we can select any period as the base period, let’s select N = 5. Then we need
to calculate the equivalent value of $2,042 today five years from now.
F = $2,04211 + 0.1025 = $3,289.
Since this amount is greater than $3,000, the change in interest rate destroys the
equivalence between the two cash flows.
Principle 3: Equivalence Calculations May Require
the Conversion of Multiple Payment Cash Flows
to a Single Cash Flow
In all the examples presented thus far, we have limited ourselves to the simplest case of
converting a single payment at one time to an equivalent single payment at another time.
Part of the task of comparing alternative cash flow series involves moving each individual cash flow in the series to the same single point in time and summing these values to
yield a single equivalent cash flow. We perform such a calculation in Example 3.6.
EXAMPLE 3.6 Equivalence Calculations with Multiple
Payments
Suppose that you borrow $1,000 from a bank for three years at 10% annual interest.
The bank offers two options: (1) repaying the interest charges for each year at the end
of that year and repaying the principal at the end of year 3 or (2) repaying the loan all
at once (including both interest and principal) at the end of year 3. The repayment
schedules for the two options are as follows:
Options
Year 1
• Option 1: End-of-year repayment
of interest, and principal repayment
at end of loan
• Option 2: One end-of-loan
repayment of both principal
and interest
Year 2
Year 3
$100
$100
$1,100
0
0
1,331
Determine whether these options are equivalent, assuming that the appropriate interest
rate for the comparison is 10%.
STRATEGY: Since we pay the principal after three years in either plan, the repayment of principal can be removed from our analysis. This is an important point: We
can ignore the common elements of alternatives being compared so that we can
focus entirely on comparing the interest payments. Notice that under Option 1, we
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70 CHAPTER 3 Interest Rate and Economic Equivalence
will pay a total of $300 interest, whereas under Option 2, we will pay a total of $331.
Before concluding that we prefer Option 2, remember that a comparison of the two
cash flows is based on a combination of payment amounts and the timing of those
payments. To make our comparison, we must compare the equivalent value of each
option at a single point in time. Since Option 2 is already a single payment at
n = 3 years, it is simplest to convert the cash flow pattern of Option 1 to a single
value at n = 3. To do this, we must convert the three disbursements of Option 1 to
their respective equivalent values at n = 3. At that point, since they share a time in
common, we can simply sum them in order to compare them with the $331 sum in
Option 2.
SOLUTION
Given: Interest payment series; i = 10% per year.
Find: A single future value F of the flows in Option 1.
Equation: F = P11 + i2N, applied to each disbursement in the cash flow diagram.
N in Eq. (3.3) is the number of interest periods during which interest is in effect, and
n is the period number (i.e., for year 1, n = 1). We determine the value of F by
finding the interest period for each payment. Thus, for each payment in the series,
N can be calculated by subtracting n from the total number of years of the loan
(3). That is, N = 3 - n. Once the value of each payment has been found, we sum
the payments:
0
Option 1
1
2
3
0
Option 2
1
2
3
⬅
$100
$100
$100
$331
F3 for $100 at n = 1 : $10011 + .1023 - 1 = $121;
F3 for $100 at n = 2 : $10011 + .1023 - 2 = $110;
F3 for $100 at n = 3 : $10011 + .1023 - 3 = $100;
Total = $331.
By converting the cash flow in Option 1 to a single future payment at year 3, we
can compare Options 1 and 2. We see that the two interest payments are equivalent. Thus, the bank would be economically indifferent to a choice between the
two plans. Note that the final interest payment in Option 1 does not accrue any
compound interest.
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Section 3.3 Development of Interest Formulas 71
Principle 4: Equivalence Is Maintained Regardless
of Point of View
As long as we use the same interest rate in equivalence calculations, equivalence can be
maintained regardless of point of view. In Example 3.6, the two options were equivalent at
an interest rate of 10% from the banker’s point of view. What about from a borrower’s
point of view? Suppose you borrow $1,000 from a bank and deposit it in another bank that
pays 10% interest annually. Then you make future loan repayments out of this savings account. Under Option 1, your savings account at the end of year 1 will show a balance of
$1,100 after the interest earned during the first period has been credited. Now you withdraw
$100 from this savings account (the exact amount required to pay the loan interest during
the first year), and you make the first-year interest payment to the bank. This leaves only
$1,000 in your savings account. At the end of year 2, your savings account will earn another interest payment in the amount of $1,00010.102 = $100, making an end-of-year balance
of $1,100. Now you withdraw another $100 to make the required loan interest payment.
After this payment, your remaining balance will be $1,000. This balance will grow again
at 10%, so you will have $1,100 at the end of year 3. After making the last loan payment
($1,100), you will have no money left in either account. For Option 2, you can keep track of
the yearly account balances in a similar fashion. You will find that you reach a zero balance
after making the lump-sum payment of $1,331. If the borrower had used the same interest
rate as the bank, the two options would be equivalent.
3.2.3 Looking Ahead
The preceding examples should have given you some insight into the basic concepts and
calculations involved in the concept of economic equivalence. Obviously, the variety of
financial arrangements possible for borrowing and investing money is extensive, as is the
variety of time-related factors (e.g., maintenance costs over time, increased productivity
over time, etc.) in alternative proposals for various engineering projects. It is important to
recognize that even the most complex relationships incorporate the basic principles we
have introduced in this section.
In the remainder of the chapter, we will represent all cash flow diagrams either in the
context of an initial deposit with a subsequent pattern of withdrawals or in an initial borrowed amount with a subsequent pattern of repayments. If we were limited to the methods
developed in this section, a comparison between the two payment options would involve a
large number of calculations. Fortunately, in the analysis of many transactions, certain cash
flow patterns emerge that may be categorized. For many of these patterns, we can derive
formulas that can be used to simplify our work. In Section 3.3, we develop these formulas.
3.3 Development of Interest Formulas
Now that we have established some working assumptions and notations and have a preliminary understanding of the concept of equivalence, we will develop a series of interest
formulas for use in more complex comparisons of cash flows.
As we begin to compare series of cash flows instead of single payments, the required
analysis becomes more complicated. However, when patterns in cash flow transactions
can be identified, we can take advantage of these patterns by developing concise expressions
for computing either the present or future worth of the series. We will classify five major
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72 CHAPTER 3 Interest Rate and Economic Equivalence
categories of cash flow transactions, develop interest formulas for them, and present several
working examples of each type. Before we give the details, however, we briefly describe the
five types of cash flows in the next subsection.
3.3.1 The Five Types of Cash Flows
Whenever we identify patterns in cash flow transactions, we may use those patterns to develop concise expressions for computing either the present or future worth of the series.
For this purpose, we will classify cash flow transactions into five categories: (1) a single
cash flow, (2) a uniform series, (3) a linear gradient series, (4) a geometric gradient series,
and (5) an irregular series. To simplify the description of various interest formulas, we
will use the following notation:
1. Single Cash Flow: The simplest case involves the equivalence of a single present
amount and its future worth. Thus, the single-cash-flow formulas deal with only
two amounts: a single present amount P and its future worth F (Figure 3.10a). You
have already seen the derivation of one formula for this situation in Section 3.1.3,
which gave us Eq. (3.3):
F = P11 + i2N.
2. Equal (Uniform) Series: Probably the most familiar category includes transactions
arranged as a series of equal cash flows at regular intervals, known as an equal payment series (or uniform series) (Figure 3.10b). For example, this category describes the
cash flows of the common installment loan contract, which arranges the repayment of
a loan in equal periodic installments. The equal-cash-flow formulas deal with the
equivalence relations P, F, and A (the constant amount of the cash flows in the series).
3. Linear Gradient Series: While many transactions involve series of cash flows, the
amounts are not always uniform; they may, however, vary in some regular way. One
common pattern of variation occurs when each cash flow in a series increases (or
decreases) by a fixed amount (Figure 3.10c). A five-year loan repayment plan might
specify, for example, a series of annual payments that increase by $500 each year.
We call this type of cash flow pattern a linear gradient series because its cash flow
diagram produces an ascending (or descending) straight line, as you will see in
Section 3.3.5. In addition to using P, F, and A, the formulas employed in such problems involve a constant amount G of the change in each cash flow.
4. Geometric Gradient Series: Another kind of gradient series is formed when the
series in a cash flow is determined not by some fixed amount like $500, but by
some fixed rate, expressed as a percentage. For example, in a five-year financial
plan for a project, the cost of a particular raw material might be budgeted to increase at a rate of 4% per year. The curving gradient in the diagram of such a series
suggests its name: a geometric gradient series (Figure 3.10d). In the formulas
dealing with such series, the rate of change is represented by a lowercase g.
5. Irregular (Mixed) Series: Finally, a series of cash flows may be irregular, in that it
does not exhibit a regular overall pattern. Even in such a series, however, one or
more of the patterns already identified may appear over segments of time in the
total length of the series. The cash flows may be equal, for example, for 5 consecutive
periods in a 10-period series. When such patterns appear, the formulas for dealing
with them may be applied and their results included in calculating an equivalent
value for the entire series.
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Section 3.3 Development of Interest Formulas 73
$100
(a) Single cash flow
0
Years
1
2
3
4
5
$100
$100
$100
$100
$100
(b) Equal (uniform) payment
series at regular intervals
0
Years
1
$50
0
Years
1
2
3
$50 2G
$50 G
2
3
4
5
$50 4G
$50 3G
(c) Linear gradient series,
where each cash flow in the
series increases or decreases
by a fixed amount G
4
5
$50(1g)4
$50
0
Years
1
$70
0
Years
1
$50(1g)3
$50(1g)2
$50(1g)
2
3
4
$100
5
$100
$60
2
(d) Geometric gradient series,
where each cash flow in the
series increases or decreases
by a fixed rate (percentage) g
3
$50
4
5
(e) Irregular payment series,
which exhibits no regular overall
pattern
Figure 3.10 Five types of cash flows: (a) Single cash flow, (b) equal
(uniform) payment series, (c) linear gradient series, (d) geometric gradient
series, and (e) irregular payment series.
3.3.2 Single-Cash-Flow Formulas
We begin our coverage of interest formulas by considering the simplest of cash flows:
single cash flows.
Compound Amount Factor
Given a present sum P invested for N interest periods at interest rate i, what sum will have accumulated at the end of the N periods? You probably noticed right away that this description
matches the case we first encountered in describing compound interest. To solve for F
(the future sum), we use Eq. (3.3):
F = P11 + i2N.
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74 CHAPTER 3 Interest Rate and Economic Equivalence
F
P
Compounding
process
F = P(1 + i)N
O
N
F
P
Discounting
process
P = F(1 + i)–N
N
O
Figure 3.11
Compounding
process:
the process of
computing the
future value of a
current sum.
Equivalence relation between P and F.
Because of its origin in the compound-interest calculation, the factor 11 + i2N is known
as the compound-amount factor. Like the concept of equivalence, this factor is one of
the foundations of engineering economic analysis. Given the compound-amount factor, all
the other important interest formulas can be derived.
This process of finding F is often called the compounding process. The cash flow
transaction is illustrated in Figure 3.11. (Note the time-scale convention: The first period
begins at n = 0 and ends at n = 1.) If a calculator is handy, it is easy enough to calculate
11 + i2N directly.
Interest Tables
Interest formulas such as the one developed in Eq. (3.3), F = P11 + i2N, allow us to
substitute known values from a particular situation into the equation and to solve for the
unknown. Before the hand calculator was developed, solving these equations was very tedious. With a large value of N, for example, one might need to solve an equation such as
F = $20,00011 + 0.12215. More complex formulas required even more involved calculations. To simplify the process, tables of compound-interest factors were developed, and
these tables allow us to find the appropriate factor for a given interest rate and the number
of interest periods. Even with hand calculators, it is still often convenient to use such tables, and they are included in this text in Appendix A. Take some time now to become familiar with their arrangement and, if you can, locate the compound-interest factor for the
example just presented, in which we know P. Remember that, to find F, we need to know
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Section 3.3 Development of Interest Formulas 75
the factor by which to multiply $20,000 when the interest rate i is 12% and the number of
periods is 15:
F = $20,00011 + 0.12215 = $109,472.
5
5.4736
Factor Notation
As we continue to develop interest formulas in the rest of this chapter, we will express the
resulting compound-interest factors in a conventional notation that can be substituted in a
formula to indicate precisely which table factor to use in solving an equation. In the
preceding example, for instance, the formula derived as Eq. (3.3) is F = P11 + i2N.
In ordinary language, this tells us that, to determine what future amount F is equivalent to
a present amount P, we need to multiply P by a factor expressed as 1 plus the interest
rate, raised to the power given by the number of interest periods. To specify how the interest tables are to be used, we may also express that factor in functional notation as
(F/P, i, N), which is read as “Find F, Given P, i, and N.” This is known as the single-payment
compound-amount factor. When we incorporate the table factor into the formula, it is
expressed as
F = P11 + i2N = P1F>P, i, N2.
Thus, in the preceding example, where we had F = $20,00011.12215, we can write
F = $20,0001F/P, 12%, 152. The table factor tells us to use the 12% interest table and
find the factor in the F/P column for N = 15. Because using the interest tables is often
the easiest way to solve an equation, this factor notation is included for each of the formulas derived in the sections that follow.
EXAMPLE 3.7 Single Amounts: Find F, Given i, N, and P
If you had $2,000 now and invested it at 10%, how much would it be worth in eight
years (Figure 3.12)?
F
i = 10%
0
1
2
3
4
Years
5
6
7
$2,000
Figure 3.12 A cash flow diagram from the investor’s
point of view (Example 3.7).
8
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76 CHAPTER 3 Interest Rate and Economic Equivalence
SOLUTION
Given: P = $2,000, i = 10% per year, and N = 8 years.
Find: F.
We can solve this problem in any of three ways:
1. Using a calculator. You can simply use a calculator to evaluate the 11 + i2N term
(financial calculators are preprogrammed to solve most future-value problems):
F = $2,00011 + 0.1028
= $4,287.18.
2. Using compound-interest tables. The interest tables can be used to locate the
compound-amount factor for i = 10% and N = 8. The number you get can be
substituted into the equation. Compound-interest tables are included as Appendix A
of this book. From the tables, we obtain
F = $2,0001F>P, 10%, 82 = $2,00012.14362 = $4,287.20.
This is essentially identical to the value obtained by the direct evaluation of the
single-cash-flow compound-amount factor. This slight difference is due to rounding errors.
3. Using Excel. Many financial software programs for solving compound-interest
problems are available for use with personal computers. Excel provides financial
functions to evaluate various interest formulas, where the future-worth calculation looks like the following:
=FV110%,8,0,-20002
Discounting
process: A
process of
calculating the
present value
of a future
amount.
Present-Worth Factor
Finding the present worth of a future sum is simply the reverse of compounding and is
known as the discounting process. In Eq. (3.3), we can see that if we were to find a present sum P, given a future sum F, we simply solve for P:
P = Fc
1
d = F1P>F, i, N2.
11 + i2N
(3.7)
The factor 1/11 + i2N is known as the single-payment present-worth factor and is designated (P/F, i, N). Tables have been constructed for P/F factors and for various values of
i and N. The interest rate i and the P/F factor are also referred to as the discount rate and
discounting factor, respectively.
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Section 3.3 Development of Interest Formulas 77
EXAMPLE 3.8 Single Amounts: Find P, Given F, i, and N
Suppose that $1,000 is to be received in five years. At an annual interest rate of 12%,
what is the present worth of this amount?
SOLUTION
Given: F = $1,000, i = 12% per year, and N = 5 years.
Find: P.
P = $1,00011 + 0.122-5 = $1,00010.56742 = $567.40.
Using a calculator may be the best way to make this simple calculation. To have $1,000
in your savings account at the end of five years, you must deposit $567.40 now.
We can also use the interest tables to find that
10.56742
4
P = $1,0001P> F, 12%, 52 = $567.40.
Again, you could use a financial calculator or a computer to find the present worth.
With Excel, the present-value calculation looks like the following:
= PV112%,5,0,-10002
Solving for Time and Interest Rates
At this point, you should realize that the compounding and discounting processes are reciprocals of one another and that we have been dealing with one equation in two forms:
Future-value form: F = P11 + i2N;
Present-value form: P = F11 + i2-N.
There are four variables in these equations: P, F, N, and i. If you know the values of any
three, you can find the value of the fourth. Thus far, we have always given you the interest rate i and the number of years N, plus either P or F. In many situations, though, you
will need to solve for i or N, as we discuss next.
EXAMPLE 3.9 Solving for i
Suppose you buy a share for $10 and sell it for $20. Then your profit is $10. If that
happens within a year, your rate of return is an impressive 100% 1$10/$10 = 12. If
it takes five years, what would be the average annual rate of return on your investment?
(See Figure 3.13.)
SOLUTION
Given: P = $10, F = $20, and N = 5.
Find: i.
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78 CHAPTER 3 Interest Rate and Economic Equivalence
$20
i?
0
1
2
3
4
5
Years
$10
Figure 3.13
Cash flow diagram (Example 3.9).
Here, we know P, F, and N, but we do not know i, the interest rate you will earn on
your investment. This type of rate of return is a lot easier to calculate, because you make
only a one-time lump-sum investment. Problems such as this are solved as follows:
F = P11 + i2N;
$20 = $1011 + i25; solve for i.
• Method 1. Go through a trial-and-error process in which you insert different
values of i into the equation until you find a value that “works” in the sense
that the right-hand side of the equation equals $20. The solution is
i = 14.87%. The trial-and-error procedure is extremely tedious and inefficient for most problems, so it is not widely practiced in the real world.
• Method 2. You can solve the problem by using the interest tables in Appendix A. Now look across the N = 5 row, under the (F/P, i, 5) column, until you
can locate the value of 2:
$20 = $1011 + i25;
2 = 11 + i25 = 1F>P, i, 52.
This value is close to the 15% interest table with 1F/P, 15%, 52 = 2.0114, so
the interest rate at which $10 grows to $20 over five years is very close to 15%.
This procedure will be very tedious for fractional interest rates or when N is
not a whole number, because you may have to approximate the solution by linear interpolation.
• Method 3. The most practical approach is to use either a financial calculator
or an electronic spreadsheet such as Excel. A financial function such as
RATE(N,0,P,F) allows us to calculate an unknown interest rate. The precise
command statement would be
= RATE15,0,-10,202=14.87%
Note that, in Excel format, we enter the present value (P) as a negative number, indicating a cash outflow.
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Section 3.3 Development of Interest Formulas 79
A
B
1
P
−10
2
F
20
3
N
5
4
i
14.87%
C
RATE(5,0,10,20)
5
EXAMPLE 3.10 Single Amounts: Find N, Given P, F, and i
You have just purchased 100 shares of General Electric stock at $60 per share. You will
sell the stock when its market price has doubled. If you expect the stock price to increase
20% per year, how long do you anticipate waiting before selling the stock (Figure 3.14)?
SOLUTION
Given: P = $6,000, F = $12,000, and i = 20% per year.
Find: N (years).
Using the single-payment compound-amount factor, we write
F = P11 + i2N = P1F>P, i, N2;
$12,000 = $6,00011 + 0.202N = $6,0001F>P, 20%, N2;
2 = 11.202N = 1F>P, 20%, N2.
Again, we could use a calculator or a computer spreadsheet program to find N.
$12,000
i = 20%
0
N=?
$6,000
Figure 3.14
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80 CHAPTER 3 Interest Rate and Economic Equivalence
1. Using a calculator. Solving for N gives
log 2 = N log 1.20,
or
log 2
N =
log 1.20
= 3.80 L 4 years.
2. Using Excel. Within Excel, the financial function NPER(i,0,P,F) computes the
number of compounding periods it will take an investment (P) to grow to a future
value (F), earning a fixed interest rate (i) per compounding period. In our example, the Excel command would look like this:
= NPER120%,0,-6000,120002
= 3.801784.
Rule of 72:
Rule giving the
approximate
number of years
that it will take
for your investment to double.
COMMENTS: A very handy rule of thumb, called the Rule of 72, estimates approximately how long it will take for a sum of money to double. The rule states that, to
find the time it takes for a present sum of money to grow by a factor of two, we divide
72 by the interest rate. In our example, the interest rate is 20%. Therefore, the Rule of
72 indicates 72/20 = 3.60, or roughly 4 years, for a sum to double. This is, in fact,
relatively close to our exact solution. Figure 3.15 illustrates the number of years
required to double an investment at various interest rates.
18
years
Imagine! If you were to invest
at a higher rate, you’d double
money that much faster!
12
years
4%
6%
9
years
8%
7.2
years
10%
6
years
12%
5.14
years
14%
Figure 3.15 Number of years required to double an
initial investment at various interest rates.
3.3.3 Uneven Payment Series
A common cash flow transaction involves a series of disbursements or receipts. Familiar
examples of series payments are payment of installments on car loans and home mortgage payments. Payments on car loans and home mortgages typically involve identical
sums to be paid at regular intervals. However, there is no clear pattern over the series; we
call the transaction an uneven cash flow series.
We can find the present worth of any uneven stream of payments by calculating the
present value of each individual payment and summing the results. Once the present worth
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Section 3.3 Development of Interest Formulas 81
is found, we can make other equivalence calculations (e.g., future worth can be calculated by using the interest factors developed in the previous section).
EXAMPLE 3.11 Present Values of an Uneven Series by
Decomposition into Single Payments
Wilson Technology, a growing machine shop, wishes to set aside money now to invest
over the next four years in automating its customer service department. The company
can earn 10% on a lump sum deposited now, and it wishes to withdraw the money in
the following increments:
• Year 1: $25,000, to purchase a computer and database software designed for customer service use;
• Year 2: $3,000, to purchase additional hardware to accommodate anticipated
growth in use of the system;
• Year 3: No expenses; and
• Year 4: $5,000, to purchase software upgrades.
How much money must be deposited now to cover the anticipated payments over the
next 4 years?
STRATEGY: This problem is equivalent to asking what value of P would make you
indifferent in your choice between P dollars today and the future expense stream of
($25,000, $3,000, $0, $5,000). One way to deal with an uneven series of cash flows
is to calculate the equivalent present value of each single cash flow and to sum the
present values to find P. In other words, the cash flow is broken into three parts as
shown in Figure 3.16.
SOLUTION
Given: Uneven cash flow in Figure 3.16, with i = 10% per year.
Find: P.
P = $25,0001P>F, 10%, 12 + $3,0001P>F, 10%, 22
+ $5,0001P>F, 10%, 42
= $28,622.
COMMENTS: To see if $28,622 is indeed sufficient, let’s calculate the balance at the
end of each year. If you deposit $28,622 now, it will grow to (1.10)($28,622), or
$31,484, at the end of year 1. From this balance, you pay out $25,000. The remaining
balance, $6,484, will again grow to (1.10)($6,484), or $7,132, at the end of year 2.
Now you make the second payment ($3,000) out of this balance, which will leave
you with only $4,132 at the end of year 2. Since no payment occurs in year 3, the
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82 CHAPTER 3 Interest Rate and Economic Equivalence
$25,000
$5,000
$3,000
0
1
$25,000
2
3
Years
4
P
$5,000
$3,000
0
1
2
3
4
0
1
2
3
4
P2
P1
P1 $25,000(P/F, 10%, 1)
$22,727
P2 $3,000(P/F, 10%, 2)
$2,479
0
1
2
3
4
P4
P4 $5,000(P/F, 10%, 4)
$3,415
P P1 P2 P4 $28,622
Figure 3.16
Decomposition of uneven cash flow series (Example 3.11).
balance will grow to $(1.10)2 ($4,132), or $5,000, at the end of year 4. The final withdrawal in the amount of $5,000 will deplete the balance completely.
EXAMPLE 3.12 Calculating the Actual Worth of a Long-Term
Contract of Michael Vick with Atlanta Falcons2
On December 23, 2004, Michael Vick became the richest player in the National
Football League by agreeing to call Atlanta home for the next decade. The Falcons’
quarterback signed a 10-year, $130 million contract extension Thursday that guarantees him an NFL-record $37 million in bonuses.
Base salaries for his new contract are $600,000 (2005), $1.4 million (2006),
$6 million (2007), $7 million (2008), $9 million (2009), $10.5 million (2010),
$13.5 million (2011), $13 million (2012), $15 million (2013), and $17 million
(2014). He received an initial signing bonus of $7.5 million. Vick also received
two roster bonuses in the new deal. The first is worth $22.5 million and is due in
March 2005. The second is worth $7 million and is due in March 2006. Both roster bonuses will be treated as signing bonuses and prorated annually. Because
2011 is an uncapped year (the league’s collective bargaining agreement (CBA)
expires after the 2010 season), the initial signing bonus and 2005 roster bonus can
be prorated only over the first six years of the contract. If the CBA is extended
2
Source: http://www.falcfans.com/players/michael_vick.html.
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Section 3.3 Development of Interest Formulas 83
prior to March 2006, then the second roster bonus of $7 million can be prorated
over the final nine seasons of the contract. If the CBA is extended prior to March
2006, then his cap hits (rounded to nearest thousand) will change to $7.178 million (2006), $11.778 million (2007), $12.778 million (2008), $14.778 million
(2009), $16.278 million (2010), $14.278 million (2011), $13.778 million (2012),
$15.778 million (2013), and $17.778 million (2014). With the salary and signing
bonus paid at the beginning of each season, the net annual payment schedule
looks like the following:
Beginning
of Season
2005
Base
Salary
$
600,000
Prorated
Signing Bonus
Total
Annual Payment
$5,000,000
$5,600,000
2006
1,400,000
5,000,000 + 778,000
7,178,000
2007
6,000,000
5,000,000 + 778,000
11,778,000
2008
7,000,000
5,000,000 + 778,000
12,778,000
2009
9,000,000
5,000,000 + 778,000
14,778,000
2010
10,500,000
778,000
16,278,000
2011
13,500,000
778,000
14,278,000
2012
13,000,000
778,000
13,778,000
2013
15,000,000
778,000
15,778,000
2014
17,000,000
778,000
17,778,000
(a) How much is Vick’s contract actually worth at the time of signing? Assume that
Vick’s interest rate is 6% per year.
(b) For the initial signing bonus and the first year’s roster bonus, suppose that the
Falcons allow Vick to take either the prorated payment option as just described
($30 million over five years) or a lump-sum payment option in the amount of
$23 million at the time he signs the contract. Should Vick take the lump-sum option instead of the prorated one?
SOLUTION
Given: Payment series given in Figure 3.17, with i = 6% per year.
Find: P.
(a) Actual worth of the contract at the time of signing:
Pcontract = $5,600,000 + $7,178,0001P>F, 6%, 12
+ $11,778,0001P>F, 6%, 22 + Á
+ $17,778,0001P>F, 6%, 92
= $97,102,827.
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84 CHAPTER 3 Interest Rate and Economic Equivalence
$17.778 M
$16.278 M
$5.6 M
04
05
06
07
08
09
10
11
12
13
14
Figure 3.17 Cash flow diagram for Michael Vick’s contract with Atlanta Falcons.
(b) Choice between the prorated payment option and the lump-sum payment: The
equivalent present worth of the prorated payment option is
Pbonus = $5,000,000 + $5,000,0001P>F, 6%, 12
+ $5,000,0001P>F, 6%, 22 + $5,000,0001P>F, 6%, 32
+ $5,000,0001P>F, 6%, 42
= $22,325,528
which is smaller than $23,000,000. Therefore, Vick would be better off taking the
lump-sum option if, and only if, his money could be invested at 6% or higher.
COMMENTS: Note that the actual contract is worth less than the published figure of
$130 million. This “brute force” approach of breaking cash flows into single
amounts will always work, but it is slow and subject to error because of the many factors that must be included in the calculation. We develop more efficient methods in
later sections for cash flows with certain patterns.
3.3.4 Equal Payment Series
As we learned in Example 3.12, the present worth of a stream of future cash flows can always be found by summing the present worth of each of the individual cash flows. However, if cash flow regularities are present within the stream (such as we just saw in the
prorated bonus payment series in Example 3.12) then the use of shortcuts, such as finding
the present worth of a uniform series, may be possible. We often encounter transactions
in which a uniform series of payments exists. Rental payments, bond interest payments,
and commercial installment plans are based on uniform payment series.
Compound-Amount Factor: Find F, Given A, i, and N
Suppose we are interested in the future amount F of a fund to which we contribute A dollars each period and on which we earn interest at a rate of i per period. The contributions
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Section 3.3 Development of Interest Formulas 85
F
F F1 F2 FN
0
1
2
N1
A1
A2
AN1
F2
F1
0 1 2 3
N
0 1 2 3
N
A1
N
AN
FN1
A2
0 1 2 N1
N
FN
0 1 2 3
AN1
N
AN
Figure 3.18 The future worth of a cash flow series obtained by summing the
future-worth figures of each of the individual flows.
are made at the end of each of N equal periods. These transactions are graphically illustrated in Figure 3.18. Looking at this diagram, we see that if an amount A is invested at
the end of each period, for N periods, the total amount F that can be withdrawn at the end
of the N periods will be the sum of the compound amounts of the individual deposits.
As shown in Figure 3.18, the A dollars we put into the fund at the end of the first period will be worth A11 + i2N - 1 at the end of N periods. The A dollars we put into the
fund at the end of the second period will be worth A11 + i2N - 2, and so forth. Finally, the
last A dollars that we contribute at the end of the Nth period will be worth exactly A dollars at that time. This means that there exists a series of the form
F = A11 + i2N - 1 + A11 + i2N - 2 + Á + A11 + i2 + A,
or, expressed alternatively,
F = A + A11 + i2 + A11 + i22 + Á + A11 + i2N - 1.
(3.8)
Multiplying Eq. (3.8) by 11 + i2 results in
11 + i2F = A11 + i2 + A11 + i22 + Á + A11 + i2N.
(3.9)
Subtracting Eq. (3.8) from Eq. (3.9) to eliminate common terms gives us
F11 + i2 - F = -A + A11 + i2N.
Solving for F yields
F = Ac
11 + i2N - 1
d = A1F>A, i, N2.
i
(3.10)
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The bracketed term in Eq. (3.10) is called the equal payment series compound-amount
factor, or the uniform series compound-amount factor; its factor notation is (F/A, i,
N). This interest factor has been calculated for various combinations of i and N in the interest tables.
EXAMPLE 3.13 Uniform Series: Find F, Given i, A, and N
Suppose you make an annual contribution of $3,000 to your savings account at the
end of each year for 10 years. If the account earns 7% interest annually, how much
can be withdrawn at the end of 10 years (Figure 3.19)?
F
i = 7%
0
1
2
3
4
Years
5
6
7
8
9
10
A = $3,000
Figure 3.19 Cash flow diagram (Example 3.13).
SOLUTION
Given: A = $3,000, N = 10 years, and i = 7% per year.
Find: F.
F = $3,0001F>A, 7%, 102
= $3,000113.81642
= $41,449.20.
To obtain the future value of the annuity with the use of Excel, we may use the following financial command:
= FV17%,10, - 3000,0,02
EXAMPLE 3.14 Handling Time Shifts in a Uniform Series
In Example 3.13, the first deposit of the 10-deposit series was made at the end of
period 1 and the remaining nine deposits were made at the end of each following
period. Suppose that all deposits were made at the beginning of each period instead.
How would you compute the balance at the end of period 10?
SOLUTION
Given: Cash flow as shown in Figure 3.20, and i = 7% per year.
Find: F10.
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Section 3.3 Development of Interest Formulas 87
F
i = 7%
First deposit occurs at n = 0
0
1
2
3
4
Years
5
6
7
8
9
10
A = $3,000
Figure 3.20
Cash Flow diagram (Example 3.14).
Compare Figure 3.20 with Figure 3.19: Each payment has been shifted to one year earlier; thus, each payment would be compounded for one extra year. Note that with the
end-of-year deposit, the ending balance (F) was $41,449.20. With the beginning-of-year
deposit, the same balance accumulates by the end of period 9. This balance can earn interest for one additional year. Therefore, we can easily calculate the resulting balance as
F10 = $41,449.2011.072 = $44,350.64.
The annuity due can be easily evaluated with the following financial command available
on Excel:
= FV17%,10, -3000,0,12
COMMENTS: Another way to determine the ending balance is to compare the two
cash flow patterns. By adding the $3,000 deposit at period 0 to the original cash flow
and subtracting the $3,000 deposit at the end of period 10, we obtain the second cash
flow. Therefore, the ending balance can be found by making the following adjustment to the $41,449.20:
F10 = $41,449.20 + $3,0001F>P, 7%, 102 - $3,000 = $44,350.64.
Sinking-Fund Factor: Find A, Given F, i, and N
If we solve Eq. (3.10) for A, we obtain
i
d = F1A>F, i, N2.
A = Fc
11 + i2N - 1
(3.11)
The term within the brackets is called the equal payment series sinking-fund
factor, or sinking-fund factor, and is referred to by the notation (A/F, i, N). A sinking
fund is an interest-bearing account into which a fixed sum is deposited each interest
period; it is commonly established for the purpose of replacing fixed assets or retiring
corporate bonds.
Sinking fund:
A means of
repaying funds
advanced
through a bond
issue.This means
that every period,
a company will
pay back a portion
of its bonds.
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88 CHAPTER 3 Interest Rate and Economic Equivalence
EXAMPLE 3.15 Combination of a Uniform Series and
a Single Present and Future Amount
To help you reach a $5,000 goal five years from now, your father offers to give you $500
now. You plan to get a part-time job and make five additional deposits, one at the end of
each year. (The first deposit is made at the end of the first year.) If all your money is
deposited in a bank that pays 7% interest, how large must your annual deposit be?
STRATEGY: If your father reneges on his offer, the calculation of the required annual
deposit is easy because your five deposits fit the standard end-of-period pattern for a
uniform series. All you need to evaluate is
A = $5,0001A>F, 7%, 52 = $5,00010.17392 = $869.50.
If you do receive the $500 contribution from your father at n = 0, you may divide
the deposit series into two parts: one contributed by your father at n = 0 and five
equal annual deposit series contributed by yourself. Then you can use the F/P factor
to find how much your father’s contribution will be worth at the end of year 5 at a 7%
interest rate. Let’s call this amount Fc. The future value of your five annual deposits
must then make up the difference, $5,000 - Fc.
SOLUTION
Given: Cash flow as shown in Figure 3.21, with i = 7% per year, and N = 5 years.
Find: A.
1$5,000 - FC21A>F, 7%, 52
[$5,000 - $5001F>P, 7%, 52]1A>F, 7%, 52
[$5,000 - $50011.40262]10.17392
$747.55.
A =
=
=
=
$5,000
Original cash flow
i 7%
0
$500
1
Years
2
3
4
A
A
A
A
5
[$5,000 $500(F/P, 7%, 5)]
A
i 7%
0
Figure 3.21
1
Years
2
3
4
A
A
A
A
Cash flow diagram (Example 3.15).
5
A
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Section 3.3 Development of Interest Formulas 89
EXAMPLE 3.16 Comparison of Three Different
Investment Plans
Consider three investment plans at an annual interest rate of 9.38% (Figure 3.22):
• Investor A. Invest $2,000 per year for the first 10 years of your career. At the end
of 10 years, make no further investments, but reinvest the amount accumulated at the end of 10 years for the next 31 years.
• Investor B. Do nothing for the first 10 years. Then start investing $2,000 per year
for the next 31 years.
• Investor C. Invest $2,000 per year for the entire 41 years.
Note that all investments are made at the beginning of each year; the first deposit will
be made at the beginning of age 25 1n = 02, and you want to calculate the balance at
the age of 65 1n = 412.
STRATEGY: Since the investments are made at the beginning of each year, we need to
use the procedure outlined in Example 3.14. In other words, each deposit has one
extra interest-earning period.
F
Investor A
0 1 2 3 4 5 6 7 8 9 10
40 41
$2,000
F
Investor B
0 1 2 3 4 5 6 7 8 9 1011 12
40 41
$2,000
F
Investor C
0 1 2 3 4 5 6 7 8 9 10
40 41
$2,000
Figure 3.22
Cash flow diagrams for three investment options (Example 3.16).
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90 CHAPTER 3 Interest Rate and Economic Equivalence
SOLUTION
Given: Three different deposit scenarios with i = 9.38% and N = 41 years.
Find: Balance at the end of 41 years (or at the age of 65).
• Investor A:
Balance at the end of 10 years
$''''''%''''''&
F65 = $2,0001F/A, 9.38%, 10211.093821F/P, 9.38%, 312
('''''')''''''*
$33,845
= $545,216.
• Investor B:
F65 = $2,0001F/P, 9.38%, 31211.09382
5
$322,159
= $352,377.
• Investor C:
F65 = $2,0001F/P, 9.38%, 41211.09382
5
$820,620
= $897,594.
If you know how your balance changes at the end of each year, you may want to construct a tableau such as the one shown in Table 3.3. Note that, due to rounding errors,
the final balance figures are slightly off from those calculated by interest formulas.
TABLE 3.3
How Time Affects the Value of Money
Investor A
Age Years Contribution
Investor B
Investor C
Year-End
Year-End
Year-End
Value
Contribution
Value
Contribution
Value
25
1
$2,000
$ 2,188
$0
$0
$2,000
$ 2,188
26
2
$2,000
$ 4,580
$0
$0
$2,000
$ 4,580
27
3
$2,000
$ 7,198
$0
$0
$2,000
$ 7,198
28
4
$2,000
$10,061
$0
$0
$2,000
$10,061
29
5
$2,000
$13,192
$0
$0
$2,000
$13,192
30
6
$2,000
$16,617
$0
$0
$2,000
$16,617
31
7
$2,000
$20,363
$0
$0
$2,000
$20,363
32
8
$2,000
$24,461
$0
$0
$2,000
$24,461
33
9
$2,000
$28,944
$0
$0
$2,000
$28,944
34
10
$2,000
$33,846
$0
$0
$2,000
$33,846
(Continued)
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Section 3.3 Development of Interest Formulas 91
TABLE 3.3
Continued
Investor A
Investor B
Investor C
Age Years Contribution
Year-End
Year-End
Year-End
Value
Contribution
Value
Contribution
Value
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
$0
$0
$0
$0
$0
$0
$0
$0
$0
$0
$0
$0
$0
$0
$0
$0
$0
$0
$0
$0
$0
$0
$0
$0
$0
$0
$0
$0
$0
$0
$0
$20,000
$ 37,021
$ 40,494
$ 44,293
$ 48,448
$ 52,992
$ 57,963
$ 63,401
$ 69,348
$ 75,854
$ 82,969
$ 90,752
$ 99,265
$108,577
$118,763
$129,903
$142,089
$155,418
$169,997
$185,944
$203,387
$222,466
$243,335
$266,162
$291,129
$318,439
$348,311
$380,985
$416,724
$455,816
$498,574
$545,344
Value at 65
Less Total Contributions
Net Earnings
$545,344
$20,000
$525,344
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
$2,000
$2,000
$2,000
$2,000
$2,000
$2,000
$2,000
$2,000
$2,000
$2,000
$2,000
$2,000
$2,000
$2,000
$2,000
$2,000
$2,000
$2,000
$2,000
$2,000
$2,000
$2,000
$2,000
$2,000
$2,000
$2,000
$2,000
$2,000
$2,000
$2,000
$2,000
$62,000
$ 2,188
$ 4,580
$ 7,198
$ 10,061
$ 13,192
$ 16,617
$ 20,363
$ 24,461
$ 28,944
$ 33,846
$ 39,209
$ 45,075
$ 51,490
$ 58,508
$ 66,184
$ 74,580
$ 83,764
$ 93,809
$104,797
$116,815
$129,961
$144,340
$160,068
$177,271
$196,088
$216,670
$239,182
$263,807
$290,741
$320,202
$352,427
$ 2,000
$ 2,000
$ 2,000
$ 2,000
$ 2,000
$ 2,000
$ 2,000
$ 2,000
$ 2,000
$ 2,000
$ 2,000
$ 2,000
$ 2,000
$ 2,000
$ 2,000
$ 2,000
$ 2,000
$ 2,000
$ 2,000
$ 2,000
$ 2,000
$ 2,000
$ 2,000
$ 2,000
$ 2,000
$ 2,000
$ 2,000
$ 2,000
$ 2,000
$ 2,000
$ 2,000
$82,000
$352,427
$ 62,000
$290,427
Source: Adapted from Making Money Work for You, UNH Cooperative Extension.
$ 39,209
$ 45,075
$ 51,490
$ 58,508
$ 66,184
$ 74,580
$ 83,764
$ 93,809
$104,797
$116,815
$129,961
$144,340
$160,068
$177,271
$196,088
$216,670
$239,182
$263,807
$290,741
$320,202
$352,427
$387,675
$426,229
$468,400
$514,527
$564,981
$620,167
$680,531
$746,557
$818,777
$897,771
$897,771
$82,000
$815,771
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92 CHAPTER 3 Interest Rate and Economic Equivalence
A
A
1
2
A
A
A
3
N–1
N
0
Years
Lender’s point of view
P
Borrower’s point of view
Years
1
2
3
N–1
N
A
A
A
A
A
0
Figure 3.23
Capital Recovery Factor (Annuity Factor): Find A, Given P, i, and N
We can determine the amount of a periodic payment A if we know P, i, and N. Figure 3.23
illustrates this situation. To relate P to A, recall the relationship between P and F in Eq. (3.3),
F = P11 + i2N. Replacing F in Eq. (3.11) by P11 + i2N, we get
A = P11 + i2N c
i
d,
11 + i2N - 1
or
A = Pc
Capital recovery
factor:
Commonly used
to determine
the revenue
requirements
needed to
address the upfront capital costs
for projects.
Annuity:
An annuity is
essentially a level
stream of cash
flows for a fixed
period of time.
i11 + i2N
11 + i2N - 1
d = P1A>P, i, N2.
(3.12)
Now we have an equation for determining the value of the series of end-of-period payments A when the present sum P is known. The portion within the brackets is called the
equal payment series capital recovery factor, or simply capital recovery factor,
which is designated (A/P, i, N). In finance, this A/P factor is referred to as the annuity
factor and indicates a series of payments of a fixed, or constant, amount for a specified
number of periods.
EXAMPLE 3.17 Uniform Series: Find A, Given P, i, and N
BioGen Company, a small biotechnology firm, has borrowed $250,000 to purchase
laboratory equipment for gene splicing. The loan carries an interest rate of 8% per
year and is to be repaid in equal installments over the next six years. Compute the
amount of the annual installment (Figure 3.24).
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Section 3.3 Development of Interest Formulas 93
$250,000
Years
1
2
3
4
5
6
A
A
A
A
A
A
0
Figure 3.24 A loan cash flow diagram from BioGen’s point of view.
SOLUTION
Given: P = $250,000, i = 8% per year, and N = 6 years.
Find: A.
A = $250,0001A>P, 8%, 62
= $250,00010.21632
= $54,075.
Here is an Excel solution using annuity function commands:
= PMT1i, N, P2
= PMT18%,6, -2500002
= $54,075
EXAMPLE 3.18 Deferred Loan Repayment
In Example 3.17, suppose that BioGen wants to negotiate with the bank to defer the
first loan repayment until the end of year 2 (but still desires to make six equal installments at 8% interest). If the bank wishes to earn the same profit, what should be the
annual installment, also known as deferred annuity (Figure 3.25)?
SOLUTION
Given: P = $250,000, i = 8% per year, and N = 6 years, but the first payment
occurs at the end of year 2.
Find: A.
By deferring the loan for year, the bank will add the interest accrued during the first
year to the principal. In other words, we need to find the equivalent worth P¿ of
$250,000 at the end of year 1:
P¿ = $250,0001F>P, 8%, 12
= $270,000.
Deferred
annuity:
A type of annuity
contract that
delays payments
of income,
installments, or
a lump sum until
the investor
elects to
receive them.
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94 CHAPTER 3 Interest Rate and Economic Equivalence
$250,000
Grace period
1
2
3
Years
4
5
6
7
A
A
A
A
Years
4
5
6
7
A
A
A
0
Grace period:
The additional
period of time a
lender provides
for a borrower
to make
payment on a
debt without
penalty.
A
A
(a) Original cash flow
P $250,000 (F/P, 8%, 1)
Grace period
1
2
3
0
A
A
(b) Equivalent cash flow
A
Figure 3.25 A deferred loan cash flow diagram
from BioGen’s point of view (Example 3.17).
In fact, BioGen is borrowing $270,000 for six years. To retire the loan with six equal
installments, the deferred equal annual payment on P¿ will be
A¿ = $270,0001A>P, 8%, 62
= $58,401.
By deferring the first payment for one year, BioGen needs to make additional payments of $4,326 in each year.
Present-Worth Factor: Find P, Given A, i, and N
What would you have to invest now in order to withdraw A dollars at the end of each of
the next N periods? In answering this question, we face just the opposite of the equal payment capital recovery factor situation: A is known, but P has to be determined. With the
capital recovery factor given in Eq. (3.12), solving for P gives us
P = Ac
11 + i2N - 1
i11 + i2N
d = A1P>A, i, N2.
(3.13)
The bracketed term is referred to as the equal payment series present-worth factor and
is designated (P/A, i, N).
EXAMPLE 3.19 Uniform Series: Find P, Given A, i, and N
Let us revisit Louise Outing’s lottery problem, introduced in the chapter opening.
Suppose that Outing were able to find an investor who was willing to buy her lottery
ticket for $2 million. Recall that after an initial gross payment of $283,770, Outing
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Section 3.3 Development of Interest Formulas 95
A $280,000
0
1 2 3
17 18 19
P?
Figure 3.26 A cash flow diagram for Louise Outing’s lottery
winnings (Example 3.19).
would be paid 19 annual gross checks of $280,000. (See Figure 3.26.) If she could
invest her money at 8% interest, what would be the fair amount to trade her 19 future
lottery receipts? (Note that she already cashed in $283,770 after winning the lottery,
so she is giving up 19 future lottery checks in the amount of $280,000.)
SOLUTION
Given: i = 8% per year, A = $280,000, and N = 19 years.
Find: P.
• Using interest factor:
P = $280,0001P>A, 8%, 192 = $280,00019.60362
= $2,689,008.
• Using Excel:
= PV18%,19,-2800002= $2,689,008
COMMENTS: Clearly, we can tell Outing that giving up $280,000 a year for 19 years
to receive $2 million today is a losing proposition if she can earn only an 8% return
on her investment. At this point, we may be interested in knowing at just what rate of
return her deal (receiving $2 million) would in fact make sense. Since we know that
P = $2,000,000, N = 19, and A = $280,000, we solve for i.
If you know the cash flows and the PV (or FV) of a cash flow stream, you can determine the interest rate. In this case, you are looking for the interest rate that caused
the P/A factor to equal 1P/A, i, 192 = 1$2,000,000/$280,0002 = 7.1429. Since we
are dealing with an annuity, we could proceed as follows:
• With a financial calculator, enter N = 19, PV = $2,000,000, PMT =
-280,000, and then press the i key to find i = 12.5086%.
• To use the interest tables, first recognize that $2,000,000 = $280,000 *
(P/A, i, 19) or 1P/A, i, 192 = 7.1429. Look up 7.1429 or a close value in
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96 CHAPTER 3 Interest Rate and Economic Equivalence
Appendix A. In the P/A column with N = 19 in the 12% interest table, you
will find that 1P/A, 12%, 192 = 7.3658. If you look up the 13% interest
table, you find that 1P/A, 12%, 192 = 6.9380, indicating that the interest rate
should be closer to 12.5%.
• To use Excel’s financial command, you simply evaluate the following command to solve the unknown interest rate problem for an annuity:
= RATE1N,A,P,F,type,guess2
= RATE119,280000, -2000000,0,0,10%2
= 12.5086%
It is not likely that Outing will find a financial investment which provides this
high rate of return. Thus, even though the deal she has been offered is not a
good one for economic reasons, she could accept it knowing that she has not
much time to enjoy the future benefits.
3.3.5 Linear Gradient Series
Engineers frequently encounter situations involving periodic payments that increase or
decrease by a constant amount (G) from period to period. These situations occur often
enough to warrant the use of special equivalence factors that relate the arithmetic gradient to other cash flows. Figure 3.27 illustrates a strict gradient series, A n = 1n - 12G.
Note that the origin of the series is at the end of the first period with a zero value. The gradient G can be either positive or negative. If G 7 0, the series is referred to as an
increasing gradient series. If G 6 0, it is a decreasing gradient series.
Unfortunately, the strict form of the increasing or decreasing gradient series does not
correspond with the form that most engineering economic problems take. A typical problem involving a linear gradient includes an initial payment during period 1 that increases
by G during some number of interest periods, a situation illustrated in Figure 3.28. This
contrasts with the strict form illustrated in Figure 3.27, in which no payment is made during period 1 and the gradient is added to the previous payment beginning in period 2.
Gradient Series as Composite Series
In order to utilize the strict gradient series to solve typical problems, we must view cash
flows as shown in Figure 3.28 as a composite series, or a set of two cash flows, each
(N 1)G
(N 2)G
Note that the first cash flow in
a strict linear gradient series is 0.
3G
2G
G
0
0
Figure 3.27
1
2
3
Years
4
N1
A cash flow diagram for a strict gradient series.
N
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Section 3.3 Development of Interest Formulas 97
A1 5G
A1 G
A1
A1
0
1
2
3
4
5
6
G
3 4 5 6 0
Years
(a) Increasing gradient series
0
1
A1
A1 G
2
1
2
2G
3G
3
4
4G
5
5G
6
A1
A1 5G
G
3 4 5 6 0
Years
(b) Decreasing gradient series
2G
3G
4G
5G
0
1
2
3
4
5
6
0
1
2
1
2
3
4
5
6
Figure 3.28 Two types of linear gradient series as composites of a uniform
series of N payments of A 1 and the gradient series of increments of constant
amount G.
corresponding to a form that we can recognize and easily solve: a uniform series of N
payments of amount A 1 and a gradient series of increments of constant amount G. The
need to view cash flows that involve linear gradient series as composites of two series is
very important in solving problems, as we shall now see.
Present-Worth Factor: Linear Gradient: Find P, Given G, N, and i
How much would you have to deposit now to withdraw the gradient amounts specified in
Figure 3.27? To find an expression for the present amount P, we apply the single-payment
present-worth factor to each term of the series and obtain
P = 0 +
or
1N - 12G
2G
G
Á +
+
+
,
11 + i2N
11 + i22
11 + i23
N
P = a 1n - 12G11 + i2-n.
(3.14)
n=1
Letting G = a and 1/11 + i2 = x yields
P = 0 + ax2 + 2ax3 + Á + 1N - 12axN
= ax[0 + x + 2x2 + Á + 1N - 12xN - 1].
Since an arithmetic–geometric series 50, x, 2x , Á , 1N - 12x
2
0 + x + 2x2 + Á + 1N - 12xN - 1 = x B
N-1
(3.15)
6 has the finite sum
1 - NxN - 1 + 1N - 12xN
11 - x22
R,
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98 CHAPTER 3 Interest Rate and Economic Equivalence
we can rewrite Eq. (3.15) as
P = ax2 B
1 - NxN - 1 + 1N - 12xN
11 - x22
R.
(3.16)
Replacing the original values for A and x, we obtain
P = GB
11 + i2N - iN - 1
i211 + i2N
R = G1P>G, i, N2.
(3.17)
The resulting factor in brackets is called the gradient series present-worth factor,
which we denote as (P/G, i, N).
EXAMPLE 3.20 Linear Gradient: Find P, Given A1, G, i, and N
A textile mill has just purchased a lift truck that has a useful life of five years. The engineer estimates that maintenance costs for the truck during the first year will be
$1,000. As the truck ages, maintenance costs are expected to increase at a rate of
$250 per year over the remaining life. Assume that the maintenance costs occur at the
end of each year. The firm wants to set up a maintenance account that earns 12% annual interest. All future maintenance expenses will be paid out of this account. How
much does the firm have to deposit in the account now?
SOLUTION
Given: A 1 = $1,000, G = $250, i = 12% per year, and N = 5 years.
Find: P.
Asking how much the firm has to deposit now is equivalent to asking what the equivalent present worth for this maintenance expenditure is if 12% interest is used. The
cash flow may be broken into two components as shown in Figure 3.29.
Equal payment series
$1,000
$1,250
$1,500
$1,750
$1,000
0
1 2 3 4 5
$2,000
0
1
2
3
Years
4
P1
Gradient series
$1,000
$750
$500
$250
5
P P1 P2
0
1 2 3 4 5
P2
Figure 3.29
Cash flow diagram (Example 3.20).
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Section 3.3 Development of Interest Formulas 99
The first component is an equal payment series 1A 12, and the second is a linear gradient series (G). We have
P = P1 + P2
P = A 11P>A, 12%, 52 + G1P>G, 12%, 52
= $1,00013.60482 + $25016.3972
= $5,204.
Note that the value of N in the gradient factor is 5, not 4. This is because, by definition of the series, the first gradient value begins at period 2.
COMMENTS: As a check, we can compute the present worth of the cash flow by using
the (P/F, 12%, n) factors:
Period (n)
Cash Flow
(P/F, 12%, n)
Present Worth
1
$1,000
0.8929
$ 892.90
2
1,250
0.7972
996.50
3
1,500
0.7118
1,067.70
4
1,750
0.6355
1,112.13
5
2,000
0.5674
Total
1,134.80
$5,204.03
The slight difference is caused by a rounding error.
Gradient-to-Equal-Payment Series Conversion Factor: Find A,
Given G, i, and N
We can obtain an equal payment series equivalent to the gradient series, as depicted in
Figure 3.30, by substituting Eq. (3.17) for P into Eq. (3.12) to obtain
A = GB
11 + i2N - iN - 1
i[11 + i2N - 1]
R = G1A>G, i, N2,
(3.18)
where the resulting factor in brackets is referred to as the gradient-to-equal-payment series
conversion factor and is designated (A/G, i, N).
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100 CHAPTER 3 Interest Rate and Economic Equivalence
(N – 1)G
Strict gradient series
Equivalent uniform series
(N – 2)G
A = G (A/G, i, N)
3G
2G
A
A
A
A
A
A
1
2
3
Years
4
N–1
N
G
⬅
0
1
2
3
Years
4
N–1
N
0
Figure 3.30
EXAMPLE 3.21 Linear Gradient: Find A, Given A1, G, i, and N
John and Barbara have just opened two savings accounts at their credit union. The
accounts earn 10% annual interest. John wants to deposit $1,000 in his account at the
end of the first year and increase this amount by $300 for each of the next five years.
Barbara wants to deposit an equal amount each year for the next six years. What
should be the size of Barbara’s annual deposit so that the two accounts will have
equal balances at the end of six years (Figure 3.31)?
0
John’s deposit plan
0
1
Years
2
3
4
Equal deposit plan
1 2 3 4 5
6
A1 $1,000
5
6
⬅
$1,000
$1,300
$1,600
$1,900
$2,200
$2,500
0
Gradient deposit series
1 2 3 4 5 6
$300
$600
$900
$1,200
$1,500
G $300
Figure 3.31 John’s deposit series viewed as a combination of uniform
and gradient series (Example 3.21).
SOLUTION
Given: A 1 = $1,000, G = $300, i = 10%, and N = 6.
Find: A.
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Section 3.3 Development of Interest Formulas 101
Since we use the end-of-period convention unless otherwise stated, this series begins
at the end of the first year and the last contribution occurs at the end of the sixth year.
We can separate the constant portion of $1,000 from the series, leaving the gradient
series of 0, 0, 300, 600, Á , 1,500.
To find the equal payment series beginning at the end of year 1 and ending at
year 6 that would have the same present worth as that of the gradient series, we may
proceed as follows:
A = $1,000 + $3001A>G, 10%, 62
= $1,000 + $30012.222362
= $1,667.08.
Barbara’s annual contribution should be $1,667.08.
COMMENTS: Alternatively, we can compute Barbara’s annual deposit by first computing the equivalent present worth of John’s deposits and then finding the equivalent
uniform annual amount. The present worth of this combined series is
P = $1,0001P>A, 10%, 62 + $3001P>G, 10%, 62
= $1,00014.35532 + $30019.68422
= $7,260.56.
The equivalent uniform deposit is
A = $7,260.561A>P, 10%, 62 = $1,667.02.
(The slight difference in cents is caused by a rounding error.)
EXAMPLE 3.22 Declining Linear Gradient: Find F,
Given A1, G, i, and N
Suppose that you make a series of annual deposits into a bank account that pays 10%
interest. The initial deposit at the end of the first year is $1,200. The deposit amounts
decline by $200 in each of the next four years. How much would you have immediately after the fifth deposit?
SOLUTION
Given: Cash flow shown in Figure 3.32, i = 10% per year, and N = 5 years.
Find: F.
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102 CHAPTER 3 Interest Rate and Economic Equivalence
Equal payment series
F1
F = F1 – F2
0
1
2
Original cash flow
Years
3
4
5
A1 = $1,200
Years
0
2
1
3
4
5
–
$400
Gradient series
$600
F2
$800
$1,000
0
$1,200
Years
2
3
1
$200
$400
4
5
$600
$800
Figure 3.32
The cash flow includes a decreasing gradient series. Recall that we derived the linear
gradient factors for an increasing gradient series. For a decreasing gradient series, the
solution is most easily obtained by separating the flow into two components: a uniform series and an increasing gradient that is subtracted from the uniform series
(Figure 3.32). The future value is
Geometric
growth:
The year-overyear growth rate
of an investment
over a specified
period of time.
Compound
growth is an
imaginary number that describes the rate
at which an investment grew
as though it had
grown at a
steady rate.
F =
=
=
=
F1 - F2
A 11F>A, 10%, 52 - $2001P>G, 10%, 521F>P, 10%, 52
$1,20016.1052 - $20016.862211.6112
$5,115.
3.3.6 Geometric Gradient Series
Many engineering economic problems—particularly those relating to construction
costs—involve cash flows that increase or decrease over time, not by a constant amount
(as with a linear gradient), but rather by a constant percentage (a geometric gradient).
This kind of cash flow is called compound growth. Price changes caused by inflation are
a good example of a geometric gradient series. If we use g to designate the percentage
change in a payment from one period to the next, the magnitude of the nth payment, A n,
is related to the first payment A 1 by the formula
A n = A 111 + g2n - 1, n = 1, 2, Á , N.
(3.19)
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Section 3.3 Development of Interest Formulas 103
A1(1 g)N1
A1
A1(1 g)
A1(1 g)2
A1(1 g)2
A1(1 g)
A1(1 g)N1
A1
0
1
2
3
N
4
1
2
3
4
g>0
g<0
Increasing geometric series
Decreasing geometric series
P
N
P
Figure 3.33 A geometrically increasing or decreasing gradient series at a
constant rate g.
The variable g can take either a positive or a negative sign, depending on the type of cash
flow. If g 7 0, the series will increase, and if g 6 0, the series will decrease. Figure 3.33
illustrates the cash flow diagram for this situation.
Present-Worth Factor: Find P, Given A1,g, i, and N
Notice that the present worth of any cash flow A n at interest rate i is
Pn = A n11 + i2-n = A 111 + g2n - 111 + i2-n.
To find an expression for the present amount P for the entire series, we apply the
single-payment present-worth factor to each term of the series:
N
P = a A 111 + g2n - 111 + i2-n.
(3.20)
n=1
Bringing the constant term A 111 + g2-1 outside the summation yields
P =
Let a =
N 1 + g n
A1
c
d .
11 + g2 na
=1 1 + i
(3.21)
1 + g
A1
and x =
. Then, rewrite Eq. (3.21) as
1 + g
1 + i
P = a1x + x2 + x3 + Á + xN2.
(3.22)
Since the summation in Eq. (3.22) represents the first N terms of a geometric series, we
may obtain the closed-form expression as follows: First, multiply Eq. (3.22) by x to get
xP = a1x2 + x3 + x4 + Á + xN + 12.
(3.23)
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104 CHAPTER 3 Interest Rate and Economic Equivalence
Then, subtract Eq. (3.23) from Eq. (3.22):
P - xP = a1x - xN + 12
P11 - x2 = a1x - xN + 12
P =
a1x - xN + 12
1 - x
1x Z 12.
(3.24)
If we replace the original values for a and x, we obtain
1 - 11 + g2N11 + i2-N
d
i - g
P = c
NA 1>11 + i2
A1 c
if i Z g
,
if i = g
(3.25)
or
P = A 11P>A 1, g, i, N2.
The factor within brackets is called the geometric-gradient-series present-worth factor
and is designated 1P/A 1, g, i, N2. In the special case where i = g, Eq. (3.21) becomes
P = [A 1/11 + i2]N.
EXAMPLE 3.23 Geometric Gradient: Find P, Given A1,
g, i, and N
Ansell, Inc., a medical device manufacturer, uses compressed air in solenoids and
pressure switches in its machines to control various mechanical movements. Over
the years, the manufacturing floor has changed layouts numerous times. With each
new layout, more piping was added to the compressed-air delivery system to accommodate new locations of manufacturing machines. None of the extra, unused old pipe was
capped or removed; thus, the current compressed-air delivery system is inefficient
and fraught with leaks. Because of the leaks, the compressor is expected to run 70%
of the time that the plant will be in operation during the upcoming year. This will
require 260 kWh of electricity at a rate of $0.05/kWh. (The plant runs 250 days a
year, 24 hours per day.) If Ansell continues to operate the current air delivery system,
the compressor run time will increase by 7% per year for the next five years because
of ever-worsening leaks. (After five years, the current system will not be able to meet
the plant’s compressed-air requirement, so it will have to be replaced.) If Ansell
decides to replace all of the old piping now, it will cost $28,570.
The compressor will still run the same number of days; however, it will run 23%
less (or will have 70%11 - 0.232 = 53.9% usage during the day) because of the reduced air pressure loss. If Ansell’s interest rate is 12%, is the machine worth fixing now?
SOLUTION
Given: Current power consumption, g = 7%, i = 12%, and N = 5 years.
Find: A 1 and P.
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Section 3.3 Development of Interest Formulas 105
Step 1: We need to calculate the cost of power consumption of the current piping
system during the first year:
Power cost = % of day operating
* days operating per year
* hours per day
* kWh * $/kWh
= 170%2 * 1250 days>year2 * 124 hours>day2
* 1260 kWh2 * 1$0.05/kWh2
= $54,440.
Step 2: Each year, the annual power cost will increase at the rate of 7% over the previous year’s cost. The anticipated power cost over the five-year period is
summarized in Figure 3.34. The equivalent present lump-sum cost at 12%
for this geometric gradient series is
POld = $54,4401P>A 1, 7%, 12%, 52
= $54,440c
1 - 11 + 0.072511 + 0.122-5
d
0.12 - 0.07
= $222,283.
Years
0
1
2
$54,440
$58,251
3
4
5
$62,328
$66,691
g = 7%
$71,360
Figure 3.34 Annual power cost series if repair is not performed.
Step 3: If Ansell replaces the current compressed-air system with the new one, the
annual power cost will be 23% less during the first year and will remain at
that level over the next five years. The equivalent present lump-sum cost at
12% is then
PNew = $54,44011 - 0.2321P>A, 12%, 52
= $41,918.8013.60482
= $151,109.
Step 4: The net cost for not replacing the old system now is $71,174
( = $222,283 - $151,1092. Since the new system costs only $28,570, the
replacement should be made now.
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106 CHAPTER 3 Interest Rate and Economic Equivalence
COMMENTS: In this example, we assumed that the cost of removing the old system
was included in the cost of installing the new system. If the removed system has
some salvage value, replacing it will result in even greater savings. We will consider
many types of replacement issues in Chapter 14.
EXAMPLE 3.24 Geometric Gradient: Find A1, Given F, g, i, and N
Jimmy Carpenter, a self-employed individual, is opening a retirement account at a
bank. His goal is to accumulate $1,000,000 in the account by the time he retires from
work in 20 years’ time. A local bank is willing to open a retirement account that pays
8% interest compounded annually throughout the 20 years. Jimmy expects that his
annual income will increase 6% yearly during his working career. He wishes to start
with a deposit at the end of year 1 1A 12 and increase the deposit at a rate of 6% each
year thereafter. What should be the size of his first deposit 1A 12? The first deposit
will occur at the end of year 1, and subsequent deposits will be made at the end of
each year. The last deposit will be made at the end of year 20.
SOLUTION
Given: F = $1,000,000, g = 6% per year, i = 8% per year, and N = 20 years.
Find: A 1 as in Figure 3.35.
We have
F = A11P>A1, 6%, 8%, 202 (F>P, 8%, 20)
= A1172.69112.
Solving for A 1 yields
A 1 = $1,000,000>72.6911 = $13,757.
$1,000,000
Years
0
1
2
3
4
5
6
7
8
9
10 11 12 13 14 15 16 17 18 19 20
A1 = $13,757
A20 = A1(1+ 0.06)19
= $41,623
Figure 3.35
Jimmy Carpenter’s retirement plan (Example 3.24).
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Section 3.4 Unconventional Equivalence Calculations 107
Table 3.4 summarizes the interest formulas developed in this section and the cash
flow situations in which they should be used. Recall that these formulas are applicable
only to situations where the interest (compounding) period is the same as the payment period (e.g., annual compounding with annual payment). Also, we present some useful
Excel financial commands in the table.
3.4 Unconventional Equivalence Calculations
In the preceding section, we occasionally presented two or more methods of attacking
problems even though we had standard interest factor equations by which to solve them.
It is important that you become adept at examining problems from unusual angles and
that you seek out unconventional solution methods, because not all cash flow problems
conform to the neat patterns for which we have discovered and developed equations.
Two categories of problems that demand unconventional treatment are composite
(mixed) cash flows and problems in which we must determine the interest rate implicit
in a financial contract. We will begin this section by examining instances of composite
cash flows.
3.4.1 Composite Cash Flows
Although many financial decisions do involve constant or systematic changes in cash
flows, others contain several components of cash flows that do not exhibit an overall pattern. Consequently, it is necessary to expand our analysis to deal with these mixed types of
cash flows.
To illustrate, consider the cash flow stream shown in Figure 3.36. We want to compute the equivalent present worth for this mixed payment series at an interest rate of 15%.
Three different methods are presented.
Method 1. A “brute force” approach is to multiply each payment by the appropriate (P/F, 10%, n) factors and then to sum these products to obtain the present worth
of the cash flows, $543.72. Recall that this is exactly the same procedure we used to
solve the category of problems called the uneven payment series, described in
Section 3.3.3. Figure 3.36 illustrates this computational method.
Method 2. We may group the cash flow components according to the type of cash
flow pattern that they fit, such as the single payment, equal payment series, and so
forth, as shown in Figure 3.37. Then the solution procedure involves the following
steps:
• Group 1: Find the present worth of $50 due in year 1:
$501P>F, 15%, 12 = $43.48.
• Group 2: Find the equivalent worth of a $100 equal payment series at year 1
1V12, and then bring this equivalent worth at year 0 again:
$1001P>A, 15%, 321P>F, 15%, 12 = $198.54.
5
V1
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108 CHAPTER 3 Interest Rate and Economic Equivalence
TABLE 3.4
Flow Type
Summary of Discrete Compounding Formulas with Discrete Payments
Factor
Notation
Formula
Compound
amount
(F/P, i, N)
Present
worth
(P/F, i, N)
Compound
amount
(F/A, i, N)
P
A
Y
M
E
N
T
Sinking
fund
(A/F, i, N)
A = Fc
Present
worth
(P/A, i, N)
P = Ac
G
R
A
D
I
E
N
T
S
E
R
I
E
S
Cash Flow
Diagram
F = P11 + i2N
S
I
N
G
L
E
E
Q
U
A
L
S
E
R
I
E
S
Excel Command
F
= FV1i, N, P,, 02
0
N
P = F11 + i2-N
F = Ac
11 + i2N - 1
i
= PV1i, N, F,, 02
d
F
= PV1i, N, A,, 02
F
0 1 2 3 N1
= PMT1i, N, P, F, 02
i
d
11 + i2N - 1
11 + i2N - 1
i11 + i2N
d
= PV1i, N, A,, 02
AAA
AAA
N
AA
AA
1 2 3 N1N
Capital
recovery
(A/P, i, N)
A = Pc
Linear
gradient
i11 + i2N
11 + i2N - 1
d
Present
worth
(P/G, i, N)
P = Gc
11 + i2N - iN - 1
Conversion factor
(A/G, i, N)
A = Gc
11 + i2N - iN - 1
Geometric
gradient
A1 c
i 11 + i2
2
N
i[11 + i2N - 1]
= PMT1i, N,, P2
d
1 23
P
d
1 - 11 + g2N11 + i2-N
i - g
Present
P = D
N
worth
A1 a
b1if i = g2
1
+ i
1P/A 1, g, i, N2
(N2)G
2G
G
d
N1N
A1(1g)N1
A3
A2
A1
123
P
N
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Section 3.4 Unconventional Equivalence Calculations 109
$200
$150
$100
$100
2
3
$150
$150
$150
6
7
8
$100
$50
1
0
5
4
Years
9
$50 (P/F,15%,1) = $43.48
$100 (P/F,15%,2) = $75.61
$100 (P/F,15%,3) = $65.75
$100 (P/F,15%,4) = $57.18
$150 (P/F,15%,5) = $74.58
$150 (P/F,15%,6) = $64.85
$150 (P/F,15%,7) = $56.39
$150 (P/F,15%,8) = $49.04
$200 (P/F,15%,9) = $57.78
$543.72
$543.72
Figure 3.36 Equivalent present worth calculation using only
P/F factors (Method 1 “Brute Force Approach”).
Group 1
Group 2
Group 4
Group 3
$200
$150
$100
$100
2
3
$150
$150
$150
6
7
8
$100
$50
0
1
5
4
Years
9
$43.48 $50(P/F, 15%, 1)
$198.54 $100(P/A, 15%, 3)(P/F, 15%, 1)
$244.85 $150(P/A, 15%, 4)(P/F, 15%, 4)
$543.72
$56.85 $200(P/F, 15%, 9)
$543.72
Figure 3.37 Equivalent present-worth calculation for an
uneven payment series, using P/F and P/A factors (Method 2:
grouping approach).
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110 CHAPTER 3 Interest Rate and Economic Equivalence
• Group 3: Find the equivalent worth of a $150 equal payment series at year 4
1V42, and then bring this equivalent worth at year 0.
$1501P>A, 15%, 421P>F, 15%, 42 = $244.85.
5
V4
• Group 4: Find the equivalent present worth of the $200 due in year 9:
$2001P>F, 15%, 92 = $56.85.
• Group total—sum the components:
P = $43.48 + $198.54 + $244.85 + $56.85 = $543.72.
A pictorial view of this computational process is given in Figure 3.37.
Method 3. In computing the present worth of the equal payment series components, we may use an alternative method.
• Group 1: Same as in Method 2.
• Group 2: Recognize that a $100 equal payment series will be received during
years 2 through 4. Thus, we could determine the value of a four-year annuity,
subtract the value of a one-year annuity from it, and have remaining the value
of a four-year annuity whose first payment is due in year 2. This result is
achieved by subtracting the (P/A, 15%, 1) for a one-year, 15% annuity from
that for a four-year annuity and then multiplying the difference by $100:
$100[1P>A, 15%, 42 - 1P>A, 15%, 12] = $10012.8550 - 0.86962
= $198.54.
Thus, the equivalent present worth of the annuity component of the uneven
stream is $198.54.
• Group 3: We have another equal payment series that starts in year 5 and ends
in year 8.
$150[1P>A, 15%, 82 - 1P>A, 15%, 42] = $15014.4873 - 2.85502
= $244.85.
• Group 4: Same as Method 2.
• Group total—sum the components:
P = $43.48 + $198.54 + $244.85 + $56.85 = $543.72.
Either the “brute force” method of Figure 3.35 or the method utilizing both (P/A, i, n)
and (P/F, i, n) factors can be used to solve problems of this type. However, Method 2 or
Method 3 is much easier if the annuity component runs for many years. For example,
the alternative solution would be clearly superior for finding the equivalent present
worth of a stream consisting of $50 in year 1, $200 in years 2 through 19, and $500 in
year 20.
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Section 3.4 Unconventional Equivalence Calculations 111
Also, note that in some instances we may want to find the equivalent value of a
stream of payments at some point other than the present (year 0). In this situation, we
proceed as before, but compound and discount to some other point in time—say, year 2,
rather than year 0. Example 3.25 illustrates the situation.
EXAMPLE 3.25 Cash Flows with Subpatterns
The two cash flows in Figure 3.38 are equivalent at an interest rate of 12% compounded annually. Determine the unknown value C.
SOLUTION
Given: Cash flows as in Figure 3.38; i = 12% per year.
Find: C.
• Method 1.
Compute the present worth of each cash flow at time 0:
P1 = $1001P>A, 12%, 22 + $3001P>A, 12%, 321P>F, 12%, 22
= $743.42;
P2 = C1P>A, 12%, 52 - C1P>F, 12%, 32
= 2.8930C.
Since the two flows are equivalent, P1 = P2, and we have
743.42 = 2.8930C.
Solving for C, we obtain C = $256.97.
• Method 2. We may select a time point other than 0 for comparison. The best
choice of a base period is determined largely by the cash flow patterns. Obviously, we want to select a base period that requires the minimum number of interest factors for the equivalence calculation. Cash flow 1 represents a
combined series of two equal payment cash flows, whereas cash flow 2 can be
viewed as an equal payment series with the third payment missing. For cash
Cash flow 2
Cash flow 1
$300
0
$100
$100
1
2
3
$300
4
$300
5
0
Years
Figure 3.38
Equivalence calculation (Example 3.25).
C
C
1
2
3
Years
C
C
4
5
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112 CHAPTER 3 Interest Rate and Economic Equivalence
flow 1, computing the equivalent worth at period 5 will require only two interest factors:
V5,1 = $1001F>A, 12%, 52 + $2001F>A, 12%, 32
= $1,310.16.
For cash flow 2, computing the equivalent worth of the equal payment series at
period 5 will also require two interest factors:
V5,2 = C1F>A, 12%, 52 - C1F>P, 12%, 22
= 5.0984C.
Therefore, the equivalence would be obtained by letting V5,1 = V5,2:
$1,310.16 = 5.0984C.
Solving for C yields C = $256.97, which is the same result obtained from
Method 1. The alternative solution of shifting the time point of comparison will
require only four interest factors, whereas Method 1 requires five interest factors.
EXAMPLE 3.26 Establishing a College Fund
A couple with a newborn daughter wants to save for their child’s college expenses in
advance. The couple can establish a college fund that pays 7% annual interest. Assuming that the child enters college at age 18, the parents estimate that an amount of
$40,000 per year (actual dollars) will be required to support the child’s college expenses
for 4 years. Determine the equal annual amounts the couple must save until they send
their child to college. (Assume that the first deposit will be made on the child’s first
birthday and the last deposit on the child’s 18th birthday. The first withdrawal will be
made at the beginning of the freshman year, which also is the child’s 18th birthday.)
SOLUTION
Given: Deposit and withdrawal series shown in Figure 3.39; i = 7% per year.
Find: Unknown annual deposit amount (X).
$40,000
0
1
2
3
4
5
6
7
Years
8 9 10 11 12 13 14 15 16 17
18 19 20 21
X?
Figure 3.39 Establishing a college fund (Example 3.26). Note
that the $40,000 figure represents the actual anticipated expenditures
considering the future inflation.
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Section 3.4 Unconventional Equivalence Calculations 113
• Method 1.
Establish economic equivalence at period 0:
Step 1: Find the equivalent single lump-sum deposit now:
PDeposit = X1P/A, 7%, 182
= 10.0591X.
Step 2: Find the equivalent single lump-sum withdrawal now:
PWithdrawal = $40,0001P>A, 7%, 421P>F, 7%, 172
= $42,892.
Step 3: Since the two amounts are equivalent, by equating PDeposit =
PWithdrawal, we obtain X:
10.0591X = $42,892
X = $4,264.
• Method 2.
Establish the economic equivalence at the child’s 18th birthday:
Step 1: Find the accumulated deposit balance on the child’s 18th birthday:
V18 = X1F>A, 7%, 182
= 33.9990X.
Step 2: Find the equivalent lump-sum withdrawal on the child’s 18th
birthday:
V18 = $40,000 + $40,0001P>A, 7%, 32
= $144,972.
Step 3: Since the two amounts must be the same, we obtain
33.9990X = $144,972
X = $4,264.
The computational steps are summarized in Figure 3.40. In general, the second
method is the more efficient way to obtain an equivalence solution to this type
of decision problem.
COMMENTS: To verify whether the annual deposits of $4,264 over 18 years would be
sufficient to meet the child’s college expenses, we can calculate the actual year-byyear balances: With the 18 annual deposits of $4,264, the balance on the child’s 18th
birthday is
$4,2641F>A, 7%, 182 = $144,972.
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114 CHAPTER 3 Interest Rate and Economic Equivalence
$144,972
$40,000
0
1
2
3
4
5
6
7
8
9 10 11 12 13 14 15 16 17
18 19 20 21
x?
33.999X
Figure 3.40
An alternative equivalence calculation (Example 3.26).
From this balance, the couple will make four annual tuition payments:
Year
N
Tuition
Payment
Ending
Balance
0
$40,000
$104,972
104,972
7,348
40,000
72,320
Junior
72,320
5,062
40,000
37,382
Senior
37,382
2,618
40,000
0
Freshman
Sophomore
Beginning
Balance
$144,972
Interest
Earned
$
3.4.2 Determining an Interest Rate to Establish
Economic Equivalence
Thus far, we have assumed that, in equivalence calculations, a typical interest rate is
given. Now we can use the same interest formulas that we developed earlier to determine interest rates that are explicit in equivalence problems. For most commercial loans,
interest rates are already specified in the contract. However, when making some investments in financial assets, such as stocks, you may want to know the rate of growth (or
rate of return) at which your asset is appreciating over the years. (This kind of calculation is the basis of rate-of-return analysis, which is covered in Chapter 7.) Although we
can use interest tables to find the rate that is implicit in single payments and annuities, it
is more difficult to find the rate that is implicit in an uneven series of payments. In such
cases, a trial-and-error procedure or computer software may be used. To illustrate, consider Example 3.27.
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Section 3.4 Unconventional Equivalence Calculations 115
EXAMPLE 3.27 Calculating an Unknown Interest Rate
with Multiple Factors
You may have already won $2 million! Just peel the game piece off the Instant Winner
Sweepstakes ticket, and mail it to us along with your order for subscriptions to your
two favorite magazines. As a grand prize winner, you may choose between a $1 million
cash prize paid immediately or $100,000 per year for 20 years—that’s $2 million! Suppose that, instead of receiving one lump sum of $1 million, you decide to accept the 20
annual installments of $100,000. If you are like most jackpot winners, you will be
tempted to spend your winnings to improve your lifestyle during the first several years.
Only after you get this type of spending “out of your system” will you save later sums
for investment purposes. Suppose that you are considering the following two options:
Option 1: You save your winnings for the first 7 years and then spend every
cent of the winnings in the remaining 13 years.
Option 2: You do the reverse, spending for 7 years and then saving for 13 years.
If you can save winnings at 7% interest, how much would you have at the end of 20
years, and what interest rate on your savings will make these two options equivalent? (Cash flows into savings for the two options are shown in Figure 3.41.)
SOLUTION
Given: Cash flows in Figure 3.41.
Find: (a) F and (b) i at which the two flows are equivalent.
(a) In Option 1, the net balance at the end of year 20 can be calculated in two
steps: Find the accumulated balance at the end of year 7 1V72 first; then find
the equivalent worth of V7 at the end of year 20. For Option 2, find the equivalent worth of the 13 equal annual deposits at the end of year 20. We thus have
F?
Option 1: Early savings series
0
1
2
3
4
5
6
7
8
Years
9 10 11 12 13 14 15 16 17 18 19 20
$100,000
F?
Option 2: Late savings series
0
1
2
3
4
5
6
7
8
Years
9 10 11 12 13 14 15 16 17 18 19 20
$100,000
Figure 3.41
Equivalence calculation (Example 3.27).
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116 CHAPTER 3 Interest Rate and Economic Equivalence
FOption 1 = $100,0001F>A, 7%, 721F>P, 7%, 132
= $2,085,485;
FOption 2 = $100,0001F>A, 7%, 132
= $2,014,064.
Option 1 accumulates $71,421 more than Option 2.
(b) To compare the alternatives, we may compute the present worth for each option
at period 0. By selecting period 7, however, we can establish the same economic equivalence with fewer interest factors. As shown in Figure 3.42, we calculate the equivalent value V7 for each option at the end of period 7, remembering
that the end of period 7 is also the beginning of period 8. (Recall from Example
3.4 that the choice of the point in time at which to compare two cash flows for
equivalence is arbitrary.)
• For Option 1,
V7 = $100,0001F>A, i, 72.
• For Option 2,
V7 = $100,0001P>A, i, 132.
We equate the two values:
$100,0001F>A, i, 72 = $100,0001P>A, i, 132;
1F>A, i, 72
= 1.
1P>A, i, 132
Option 1: Early savings series
V7 $100,000(F/A, i, 7)
V7
0
1
2
3
4
5
6
7
8
$100,000
Option 2: Late savings series
V7 $100,000(P/A, i, 13)
V7
0
1
2
3
4
5
6
Years
9 10 11 12 13 14 15 16 17 18 19 20
7
8
Years
9 10 11 12 13 14 15 16 17 18 19 20
$100,000
Figure 3.42 Establishing an economic equivalence at period 7
(Example 3.27).
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Section 3.4 Unconventional Equivalence Calculations 117
Here, we are looking for an interest rate that gives a ratio of unity. When using the interest tables, we need to resort to a trial-and-error method. Suppose that we guess the
interest rate to be 6%. Then
1F>A, 6%, 72
8.3938
=
= 0.9482.
1P>A, 6%, 132
8.8527
This is less than unity. To increase the ratio, we need to use a value of i such that it
increases the (F/A, i, 7) factor value, but decreases the (P/A, i, 13) value. This will
happen if we use a larger interest rate. Let’s try i = 7%:
1F>A, 7%, 72
8.6540
=
= 1.0355.
1P>A, 7%, 132
8.3577
Now the ratio is greater than unity.
Interest Rate
(F/A, i, 7)/(P/A, i, 13)
6%
0.9482
?
1.0000
7%
1.0355
As a result, we find that the interest rate is between 6% and 7% and may be approximated by linear interpolation as shown in Figure 3.43:
i = 6% + 17% - 6%2c
= 6% + 1% c
1 - 0.9482
d
1.0355 - 0.9482
0.0518
d
0.0873
= 6.5934%.
a:b=c:d
bc=ad
(F/A, i, 7)
(P/A, i, 13)
1.0355
1.0000
a
c
0.9482
d
b
0.9000
0
6%
6.5934%
i
7%
Figure 3.43 Linear interpolation to find an unknown interest rate
(Example 3.27).
Interpolation
is a method of
constructing
new data points
from a discrete
set of known
data points.
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118 CHAPTER 3 Interest Rate and Economic Equivalence
At 6.5934% interest, the two options are equivalent, and you may decide to indulge
your desire to spend like crazy for the first 7 years. However, if you could obtain a
higher interest rate, you would be wiser to save for 7 years and spend for the next 13.
COMMENTS: This example demonstrates that finding an interest rate is an iterative
process that is more complicated and generally less precise than finding an equivalent worth at a known interest rate. Since computers and financial calculators can
speed the process of finding unknown interest rates, such tools are highly recommended for these types of problem solving. With Excel, a more precise break-even
value of 6.60219% is found by using the Goal Seek function.3
In Figure 3.44, the cell that contains the formula that you want to settle is called
the Set cell 1$F$11 F7-F92. The value you want the formula to change to is called
To value (0) and the part of the formula that you wish to change is called By changing cell ($F$5, interest rate). The Set cell MUST always contain a formula or a function, whereas the Changing cell must contain a value only, not a formula or function.
Figure 3.44 Using the Goal Seek function in Excel to find the break-even interest rate
(Example 3.27). As soon as you select OK you will see that Goal Seek recalculates your formula. You then have two options, OK or Cancel. If you select OK the new term will be inserted
into your worksheet. If you select Cancel, the Goal Seek box will disappear, and your worksheet will be in its original state.
3
Goal Seek can be used when you know the result of a formula, but not the input value required by the formula
to decide the result. You can change the value of a specified cell until the formula that is dependent on the
changed cell returns the result you want. Goal Seek is found under the Tools menu.
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Problems 119
SUMMARY
Money has a time value because it can earn more money over time. A number of terms
involving the time value of money were introduced in this chapter:
Interest is the cost of money. More specifically, it is a cost to the borrower and an
earning to the lender, above and beyond the initial sum borrowed or loaned.
Interest rate is a percentage periodically applied to a sum of money to determine the
amount of interest to be added to that sum.
Simple interest is the practice of charging an interest rate only to an initial sum.
Compound interest is the practice of charging an interest rate to an initial sum and to
any previously accumulated interest that has not been withdrawn from the initial sum.
Compound interest is by far the most commonly used system in the real world.
Economic equivalence exists between individual cash flows and/or patterns of cash
flows that have the same value. Even though the amounts and timing of the cash flows
may differ, the appropriate interest rate makes them equal.
The compound-interest formula, perhaps the single most important equation in this text, is
F = P11 + i2N,
where P is a present sum, i is the interest rate, N is the number of periods for which interest is compounded, and F is the resulting future sum. All other important interest
formulas are derived from this one.
Cash flow diagrams are visual representations of cash inflows and outflows along a
timeline. They are particularly useful for helping us detect which of the following five
patterns of cash flow is represented by a particular problem:
1. Single payment. A single present or future cash flow.
2. Uniform series. A series of flows of equal amounts at regular intervals.
3. Linear gradient series. A series of flows increasing or decreasing by a fixed
amount at regular intervals.
4. Geometric gradient series. A series of flows increasing or decreasing by a fixed
percentage at regular intervals.
5. Irregular series. A series of flows exhibiting no overall pattern. However, patterns
might be detected for portions of the series.
Cash flow patterns are significant because they allow us to develop interest formu-
las, which streamline the solution of equivalence problems. Table 3.4 summarizes the
important interest formulas that form the foundation for all other analyses you will
conduct in engineering economic analysis.
PROBLEMS
Types of Interest
3.1 You deposit $5,000 in a savings account that earns 8% simple interest per year.
What is the minimum number of years you must wait to double your balance?
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120 CHAPTER 3 Interest Rate and Economic Equivalence
Suppose instead that you deposit the $5,000 in another savings account that earns
7% interest compounded yearly. How many years will it take now to double your
balance?
3.2 Compare the interest earned by $1,000 for five years at 8% simple interest with
that earned by the same amount for five years at 8% compounded annually.
3.3 You are considering investing $3,000 at an interest rate of 8% compounded annually for five years or investing the $3,000 at 9% per year simple interest for five
years. Which option is better?
3.4 You are about to borrow $10,000 from a bank at an interest rate of 9% compounded annually. You are required to make five equal annual repayments in the
amount of $2,571 per year, with the first repayment occurring at the end of year 1.
Show the interest payment and principal payment in each year.
Equivalence Concept
3.5 Suppose you have the alternative of receiving either $12,000 at the end of five
years or P dollars today. Currently you have no need for money, so you would deposit the P dollars in a bank that pays 5% interest. What value of P would make
you indifferent in your choice between P dollars today and the promise of $12,000
at the end of five years?
3.6 Suppose that you are obtaining a personal loan from your uncle in the amount of
$20,000 (now) to be repaid in two years to cover some of your college expenses.
If your uncle usually earns 8% interest (annually) on his money, which is invested
in various sources, what minimum lump-sum payment two years from now would
make your uncle happy?
Single Payments (Use of F/P or P/F Factors)
3.7 What will be the amount accumulated by each of these present investments?
(a) $5,000 in 8 years at 5% compounded annually
(b) $2,250 in 12 years at 3% compounded annually
(c) $8,000 in 31 years at 7% compounded annually
(d) $25,000 in 7 years at 9% compounded annually
3.8 What is the present worth of these future payments?
(a) $5,500 6 years from now at 10% compounded annually
(b) $8,000 15 years from now at 6% compounded annually
(c) $30,000 5 years from now at 8% compounded annually
(d) $15,000 8 years from now at 12% compounded annually
3.9 For an interest rate of 13% compounded annually, find
(a) How much can be lent now if $10,000 will be repaid at the end of five years?
(b) How much will be required in four years to repay a $25,000 loan received now?
3.10 How many years will it take an investment to triple itself if the interest rate is 12%
compounded annually?
3.11 You bought 300 shares of Microsoft (MSFT) stock at $2,600 on December 31,
2005. Your intention is to keep the stock until it doubles in value. If you expect
15% annual growth for MSFT stock, how many years do you anticipate holding
onto the stock? Compare your answer with the solution obtained by the Rule of 72
(discussed in Example 3.10).
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Problems 121
3.12 From the interest tables in the text, determine the values of the following factors
by interpolation, and compare your answers with those obtained by evaluating the
F/P factor or the P/F factor:
(a) The single-payment compound-amount factor for 38 periods at 9.5% interest
(b) The single-payment present-worth factor for 47 periods at 8% interest
Uneven Payment Series
3.13 If you desire to withdraw the following amounts over the next five years from a
savings account that earns 8% interest compounded annually, how much do you
need to deposit now?
N
Amount
2
$32,000
3
43,000
4
46,000
5
28,000
3.14 If $1,500 is invested now, $1,800 two years from now, and $2,000 four years from
now at an interest rate of 6% compounded annually, what will be the total amount
in 15 years?
3.15 A local newspaper headline blared, “Bo Smith Signed for $30 Million.” A reading
of the article revealed that on April 1, 2005, Bo Smith, the former record-breaking
running back from Football University, signed a $30 million package with the Dallas Rangers. The terms of the contract were $3 million immediately, $2.4 million
per year for the first five years (with the first payment after 1 year) and $3 million
per year for the next five years (with the first payment at year 6). If Bo’s interest
rate is 8% per year, what would his contract be worth at the time he signs it?
3.16 How much invested now at 6% would be just sufficient to provide three payments,
with the first payment in the amount of $7,000 occurring two years hence, then
$6,000 five years hence, and finally $5,000 seven years hence?
Equal Payment Series
3.17 What is the future worth of a series of equal year-end deposits of $1,000 for
10 years in a savings account that earns 7%, annual interest if
(a) All deposits are made at the end of each year?
(b) All deposits are made at the beginning of each year?
3.18 What is the future worth of the following series of payments?
(a) $3,000 at the end of each year for 5 years at 7% compounded annually
(b) $4,000 at the end of each year for 12 years at 8.25% compounded annually
(c) $5,000 at the end of each year for 20 years at 9.4% compounded annually
(d) $6,000 at the end of each year for 12 years at 10.75% compounded annually
3.19 What equal annual series of payments must be paid into a sinking fund to accumulate the following amounts?
(a) $22,000 in 13 years at 6% compounded annually
(b) $45,000 in 8 years at 7% compounded annually
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122 CHAPTER 3 Interest Rate and Economic Equivalence
(c) $35,000 in 25 years at 8% compounded annually
(d) $18,000 in 8 years at 14% compounded annually
3.20 Part of the income that a machine generates is put into a sinking fund to replace
the machine when it wears out. If $1,500 is deposited annually at 7% interest, how
many years must the machine be kept before a new machine costing $30,000 can
be purchased?
3.21 A no-load (commission-free) mutual fund has grown at a rate of 11% compounded annually since its beginning. If it is anticipated that it will continue to
grow at that rate, how much must be invested every year so that $15,000 will be
accumulated at the end of five years?
3.22 What equal annual payment series is required to repay the following present
amounts?
(a) $10,000 in 5 years at 5% interest compounded annually
(b) $5,500 in 4 years at 9.7% interest compounded annually
(c) $8,500 in 3 years at 2.5% interest compounded annually
(d) $30,000 in 20 years at 8.5% interest compounded annually
3.23 You have borrowed $25,000 at an interest rate of 16%. Equal payments will be
made over a three-year period. (The first payment will be made at the end of the
first year.) What will the annual payment be, and what will the interest payment be
for the second year?
3.24 What is the present worth of the following series of payments?
(a) $800 at the end of each year for 12 years at 5.8% compounded annually
(b) $2,500 at the end of each year for 10 years at 8.5% compounded annually
(c) $900 at the end of each year for 5 years at 7.25% compounded annually
(d) $5,500 at the end of each year for 8 years at 8.75% compounded annually
3.25 From the interest tables in Appendix B, determine the values of the following factors by interpolation and compare your results with those obtained from evaluating the A/P and P/A interest formulas:
(a) The capital recovery factor for 38 periods at 6.25% interest
(b) The equal payment series present-worth factor for 85 periods at 9.25% interest
Linear Gradient Series
3.26 An individual deposits an annual bonus into a savings account that pays 8% interest compounded annually. The size of the bonus increases by $2,000 each year,
and the initial bonus amount was $5,000. Determine how much will be in the account immediately after the fifth deposit.
3.27 Five annual deposits in the amounts of $3,000, $2,500, $2,000, $1,500, and
$1,000, in that order, are made into a fund that pays interest at a rate of 7% compounded annually. Determine the amount in the fund immediately after the fifth
deposit.
3.28 Compute the value of P in the accompanying cash flow diagram, assuming that
i = 9%.
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Problems 123
$250 $250
$200 $200
$150 $150
$100 $100
0
1
2
3
4
5
6
7
Years
P
3.29 What is the equal payment series for 12 years that is equivalent to a payment series of $15,000 at the end of the first year, decreasing by $1,000 each year over
12 years? Interest is 8% compounded annually.
Geometric Gradient Series
3.30 Suppose that an oil well is expected to produce 100,000 barrels of oil during its
first year in production. However, its subsequent production (yield) is expected to
decrease by 10% over the previous year’s production. The oil well has a proven
reserve of 1,000,000 barrels.
(a) Suppose that the price of oil is expected to be $60 per barrel for the next
several years. What would be the present worth of the anticipated revenue
stream at an interest rate of 12% compounded annually over the next seven
years?
(b) Suppose that the price of oil is expected to start at $60 per barrel during the
first year, but to increase at the rate of 5% over the previous year’s price. What
would be the present worth of the anticipated revenue stream at an interest rate
of 12% compounded annually over the next seven years?
(c) Consider part (b) again. After three years’ production, you decide to sell the oil
well. What would be a fair price?
3.31 A city engineer has estimated the annual toll revenues from a newly proposed
highway construction over 20 years as follows:
A n = 1$2,000,00021n211.062n - 1,
n = 1, 2, Á , 20.
To validate the bond, the engineer was asked to present the estimated total present
value of toll revenue at an interest rate of 6%. Assuming annual compounding,
find the present value of the estimated toll revenue.
3.32 What is the amount of 10 equal annual deposits that can provide five annual withdrawals when a first withdrawal of $5,000 is made at the end of year 11 and subsequent withdrawals increase at the rate of 8% per year over the previous year’s
withdrawal if
(a) The interest rate is 9% compounded annually?
(b) The interest rate is 6% compounded annually?
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124 CHAPTER 3 Interest Rate and Economic Equivalence
Various Interest Factor Relationships
3.33 By using only those factors given in interest tables, find the values of the factors
that follow, which are not given in your tables. Show the relationship between the
factors by using factor notation, and calculate the value of the factor. Then compare the solution you obtained by using the factor formulas with a direct calculation of the factor values.
Example: 1F/P, 8%, 382 = 1F/P, 8%, 3021F/P, 8%, 82 = 18.6253
(a) (P/F, 8%, 67)
(b) (A/P, 8%, 42)
(c) (P/A, 8%, 135)
3.34 Prove the following relationships among interest factors:
(a) 1F/P, i, N2 = i1F/A, i, N2 + 1
(b) 1P/F, i, N2 = 1 - 1P/A, i, N2i
(c) 1A/F, i, N2 = 1A/P, i, N2 - i
(d) 1A/P, i, N2 = i/[1 - 1P/F, i, N2]
(e) 1P/A, i, N : q 2 = 1/i
(f) 1A/P, i, N : q 2 = i
(g) 1P/G, i, N : q 2 = 1/i2
Equivalence Calculations
3.35 Find the present worth of the cash receipts where i = 12% compounded annually with only four interest factors.
$200
$150
$100
0
2
1
4
3
5 6
Years
7
8
9
10
P
3.36 Find the equivalent present worth of the cash receipts where i = 8%. In other
words, how much do you have to deposit now (with the second deposit in the
amount of $200 at the end of the first year) so that you will be able to withdraw
$200 at the end of second year, $120 at the end of third year, and so forth if the
bank pays you a 8% annual interest on your balance?
$300
$200
0
$120
$120
3
4
1
2
Years
$200
P
5
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Problems 125
3.37 What value of A makes two annual cash flows equivalent at 13% interest compounded annually?
0
$100
$100
1
2
$120
$120
$120
3
4
5
A
A
A
3
4
5
Years
0
A
A
1
2
Years
3.38 The two cash flow transactions shown in the accompanying cash flow diagram are
said to be equivalent at 6% interest compounded annually. Find the unknown
value of X that satisfies the equivalence.
$200
$200
$150
$150
$100
$100
2
3
4
5
X
X
X
3
4
5
1
0
Years
0
X
X
1
2
Years
3.39 Solve for the present worth of this cash flow using at most three interest factors at
10% interest compounded annually.
$60
$40
$40
$20
Years
0
1
2
6
7
8
9
10 11 12
$20
$20
$60
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126 CHAPTER 3 Interest Rate and Economic Equivalence
3.40 From the accompanying cash flow diagram, find the value of C that will establish
the economic equivalence between the deposit series and the withdrawal series at
an interest rate of 8% compounded annually.
$6000
Years
0
1
2
3
4
5
6
7
8
10
9
C
C
C
C
C
C
(a) $1,335
(b) $862
(c) $1,283
(d) $828
3.41 The following equation describes the conversion of a cash flow into an equivalent
equal payment series with N = 10:
A = [800 + 201A>G, 6%, 72]
* 1P>A, 6%, 721A>P, 6%, 102
+ [3001F>A, 6%, 32 - 500]1A>F, 6%, 102.
Reconstruct the original cash flow diagram.
3.42 Consider the cash flow shown in the accompanying diagram. What value of C
makes the inflow series equivalent to the outflow series at an interest rate of 10%?
$500
$300
0
1
2
Years
3
C
4
C
C
5
6
C
C
7
8
C
C
2C
3.43 Find the value of X so that the two cash flows shown in the diagram are equivalent
for an interest rate of 8%.
$200
$200
$200
$200
$150
0
1
2
3
4
5
Years
0
X
X
1
2
X
3
4
$200
$200
Years
5
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Problems 127
3.44 What single amount at the end of the fifth year is equivalent to a uniform annual series
of $3,000 per year for 10 years if the interest rate is 9% compounded annually?
3.45 From the following list, identify all the correct equations used in computing either
the equivalent present worth (P) or future worth (F) for the cash flow shown at
i = 10%.
R
R
R
0
1
2
R
R
R
3
4
5
Years
(1) P = R1P/A, 10%, 62
(2) P = R + R1P/A, 10%, 52
(3) P = R1P/F, 10%, 52 + R1P/A, 10%, 52
(4) F = R1F/A, 10%, 52 + R1F/P, 10%, 52
(5) F = R + R1F/A, 10%, 52
(6) F = R1F/A, 10%, 62
(7) F = R1F/A, 10%, 62 - R
3.46 On the day his baby was born, a father decided to establish a savings account for
the child’s college education. Any money that is put into the account will earn an
interest rate of 8% compounded annually. The father will make a series of annual
deposits in equal amounts on each of his child’s birthdays from the 1st through the
18th, so that the child can make four annual withdrawals from the account in the
amount of $30,000 on each birthday. Assuming that the first withdrawal will be
made on the child’s 18th birthday, which of the following equations are correctly
used to calculate the required annual deposit?
(1) A = 1$30,000 * 42/18
(2) A = $30,0001F/A, 8%, 42 * 1P/F, 8%, 2121A/P, 8%, 182
(3) A = $30,0001P/A, 8%, 182 * 1F/P, 8%, 2121A/F, 8%, 42
(4) A = [$30,0001P/A, 8%, 32 + $30,000]1A/F, 8%, 182
(5) A = $30,000[1P/F, 8%, 182 + 1P/F, 8%, 192 +
1P/F, 8%, 202 + 1P/F, 8%, 212]1A/P, 8%, 182
3.47 Find the equivalent equal payment series (A) using an A/G factor such that the two
cash flows are equivalent at 10% compounded annually.
$250
Cash flow diagram
$200
$150
$100
0
1
2
3
4
5
A
A
A
3
4
5
Years
$50
A
A
A
0
1
2
Years
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128 CHAPTER 3 Interest Rate and Economic Equivalence
3.48 Consider the following cash flow:
Year End
Payment
0
$500
1–5
$1,000
In computing F at the end of year 5 at an interest rate of 12%, which of the following equations is incorrect?
(a) F = $1,0001F/A, 12%, 52 - $5001F/P, 12%, 52
(b) F = $5001F/A, 12%, 62 + $5001F/A, 12%, 52
(c) F = [$500 + $1,0001P/A, 12%, 52] * 1F/P, 12%, 52
(d) F = [$5001A/P, 12%, 52 + $1,000] * 1F/A, 12%, 52
3.49 Consider the cash flow series given. In computing the equivalent worth at n = 4,
which of the following equations is incorrect?
Cash flow diagram
V4
$100 $100 $100
0
1
2
3
Years
$100 $100
4
5
6
(a) V4 = [$1001P/A, i, 62 - $1001P/F, i, 42]1F/P, i, 42
(b) V4 = $1001F/A, i, 32 + $1001P/A, i, 22
(c) V4 = $1001F/A, i, 42 - $100 + $1001P/A, i, 22
(d) V4 = [$1001F/A, i, 62 - $1001F/P, i, 22]1P/F, i, 22
3.50 Henry Cisco is planning to make two deposits: $25,000 now and $30,000 at the end
of year 6. He wants to withdraw C each year for the first six years and 1C + $1,0002
each year for the next six years. Determine the value of C if the deposits earn 10% interest compounded annually.
(a) $7,711
(b) $5,794
(c) $6,934
(d) $6,522
Solving for an Unknown Interest Rate or Unknown Interest Periods
3.51 At what rate of interest compounded annually will an investment double itself in
five years?
3.52 Determine the interest rate (i) that makes the pairs of cash flows shown economically equivalent.
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$2000 $2000 $2000 $2000 $2000 $2000
0
1
$2500
0
1
2
$1875
2
3
Years
i=?
$1406
3
Years
4
5
$1055 $791
4
5
6
$593
6
3.53 You have $10,000 available for investment in stock. You are looking for a growth
stock whose value can grow to $35,000 over five years. What kind of growth rate
are you looking for?
3.54 How long will it take to save $1 million if you invest $2,000 each year at 6%?
Short Case Studies
ST3.1 Read the following letter from a magazine publisher:
Dear Parent:
Currently your Growing Child/Growing Parent subscription will expire with your
24-month issue. To renew on an annual basis until your child reaches 72 months
would cost you a total of $63.84 ($15.96 per year). We feel it is so important for
you to continue receiving this material until the 72nd month, that we offer you an
opportunity to renew now for $57.12. Not only is this a savings of 10% over the
regular rate, but it is an excellent inflation hedge for you against increasing rates
in the future. Please act now by sending $57.12.
(a) If your money is worth 6% per year, determine whether this offer can be of any
value.
(b) What rate of interest would make you indifferent between the two renewal
options?
ST3.2 The State of Florida sold a total of 36.1 million lottery tickets at $1 each during
the first week of January 2006. As prize money, a total of $41 million will be distributed ($1,952,381 at the beginning of each year) over the next 21 years. The
distribution of the first-year prize money occurs now, and the remaining lottery
proceeds will be put into the state’s educational reserve fund, which earns interest
at the rate of 6% compounded annually. After making the last prize distribution (at
the beginning of year 21), how much will be left over in the reserve account?
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130 CHAPTER 3 Interest Rate and Economic Equivalence
ST3.3 A local newspaper carried the following story:
Texas Cowboys wide receiver John Young will earn either $11,406,000 over
12 years or $8,600,000 over 6 years. Young must declare which plan he prefers.
The $11 million package is deferred through the year 2017, while the nondeferred arrangement ends after the 2011 season. Regardless of which plan is
chosen, Young will be playing through the 2011 season. Here are the details of
the two plans:
Deferred Plan
Nondeferred Plan
2006
$2,000,000
2006
$2,000,000
2007
566,000
2007
900,000
2008
920,000
2008
1,000,000
2009
930,000
2009
1,225,000
2010
740,000
2010
1,500,000
2011
740,000
2011
1,975,000
2012
740,000
2013
790,000
2014
540,000
2015
1,040,000
2016
1,140,000
2017
1,260,000
Total
$11,406,000
Total
$8,600,000
(a) As it happened, Young ended up with the nondeferred plan. In retrospect, if
Young’s interest rate were 6%, did he make a wise decision in 2006?
(b) At what interest rate would the two plans be economically equivalent?
ST3.4 Fairmont Textile has a plant in which employees have been having trouble with
carpal tunnel syndrome (CTS, an inflammation of the nerves that pass through
the carpal tunnel, a tight space at the base of the palm), resulting from long-term
repetitive activities, such as years of sewing. It seems as if 15 of the employees
working in this facility developed signs of CTS over the last five years. DeepSouth, the company’s insurance firm, has been increasing Fairmont’s liability
insurance steadily because of this problem. DeepSouth is willing to lower the
insurance premiums to $16,000 a year (from the current $30,000 a year) for the
next five years if Fairmont implements an acceptable CTS-prevention program
that includes making the employees aware of CTS and how to reduce the
chances of it developing. What would be the maximum amount that Fairmont
should invest in the program to make it worthwhile? The firm’s interest rate is
12% compounded annually.
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ST3.5 Kersey Manufacturing Co., a small fabricator of plastics, needs to purchase an
extrusion molding machine for $120,000. Kersey will borrow money from a bank
at an interest rate of 9% over five years. Kersey expects its product sales to be
slow during the first year, but to increase subsequently at an annual rate of 10%.
Kersey therefore arranges with the bank to pay off the loan on a “balloon scale,”
which results in the lowest payment at the end of the first year and each subsequent payment being just 10% over the previous one. Determine the five annual
payments.
ST3.6 Adidas will put on sale what it bills as the world’s first computerized “smart
shoe.” But consumers will decide whether to accept the bionic running shoe’s
$250 price tag—four times the average shoe price at stores such as Foot Locker.
Adidas uses a sensor, a microprocessor, and a motorized cable system to automatically adjust the shoe’s cushioning. The sensor under the heel measures compression and decides whether the shoe is too soft or firm. That information is sent
to the microprocessor and, while the shoe is in the air, the cable adjusts the heel
cushion. The whole system weighs less than 40 grams. Adidas’s computer-driven
shoe—three years in the making—is the latest innovation in the $16.4 billion
U.S. sneaker industry. The top-end running shoe from New Balance lists for
$199.99. With runners typically replacing shoes by 500 miles, the $250 Adidas
could push costs to 50 cents per mile. Adidas is spending an estimated $20 million on the rollout.4
The investment required to develop a full-scale commercial rollout cost Adidas
$70 million (including the $20 million ad campaign), which will be financed at an
interest rate of 10%. With a price tag of $250, Adidas will have about $100 net
cash profit from each sale. The product will have a five-year market life. Assuming that the annual demand for the product remains constant over the market life,
how many units does Adidas have to sell each year to pay off the initial investment
and interest?
ST3.7 Millionaire Babies: How to Save Our Social Security System. It sounds a little
wild, but that is probably the point. Former Senator Bob Kerrey, D-Nebraska, had
proposed giving every newborn baby a $1,000 government savings account at
birth, followed by five annual contributions of $500 each. If the money is then left
untouched in an investment account, Kerrey said, by the time the baby reaches age
65, the $3,500 contribution per child would grow to $600,000, even at medium returns for a thrift savings plan. At about 9.4%, the balance would grow to be
$1,005,132. (How would you calculate this number?) With about 4 million babies
born each year, the proposal would cost the federal government $4 billion annually. Kerrey offered this idea in a speech devoted to tackling Social Security reform.
About 90% of the total annual Social Security tax collections of more than $300
billion are used to pay current beneficiaries in the largest federal program. The remaining 10% is invested in interest-bearing government bonds that finance the
day-to-day expenses of the federal government. Discuss the economics of Kerrey’s
Social Security savings plan.
4
Source: “Adidas puts computer on new footing,” by Michael McCarthy, USA Today—Thursday, March 3,
2006, section 5B.
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132 CHAPTER 3 Interest Rate and Economic Equivalence
ST3.8 Recently an NFL quarterback agreed to an eight-year, $50 million contract that at
the time made him one of the highest paid players in professional football history.
The contract included a signing bonus of $11 million. The agreement called for annual salaries of $2.5 million in 2005, $1.75 million in 2006, $4.15 million in 2007,
$4.90 million in 2008, $5.25 million in 2009, $6.2 million in 2010, $6.75 million in
2011, and $7.5 million in 2012. The $11 million signing bonus was prorated over
the course of the contract, so that an additional $1.375 million was paid each year
over the eight-year contract period. Table ST3.8 shows the net annual payment
schedule, with the salary paid at the beginning of each season.
(a) How much was the quarterback’s contract actually worth at the time of
signing?
(b) For the signing bonus portion, suppose that the quarterback was allowed to
take either the prorated payment option as just described or a lump-sum payment option in the amount of $8 million at the time he signed the contract.
Should he have taken the lump-sum option instead of the prorated one? Assume that his interest rate is 6%.
TABLE ST3.8
Net Annual Payment Schedule
Beginning
of Season
Prorated
Contract
Salary
Actual
Signing
Bonus
Annual
Payment
2005
$2,500,000
$1,375,000
$3,875,000
2006
1,750,000
1,375,000
3,125,000
2007
4,150,000
1,375,000
5,525,000
2008
4,900,000
1,375,000
6,275,000
2009
5,250,000
1,375,000
6,625,000
2010
6,200,000
1,375,000
7,575,000
2011
6,750,000
1,375,000
8,125,000
2012
7,500,000
1,375,000
8,875,000
ST3.9 Yuma, Arizona, resident Rosalind Setchfield won $1.3 million in a 1987 Arizona
lottery drawing, to be paid in 20 annual installments of $65,277. However, in
1989, her husband, a construction worker, suffered serious injuries when a crane
dropped a heavy beam on him. The couple’s medical expenses and debt mounted.
Six years later, in early 1995, a prize broker from Singer Asset Finance Co. telephoned Mrs. Setchfield with a promising offer. Singer would immediately give
her $140,000 for one-half of each of her next 9 prize checks, an amount equal to
48% of the $293,746 she had coming over that period. A big lump sum had obvious appeal to Mrs. Setchfield at that time, and she ended up signing a contract
with Singer. Did she make the right decision? The following table gives the details
of the two options:
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Short Case Studies 133
Reduced
Payment
Year
Installment
Year
Installment
1988
$65,277
1995
$65,277
$32,639
1989
65,277
1996
65,277
32,639
1990
65,277
1997
65,277
32,639
1991
65,277
1998
65,277
32,639
1992
65,277
1999
65,277
32,639
1993
65,277
2000
65,277
32,639
1994
65,277
2001
65,277
32,639
2002
65,277
32,639
2003
65,277
32,639
2004
65,277
2005
65,277
2006
65,277
2007
65,277
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FOUR
CHAPTER
Understanding Money
and Its Management
Hybrid Mortgages Can Cause Pain Should Rates Start to
Rise1 Are you shopping for a mortgage to finance a home that you expect to own for no more than a few years? If so, you should know about
a hybrid mortgage. Hybrid loans give prospective home buyers the ability to buy a lot more home than they can afford, thanks to the initially
lower interest rate. But with such flexibility comes greater risk. Since
lenders are free to design loans to fit borrowers’ needs, the terms and
fees vary widely and homeowners can get burned if rates turn higher.
Hybrid mortgages allow homeowners to benefit from the best aspects of both fixed-rate and adjustable-rate mortgages (ARMs). With
hybrids, borrowers choose to accept a fixed interest rate over a number of years—usually, 3, 5, 7, or 10 years—and afterward the loan
Rates Rising
Mortgages are more costly than they were a year
ago.
2006
2005
1
30-year fixedrate mortgage
6.48%
5.85%
Hybrid ARM with
fixed rate for first
10 years
6.32
5.58
Hybrid ARM with
fixed rate for first
5 years
6.02
5.01
ARM with rate that
adjusts annually
5.29
4.21
“Hybrid mortgages can cause pain should rates start to rise,” by Terri Cullen, Wall Street Journal Online,
December 5, 2002. @2002 Dow Jones & Company, Inc.
134
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converts to an ARM. But therein lies the danger: While you’re getting an extraordinarily low rate up front for a few years, when the fixed-rate period
expires you could very well end up paying more than double your current
rate of interest.
At today’s rate of 6.16% for a 30-year mortgage, a person borrowing
$200,000 would pay $1,220 a month.With a 7-year hybrid, more commonly
called a 7/1 loan, at the going rate of 5.61% that monthly payment drops to
$1,150. By the end of the seventh year, the homeowner would save $7,700
in interest charges by going with a 7-year hybrid.
To say that there are drawbacks is an understatement. Despite the surge
in popularity, a hybrid loan can be a ticking time bomb for borrowers who
plan on holding the loan for the long term.
Let’s say a borrower takes out a 30-year $200,000 hybrid loan that will
remain at a fixed rate of 5.19% for 5 years and then will switch to an adjustable-rate mortgage for the remaining period.The lender agrees to set a
cap on the adjustable-rate portion of the loan, so that the rate will climb no
more than 5 percentage points. Conceivably, then, if rates are sharply higher
after that initial 7-year period, a borrower could be looking at a mortgage
rate of more than 10%. Under this scenario, the homeowner’s monthly payment on a $200,000 mortgage would jump to $1,698 from $1,097 after the
5-year term expires—a 55% increase! But if there’s a very real chance you’ll
be looking to sell your home over the next 10 years, hybrids can make a lot
of sense, since shorter term loans usually carry the lowest rates.
In this chapter, we will consider several concepts crucial to managing
money. In Chapter 3, we examined how time affects the value of money, and
we developed various interest formulas for that purpose. Using these basic
formulas, we will now extend the concept of equivalence to determine interest rates that are implicit in many financial contracts. To this end, we will
introduce several examples in the area of loan transactions. For example,
many commercial loans require that interest compound more frequently
than once a year—for instance, monthly or quarterly. To consider the effect
of more frequent compounding, we must begin with an understanding of the
concepts of nominal and effective interest.
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136 CHAPTER 4 Understanding Money and Its Management
CHAPTER LEARNING OBJECTIVES
After completing this chapter, you should understand the following concepts:
The difference between the nominal interest rate and the effective
interest rate.
The procedure for computing the effective interest rate, based on a
payment period.
How commercial loans and mortgages are structured in terms of
interest and principal payments.
The basics of investing in financial assets.
4.1 Nominal and Effective Interest Rates
n all the examples in Chapter 3, the implicit assumption was that payments are received once a year, or annually. However, some of the most familiar financial
transactions, both personal and in engineering economic analysis, involve payments that are not based on one annual payment—for example, monthly mortgage
payments and quarterly earnings on savings accounts. Thus, if we are to compare different cash flows with different compounding periods, we need to evaluate them on a
common basis. This need has led to the development of the concepts of the nominal interest rate and the effective interest rate.
I
4.1.1 Nominal Interest Rates
Take a closer look at the billing statement from any of your credit cards. Or if you financed a new car recently, examine the loan contract. You will typically find the interest
that the bank charges on your unpaid balance. Even if a financial institution uses a unit of
time other than a year—say, a month or a quarter (e.g., when calculating interest payments)—the institution usually quotes the interest rate on an annual basis. Many banks,
for example, state the interest arrangement for credit cards in this way:
18% compounded monthly.
Annual
percentage rate
(APR) is the
yearly cost of a
loan, including
interest,
insurance, and
the origination
fee, expressed as
a percentage.
This statement simply means that each month the bank will charge 1.5% interest on
an unpaid balance. We say that 18% is the nominal interest rate or annual percentage
rate (APR), and the compounding frequency is monthly (12). As shown in Figure 4.1, to
obtain the interest rate per compounding period, we divide, for example, 18% by 12, to
get 1.5% per month.
Although the annual percentage rate, or APR, is commonly used by financial institutions and is familiar to many customers, the APR does not explain precisely the amount
of interest that will accumulate in a year. To explain the true effect of more frequent
compounding on annual interest amounts, we will introduce the term effective interest
rate, commonly known as annual effective yield, or annual percentage yield (APY).
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Section 4.1 Nominal and Effective Interest Rates
Month
Interest
rate (%)
1
2
3
4
5
6
7
8
9
10
11
12
1.5
1.5
1.5
1.5
1.5
1.5
1.5
1.5
1.5
1.5
1.5
1.5
Nominal interest rate.
1.5% 12 18%
Figure 4.1 The nominal interest rate is determined by summing the individual interest
rates per period.
4.1.2 Effective Annual Interest Rates
The effective annual interest rate is the one rate that truly represents the interest earned in
a year. For instance, in our credit card example, the bank will charge 1.5% interest on any
unpaid balance at the end of each month. Therefore, the 1.5% rate represents the effective
interest rate per month. On a yearly basis, you are looking for a cumulative rate—1.5% each
month for 12 months. This cumulative rate predicts the actual interest payment on your outstanding credit card balance.
Suppose you deposit $10,000 in a savings account that pays you at an interest rate of
9% compounded quarterly. Here, 9% represents the nominal interest rate, and the interest
rate per quarter is 2.25% (9%/4). The following is an example of how interest is compounded when it is paid quarterly:
End of
Period
Base amount
Interest Earned
2.25% : 1Base amount2
First
quarter
$10,000.00
2.25% * $10,000.00 = $225.00
$10,225.00
Second
quarter
$10,225.00
2.25% * $10,225.00 = $230.06
$10,455.06
Third
quarter
$10,455.06
2.25% * $10,455.06 = $225.24
$10,690.30
Fourth
quarter
$10,690.30
2.25% * $10,690.30 = $240.53
$10,930.83
New Base
Clearly, you are earning more than 9% on your original deposit. In fact, you are earning
9.3083% ($930.83/$10,000). We could calculate the total annual interest payment for a principal amount of $10,000 with the formula given in Eq. (3.3). If P = $10,000, i = 2.25%,
and N = 4, we obtain
F = P11 + i2N
= $10,00011 + 0.022524
= $10,930.83.
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138 CHAPTER 4 Understanding Money and Its Management
The implication is that, for each dollar deposited, you are earning an equivalent annual
interest of 9.38 cents. In terms of an effective annual interest rate 1ia2, the interest payment can be rewritten as a percentage of the principal amount:
ia = $930.83> $10,000 = 0.093083, or 9.3083%.
In other words, earning 2.25% interest per quarter for four quarters is equivalent to earning
9.3083% interest just one time each year.
Table 4.1 shows effective interest rates at various compounding intervals for 4%–12%
APR. As you can see, depending on the frequency of compounding, the effective interest
earned or paid by the borrower can differ significantly from the APR. Therefore, truthin-lending laws require that financial institutions quote both the nominal interest rate
and the compounding frequency (i.e., the effective interest) when you deposit or borrow
money.
Certainly, more frequent compounding increases the amount of interest paid over a
year at the same nominal interest rate. Assuming that the nominal interest rate is r, and
M compounding periods occur during the year, we can calculate the effective annual interest rate
ia = a1 +
r M
b - 1.
M
(4.1)
When M = 1, we have the special case of annual compounding. Substituting M = 1
in Eq. (4.1) reduces it to ia = r. That is, when compounding takes place once annually,
the effective interest is equal to the nominal interest. Thus, in the examples in Chapter 3,
in which only annual interest was considered, we were, by definition, using effective interest rates.
TABLE 4.1
Nominal and Effective Interest Rates with Different Compounding
Periods
Effective Rates
Nominal Compounding Compounding Compounding Compounding Compounding
Rate
Annually
Semiannually
Quarterly
Monthly
Daily
4%
4.00%
4.04%
4.06%
4.07%
4.08%
5
5.00
5.06
5.09
5.12
5.13
6
6.00
6.09
6.14
6.17
6.18
7
7.00
7.12
7.19
7.23
7.25
8
8.00
8.16
8.24
8.30
8.33
9
9.00
9.20
9.31
9.38
9.42
10
10.00
10.25
10.38
10.47
10.52
11
11.00
11.30
11.46
11.57
11.62
12
12.00
12.36
12.55
12.68
12.74
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Section 4.1 Nominal and Effective Interest Rates 139
EXAMPLE 4.1 Determining a Compounding Period
The following table summarizes interest rates on certificates of deposit (CDs) offered
by various lending institutions during November 2005:
Product
Bank
3-Month CD
*
Minimum
Rate
APY
Imperial Capital Bank
$2,000
4.03%
4.10%
6-Month
Jumbo CD
IndyMac Bank
$5,000
4.21%
4.30%
1-Year
Jumbo CD
VirtualBank
$10,000
4.50%
4.60%
1.5-Year CD
AmTrust Bank
$1,000
4.50%
4.60%
2-Year CD
Ohio Savings Bank
$1,000
4.59%
4.70%
2.5-Year
Jumbo CD
Countrywide Bank
$98,000
4.66%
4.77%
3-Year CD
ING Direct
5-Year CD
Citizens & Northern
Bank
*
$0
4.70%
4.70%
$500
4.70%
4.78%
Annual percentage yield = effective annual interest rate 1ia2
In the table, no mention is made of specific interest compounding frequencies.
(a) Find the interest periods assumed for the 2.5-year Jumbo CD offered by Countrywide Bank. (b) Find the total balance for a deposit amount of $100,000 at the end of
2.5 years.
SOLUTION
Given: r = 4.66% per year, ia 1APY2 = 4.77%, P = $100,000, and N = 2.5 years.
Find: M and the balance at the end of 2.5 years.
(a) The nominal interest rate is 4.66% per year, and the effective annual interest rate
(yield) is 4.77%. Using Eq. (4.1), we obtain the expression
0.0477 = a 1 +
0.0466 M
b - 1,
M
1.0477 = a 1 +
0.0466 M
b .
M
or
By trial and error, we find that M = 365, which indicates daily compounding. Thus, the 2.5-year Jumbo CD earns 4.66% interest compounded daily.
Annual
percentage yield
(APY) is the rate
actually earned
or paid in one
year, taking into
account the
affect of
compounding.
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140 CHAPTER 4 Understanding Money and Its Management
Normally, if the CD is not cashed at maturity, it will be renewed automatically
at the original interest rate. Similarly, we can find the interest periods for the
other CDs.
(b) If you purchase the 2.5-year Jumbo CD, it will earn 4.66% interest compounded
daily. This means that your CD earns an effective annual interest of 4.77%:
F = P11 + ia2N
= $100,00011 + 0.047722.5
= $100,000 a1 +
0.0466 365 * 2.5
b
365
= $112,355.
4.1.3 Effective Interest Rates per Payment Period
We can generalize the result of Eq. (4.1) to compute the effective interest rate for periods
of any duration. As you will see later, the effective interest rate is usually computed on
the basis of the payment (transaction) period. For example, if cash flow transactions
occur quarterly, but interest is compounded monthly, we may wish to calculate the effective interest rate on a quarterly basis. To do this, we may redefine Eq. (4.1) as
i = a1 +
= a1 +
r C
b - 1
M
r C
b - 1,
CK
(4.2)
where
M = the number of interest periods per year,
C = the number of interest periods per payment period, and
K = the number of payment periods per year.
Note that M = CK in Eq. (4.2). For the special case of annual payments with annual
compounding, we obtain i = ia with C = M and K = 1.
EXAMPLE 4.2 Effective Rate per Payment Period
Suppose that you make quarterly deposits in a savings account that earns 9% interest
compounded monthly. Compute the effective interest rate per quarter.
SOLUTION
Given: r = 9%, C = three interest periods per quarter, K = four quarterly payments per year, and M = 12 interest periods per year.
Find: i.
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Section 4.1 Nominal and Effective Interest Rates 141
Using Eq. (4.2), we compute the effective interest rate per quarter as
i = a1 +
0.09 3
b - 1
12
= 2.27%.
COMMENTS: Figure 4.2 illustrates the relationship between the nominal and effective
interest rates per payment period.
9%/12 or 0.75% per month
First
quarter
Second
quarter
Third
quarter
Fourth
quarter
0.75% 0.75% 0.75% 0.75% 0.75% 0.75% 0.75% 0.75% 0.75% 0.75% 0.75% 0.75%
2.27%
2.27%
2.27%
2.27%
i = (1 + 0.0075)3 –1 = 2.27%
Effective interest rate per quarter
ia = (1 + 0.0075)12 –1 = 9.38%
Effective annual interest rate
Figure 4.2 Functional relationships among r, i, and i a, where interest is calculated based
on 9% compounded monthly and payments occur quarterly (Example 4.2).
4.1.4 Continuous Compounding
To be competitive on the financial market or to entice potential depositors, some financial
institutions offer frequent compounding. As the number of compounding periods (M) becomes very large, the interest rate per compounding period (r/M) becomes very small. As
M approaches infinity and r/M approaches zero, we approximate the situation of continuous compounding.
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142 CHAPTER 4 Understanding Money and Its Management
By taking limits on both sides of Eq. (4.2), we obtain the effective interest rate per
payment period as
r C
b - 1d
CK: q
CK
r C
= lim a 1 +
b - 1
CK: q
CK
= 1er21>K - 1.
i =
Continuous
compounding:
The process of
calculating
interest and
adding it to
existing principal
and interest at
infinitely short
time intervals.
lim c a1 +
Therefore, the effective interest rate per payment period is
i = er>K - 1.
(4.3)
To calculate the effective annual interest rate for continuous compounding, we set K
equal to unity, resulting in
ia = er - 1.
(4.4)
As an example, the effective annual interest rate for a nominal interest rate of 12% compounded continuously is ia = e0.12 - 1 = 12.7497%.
EXAMPLE 4.3 Calculating an Effective Interest
Rate with Quarterly Payment
Find the effective interest rate per quarter at a nominal rate of 8% compounded
(a) quarterly, (b) monthly, (c) weekly, (d) daily, and (e) continuously.
SOLUTION
Given: r = 8%, M, C, and K = 4 quarterly payments per year.
Find: i.
(a) Quarterly compounding:
r = 8%, M = 4, C = 1 interest period per quarter, and K = 4 payments per
year. Then
0.08 1
b - 1 = 2.00%.
i = a1 +
4
1st Q
4th Q
3rd Q
2nd Q
3
1 interest period
(b) Monthly compounding:
r = 8%, M = 12, C = 3 interest periods per quarter, and K = 4 payments per
year. Then
0.08 3
i = a1 +
b - 1 = 2.013%.
12
1st Q
2nd Q
3
3 interest periods
3rd Q
4th Q
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Section 4.2 Equivalence Calculations with Effective Interest Rates 143
(c) Weekly compounding:
r = 8%, M = 52, C = 13 interest periods per quarter, and K = 4 payments
per year. Then
0.08 13
b - 1 = 2.0186%.
52
i = a1 +
1st Q
2nd Q
3rd Q
4th Q
3
13 interest periods
(d) Daily compounding:
r = 8%, M = 365, C = 91.25 days per quarter, and K = 4. Then
i = a1 +
0.08 91.25
b
- 1 = 2.0199%.
365
1st Q
3
2nd Q
3rd Q
4th Q
91.25 interest periods
(e) Continuous compounding:
r = 8%, M : q , C : q , and K = 4. Then, from Eq. 14.32,
i = e0.08>4 - 1 = 2.0201%.
1st Q
2nd Q
3rd Q
4th Q
3
∞ interest periods
COMMENTS: Note that the difference between daily compounding and continuous
compounding is often negligible. Many banks offer continuous compounding to entice deposit customers, but the extra benefits are small.
4.2 Equivalence Calculations with Effective
Interest Rates
All the examples in Chapter 3 assumed annual payments and annual compounding. However, a number of situations involve cash flows that occur at intervals that are not the same
as the compounding intervals often used in practice. Whenever payment and compounding
periods differ from each other, one or the other must be transformed so that both conform
to the same unit of time. For example, if payments occur quarterly and compounding occurs monthly, the most logical procedure is to calculate the effective interest rate per quarter. By contrast, if payments occur monthly and compounding occurs quarterly, we may be
able to find the equivalent monthly interest rate. The bottom line is that, to proceed with
equivalency analysis, the compounding and payment periods must be in the same order.
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144 CHAPTER 4 Understanding Money and Its Management
4.2.1 When Payment Period Is Equal to Compounding Period
Whenever the compounding and payment periods are equal 1M = K2, whether the interest is compounded annually or at some other interval, the following solution method can
be used:
1. Identify the number of compounding periods (M) per year.
2. Compute the effective interest rate per payment period—that is, using Eq. (4.2).
Then, with C = 1 and K = M, we have
i =
r
.
M
3. Determine the number of compounding periods:
N = M * 1number of years2.
EXAMPLE 4.4 Calculating Auto Loan Payments
Suppose you want to buy a car. You have surveyed the dealers’ newspaper advertisements, and the following one has caught your attention:
College Graduate Special: New 2006 Nissan Altima 2.55 with automatic transmission, A/C, power package, and cruise control
MSRP:
$20,870
Dealer’s discount:
$1,143
Manufacturer rebate
$800
College graduate cash:
$500
Sale price:
$18,427
Price and payment is plus tax, title, customer service fee, with approved credit for
72 months at 6.25% APR.
You can afford to make a down payment of $3,427 (and taxes and insurance as
well), so the net amount to be financed is $15,000. What would the monthly payment
be (Figure 4.3)?
$15,000
APR 6.25%
1
2
3
4
72 months
0
A
Figure 4.3
A car loan cash transaction (Example 4.4).
DISCUSSION: The advertisement does not specify a compounding period, but in automobile financing, the interest and the payment periods are almost always monthly.
Thus, the 6.25% APR means 6.25% compounded monthly.
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Section 4.2 Equivalence Calculations with Effective Interest Rates 145
SOLUTION
Given: P = $25,000, r = 6.25% per year, K = 12 payments per year, N =
72 months, and M = 12 interest periods per year.
Find: A.
In this situation, we can easily compute the monthly payment with Eq. (3.12):
i = 6.25%>12 = 0.5208% per month,
N = 1122162 = 72 months,
A = $15,0001A>P, 0.5208%, 722 = $250.37.
4.2.2 Compounding Occurs at a Different Rate than That
at Which Payments Are Made
We will consider two situations: (1) compounding is more frequent than payments and
(2) compounding is less frequent than payments.
Compounding Is More Frequent than Payments
The computational procedure for compounding periods and payment periods that cannot
be compared is as follows:
1. Identify the number of compounding periods per year (M), the number of payment
periods per year (K), and the number of interest periods per payment period (C).
2. Compute the effective interest rate per payment period:
• For discrete compounding, compute
i = a1 +
r
b - 1.
M
• For continuous compounding, compute
i = er>K - 1.
3. Find the total number of payment periods:
N = K * 1number of years2.
4. Use i and N in the appropriate formulas in Table 3.4.
EXAMPLE 4.5 Compounding Occurs More Frequently
than Payments Are Made (DiscreteCompounding Case)
Suppose you make equal quarterly deposits of $1,500 into a fund that pays interest at
a rate of 6% compounded monthly, as shown in Figure 4.4. Find the balance at the
end of year 2.
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146 CHAPTER 4 Understanding Money and Its Management
F?
i (1 0.06/12)3 1
1.5075% per quarter
0
Year 1
1
2
Year 2
3
4
5
6
7
8
Quarters
A $1,500
Figure 4.4 Quarterly deposits with monthly compounding
(Example 4.5).
SOLUTION
Given: A = $1,500 per quarter, r = 6% per year, M = 12 compounding periods
per year, and N = 8 quarters.
Find: F.
We follow the aforementioned procedure for noncomparable compounding and
payment periods:
1. Identify the parameter values for M, K, and C, where
M = 12 compounding periods per year,
K = 4 payment periods per year,
C = 3 interest periods per payment period.
2. Use Eq. (4.2) to compute the effective interest:
i = a1 +
0.06 3
b - 1
12
= 1.5075% per quarter.
3. Find the total number of payment periods, N:
N = K1number of years2 = 4122 = 8 quarters.
4. Use i and N in the appropriate equivalence formulas:
F = $1,5001F>A, 1.5075%, 82 = $12,652.60.
COMMENT: No 1.5075% interest table appears in Appendix A, but the interest factor can still be evaluated by F = $1,5001A>F, 0.5%, 321F>A, 0.5%, 242, where
the first interest factor finds its equivalent monthly payment and the second interest factor converts the monthly payment series to an equivalent lump-sum future
payment.
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EXAMPLE 4.6 Compounding Occurs More Frequently
than Payments Are Made (ContinuousCompounding Case)
A series of equal quarterly receipts of $500 extends over a period of five years as
shown in Figure 4.5. What is the present worth of this quarterly payment series at 8%
interest compounded continuously?
Effective interest rate
i = 2.02% per quarter
A = $500
0
4
8
12
16
20
Quarters
P=?
Figure 4.5 A present-worth calculation for an equal payment series with an interest rate
of 8% compounded continuously (Example 4.6).
Discussion: A question that is equivalent to the preceding one is “How much do you
need to deposit now in a savings account that earns 8% interest compounded continuously so that you can withdraw $500 at the end of each quarter for five years?”
Since the payments are quarterly, we need to compute i per quarter for the equivalence calculations:
i = er>K - 1 = e0.08>4 - 1
= 2.02% per quarter
N = 14 payment periods per year215 years2
= 20 quarterly periods.
SOLUTION
Given: i = 2.02% per quarter, N = 20 quarters, and A = $500 per quarter.
Find: P.
Using the (P/A, i, N) factor with i = 2.02% and N = 20, we find that
P = $5001P>A, 2.02%, 202
= $500116.31992
= $8,159.96.
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148 CHAPTER 4 Understanding Money and Its Management
Compounding Is Less Frequent than Payments
The next two examples contain identical parameters for savings situations in which
compounding occurs less frequently than payments. However, two different underlying assumptions govern how interest is calculated. In Example 4.7, the assumption is
that, whenever a deposit is made, it starts to earn interest. In Example 4.8, the assumption is that the deposits made within a quarter do not earn interest until the end of
that quarter. As a result, in Example 4.7 we transform the compounding period to conform to the payment period, and in Example 4.8 we lump several payments together to
match the compounding period. In the real world, which assumption is applicable depends on the transactions and the financial institutions involved. The accounting
methods used by many firms record cash transactions that occur within a compounding period as if they had occurred at the end of that period. For example, when cash
flows occur daily, but the compounding period is monthly, the cash flows within each
month are summed (ignoring interest) and treated as a single payment on which interest is calculated.
Note: In this textbook, we assume that whenever the time point of a cash flow is specified, one cannot move it to another time point without considering the time value of
money (i.e., the practice demonstrated in Example 4.7 should be followed).
EXAMPLE 4.7 Compounding Is Less Frequent than Payments:
Effective Interest Rate per Payment Period
Suppose you make $500 monthly deposits to a tax-deferred retirement plan that pays
interest at a rate of 10% compounded quarterly. Compute the balance at the end of
10 years.
SOLUTION
Given: r = 10% per year, M = 4 quarterly compounding periods per year, K = 12
payment periods per year, A = $500 per month, N = 120 months, and interest is
accrued on deposits made during the compounding period.
Find: i, F.
As in the case of Example 4.5, the procedure for noncomparable compounding and
payment periods is followed:
1. The parameter values for M, K, and C are
M = 4 compounding periods per year,
K = 12 payment periods per year,
C =
1
3
interest period per payment period.
2. As shown in Figure 4.6, the effective interest rate per payment period is calculated
with Eq. (4.2):
i = 0.826% per month.
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Section 4.2 Equivalence Calculations with Effective Interest Rates 149
Equivalent
monthly interest
i = (1 + 0.025)1/3 – 1
= 0.826% per month
Months
0
1
2
3
4
5
6
7
8
9
10
11
12
Monthly payments
i = 10%/4 = 2.5%
per quarter
Actual
compounding
period
(quarterly)
Figure 4.6 Calculation of equivalent monthly interest when the quarterly interest
rate is specified (Example 4.7).
3. Find N:
N = 11221102 = 120 payment periods.
4. Use i and N in the appropriate equivalence formulas (Figure 4.7):
F = $5001F>A, 0.826%, 1202
= $101,907.89.
F
F = $500 (F/A 0.826%,120)
= $101,907.89
i = 0.826% per month
Months
0
6
120
A = $500
Figure 4.7
Cash flow diagram (Example 4.7).
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150 CHAPTER 4 Understanding Money and Its Management
EXAMPLE 4.8 Compounding Is Less Frequent than Payment:
Summing Cash Flows to the End of the
Compounding Period
Some financial institutions will not pay interest on funds deposited after the start of
the compounding period. To illustrate, consider Example 4.7 again. Suppose that
money deposited during a quarter (the compounding period) will not earn any interest (Figure 4.8). Compute what F will be at the end of 10 years.
F
Months
0
3
6
9
12
15
18
111
114
117
120
A = $500
F
Quarters
0
1
2
3
4
5
6
37
38
39
40
A = $1,500
Figure 4.8 Transformed cash flow diagram created by summing monthly cash flows to
the end of the quarterly compounding period (Example 4.8).
SOLUTION
Given: Same as for Example 4.7; however, no interest on flow during the compounding period.
Find: F.
In this case, the three monthly deposits during each quarterly period will be placed at
the end of each quarter. Then the payment period coincides with the interest period,
and we have
10%
= 2.5% per quarter,
4
A = 31$5002 = $1,500 per quarter,
i =
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Section 4.2 Equivalence Calculations with Effective Interest Rates 151
N = 41102 = 40 payment periods,
F = $1,5001F>A, 2.5%, 402 = $101,103.83.
COMMENTS: In Example 4.8, the balance will be $804.06 less than in Example 4.7,
a fact that is consistent with our understanding that increasing the frequency of
compounding increases the future value of money. Some financial institutions follow the practice illustrated in Example 4.7. As an investor, you should reasonably
ask yourself whether it makes sense to make deposits in an interest-bearing account more frequently than interest is paid. In the interim between interests compounding, you may be tying up your funds prematurely and forgoing other
opportunities to earn interest.
Figure 4.9 is a decision chart that allows you to sum up how you can proceed to
find the effective interest rate per payment period, given the various possible compounding and interest arrangements.
No
Annual compounding/
annual payments?
Compounding same as payments?
Yes
No
Yes
i=r
Yes
(Examples in Chapter 3)
Compounding more frequent
than payments?
No
i = r/M
Payments start to earn interest
immediately (recommended procedure)
(Example 4.4)
Case 1
Continuous
compounding?
i = (1 + r/M)M/K – 1
(Example 4.7)
Payments do not earn interest
until end of compounding period
No
Yes
Case 2
i = r/K
(Example 4.8)
i=
er/K
–1
(Example 4.6)
i = [1 + r/ (CK)]c – 1
(Example 4.5)
Figure 4.9 A decision flowchart demonstrating how to compute the effective interest rate i
per payment period.
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152 CHAPTER 4 Understanding Money and Its Management
4.3 Equivalence Calculations with Continuous Payments
As we have seen so far, interest can be compounded annually, semiannually, monthly, or
even continuously. Discrete compounding is appropriate for many financial transactions;
mortgages, bonds, and installment loans, which require payments or receipts at discrete
times, are good examples. In most businesses, however, transactions occur continuously
throughout the year. In these circumstances, we may describe the financial transactions as
having a continuous flow of money, for which continuous compounding and discounting
are more realistic. This section illustrates how one establishes economic equivalence between cash flows under continuous compounding.
Continuous cash flows represent situations in which money flows continuously and
at a known rate throughout a given period. In business, many daily cash flow transactions
can be viewed as continuous. An advantage of the continuous-flow approach is that it
more closely models the realities of business transactions. Costs for labor, for carrying inventory, and for operating and maintaining equipment are typical examples. Others include capital improvement projects that conserve energy or water or that process steam.
Savings on these projects can occur continuously.
4.3.1 Single-Payment Transactions
First we will illustrate how single-payment formulas for continuous compounding and
discounting are derived. Suppose that you invested P dollars at a nominal rate of r % interest for N years. If interest is compounded continuously, the effective annual interest is
i = er - 1. The future value of the investment at the end of N years is obtained with the
F/P factor by substituting er - 1 for i:
F = P11 + i2N
= P11 + er - 12N
= PerN.
This implies that $1 invested now at an interest rate of r% compounded continuously accumulates to erN dollars at the end of N years. Correspondingly, the present
value of F due N years from now and discounted continuously at an interest rate of r%
is equal to
P = Fe -rN.
We can say that the present value of $1 due N years from now and discounted continuously at an annual interest rate of r% is equal to e -rN dollars.
4.3.2 Continuous-Funds Flow
Suppose that an investment’s future cash flow per unit of time (e.g., per year) can be
expressed by a continuous function (f(t)) that can take any shape. Suppose also that the
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Section 4.3 Equivalence Calculations with Continuous Payments 153
f(t)
f(t)t ert
f(t)t
P
f(t)
0
t
t
N
0
t
Equivalent present worth at t = 0
Figure 4.10 Finding an equivalent present worth of a continuous-flow payment function
f(t) at a nominal rate of r%.
investment promises to generate cash of f1t2¢t dollars between t and t + ¢t, where t
is a point in the time interval 0 … t … N (Figure 4.10). If the nominal interest rate is a
constant r during this interval, the present value of the cash stream is given approximately
by the expression
g1f1t2¢t2e -rt,
where e -rt is the discounting factor that converts future dollars into present dollars. With
the project’s life extending from 0 to N, we take the summation over all subperiods (compounding periods) in the interval from 0 to N. As the interval is divided into smaller and
smaller segments (i.e., as ¢t approaches zero), we obtain the expression for the present
value by the integral
N
P =
L0
f1t2e - rt dt.
(4.5)
Similarly, the expression for the future value of the cash flow stream is given by the
equation
N
F = Pe
rN
=
L0
f1t2er1N - t2 dt,
(4.6)
where er1N - t2 is the compounding factor that converts present dollars into future dollars.
It is important to observe that the time unit is the year, because the effective interest rate
is expressed in terms of a year. Therefore, all time units in equivalence calculations must
be converted into years. Table 4.2 summarizes some typical continuous cash functions
that can facilitate equivalence calculations.2
2
Chan S. Park and Gunter P. Sharp-Bette, Advanced Engineering Economics. New York: John Wiley & Sons,
1990. (Reprinted by permission of John Wiley & Sons, Inc.)
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154 CHAPTER 4 Understanding Money and Its Management
TABLE 4.2
Type of
Cash Flow
Summary of Interest Factors for Typical Continuous Cash Flows with
Continuous Compounding
Cash Flow Function
f (t) = A
A
Uniform
(step)
0
Parameters
Find Given
Algebraic
Notation
Factor
Notation
P
A
Ac
erN - 1
d
rerN
1P>A, r, N2
A
P
Pc
rerN
d
e - 1
1A>P, r, N2
F
A
Ac
erN - 1
d
r
1F>A, r, N2
A
F
Fc
r
d
erN - 1
1A>P, r, N2
P
G
P
c, j
rN
N
f (t) = Gt
Gradient
(ramp)
0
G
G
r2
11 - e -rN2 -
G
1Ne -rN2
r
N
f (t) = cejt
j t = decay rate
with time
Decay
0
c
11 - e -1r + j2N2
r + j
N
EXAMPLE 4.9 Comparison of Daily Flows and Daily
Compounding with Continuous Flows
and Continuous Compounding
Consider a situation in which money flows daily. Suppose you own a retail shop and
generate $200 cash each day. You establish a special business account and deposit
your daily cash flows in an account for 15 months. The account earns an interest rate
of 6%. Compare the accumulated cash values at the end of 15 months, assuming
(a) Daily compounding and
(b) Continuous compounding.
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Section 4.3 Equivalence Calculations with Continuous Payments 155
SOLUTION
(a) With daily compounding:
Given: A = $200 per day r = 6% per year, M = 365 compounding periods
per year, and N = 455 days.
Find: F.
Assuming that there are 455 days in the 15-month period, we find that
i = 6%>365
= 0.01644% per day,
N = 455 days.
The balance at the end of 15 months will be
F = $2001F>A, 0.01644%, 4552
= $2001472.40952
= $94,482.
(b) With continuous compounding:
Now we approximate this discrete cash flow series by a uniform continuous cash
flow function as shown in Figure 4.11. In this situation, an amount flows at the rate
of A per year for N years.
A = $200
455
0
Days
(a) Daily transaction
A = $73,000
f(t)
0
1.25
Years
(b) Continuous flow
Figure 4.11 Comparison between daily transaction and continuous-funds
flow transaction (Example 4.9).
t
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156 CHAPTER 4 Understanding Money and Its Management
Note: Our time unit is a year. Thus, a 15-month period is 1.25 years. Then the
cash flow function is expressed as
f1t2 = A, 0 … t … 1.25
= $20013652
= $73,000 per year.
Given: A = $73,000 per year, r = 6% per year, compounded continuously, and
N = 1.25 years.
Find: F.
Substituting these values back into Eq. (4.6) yields
1.25
F =
L0
73,000e0.0611.25 - t2 dt
= $73,000 c
e0.075 - 1
d
0.06
= $94,759.
The factor in the bracket is known as the funds flow compound amount factor and
is designated 1F>A, r, N2 as shown in Table 4.2. Notice that the difference between
the two methods is only $277 (less than 0.3%).
COMMENTS: As shown in this example, the differences between discrete daily compounding and continuous compounding have no practical significance in most cases.
Consequently, as a mathematical convenience, instead of assuming that money flows
in discrete increments at the end of each day, we could assume that money flows continuously at a uniform rate during the period in question. This type of cash flow assumption is common practice in the chemical industry.
4.4 Changing Interest Rates
Up to this point, we have assumed a constant interest rate in our equivalence calculations.
When an equivalence calculation extends over several years, more than one interest rate
may be applicable to properly account for the time value of money. That is to say, over
time, interest rates available in the financial marketplace fluctuate, and a financial institution that is committed to a long-term loan may find itself in the position of losing the opportunity to earn higher interest because some of its holdings are tied up in a lower
interest loan. The financial institution may attempt to protect itself from such lost earning
opportunities by building gradually increasing interest rates into a long-term loan at the
outset. Adjustable-rate mortgage (ARM) loans are perhaps the most common examples
of variable interest rates. In this section, we will consider variable interest rates in both
single payments and a series of cash flows.
4.4.1 Single Sums of Money
To illustrate the mathematical operations involved in computing equivalence under
changing interest rates, first consider the investment of a single sum of money, P, in a
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Section 4.4 Changing Interest Rates
savings account for N interest periods. If in denotes the interest rate appropriate during
period n, then the future worth equivalent for a single sum of money can be expressed as
F = P11 + i1211 + i22 Á 11 + iN - 1211 + iN2,
(4.7)
and solving for P yields the inverse relation
P = F[11 + i1211 + i22 Á 11 + iN - 1211 + iN2]-1.
(4.8)
EXAMPLE 4.10 Changing Interest Rates with a Lump-Sum
Amount
Suppose you deposit $2,000 in an individual retirement account (IRA) that pays interest at 6% compounded monthly for the first two years and 9% compounded
monthly for the next three years. Determine the balance at the end of five years
(Figure 4.12).
F
6% compounded
monthly
0
1
2
3
4
5
Years
9% compounded
monthly
$2,000
Figure 4.12
Changing interest rates (Example 4.10).
SOLUTION
Given: P = $2,000, r = 6% per year for first two years, 9% per year for last three
years, M = 4 compounding periods per year, N = 20 quarters.
Find: F.
We will compute the value of F in two steps. First we will compute the balance B2 in
the account at the end of two years. With 6% compounded quarterly, we have
i = 6%>12 = 0.5%
N = 12122 = 24 months
B2 = $2,0001F>P, 0.5%, 242
= $2,00011.127162
= $2,254.
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Since the fund is not withdrawn, but reinvested at 9% compounded monthly, as a
second step we compute the final balance as follows:
i = 9%>12 = 0.75%
N = 12132 = 36 months
F = B21F>P, 0.75%, 362
= $2,25411.30862
= $2,950.
4.4.2 Series of Cash Flows
The phenomenon of changing interest rates can easily be extended to a series of cash
flows. In this case, the present worth of a series of cash flows can be represented as
P = A 111 + i12-1 + A 2[11 + i12-111 + i22-1] + . . .
+ A N[11 + i12-111 + i22-1 . . . 11 + iN2-1].
(4.9)
The future worth of a series of cash flows is given by the inverse of Eq. (4.9):
F = A 1[11 + i2211 + i32 . . . 11 + iN2]
+ A 2[11 + i3211 + i42 . . . 11 + iN2] + . . . + A N.
(4.10)
The uniform series equivalent is obtained in two steps. First, the present-worth equivalent
of the series is found from Eq. (4.9). Then A is obtained after establishing the following
equivalence equation:
P = A11 + i12-1 + A[11 + i12-111 + i22-1] + . . .
+ A[11 + i12-111 + i22-1 . . . 11 + iN2-1].
(4.11)
EXAMPLE 4.11 Changing Interest Rates with Uneven Cash
Flow Series
Consider the cash flow in Figure 4.13 with the interest rates indicated, and determine
the uniform series equivalent of the cash flow series.
DISCUSSION: In this problem and many others, the easiest approach involves collapsing the original flow into a single equivalent amount, for example, at time zero, and
then converting the single amount into the final desired form.
SOLUTION
Given: Cash flows and interest rates as shown in Figure 4.13; N = 3.
Find: A.
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$250
$200
$100
i1 = 5%
0
i2 = 7%
1
2
3
A
A
A
5%
7%
1
0
i3 = 9%
9%
2
3
Figure 4.13 Equivalence calculation with changing
interest rates (Example 4.11).
Using Eq. (4.9), we find the present worth:
P = $1001P>F, 5%, 12 + $2001P>F, 5%, 121P>F, 7%, 12
+ $2501P>F, 5%, 121P>F, 7%, 121P>F, 9%, 12
= $477.41.
Then we obtain the uniform series equivalent as follows:
$477.41 = A1P>F, 5%, 12 + A1P>F, 5%, 121P>F, 7%, 12
+ A1P>F, 5%, 121P>F, 7%, 121P>F, 9%, 12
= 2.6591A
A = $179.54.
4.5 Debt Management
Credit card debt and commercial loans are among the most significant financial transactions involving interest. Many types of loans are available, but here we will focus on
those most frequently used by individuals and in business.
4.5.1 Commercial Loans
One of the most important applications of compound interest involves loans that are paid
off in installments over time. If the loan is to be repaid in equal periodic amounts (weekly,
monthly, quarterly, or annually), it is said to be an amortized loan.
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Examples of installment loans include automobile loans, loans for appliances, home
mortgage loans, and the majority of business debts other than very short-term loans. Most
commercial loans have interest that is compounded monthly. With an auto loan, a local
bank or a dealer advances you the money to pay for the car, and you repay the principal
plus interest in monthly installments, usually over a period of three to five years. The car
is your collateral. If you don’t keep up with your payments, the lender can repossess the
car and keep all the payments you have made.
Two things determine what borrowing will cost you: the finance charge and the
length of the loan. The cheapest loan is not the one with the lowest payments or even the
one with the lowest interest rate. Instead, you have to look at (1) the total cost of borrowing, which depends on the interest rate plus fees, and (2) the term, or length of time it
takes you to repay the loan. While you probably cannot influence the rate or the fees, you
may be able to arrange for a shorter term.
• The annual percentage rate (APR) is set by lenders, who are required to tell you
what a loan will actually cost per year, expressed as an APR. Some lenders charge
lower interest, but add high fees; others do the reverse. Combining the fees with a
year of interest charges to give you the true annual interest rate, the APR allows you
to compare these two kinds of loans on equal terms.
• Fees are the expenses the lender will charge to lend the money. The application fee
covers processing expenses. Attorney fees pay the lender’s attorney. Credit search
fees cover researching your credit history. Origination fees cover administrative costs
and sometimes appraisal fees. All these fees add up very quickly and can substantially increase the cost of your loan.
• Finance charges are the cost of borrowing. For most loans, they include all the interest, fees, service charges, points, credit-related insurance premiums, and any other
charges.
• The periodic interest rate is the interest the lender will charge on the amount you borrow. If lender also charges fees, the periodic interest rate will not be the true interest rate.
• The term of your loan is crucial in determining its cost. Shorter terms mean
squeezing larger amounts into fewer payments. However, they also mean paying interest for fewer years, saving a lot of money.
Amortized Installment Loans
So far, we have considered many instances of amortized loans in which we calculated present
or future values of the loans or the amounts of the installment payments. An additional
aspect of amortized loans that will be of great interest to us is calculating the amount of
interest versus the portion of the principal that is paid off in each installment. As we shall
explore more fully in Chapter 10, the interest paid on a loan is an important element in
calculating taxable income and has repercussions for both personal and business loan
transactions. For now, we will focus on several methods of calculating interest and principal paid at any point in the life of the loan.
In calculating the size of a monthly installment, lending institutions may use two
types of schemes. The first is the conventional amortized loan, based on the compoundinterest method, and the other is the add-on loan, based on the simple-interest concept.
We explain each method in what follows, but it should be understood that the amortized
loan is the most common in various commercial lending. The add-on loan is common in
financing appliances as well as furniture.
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In a typical amortized loan, the amount of interest owed for a specified period is calculated on the basis of the remaining balance on the loan at the beginning of the period.
A set of formulas has been developed to compute the remaining loan balance, interest
payment, and principal payment for a specified period. Suppose we borrow an amount P
at an interest rate i and agree to repay this principal sum P, including interest, in equal
payments A over N periods. Then the size of the payment is A = P1A>P, i, N2, and each
payment is divided into an amount that is interest and a remaining amount that goes toward paying the principal.
Let
Bn = Remaining balance at the end of period n, with B0 = P,
In = Interest payment in period n, where In = Bn - 1i,
Pn = Principal payment in period n.
Then each payment can be defined as
A = Pn + In.
(4.12)
The interest and principal payments for an amortized loan can be determined in several ways; two are presented here. No clear-cut reason is available to prefer one method
over the other. Method 1, however, may be easier to adopt when the computational process
is automated through a spreadsheet application, whereas Method 2 may be more suitable
for obtaining a quick solution when a period is specified. You should become comfortable
with at least one of these methods; pick the one that comes most naturally to you.
Method 1: Tabular Method. The first method is tabular. The interest charge for a
given period is computed progressively on the basis of the remaining balance at the beginning of that period. Example 4.12 illustrates the process of creating a loan repayment
schedule based on an iterative approach.
EXAMPLE 4.12 Loan Balance, Principal, and Interest:
Tabular Method
Suppose you secure a home improvement loan in the amount of $5,000 from a local
bank. The loan officer computes your monthly payment as follows:
Contract amount
Contract period
Annual percentage rate
Monthly installments
=
=
=
=
$5,000,
24 months,
12%,
$235.37.
Figure 4.14 is the cash flow diagram. Construct the loan payment schedule by showing the remaining balance, interest payment, and principal payment at the end of
each period over the life of the loan.
SOLUTION
Given: P = $5,000, A = $235.37 per month, r = 12% per year, M = 12 compounding periods per year, and N = 24 months.
Find: Bn and In for n = 1 to 24.
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$5,000
1% per month
Months
6
12
24
18
0
A = $235.37
Figure 4.14 Cash flow diagram of the home improvement loan with an
APR of 12% (Example 4.12).
TABLE 4.3
Creating a Loan Repayment Schedule with Excel (Example 4.12)
Payment
No.
Size of
Payment
Principal
Payment
Interest
Payment
Loan
Balance
1
$235.37
$185.37
$50.00
$4,814.63
2
235.37
187.22
48.15
4,627.41
3
235.37
189.09
46.27
4,438.32
4
235.37
190.98
44.38
4,247.33
5
235.37
192.89
42.47
4,054.44
6
235.37
194.83
40.54
3,859.62
7
235.37
196.77
38.60
3,662.85
8
235.37
198.74
36.63
3,464.11
9
235.37
200.73
34.64
3,263.38
10
235.37
202.73
32.63
3,060.65
11
235.37
204.76
30.61
2,855.89
12
235.37
206.81
28.56
2,649.08
13
235.37
208.88
26.49
2,440.20
14
235.37
210.97
24.40
2,229.24
15
235.37
213.08
22.29
2,016.16
16
235.37
215.21
20.16
1,800.96
17
235.37
217.36
18.01
1,583.60
18
235.37
219.53
15.84
1,364.07
19
235.37
221.73
13.64
1,142.34
20
235.37
223.94
11.42
918.40
21
235.37
226.18
9.18
692.21
22
235.37
228.45
6.92
463.77
23
235.37
230.73
4.64
233.04
24
235.37
233.04
2.33
0.00
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We can easily see how the bank calculated the monthly payment of $235.37. Since
the effective interest rate per payment period on this loan is 1% per month, we establish the following equivalence relationship:
$235.371P>A, 1%, 242 = $235.37121.24312 = $5,000.
The loan payment schedule can be constructed as in Table 4.3. The interest due
at n = 1 is $50.00, 1% of the $5,000 outstanding during the first month. The $185.37
left over is applied to the principal, reducing the amount outstanding in the second
month to $4,814.63. The interest due in the second month is 1% of $4,814.63, or
$48.15, leaving $187.22 for repayment of the principal. At n = 24, the last $235.37
payment is just sufficient to pay the interest on the unpaid principal of the loan and to
repay the remaining principal. Figure 4.15 illustrates the ratios between the interest
and principal payments over the life of the loan.
COMMENTS: Certainly, constructing a loan repayment schedule such as that in Table 4.3
can be tedious and time consuming, unless a computer is used. As you can see in the
website for this book, you can download an Excel file that creates the loan repayment
schedule, on which you can make any adjustment to solve a typical loan problem of
your choice.
30
Principal payment
Interest payment
25
Payment period (month)
20
15
10
5
0
250
200
150
100
50
0
50
100
$
Figure 4.15 The proportions of principal and interest payments over the life of the loan
1monthly payment = $235.372 (Example 4.12).
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Current point (n)
P
Bn = A(P/A, i, N – n)
n
N
0
A
Payments made
Figure 4.16
Payments to be made
Calculating the remaining loan balance on the basis of Method 2.
Method 2: Remaining-Balance Method. Alternatively, we can derive Bn by computing the equivalent payments remaining after the nth payment. Thus, the balance with
N - n payments remaining is
Bn = A1P>A, i, N - n2,
(4.13)
and the interest payment during period n is
In = 1Bn - 12i = A1P>A, i, N - n + 12i,
(4.14)
where A1P>A, i, N - n + 12 is the balance remaining at the end of period
n - 1 and
Pn = A - In = A - A1P>A, i, N - n + 12i
= A[1 - 1P>A, i, N - n + 12i].
Knowing the interest factor relationship 1P>F, i, n2 = 1 - 1P>A, i, n2i from Table 3.4,
we obtain
Pn = A1P>F, i, N - n + 12.
(4.15)
As we can see in Figure 4.16, this method provides more concise expressions for
computing the balance of the loan.
EXAMPLE 4.13 Loan Balances, Principal, and Interest:
Remaining-Balance Method
Consider the home improvement loan in Example 4.12, and
(a) For the sixth payment, compute both the interest and principal payments.
(b) Immediately after making the sixth monthly payment, you would like to pay off
the remainder of the loan in a lump sum. What is the required amount?
SOLUTION
(a) Interest and principal payments for the sixth payment.
Given: (as for Example 4.12)
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Find: I6 and P6.
Using Eqs. (4.14) and (4.15), we compute
I6 = $235.371P>A, 1%, 19210.012
= 1$4,054.44210.012
= $40.54.
P6 = $235.371P>F, 1%, 192 = $194.83,
or we simply subtract the interest payment from the monthly payment:
P6 = $235.37 - $40.54 = $194.83.
(b) Remaining balance after the sixth payment.
The lower half of Figure 4.17 shows the cash flow diagram that applies to this part of
the problem. We can compute the amount you owe after you make the sixth payment
by calculating the equivalent worth of the remaining 18 payments at the end of the
sixth month, with the time scale shifted by 6:
Given: A = $235.37, i = 1% per month, and N = 18 months.
Find: Balance remaining after six months 1B62.
B6 = $235.371P>A, 1%, 182 = $3,859.62.
If you desire to pay off the remainder of the loan at the end of the sixth payment, you
must come up with $3,859.62. To verify our results, compare this answer with the
value given in Table 4.3.
$5,000
B6 = $235.37 (P/A, 1%, 18) = $3,859.62
I6 = $40.54
P6 = $194.83
1% per month
Months
6
12
18
24
0
6 payments made
18 monthly payments outstanding
A = $235.37
Figure 4.17 Computing the outstanding loan balance after making the sixth payment on the home improvement loan (Example 4.13).
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Add-on Interest:
A method of
computing
interest whereby
interest charges
are made for the
entire principal
amount for the
entire term,
regardless of any
repayments of
principal made.
Add-On Interest Loans
The add-on loan is totally different from the popular amortized loan. In the add-on loan,
the total interest to be paid is precalculated and added to the principal. The principal and
this precalculated interest amount are then paid together in equal installments. In such a
case, the interest rate quoted is not the effective interest rate, but what is known as addon interest. If you borrow P for N years at an add-on rate of i, with equal payments due
at the end of each month, a typical financial institution might compute the monthly installment payments as follows:
Total add-on interest = P1i21N2,
Principal plus add-on interest = P + P1i21N2 = P11 + iN2
Monthly installments =
P11 + iN2
.
112 * N2
(4.16)
Notice that the add-on interest is simple interest. Once the monthly payment is determined, the financial institution computes the APR on the basis of this payment, and you
will be told what the value will be. Even though the add-on interest is specified along
with the APR, many ill-informed borrowers think that they are actually paying the add-on
rate quoted for this installment loan. To see how much interest you actually pay under a
typical add-on loan arrangement, consider Example 4.14.
EXAMPLE 4.14 Effective Interest Rate for an Add-On
Interest Loan
Consider again the home improvement loan problem in Example 4.12. Suppose that
you borrow $5,000 with an add-on rate of 12% for two years. You will make 24 equal
monthly payments.
(a) Determine the amount of the monthly installment.
(b) Compute the nominal and the effective annual interest rate on the loan.
SOLUTION
Given: Add-on rate = 12% per year, loan amount 1P2 = $5,000, and N = 2 years.
Find: (a) A and (b) ia and i.
(a) First we determine the amount of add-on interest:
iPN = 10.1221$5,0002122 = $1,200.
Then we add this simple-interest amount to the principal and divide the total
amount by 24 months to obtain A:
A =
1$5,000 + $1,2002
= $258.33.
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Section 4.5 Debt Management
(b) Putting yourself in the lender’s position, compute the APR value of the loan
just described. Since you are making monthly payments with monthly compounding, you need to find the effective interest rate that makes the present
$5,000 sum equivalent to 24 future monthly payments of $258.33. In this
situation, we are solving for i in the equation
$258.33 = $5,0001A>P, i, 242,
or
1A>P, i, 242 = 0.0517.
You know the value of the A/P factor, but you do not know the interest rate i. As
a result, you need to look through several interest tables and determine i by interpolation. A more effective approach is to use Excel’s RATE function with the
following parameters:
=RATE1N,A,P,F,type,guess)
=RATE124,258.33, -5000,0,0,1%2:1.7975%
The nominal interest rate for this add-on loan is 1.7975 * 12 = 21.57%, and
the effective annual interest rate is 11 + 0.01975212 - 1 = 26.45%, rather
than the 12% quoted add-on interest. When you take out a loan, you should not
confuse the add-on interest rate stated by the lender with the actual interest cost
of the loan.
COMMENTS: In the real world, truth-in-lending laws require that APR information always be provided in mortgage and other loan situations, so you would not have to
calculate nominal interest as a prospective borrower (although you might be interested in calculating the actual or effective interest). However, in later engineering
economic analyses, you will discover that solving for implicit interest rates, or rates
of return on investment, is performed regularly. Our purpose in this text is to periodically give you some practice with this type of problem, even though the scenario
described does not exactly model the real-world information you would be given.
4.5.2 Loan versus Lease Financing
When, for example, you choose a car, you also choose how to pay for it. If you do not
have the cash on hand to buy a new car outright—and most of us don’t—you can consider taking out a loan or leasing the car to spread the payments over time. Deciding
whether to pay cash, take a loan, or sign a lease depends on a number of personal as well
as economic factors. Leasing is an option that lets you pay for the portion of a vehicle you
expect to use over a specified term, plus a charge for rent, taxes, and fees. For instance,
you might want a $20,000 vehicle. Suppose that vehicle might be worth about $9,000 at
the end of your three-year lease. (This is called the residual value.)
• If you have enough money to buy the car, you could purchase it in cash. If you pay
cash, however, you will lose the opportunity to earn interest on the amount you spend.
That could be substantial if you know of an investment that is paying a good return.
• If you purchase the vehicle using debt financing, your monthly payments will be
based on the entire $20,000 value of the vehicle. You will continue to own the vehicle
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at the end of your financing term, but the interest you will pay on the loan will drive
up the real cost of the car.
• If you lease the same vehicle, your monthly payments will be based on the amount of
the vehicle you expect to “use up” over the term of the lease. This value ($11,000 in
our example) is the difference between the original cost ($20,000) and the estimated
value at the end of the lease ($9,000). With leasing, the length of your lease agreement, the monthly payments, and the yearly mileage allowance can be tailored to
your driving needs. The greatest financial appeal for leasing is low initial outlay
costs: Usually you pay a leasing administrative fee, one month’s lease payment, and
a refundable security deposit. The terms of your lease will include a specific mileage
allowance; if you put additional miles on your car, you will have to pay more for each
extra mile.
Discount Rate to Use in Comparing Different Financing Options
In calculating the net cost of financing a car, we need to decide which interest rate to use
in discounting the loan repayment series. The dealer’s (bank’s) interest rate is supposed to
reflect the time value of money of the dealer (or the bank) and is factored into the required payments. However, the correct interest rate to use in comparing financing options
is the interest rate that reflects your time value of money. For most individuals, this interest rate might be equivalent to the savings rate from their deposits. To illustrate, consider
Example 4.15, in which we compare three auto financing options.
EXAMPLE 4.15 Financing your Vehicle: Paying Cash, Taking
a Loan, or Leasing
Suppose you intend to own or lease a vehicle for 42 months. Consider the following
three ways of financing the vehicle—say, a 2006 BMW 325 Ci 2-D coupe:
• Option A: Purchase the vehicle at the normal price of $32,508, and pay for the
vehicle over 42 months with equal monthly payments at 5.65% APR
financing.
• Option B: Purchase the vehicle at a discount price of $31,020 to be paid immediately.
• Option C: Lease the vehicle for 42 months.
The accompanying chart lists the items of interest under each option. For each option, license, title, and registration fees, as well as taxes and insurance, are extra.
For the lease option, the lessee must come up with $1,507.76 at signing. This
cash due at signing includes the first month’s lease payment of $513.76 and a $994
administrative fee. The lease rate is based on 60,000 miles over the life of the contract. There will be a surcharge at the rate of 18 cents per mile for any additional
miles driven over 60,000. No security deposit is required; however, a $395 disposition fee is due at the end of the lease, at which time the lessee has the option to
purchase the car for $17,817. The lessee is also responsible for excessive wear and
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Section 4.5 Debt Management
Item
Option A
Option B
Option C
Debt Paying Lease
Financing
Cash Financing
Price
$32,508
$32,508
$32,508
Down payment
$4,500
$0
$0
APR(%)
5.65%
Monthly payment
$736.53
$513.76
Length
42 months
42 months
Fees
$994
Cash due at end of lease
$395
Purchase option at end of lease
$17,817
Cash due at signing
$4,500
$31,020
$1,507.76
use. If the funds that would be used to purchase the vehicle are presently earning
4.5% annual interest compounded monthly, which financing option is a better
choice?
DISCUSSION: With a lease payment, you pay for the portion of the vehicle you expect
to use. At the end of the lease, you simply return the vehicle to the dealer and pay the
agreed-upon disposal fee. With traditional financing, your monthly payment is based
on the entire $32,508 value of the vehicle, and you will own the vehicle at the end of
your financing terms. Since you are comparing the options over 42 months, you must
explicitly consider the unused portion (resale value) of the vehicle at the end of the
term. In other words, you must consider the resale value of the vehicle in order to figure out the net cost of owning it. As the resale value, you could use the $17,817 quoted
by the dealer in the lease option. Then you have to ask yourself if you can get that kind
of resale value after 42 months’ ownership.
Note that the 5.65% APR represents the dealer’s interest rate used in calculating the
loan payments. With 5.65% interest, your monthly payments will be A = 1$32,508 $4,50021A>P, 5.65%>12, 422 = $736.53. Note also, however, that the 4.5% APR
represents your earning opportunity rate. In other words, if you do not buy the car, your
money continues to earn 4.5% APR. Therefore, 4.5% represents your opportunity cost
of purchasing the car. So which interest rate do you use in your analysis? Clearly, the
4.5% rate is the appropriate one to use.
SOLUTION
Given: Financial facts shown in Figure 4.18, r = 4.5%, payment period = monthly,
and compounding period = monthly.
Find: The most economical financing option, under the assumption that you will be
able to sell the vehicle for $17,817 at the end of 42 months.
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$17,819
Option A: Debt Financing
1
2
3
0
42
A $736.53
$17,819
$4,500
Option B: Cash Financing
0
42
$31,020
Option C: Lease Financing
40
41
A $513.76
42
$395
$1,507.76
Figure 4.18
Comparing different financing options.
For each option, we will calculate the net equivalent total cost at n = 0. Since
the loan payments occur monthly, we need to determine the effective interest rate per
month, which is 4.5%/12.
• Option A: Conventional debt financing
The equivalent present cost of the total loan payments is
P1 = $4,500 + $736.531P>A, 4.5%>12, 422
= $33,071.77.
The equivalent present worth of the resale value of the car is
P2 = $17,8171P>F, 4.5%>12, 422 = $15,225.13.
The equivalent net financing cost is
POption A = $33,071.77 - $15,225.13
= $17,846.64.
• Option B: Cash financing
POption B = $31,020 - $17,8171P>F, 4.5%>12, 422
= $31,020 - $15,225.13
= $15,844.87.
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171
• Option C: Lease financing
The equivalent present cost of the total lease payments is
POption C = $1,507.76 + $513.761P>A, 4.5%>12, 412
+ $3951P>F, 4.5%>12, 422
= $1,507.76 + $19,490.96 + $337.54
= $21,336.26
It appears that, at 4.5% interest compounded monthly, the cash financing option
is the most economical one.
4.5.3 Home Mortgage
The term mortgage refers to a special type of loan for buying a piece of property,
such as a house. The most common mortgages are fixed-rate amortized loans, adjustable-rate loans, and graduated-payment loans. As with many of the amortized
loans we have considered so far, most home mortgages involve monthly payments
and monthly compounding. However, as the names of these types of mortgages
suggest, a number of ways determine how monthly payments and interest rates can
be structured over the life of the mortgage.
The Cost of a Mortgage
The cost of a mortgage depends on the amount you borrow, the interest you pay,
and how long you take to repay the loan. Since monthly payments spread the cost
of a mortgage over a long period, it is easy to forget the total expense. For example, if you borrow $100,000 for 30 years at 8.5% interest, your total repayment
will be around $277,000, more than two-and-a-half times the original loan! Minor
differences in the interest rate—8.5% versus 8%—can add up to a lot of money
over 30 years. At 8%, the total repaid would be $264,240, almost $13,000 less than
at the 8.5% rate. Other than the interest rate, any of the following factors will increase the overall cost, but a higher interest rate and longer term will have the
greatest impact:
• Loan amount. This is the amount you actually borrow after fees and points
are deducted. It is the basis for figuring the real interest, or APR, on the money
you are borrowing.
• Loan term. With a shorter term, you will pay less interest overall, and your
monthly payments will be somewhat larger. A 15-year mortgage, as opposed to
a 30-year mortgage for the same amount, can cut your costs by more than 55%.
• Payment frequency. You can pay your mortgage biweekly instead of monthly,
or you can make an additional payment each month. With biweekly payments,
you make 26 regular payments instead of 12 every year. The mortgage is paid
off in a little more than half the time, and you pay a little more than half the interest. With an additional payment each month, you can reduce your principal.
With a fixed-rate mortgage, you pay off the loan more quickly, but regular
monthly payments remain the same.
• Points (prepaid interest). Points are interest that you prepay at the closing on
your home. Each point is 1% of the loan amount. For example, on a $100,000
Mortgage: A
loan to finance
the purchase of
real estate,
usually with
specified payment
periods and
interest rates.
ARM: A mortgage
with an interest
rate that may
change, usually
in response to
changes in the
Treasury Bill rate
or the prime
rate.
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Origination fee:
A fee charged
by a lender for
processing a loan
application,
expressed as a
percentage of
the mortgage
amount.
loan, three points represents a prepayment of $3,000. This is equivalent to financing
$97,000, but your payments are based on a $100,000 loan.
• Fees. Fees include application fees, loan origination fees, and other initial costs imposed by the lender.
Lenders might be willing to raise the rate by a fraction (say, 18% or 14%) and lower the
points—or the reverse—as long as they make the same profit. The advantages of fewer
points are lower closing costs and laying out less money when you are apt to need it most.
However, if you plan to keep the house longer than five to seven years, paying more
points to get a lower interest rate will reduce your long-term cost. Example 4.16 examines the effect of points on the cost of borrowing.
EXAMPLE 4.16 Points or No Points?
When you are shopping for a home mortgage loan, you frequently encounter various
types of borrowing options, including paying points up front and paying no points
but accepting a slightly higher interest rate. Suppose you want to finance a home
mortgage of $100,000 at a 15-year fixed interest rate. Countrywide, a leading independent home lender offers the following two options with no origination fees:
• Option 1: Pay one point with 6.375% interest.
• Option 2: Pay no points with 6.75% interest.
A point is equivalent to 1% of the face value of the mortgage. In other words,
with Option 1, you are borrowing only $99,000, but your lender will calculate your
monthly payments on the basis of $100,000. Compute the APR for Option 1.
DISCUSSION: Discount fees or points are a fact of life with mortgages. A point is a fee
charged by a lender to increase the lender’s effective yield on the money borrowed.
Points are charged in lieu of interest; the more points paid, the lower is the rate of interest required on the loan to provide the same yield or return to the lender. One point
equals 1% of the loan amount. Origination fees are the fees charged by a lender to prepare a loan document, make credit checks, and inspect and sometimes appraise a property. These fees are usually computed as a percentage of the face value of the mortgage.
SOLUTION
Given: P = $100,000, r = 6.375%, N = 180 months, and discount point = 1 point.
Find: APR.
We first need to find out how much the lender will calculate your monthly payments
to be. Since the mortgage payments will be based on the face value of the loan, your
monthly payment will be
A = $100,0001A>P, 6.375%>12, 1802 = $863.98.
Because you have to pay $1,000 to borrow $100,000, you are actually borrowing
$99,000. Therefore, you need to find out what kind of interest you are actually paying. In other words, you borrow $99,000 and you make $863.98 monthly payments
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Section 4.5 Debt Management
over 15 years. To find the interest rate, you can set up the following equivalence
equation and solve for the unknown interest rate.
$99,000 = $863.981P>A, i, 1802,
i = 0.545% per month,
APR = 0.545% * 12 = 6.54%.
Note that with a one-point fee, the lender was able to raise its effective yield from
6.375% to 6.54%. However, the lender is still earning less than Option 2 affords.
Variable-Rate Mortgages
Mortgages can have either fixed or adjustable rates (or both, in which case they are
known as hybrid mortgages). As we mentioned earlier, interest rates in the financial marketplace rise and fall over time. If there is a possibility that market rates will rise above
the fixed rate, some lenders may be reluctant to lock a loan into a fixed interest rate over
a long term. If they did, they would have to forgo the opportunity to earn better rates because their assets are tied up in a loan that earns a lower rate of interest. The variable-rate
mortgage is meant to encourage lending institutions to commit funds over long periods.
Typically, the rate rises gradually each year for the first several years of the mortgage and
then settles into a single rate for the remainder of the loan. By contrast, a hybrid loan offers a plan with a fixed interest rate for the first few years and then converts to a variablerate schedule for the remaining periods. Example 4.17 illustrates how a lender would
calculate the monthly payments with varying interest rates.
EXAMPLE 4.17 A 5/1 Hybrid Mortgage Plan
Consider again the hybrid mortgage issue discussed in the chapter opening story.
Suppose that you finance a home on the basis of a 5/1 hybrid mortgage (five-year
fixed/adjustable) plan over 30 years. The $100,000 hybrid loan plan offers an initial
rate of 6.02% fixed for 60 months. The loan rate would be adjusted thereafter every
12 months to the lowest of three options: the then-current rate on one-year Treasury
bills plus 2.75 percent, the previous rate plus a maximum annual cap of 2.0 percent,
or a lifetime cap of 11.02 percent. There is no prepayment penalty for this type of
loan. The projected interest rates by the lender after 5 years are as follows:
Period
Projected APR
Years 1–5
6.02%
Year 6
6.45%
Year 7
6.60%
Year 8
6.80%
Year 9
7.15%
Year 10
7.30%
173
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174 CHAPTER 4 Understanding Money and Its Management
(a) Develop the payment schedule for the first 10 years.
(b) Determine the total interest paid over a 10-year ownership.
SOLUTION
Given: Varying annual mortgage rates and N = 30 years.
Find: (a) the monthly payment; (b) the total interest payment over the 10-year ownership of the home.
(a) Monthly payment calculation. During the first 5 years, you borrow $100,000 for 30
years at 6.02%. To compute the monthly payment, use i = 6.02%>12 = 0.5017%
per month and N = 360 months:
A 1 - 60 = $100,0001A>P, 0.5017%, 3602 = $600.84.
The balance remaining on the loan after you make the 60th payment will be
B60 = $600.841P>A, 0.5017%, 3002 = $93,074.
During the 6th year, the interest rate changes to 6.15%, or 0.5125% per month,
but the remaining term is only 300 months. Therefore, the new monthly payment would be
A 60 - 72 = $93,0741A>P, 0.5375%, 3002 = $625.54.
After you make the 72nd payment, the balance remaining on the mortgage is
B72 = $625.541P>A, 0.5375%, 2882 = $91,526.
During the 7th year, the interest rate changes to 6.60%, or 0.5500% per month.
The new monthly payment and the remaining balance after you make the 84th
payment are then
A 73 - 84 = $91,5261A>P, 0.5500%, 2882 = $634.03
and
B84 = $634.031P>A, 0.5500%, 2762 = $89,908.
You can compute the monthly payments in the same fashion for the remaining
years. The accompanying table gives the details over the life of the loan.
(b) To determine the total interest paid over 10 years, we first determine the total
monthly mortgage payments over 10 years. Since we know the ending balance
at the end of 10 years, we can easily calculate the interest payments during this
home ownership period:
Total mortgage payment = 60 * $600.84
+ 12 a
$625.54 + $634.03 + $645.09
b
+ $664.09 + $672.05
= $74,940.
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Section 4.6 Investing in Financial Assets 175
Month
Monthly
Payment
1
1–12
6.02%
$600.84
$98,777
2
13–24
6.02%
$600.84
$97,477
3
25–36
6.02%
$600.84
$96,098
4
37–48
6.02%
$600.84
$94,633
5
49–60
6.02%
$600.84
$93,704
6
61–72
6.45%
$625.54
$91,526
7
73–84
6.60%
$634.03
$89,908
8
85–96
6.80%
$645.09
$88,229
9
97–108
7.15%
$664.09
$86,513
10
109–120
7.30%
$672.05
$84,704
Year
Fixed rate
Variable rate
Ending
Loan
Balance
Forecast
Rate
Interest payment = Ending balance + Total mortgage payment
- $100,000
= $84,704 + $74,940 - $100,000
= $59,644.
4.6 Investing in Financial Assets
Most individual investors have three basic investment opportunities in financial assets:
stocks, bonds, and cash. Cash investments include money in bank accounts, certificates
of deposit (CDs), and U.S. Treasury bills. You can invest directly in any or all of the
three, or indirectly, by buying mutual funds that pool your money with money from
other people and then invest it. If you want to invest in financial assets, you have plenty
of opportunities. In the United States alone, there are more than 9,000 stocks, 7,500
mutual funds, and thousands of corporate and government bonds to choose from. Even
though we will discuss investment basics in the framework of financial assets in this
chapter, the same economic concepts are applicable to any business assets examined in
later chapters.
4.6.1 Investment Basics
Selecting the best investment for you depends on your personal circumstances as well
as general market conditions. For example, a good investment strategy for a long-term
retirement plan may not be a good strategy for a short-term college savings plan. In
each case, the right investment is a balance of three things: liquidity, safety, and
return.
• Liquidity: How accessible is your money? If your investment money must be available to cover financial emergencies, you will be concerned about liquidity: how
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easily you can convert it to cash. Money-market funds and savings accounts are very
liquid; so are investments with short maturity dates, such as CDs. However, if you
are investing for longer term goals, liquidity is not a critical issue. What you are after
in that case is growth, or building your assets. We normally consider certain stocks
and stock mutual funds as growth investments.
• Risk: How safe is your money? Risk is the chance you take of making or losing
money on your investment. For most investors, the biggest risk is losing money, so
they look for investments they consider safe. Usually, that means putting money
into bank accounts and U.S. Treasury bills, as these investments are either insured
or default free. The opposite, but equally important, risk is that your investments
will not provide enough growth or income to offset the impact of inflation, the gradual increase in the cost of living. There are additional risks as well, including how
the economy is doing. However, the biggest risk is not investing at all.
• Return: How much profit will you be able to expect from your investment? Safe
investments often promise a specific, though limited, return. Those which involve more
risk offer the opportunity to make—or lose—a lot of money. Both risk and reward are
time dependent. On the one hand, as time progresses, low-yielding investments become more risky because of inflation. On the other hand, the returns associated with
higher risk investments could become more stable and predictable over time, thereby
reducing the perceived level of risk.
4.6.2 How to Determine Your Expected Return
Risk-free return:
A theoretical
interest rate
that would be
returned on an
investment which
was completely
free of risk.
The 3-month
Treasury Bill
is a close
approximation,
since it is virtually
risk-free.
Risk premium:
The reward for
holding a risky
investment
rather that a
risk-free one.
Return is what you get back in relation to the amount you invested. Return is one way to
evaluate how your investments in financial assets are doing in relation to each other and
to the performance of investments in general. Let us look first at how we may derive rates
of return.
Basic Concepts
Conceptually, the rate of return that we realistically expect to earn on any investment is a
function of three components: (1) risk-free real return, (2) an inflation factor, and (3) a
risk premium.
Suppose you want to invest in stock. First, you should expect to be rewarded in some
way for not being able to use your money while you are holding the stock. Then, you
would be compensated for decreases in purchasing power between the time you invest
the money and the time it is returned to you. Finally, you would demand additional rewards for any chance that you would not get your money back or that it will have declined in value while invested.
For example, if you were to invest $1,000 in risk-free U.S. Treasury bills for a year,
you would expect a real rate of return of about 2%. Your risk premium would be also
zero. You probably think that the 2% does not sound like much. However, to that you
have to add an allowance for inflation. If you expect inflation to be about 4% during the
investment period, you should realistically expect to earn 6% during that interval (2%
real return + 4% inflation factor + 0% for risk premium). Here is what the situation
looks like in tabular form:
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Section 4.6 Investing in Financial Assets 177
Real return
2%
Inflation (loss of purchasing power)
4%
Risk premium (U.S. Treasury Bills)
0%
Total expected return
6%
How would it work out for a riskier investment, say, in an Internet stock such as
Google.com? Because you consider this stock to be a very volatile one, you would increase the risk premium to something like the following:
Real return
2%
Inflation (loss of purchasing power)
4%
Risk premium (Google.com)
20%
Total expected return
26%
So you will not invest your money in Google.com unless you are reasonably confident of having it grow at an annual rate of 26%. Again, the risk premium of 20% is a perceived value that can vary from one investor to another.
Return on Investment over Time
If you start out with $1,000 and end up with $2,000, your return is $1,000 on that investment, or 100%. If a similar investment grows to $1,500, your return is $500, or 50%.
However, unless you held those investments for the same period, you cannot determine
which has a better performance. What you need in order to compare your return on one
investment with the return on another investment is the compound annual return, the
average percentage that you have gained on each investment over a series of one-year periods. For example, if you buy a share for $15 and sell it for $20, your profit is $5. If that
happens within a year, your rate of return is an impressive 33.33%. If it takes five years,
your return (compounded) will be closer to 5.92%, since the profit is spread over a fiveyear period. Mathematically, you are solving the following equivalence equation for i:
$20 = $1511 + i25,
i = 5.92%.
Figuring out the actual return on your portfolio investment is not always that simple.
There are several reasons:
1. The amount of your investment changes. Most investment portfolios are active,
with money moving in and out.
2. The method of computing the return can vary. For example, the performance of a
stock can be averaged or compounded, which changes the rate of return significantly,
as we will demonstrate in Example 4.18.
3. The time you hold specific investments varies. When you buy or sell can have a dramatic effect on the overall return.
Compound
annual return:
The year-overyear growth rate
of an investment
over a specified
period of time.
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178 CHAPTER 4 Understanding Money and Its Management
EXAMPLE 4.18 Figuring Average versus Compound Return
Consider the following six different cases of the performance of a $1,000 investment
over a three-year holding period:
Investment
Annual Investment Yield
Case 2
Case 3
Case 4
Case 1
Case 5
Case 6
Year 1
9%
5%
0%
0%
-1%
-5%
Year 2
9%
10%
7%
0%
-1%
-8%
Year 3
9%
12%
20%
27%
29%
40%
Compute the average versus compound return for each case.
SOLUTION
Given: Three years’ worth of annual investment yield data.
Find: Compound versus average rate of return.
Average annual
return: A figure
used when
reporting the
historical return
of a mutual fund.
As an illustration, consider Case 6 for an investment of $1,000. At the end of the
first year, the value of the investment decreases to $950; at the end of second year, it
decreases again, to $95011 - 0.082 = $874; at the end of third year, it increases to
$87411 + 0.402 = $1,223.60. Therefore, one way you can look at the investment
is to ask, “At what annual interest rate would the initial $1,000 investment grow to
$1,223.60 over three years?” This is equivalent to solving the following equivalence
problem:
$1,223.60 = $1,00011 + i23,
i = 6.96%.
If someone evaluates the investment on the basis of the average annual rate of return, he or she might proceed as follows:
i =
1-5% - 8% + 40%2
= 9%.
3
If you calculate the remaining cases, you will observe that all six cases have the
same average annual rate of return, although their compound rates of return vary
from 6.96% to 9%:
Case 1
Compound versus Average Rate of Return
Case 2
Case 3
Case 4
Case 5
Case 6
Average return
9%
9%
9%
9%
9%
9%
Balance at end
of three years
$1,295
$1,294
$1,284
$1,270
$1,264
$1,224
Compound rate
of return
9.00%
8.96%
8.69%
8.29%
8.13%
6.96%
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Section 4.6 Investing in Financial Assets 179
Your immediate question is “Are they the same indeed?” Certainly not: You will
have the most money with Case 1, which also has the highest compound rate of return.
The average rate of return is easy to calculate, but it ignores the basic principle of the
time value of money. In other words, according to the average-rate-of-return concept,
we may view all six cases as indifferent. However, the amount of money available at
the end of year 3 would be different in each case. Although the average rate of return is
popular for comparing investments in terms of their yearly performance, it is not a correct measure in comparing the performance for investments over a multiyear period.
COMMENTS: You can evaluate the performance of your portfolio by comparing it
against standard indexes and averages that are widely reported in the financial press.
If you own stocks, you can compare their performance with the performance of the
Dow Jones Industrial Average (DJIA), perhaps one of the best-known measures of
stock market performance in the world. If you own bonds, you can identify an index
that tracks the type you own: corporate, government, or agency bonds. If your investments are in cash, you can follow the movement of interest rates on Treasury bills,
CDs, and similar investments. In addition, total-return figures for the performance of
mutual funds are reported regularly. You can compare how well your investments are
doing against those numbers. Another factor to take into account in evaluating your
return is the current inflation rate. Certainly, your return needs to be higher than the
inflation rate if your investments are going to have real growth.
4.6.3 Investing in Bonds
Bonds are loans that investors make to corporations and governments. As shown in
Figure 4.19, the borrowers get the cash they need while the lenders earn interest. Americans have more money invested in bonds than in stocks, mutual funds, or any other types
of securities. One of the major appeals is that bonds pay a set amount of interest on a regular basis. That is why they are called fixed-income securities. Another attraction is that
the issuer promises to repay the loan in full and on time.
Bond versus Loan
A bond is similar to a loan. For example, say you lend out $1,000 for 10 years in return
for a yearly payment of 7% interest. Here is how that arrangement translates into bond
terminology. You did not make a loan; you bought a bond. The $1,000 of principal is the
face value of the bond, the yearly interest payment is its coupon, and the length of the
loan, 10 years, is the bond’s maturity. If you buy a bond at face value, or par, and hold
it until it matures, you will earn interest at the stated, or coupon, rate. For example, if you
buy a 20-year $1,000 bond paying 8%, you will earn $80 a year for 20 years. The yield,
or return on your investment, will also be 8%, and you get your $1,000 back. You can also
buy and sell bonds through a broker after their date of issue. This is known as the
secondary market. There the price fluctuates, with a bond sometimes selling at more
than par value, at a premium price (premium bonds), and sometimes below, at a discount.
Changes in price are tied directly to the interest rate of the bond. If its rate is higher than
the rate being paid on similar bonds, buyers are willing to pay more to get the higher interest. If its rate is lower, the bond will sell for less in order to attract buyers. However, as
the price goes up, the yield goes down, and when the price goes down, the yield goes up.
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The Individual as Lender
The Institution as Borrower
Corporate bonds
Corporation
Investors
willing to
lend money
Corporations use bonds
• to raise capital to pay for expansion, modernization
• to cover operating expenses
• to finance corporate take-overs or other changes in
management structure
U.S. Treasury bonds
U.S. Treasury
Investor
Investor
gets interest
gets par
payment at
value at
specific
maturity
intervals
Municipal bonds
Municipal
Governments
Figure 4.19
The U.S. Treasury floats debt issues
• to pay for a wide range of government activities Bond
• to pay off the national debt
matures
Corporations use bonds
• to pay for a wide variety of public projects:
schools, highways, stadiums, sewage systems, bridges
• to supplement their operating budget
Types of bonds and how they are issued in the financial market.
(Source: Kenneth M. Morris and Alan M. Stegel, The Wall Street Guide to Understanding Money and Investing,
© 1993 by Lightbulb Press, Inc.)
Types of Bonds
You can choose different types of bonds to fit your financial needs—be they investing for
college, finding tax-free income, or operating over a range of other possibilities. That is
why it is important to have a sense of how the various types work. Several types of bonds
are available in the financial market:
• Corporate bonds. Bonds are the major source of corporate borrowing. Debentures
are backed by the general credit of the corporation. Specific corporate assets, such as
property or equipment, may back bonds.
• Municipal bonds. State and local governments issue bonds to finance many projects that could not be funded through tax revenues. General-obligation bonds are
backed by the full faith and credit of the issuer, revenue bonds by the earnings of the
particular project being financed.
• Treasury notes and bonds. The U.S. Treasury floats debt issues to pay for a wide
range of government activities, including financing the national debt. Intermediate (2
to 10 years) and long-term (10 to 30 years) government bonds are a major source of
government funding.
• Treasury bills. These are the largest components of the money market, where shortterm (13 weeks to 52 weeks) securities are bought and sold. Investors use T-bills for part
of their cash reserve or as an interim holding place. Interest is the difference between the
buying price and the amount paid at maturity.
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• Zero-coupon bonds. Corporations and governments sell zero-coupon bonds at a deep
discount. Investors do not collect interest; instead, the value of the bond increases to its
full value when it matures. In this way, zero-coupon bonds are similar to old Series savings bonds that you bought for $37.50 and could cash in for $50 after seven years.
Organizations like to issue zeros because they can continue to use the loan money without paying periodic interest. Investors like zeros because they can buy more bonds for
their money and then time the maturities to coincide with anticipated expenses.
• Callable bonds. Callable bonds do not run their full term. In a process called redemption, the issuer may call the bond—pay off the debt—before the maturity date.
Issuers will sometimes call bonds when interest rates drop, so that they can reduce
their debt. If they pay off their outstanding bonds, they can float another bond at the
lower rate. It is the same idea as that of refinancing a mortgage to get a lower interest
rate and make lower monthly payments. Callable bonds are more risky for investors
than noncallable bonds, because the investors are often faced with reinvesting the
money at a lower, less attractive rate. To protect bondholders who expect long-term
steady income, call provisions often specify that a bond cannot be called before a
certain number of years, usually 5 or 10 years.
• Floating-rate bonds. These bonds promise periodic adjustments of the interest rate
to persuade investors that they are not locked into what seems like an unattractively
low rate.
Under the U.S. Constitution, any type of government bond must include a tax break.
Because federal and local governments cannot interfere with each other’s affairs, income
from local government bonds (municipals, or simply munis) is immune from federal
taxes, and income from U.S. Treasury bonds is free from local taxes.
Understanding Bond Prices
Corporate bond prices are quoted either by a percentage of the bond’s face value or in
increments of points and eight fractions of a point, with a par of $1,000 as the base. The
value of each point is $10 and of each fraction $1.25. For example, a bond quoted at
86 12 would be selling for $865, and one quoted at $100 34 would be selling for $1,007.50.
Treasury bonds are measured in thirty-seconds, rather than hundredths, of a point. Each
1
32 equals 31.25 cents, and we normally drop the fractional part of the cent when stating a
2
, the price translates to
price. For example, if a bond is quoted at 100.2, or 100 + 32
$1,000.62.
Are Bonds Safe?
Just because bonds have a reputation as a conservative investment does not mean that
they are always safe. There are sources of risk in holding or trading bonds:
• To begin with, not all loans are paid back. Companies, cities, and counties occasionally do go bankrupt. Only U.S. Treasury bonds are considered rock solid.
• Another source of risk for certain bonds is that your loan may be called, or paid back
early. While that is certainly better than its not being paid back at all, it forces you to
find another, possibly less lucrative, place to put your money.
• The main danger for buy-and-hold investors, however, is a rising inflation rate.
Since the dollar amount they earn on a bond investment does not change, the value
of that money can be eroded by inflation. Worse yet, with your money locked
away in the bond, you will not be able to take advantage of the higher interest
rates that are usually available in an inflationary economy. Now you know why
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182 CHAPTER 4 Understanding Money and Its Management
bond investors cringe at cheerful headlines about full employment and strong economic growth: These traditional signs of inflation hint that bondholders may soon
lose their shirts.
Yield to
maturity (YTM):
Yield that would
be realized on a
bond or other
fixed income
security if the
bond was held
until the maturity
date.
Current yield:
Annual income
(interest) divided
by the current
market price of
the bond.
How Do Prices and Yields Work?
You can trade bonds on the market just as you do stocks. Once a bond is purchased, you
may keep it until it matures or for a variable number of interest periods before selling it. You
can purchase or sell bonds at prices other than face value, depending on the economic environment. Furthermore, bond prices change over time because of the risk of nonpayment of
interest or par value, supply and demand, and the economic outlook. These factors affect
the yield to maturity (or return on investment) of the bond:
• The yield to maturity represents the actual interest earned from a bond over the holding
period. In other words, the yield to maturity on a bond is the interest rate that establishes the equivalence between all future interest and face-value receipts and the market
price of the bond.
• The current yield of a bond is the annual interest earned, as a percentage of the current
market price. The current yield provides an indication of the annual return realized
from investment in the bond. To illustrate, we will explain these values with numerical
examples shortly.
Bond quotes follow a few unique conventions. As an example, let’s take a look at a
corporate bond with a face value of $1,000 (issued by Ford Motor Company) and traded
on November 18, 2005:
Price:
94.50
Coupon (%):
6.625
Maturity Date:
16-Jun-2008
Yield to Maturity (%):
9.079
Current Yield (%):
7.011
Debt Rating:
BBB
Coupon Payment Frequency:
Semiannual
First Coupon Date:
16-Dec-2005
Type:
Corporate
Industry:
Industrial
• Prices are given as percentages of face value, with the last digits not decimals, but in
eighths. The bond on the list, for instance, has just fallen to sell for 94.50, or 94.5%
of its $1,000 face value. In other words, an investor who bought the bond when it was
issued (at 100) could now sell it for a 5.5% discount.
• The discount over face value is explained by examining the bond’s coupon rate,
6.625%, and its current yield, 7.011%. The current yield will be higher than the
coupon rate whenever the bond is selling for less than its par value.
• The coupon rate is 6.625% paid semiannually, meaning that, for every six-month period, you will receive $66.25>2 = $33.13.
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Section 4.6 Investing in Financial Assets 183
• The bond rating system helps investors determine a company’s credit risk. Think of a
bond rating as the report card on a company’s credit rating. Blue-chip firms, which are
safer investments, have a high rating, while risky companies have a low rating. The
following chart illustrates the different bond rating scales from Moody’s, Standard
and Poor’s (S&P), and Fitch Ratings—the major rating agencies in the United States.
Bond Rating
Moody’s
S&P/Fitch
Grade
Risk
Aaa
AAA
Investment
Highest quality
Aa
AA
Investment
High quality
A
A
Investment
Strong
Baa
BBB
Investment
Medium grade
Ba, B
BB, B
Junk
Speculative
Caa/Ca/C
CCC/CC/C
Junk
Highly
speculative
C
D
Junk
In default
Notice that if the company falls below a certain credit rating, its grade changes from
investment quality to junk status. Junk bonds are aptly named: They are the debt of
companies that are in some sort of financial difficulty. Because they are so risky, they
have to offer much higher yields than any other debt. This brings up an important
point: Not all bonds are inherently safer than stocks. Certain types of bonds can be
just as risky, if not riskier, than stocks.
• If you buy a bond at face value, its rate of return, or yield, is just the coupon rate.
However, a glance at a table of bond quotes (like the preceding one) will tell you
that after they are first issued, bonds rarely sell for exactly face value. So how
much is the yield then? In our example, if you can purchase the bond for $945, you
are getting two bonuses. First, you have effectively bought a bond with a 6.625%
coupon, since the $66.25 coupon is 7.011% of your $945 purchase price. (Recall
that the coupon rate, adjusted for the current price, is the current yield of the bond.)
However, there’s more: Although you paid $945, in 2008 you will receive the full
$1,000 face value.
Example 4.19 illustrates how you calculate the yield to maturity and the current
yield, considering both the purchase price and the capital gain for a new issue.
EXAMPLE 4.19 Yield to Maturity and Current Yield
Consider buying a $1,000 corporate (Delta Corporation) bond at the market price of
$996.25. The interest will be paid semiannually, the interest rate per payment period
will be simply 4.8125%, and 20 interest payments over 10 years are required. We
show the resulting cash flow to the investor in Figure 4.20. Find (a) the yield to maturity and (b) the current yield.
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184 CHAPTER 4 Understanding Money and Its Management
$1,000
Face value
20 bond interest payments
$48.13
0
5
10
Semi-annual periods
15
20
$996.25
Figure 4.20 A typical cash flow transaction associated with an investment in Delta’s corporate bond.
DISCUSSION
• Debenture bond: Delta wants to issue bonds totaling $100 million, and the company will back the bonds with specific pieces of property, such
as buildings.
• Par value:
Delta’s bond has a par value of $1,000.
• Maturity date: Delta’s bonds, which were issued on January 30, 2006, will
mature on January 31, 2016; thus, they have a 10-year maturity
at the time of their issue.
• Coupon rate:
Delta’s coupon rate is 9.625%, and interest is payable semiannually. For example, Delta’s bonds have a $1,000 par value, and
they pay $96.25 in simple interest A 9 58% B each year ($48.13
every six months).
• Discount bond: At 99.625%, or a 0.375% discount, Delta’s bonds are offered at
less than their par value.
SOLUTION
Given: Initial purchase price = $996.25, coupon rate = 9.625% per year paid semiannually, and 10-year maturity with a par value of $1,000.
Find: (a) Yield to maturity and (b) current yield.
(a) Yield to maturity. We find the yield to maturity by determining the interest rate
that makes the present worth of the receipts equal to the market price of the
bond:
$996.25 = $48.131P>A, i, 202 + $1,0001P>F, i, 202.
The value of i that makes the present worth of the receipts equal to $996.25 lies
between 4.5% and 5%. We could use a linear interpolation to find the yield to
maturity, but it is much easier to obtain with Excel’s Goal Seek function. As
shown in Figure 4.21, solving for i yields i = 4.8422%.
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Section 4.6 Investing in Financial Assets 185
Figure 4.21 Finding the yield to maturity of the Delta bond with
Excel’s Goal Seek function.
Note that this 4.8422% is the yield to maturity per semiannual period. The
nominal (annual) yield is 214.84222 = 9.6844%, compounded semiannually.
Compared with the coupon rate of A 9 58% B (or 9.625%), purchasing the bond with
the price discounted at 0.375% brings about an additional 0.0594% yield. The
effective annual interest rate is then
ia = 11 + 0.04842222 - 1 = 9.92%.
The 9.92% represents the effective annual yield to maturity on the bond.
Notice that when you purchase a bond at par value and sell at par value, the yield
to maturity will be the same as the coupon rate of the bond.
Until now, we have observed differences in nominal and effective interest
rates because of the frequency of compounding. In the case of bonds, the reason
is different: The stated (par) value of the bond and the actual price for which it
is sold are not the same. We normally state the nominal interest as a percentage
of par value. However, when the bond is sold at a discount, the same nominal
interest on a smaller initial investment is earned; hence, your effective interest
earnings are greater than the stated nominal rate.
(b) Current yield. For our example of Delta, we compute the current yield as follows:
$48.13
= 4.83% per semiannual period,
996.25
4.83% * 2 = 9.66% per year 1nominal current yield2,
ia = 11 + 0.048322 - 1 = 9.90%.
This effective current yield is 0.02% lower than the 9.92% yield to maturity we just
computed. If the bond is selling at a discount, the current yield is smaller than the
yield to maturity. If the bond is selling at a premium, the current yield is larger than
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186 CHAPTER 4 Understanding Money and Its Management
the yield to maturity. A significant difference between the yield to maturity and the
current yield of a bond can exist because the market price of a bond may be more
or less than its face value. Moreover, both the current yield and the yield to maturity may differ considerably from the stated coupon value of the bond.
EXAMPLE 4.20 Bond Value over Time
Consider again the Delta bond investment introduced in Example 4.19. If the yield to
maturity remains constant at 9.68%, (a) what will be the value of the bond one year
after it was purchased? (b) If the market interest rate drops to 9% a year later, what
would be the market price of the bond?
SOLUTION
Given: The same data as in Example 4.19.
Find: (a) The value of the bond one year later and (b) the market price of the bond a
year later at the going rate of 9% interest.
(a) We can find the value of the bond one year later by using the same valuation procedure as in Example 4.19, but now the term to maturity is only nine years:
$48.131P>A, 4.84%, 182 + $1,0001P>F, 4.84%, 182 = $996.80.
The value of the bond will remain at $996.80 as long as the yield to maturity remains constant at 9.68% over nine years.
(b) Now suppose interest rates in the economy have fallen since the Delta bonds
were issued, and consequently, the going rate of interest is 9%. Then both the
coupon interest payments and the maturity value remain constant, but now 9%
values have to be used to calculate the value of the bond. The value of the bond
at the end of the first year would be
$48.131P>A, 4.5%, 182 + $1,0001P>F, 4.5%, 182 = $1,038.06.
Thus, the bond would sell at a premium over its par value.
COMMENTS: The arithmetic of the bond price increase should be clear, but what is the
logic behind it? We can explain the reason for the increase as follows: Because the
going market interest rate for the bond has fallen to 9%, if we had $1,000 to invest,
we could buy new bonds with a coupon rate of 9%. These would pay $90 interest
each year, rather than $96.80. We would prefer $96.80 to $90; therefore, we would be
willing to pay more than $1,000 for Delta’s bonds to obtain higher coupons. All investors would recognize these facts, and hence the Delta’s bonds would be bid up in
price to $1,038.06. At that point, they would provide the same yield to maturity (rate
of return) to a potential investor as would the new bonds, namely, 9%.
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Summary 187
SUMMARY
Interest is most frequently quoted by financial institutions as an annual percentage
rate, or APR. However, compounding frequently occurs more often than once annually,
and the APR does not account for the effect of this more frequent compounding. The
situation leads to the distinction between nominal and effective interest:
Nominal interest is a stated rate of interest for a given period (usually a year).
Effective interest is the actual rate of interest, which accounts for the interest amount
accumulated over a given period. The effective rate is related to the APR by the equation
i = a1 +
r M
b - 1,
M
where r is the APR, M is the number of compounding periods, and i is the effective interest rate.
In any equivalence problem, the interest rate to use is the effective interest rate per
payment period, or
i = c1 +
r C
d - 1,
CK
where C is the number of interest periods per payment period, K is the number of payment periods per year, and r/K is the nominal interest rate per payment period. Figure
4.9 outlines the possible relationships between compounding and payment periods and
indicates which version of the effective-interest formula to use.
The equation for determining the effective interest of continuous compounding is
i = er>K - 1.
The difference in accumulated interest between continuous compounding and very
frequent compounding 1M 7 502 is minimal.
Cash flows, as well as compounding, can be continuous. Table 4.2 shows the interest
factors to use for continuous cash flows with continuous compounding.
Nominal (and hence effective) interest rates may fluctuate over the life of a cash flow
series. Some forms of home mortgages and bond yields are typical examples.
Amortized loans are paid off in equal installments over time, and most of these loans
have interest that is compounded monthly.
Under a typical add-on loan, the lender precalculates the total simple interest amount
and adds it to the principal. The principal and this precalculated interest amount are
then paid together in equal installments.
The term mortgage refers to a special type of loan for buying a piece of property, such
as a house or a commercial building. The cost of the mortgage will depend on many
factors, including the amount and term of the loan and the frequency of payments, as
well as points and fees.
Two types of mortgages are common: fixed-rate mortgages and variable-rate mort-
gages. Fixed-rate mortgages offer loans whose interest rates are fixed over the period
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188 CHAPTER 4 Understanding Money and Its Management
of the contract, whereas variable-rate mortgages offer interest rates that fluctuate with
market conditions. In general, the initial interest rate is lower for variable-rate mortgages, as the lenders have the flexibility to adjust the cost of the loans over the period
of the contract.
Allocating one’s assets is simply a matter of answering the following question: “Given
my personal tolerance for risk and my investment objectives, what percentage of my
assets should be allocated for growth, what percentage for income, and what percentage for liquidity?”
You can determine the expected rate of return on a portfolio by computing the weighted
average of the returns on each investment.
You can determine the expected risk of a portfolio by computing the weighted average
of the volatility of each investment.
All other things being equal, if the expected returns are approximately the same, choose
the portfolio with the lowest expected risk.
All other things being equal, if the expected risk is about the same, choose the portfolio
with the highest expected return.
Asset-backed bonds: If a company backs its bonds with specific pieces of property,
such as buildings, we call these types of bonds mortgage bonds, which indicate the
terms of repayment and the particular assets pledged to the bondholders in case of
default. It is much more common, however, for a corporation simply to pledge its overall
assets. A debenture bond represents such a promise.
Par value: Individual bonds are normally issued in even denominations of $1,000 or
multiples of $1,000. The stated face value of an individual bond is termed the par value.
Maturity date: Bonds generally have a specified maturity date on which the par
value is to be repaid.
Coupon rate: We call the interest paid on the par value of a bond the annual coupon
rate. The time interval between interest payments could be of any duration, but a
semiannual period is the most common.
Discount or premium bond: A bond that sells below its par value is called a discount
bond. When a bond sells above its par value, it is called a premium bond.
PROBLEMS
Nominal and Effective Interest Rates
4.1 If your credit card calculates interest based on 12.5% APR,
(a) What are your monthly interest rate and annual effective interest rate?
(b) If you current outstanding balance is $2,000 and you skip payments for two
months, what would be the total balance two months from now?
4.2 A department store has offered you a credit card that charges interest at 1.05% per
month, compounded monthly. What is the nominal interest (annual percentage)
rate for this credit card? What is the effective annual interest rate?
4.3 A local bank advertised the following information: Interest 6.89%—effective annual yield 7.128%. No mention was made of the interest period in the advertisement.
Can you figure out the compounding scheme used by the bank?
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Problems 189
4.4 College Financial Sources, which makes small loans to college students, offers to
lend $500. The borrower is required to pay $400 at the end of each week for 16
weeks. Find the interest rate per week. What is the nominal interest rate per year?
What is the effective interest rate per year?
4.5 A financial institution is willing to lend you $40. However, $450 is repaid at the
end of one week.
(a) What is the nominal interest rate?
(b) What is the effective annual interest rate?
4.6 The Cadillac Motor Car Company is advertising a 24-month lease of a Cadillac
Deville for $520, payable at the beginning of each month. The lease requires a
$2,500 down payment, plus a $500 refundable security deposit. As an alternative,
the company offers a 24-month lease with a single up-front payment of $12,780,
plus a $500 refundable security deposit. The security deposit will be refunded at
the end of the 24-month lease. Assuming an interest rate of 6%, compounded
monthly, which lease is the preferred one?
4.7 As a typical middle-class consumer, you are making monthly payments on your
home mortgage (9% annual interest rate), car loan (12%), home improvement
loan (14%), and past-due charge accounts (18%). Immediately after getting a
$100 monthly raise, your friendly mutual fund broker tries to sell you some investment funds with a guaranteed return of 10% per year. Assuming that your only
other investment alternative is a savings account, should you buy?
Compounding More Frequent than Annually
4.8 A loan company offers money at 1.8% per month, compounded monthly.
(a) What is the nominal interest rate?
(b) What is the effective annual interest rate?
(c) How many years will it take an investment to triple if interest is compounded
monthly?
(d) How many years will it take an investment to triple if the nominal rate is compounded continuously?
4.9 Suppose your savings account pays 9% interest compounded quarterly. If you deposit $10,000 for one year, how much would you have?
4.10 What will be the amount accumulated by each of these present investments?
(a) $5,635 in 10 years at 5% compounded semiannually.
(b) $7,500 in 15 years at 6% compounded quarterly.
(c) $38,300 in 7 years at 9% compounded monthly.
4.11 How many years will it take an investment to triple if the interest rate is 9%
compounded
(a) Quarterly?
(b) Monthly?
(c) Continuously?
4.12 A series of equal quarterly payments of $5,000 for 12 years is equivalent to what
present amount at an interest rate of 9% compounded
(a) Quarterly?
(b) Monthly?
(c) Continuously?
4.13 What is the future worth of an equal payment series of $3,000 each quarter for five
years if the interest rate is 8% compounded continuously?
4.14 Suppose that $2,000 is placed in a bank account at the end of each quarter over the
next 15 years. What is the future worth at the end of 15 years when the interest rate
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190 CHAPTER 4 Understanding Money and Its Management
is 6% compounded
(a) Quarterly?
(b) Monthly?
(c) Continuously?
4.15 A series of equal quarterly deposits of $1,000 extends over a period of three years.
It is desired to compute the future worth of this quarterly deposit series at 12%
compounded monthly. Which of the following equations is correct?
(a) F = 41$1,00021F>A, 12%, 32.
(b) F = $1,0001F>A, 3%, 122.
(c) F = $1,0001F>A, 1%, 122.
(d) F = $1,0001F>A, 3.03%, 122.
4.16 If the interest rate is 8.5% compounded continuously, what is the required quarterly payment to repay a loan of $12,000 in five years?
4.17 What is the future worth of a series of equal monthly payments of $2,500 if the
series extends over a period of eight years at 12% interest compounded
(a) Quarterly?
(b) Monthly?
(c) Continuously?
4.18 Suppose you deposit $500 at the end of each quarter for five years at an interest rate
of 8% compounded monthly. What equal end-of-year deposit over the five years
would accumulate the same amount at the end of the five years under the same interest compounding? To answer the question, which of the following is correct?
(a) A = [$5001F>A, 2%, 202] * 1A>F, 8%, 52.
(b) A = $5001F>A, 2.013%, 42.
8%
, 20 b * 1A>F, 8%, 52.
12
(d) None of the above.
A series of equal quarterly payments of $2,000 for 15 years is equivalent to what
future lump-sum amount at the end of 10 years at an interest rate of 8% compounded continuously?
What will be the required quarterly payment to repay a loan of $32,000 in five years,
if the interest rate is 7.8% compounded continuously?
A series of equal quarterly payments of $4,000 extends over a period of three
years. What is the present worth of this quarterly payment series at 8.75% interest
compounded continuously?
What is the future worth of the following series of payments?
(a) $6,000 at the end of each six-month period for 6 years at 6% compounded
semiannually.
(b) $42,000 at the end of each quarter for 12 years at 8% compounded quarterly.
(c) $75,000 at the end of each month for 8 years at 9% compounded monthly.
What equal series of payments must be paid into a sinking fund to accumulate the
following amount?
(a) $21,000 in 10 years at 6.45% compounded semiannually when payments are
semiannual.
(b) $9,000 in 15 years at 9.35% compounded quarterly when payments are quarterly.
(c) $24,000 in 5 years at 6.55% compounded monthly when payments are monthly.
(c) A = $500 aF>A,
4.19
4.20
4.21
4.22
4.23
4.24 You have a habit of drinking a cup of Starbucks coffee ($2.50 a cup) on the way to
work every morning. If, instead, you put the money in the bank for 30 years, how
much would you have at the end of that time, assuming that your account earns
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Problems 191
5% interest compounded daily? Assume also that you drink a cup of coffee every
day, including weekends.
4.25 John Jay is purchasing a $24,000 automobile, which is to be paid for in 48 monthly
installments of $543.35. What effective annual interest is he paying for this financing
arrangement?
4.26 A loan of $12,000 is to be financed to assist in buying an automobile. On the basis
of monthly compounding for 42 months, the end-of-the-month equal payment is
quoted as $445. What nominal interest rate in percentage is being charged?
4.27 Suppose a young newlywed couple is planning to buy a home two years from
now. To save the down payment required at the time of purchasing a home worth
$220,000 (let’s assume that the down payment is 10% of the sales price, or
$22,000), the couple decides to set aside some money from each of their salaries
at the end of every month. If each of them can earn 6% interest (compounded
monthly) on his or her savings, determine the equal amount this couple must deposit each month until the point is reached where the couple can buy the home.
4.28 What is the present worth of the following series of payments?
(a) $1,500 at the end of each six-month period for 12 years at 8% compounded
semiannually.
(b) $2,500 at the end of each quarter for 8 years at 8% compounded quarterly.
(c) $3,800 at the end of each month for 5 years at 9% compounded monthly.
4.29 What is the amount of the quarterly deposits A such that you will be able to withdraw the amounts shown in the cash flow diagram if the interest rate is 8% compounded quarterly?
$2,500
8% Compounded quarterly
$1,500
Quarters
0
1
2
3
4
5
6
7
8
A
(Deposit)
4.30 Georgi Rostov deposits $15,000 in a savings account that pays 6% interest compounded monthly. Three years later, he deposits $14,000. Two years after the
$14,000 deposit, he makes another deposit in the amount of $12,500. Four years
after the $12,500 deposit, half of the accumulated funds is transferred to a fund
that pays 8% interest compounded quarterly. How much money will be in each
account six years after the transfer?
4.31 A man is planning to retire in 25 years. He wishes to deposit a regular amount
every three months until he retires, so that, beginning one year following his retirement, he will receive annual payments of $60,000 for the next 10 years. How
much must he deposit if the interest rate is 6% compounded quarterly?
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4.32 You borrowed $15,000 for buying a new car from a bank at an interest rate of
12% compounded monthly. This loan will be repaid in 48 equal monthly installments over four years. Immediately after the 20th payment, you desire to pay the
remainder of the loan in a single payment. Compute this lump-sum amount of
that time.
4.33 A building is priced at $125,000. If a down payment of $25,000 is made and a payment of $1,000 every month thereafter is required, how many months will it take to
pay for the building? Interest is charged at a rate of 9% compounded monthly.
4.34 You obtained a loan of $20,000 to finance an automobile. Based on monthly compounding over 24 months, the end-of-the-month equal payment was figured to be
$922.90. What APR was used for this loan?
4.35 The Engineering Economist (a professional journal) offers three types of subscriptions, payable in advance: one year at $66, two years at $120, and three years at $160.
If money can earn 6% interest compounded monthly, which subscription should you
take? (Assume that you plan to subscribe to the journal over the next three years.)
4.36 A couple is planning to finance its three-year-old son’s college education. Money
can be deposited at 6% compounded quarterly. What quarterly deposit must be
made from the son’s 3rd birthday to his 18th birthday to provide $50,000 on each
birthday from the 18th to the 21st? (Note that the last deposit is made on the date
of the first withdrawal.)
4.37 Sam Salvetti is planning to retire in 15 years. Money can be deposited at 8% compounded quarterly. What quarterly deposit must be made at the end of each quarter until Sam retires so that he can make a withdrawal of $25,000 semiannually
over the first five years of his retirement? Assume that his first withdrawal occurs
at the end of six months after his retirement.
4.38 Michelle Hunter received $250,000 from an insurance company after her husband’s death. Michelle wants to deposit this amount in a savings account that
earns interest at a rate of 6% compounded monthly. Then she would like to make
120 equal monthly withdrawals over the 10-year period such that, when she makes
the last withdrawal, the savings account will have a balance of zero. How much
can she withdraw each month?
4.39 Anita Tahani, who owns a travel agency, bought an old house to use as her business office. She found that the ceiling was poorly insulated and that the heat
loss could be cut significantly if 6 inches of foam insulation were installed. She
estimated that with the insulation, she could cut the heating bill by $40 per
month and the air-conditioning cost by $25 per month. Assuming that the summer season is three months (June, July, and August) of the year and that the
winter season is another three months (December, January, and February) of
the year, how much can Anita spend on insulation if she expects to keep the
property for five years? Assume that neither heating nor air-conditioning would
be required during the fall and spring seasons. If she decides to install the insulation, it will be done at the beginning of May. Anita’s interest rate is 9% compounded monthly.
Continuous Payments with Continuous Compounding
4.40 A new chemical production facility that is under construction is expected to be in
full commercial operation 1 year from now. Once in full operation, the facility
will generate $63,000 cash profit daily over the plant’s service life of 12 years.
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Problems 193
Determine the equivalent present worth of the future cash flows generated by the
facility at the beginning of commercial operation, assuming
(a) 12% interest compounded daily, with the daily flows.
(b) 12% interest compounded continuously, with the daily flow series approximated by a uniform continuous cash flow function.
Also, compare the difference between (a) discrete (daily) and (b) continuous
compounding.
4.41 Income from a project is expected to decline at a constant rate from an initial value
of $500,000 at time 0 to a final value of $40,000 at the end of year 3. If interest is
compounded continuously at a nominal annual rate of 11%, determine the present
value of this continuous cash flow.
4.42 A sum of $80,000 will be received uniformly over a five-year period beginning
two years from today. What is the present value of this deferred-funds flow if interest is compounded continuously at a nominal rate of 9%?
4.43 A small chemical company that produces an epoxy resin expects its production
volume to decay exponentially according to the relationship
yt = 5e -0.25t,
where yt is the production rate at time t. Simultaneously, the unit price is expected
to increase linearly over time at the rate
ut = $5511 + 0.09t2.
What is the expression for the present worth of sales revenues from t = 0 to
t = 20 at 12% interest compounded continuously?
Changing Interest Rates
4.44 Consider the accompanying cash flow diagram, which represents three different
interest rates applicable over the five-year time span shown.
$3,000
$2,000 $2,000
$2,000 $2,000
0
1
6%
P Compounded
quarterly
2
3
4
5
Years
8%
10%
Compounded Compounded
quarterly
quarterly
(a) Calculate the equivalent amount P at the present time.
(b) Calculate the single-payment equivalent to F at n = 5.
(c) Calculate the equal-payment-series cash flow A that runs from n = 1 to n = 5.
4.45 Consider the cash flow transactions depicted in the accompanying cash flow diagram, with the changing interest rates specified.
(a) What is the equivalent present worth? (In other words, how much do you
have to deposit now so that you can withdraw $300 at the end of year 1, $300
at the end of year 2, $500 at the end of year 3, and $500 at the end of year 4?)
(b) What is the single effective annual interest rate over four years?
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194 CHAPTER 4 Understanding Money and Its Management
$300
$500
$500
3
4
$300
0
1
6%
Interest
P compounded
monthly
2
Years
9%
Interest
compounded
monthly
6%
Interest
compounded
monthly
4.46 Compute the future worth of the cash flows with the different interest rates specified. The cash flows occur at the end of each year over four years.
$500
9%
9%
Interest
Interest
compounded
compounded
Years
daily
continuously
1
2
3
0
$100
F
4
$100
$250
$250
$400
Amortized Loans
4.47 An automobile loan of $20,000 at a nominal rate of 9% compounded monthly for 48
months requires equal end-of-month payments of $497.70. Complete the following
table for the first six payments, as you would expect a bank to calculate the values:
End of Month
(n)
1
Interest
Payment
Repayment of
Principal
Remaining
Loan Balance
$19,652.30
2
3
$352.94
4
$142.12
5
$139.45
6
$17,874.28
4.48 Mr. Smith wants to buy a new car that will cost $18,000. He will make a down
payment in the amount of $8,000. He would like to borrow the remainder from a
bank at an interest rate of 9% compounded monthly. He agrees to pay off the loan
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Problems 195
monthly for a period of two years. Select the correct answer for the following
questions:
(a) What is the amount of the monthly payment A?
(i) A = $10,0001A>P, 0.75%, 242.
(ii) A = $10,0001A>P, 9%, 22>12.
(iii) A = $10,0001A>F, 0.75%, 242.
(iv) A = $12,5001A>F, 9%, 22>12.
(b) Mr. Smith has made 12 payments and wants to figure out the balance remaining immediately after 12th payment. What is that balance?
(i) B12 = 12A.
(ii) B12 = A1P>A, 9%, 12>12.
(iii) B12 = A1P>A, 0.75%, 122.
(iv) B12 = 10,000 - 12A.
4.49 Tony Wu is considering purchasing a used automobile. The price, including the
title and taxes, is $12,345. Tony is able to make a $2,345 down payment. The balance, $10,000, will be borrowed from his credit union at an interest rate of 8.48%
compounded daily. The loan should be paid in 36 equal monthly payments. Compute the monthly payment. What is the total amount of interest Tony has to pay
over the life of the loan?
4.50 Suppose you are in the market for a new car worth $18,000. You are offered a deal to
make a $1,800 down payment now and to pay the balance in equal end-of-month
payments of $421.85 over a 48-month period. Consider the following situations:
(a) Instead of going through the dealer’s financing, you want to make a down payment of $1,800 and take out an auto loan from a bank at 11.75% compounded
monthly. What would be your monthly payment to pay off the loan in four years?
(b) If you were to accept the dealer’s offer, what would be the effective rate of interest per month the dealer charges on your financing?
4.51 Bob Pearson borrowed $25,000 from a bank at an interest rate of 10% compounded
monthly. The loan will be repaid in 36 equal monthly installments over three
years. Immediately after his 20th payment, Bob desires to pay the remainder of
the loan in a single payment. Compute the total amount he must pay.
4.52 You plan to buy a $200,000 home with a 10% down payment. The bank you want
to finance the loan suggests two options: a 20-year mortgage at 9% APR and a 30year mortgage at 10% APR. What is the difference in monthly payments (for the
first 20 years) between these two options?
4.53 David Kapamagian borrowed money from a bank to finance a small fishing boat.
The bank’s terms allowed him to defer payments (including interest) on the loan
for six months and to make 36 equal end-of-month payments thereafter. The original bank note was for $4,800, with an interest rate of 12% compounded monthly.
After 16 monthly payments, David found himself in a financial bind and went to a
loan company for assistance in lowering his monthly payments. Fortunately, the
loan company offered to pay his debts in one lump sum if he would pay the company $104 per month for the next 36 months. What monthly rate of interest is the
loan company charging on this transaction?
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4.54 You are buying a home for $250,000.
(a) If you make a down payment of $50,000 and take out a mortgage on the rest of
the money at 8.5% compounded monthly, what will be your monthly payment
to retire the mortgage in 15 years?
(b) Consider the seventh payment. How much will the interest and principal payments be?
4.55 With a $350,000 home mortgage loan with a 20-year term at 9% APR compounded
monthly, compute the total payments on principal and interest over the first 5 years
of ownership.
4.56 A lender requires that monthly mortgage payments be no more than 25% of gross
monthly income with a maximum term of 30 years. If you can make only a 15%
down payment, what is the minimum monthly income needed to purchase a
$400,000 house when the interest rate is 9% compounded monthly?
4.57 To buy a $150,000 house, you take out a 9% (APR) mortgage for $120,000. Five
years later, you sell the house for $185,000 (after all other selling expenses). What
equity (the amount that you can keep before tax) would you realize with a 30-year
repayment term?
4.58 Just before their 15th payment,
• Family A had a balance of $80,000 on a 9%, 30-year mortgage;
• Family B had a balance of $80,000 on a 9%, 15-year mortgage; and
• Family C had a balance of $80,000 on a 9%, 20-year mortgage.
How much interest did each family pay on the 15th payment?
4.59 Home mortgage lenders usually charge points on a loan to avoid exceeding a legal
limit on interest rates or to be competitive with other lenders. As an example, for a
two-point loan, the lender would lend only $98 for each $100 borrowed. The borrower would receive only $98, but would have to make payments just as if he or
she had received $100. Suppose that you receive a loan of $130,000, payable at
the end of each month for 30 years with an interest rate of 9% compounded
monthly, but you have been charged three points. What is the effective interest rate
on this home mortgage loan?
4.60 A restaurant is considering purchasing a lot adjacent to its business to provide adequate parking space for its customers. The restaurant needs to borrow $35,000 to
secure the lot. A deal has been made between a local bank and the restaurant so
that the restaurant would pay the loan back over a five-year period with the following payment terms: 15%, 20%, 25%, 30%, and 35% of the initial loan at the
end of first, second, third, fourth, and fifth years, respectively.
(a) What rate of interest is the bank earning from this loan?
(b) What would be the total interest paid by the restaurant over the five-year period?
4.61 Don Harrison’s current salary is $60,000 per year, and he is planning to retire 25
years from now. He anticipates that his annual salary will increase by $3,000 each
year (to $60,000 the first year, $63,000 the second year, $66,000 the third year,
and so forth), and he plans to deposit 5% of his yearly salary into a retirement
fund that earns 7% interest compounded daily. What will be the amount accumulated at the time of Don’s retirement?
4.62 Consider the following two options for financing a car:
• Option A. Purchase the vehicle at the normal price of $26,200 and pay for the
vehicle over three years with equal monthly payments at 1.9% APR financing.
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Problems 197
• Option B. Purchase the vehicle for a discount price of $24,048, to be paid immediately. The funds that would be used to purchase the vehicle are presently
earning 5% annual interest compounded monthly.
(a) What is the meaning of the APR of 1.9% quoted by the dealer?
(b) Under what circumstances would you prefer to go with the dealer’s financing?
(c) Which interest rate (the dealer’s interest rate or the savings rate) would you
use in comparing the two options?
Add-On Loans
4.63 Katerina Unger wants to purchase a set of furniture worth $3,000. She plans to finance the furniture for two years. The furniture store tells Katerina that the interest
rate is only 1% per month, and her monthly payment is computed as follows:
• Installment period = 24 months.
• Interest = 2410.0121$3,0002 = $720.
• Loan processing fee = $25.
• Total amount owed = $3,000 + $720 + $25 = $3,745.
• Monthly payment = $3,745>24 = $156.04 per month.
(a) What is the annual effective interest rate that Katerina is paying for her loan
transaction? What is the nominal interest (annual percentage rate) for the loan?
(b) Katerina bought the furniture and made 12 monthly payments. Now she wants
to pay off the remaining installments in one lump sum (at the end of 12
months). How much does she owe the furniture store?
4.64 You purchase a piece of furniture worth $5,000 on credit through a local furniture store. You are told that your monthly payment will be $146.35, including an
acquisition fee of $25, at a 10% add-on interest rate over 48 months. After making
15 payments, you decide to pay off the balance. Compute the remaining balance,
based on the conventional amortized loan.
Loans with Variable Payments
4.65 Kathy Stonewall bought a new car for $15,458. A dealer’s financing was available
through a local bank at an interest rate of 11.5% compounded monthly. Dealer financing required a 10% down payment and 60 equal monthly payments. Because
the interest rate was rather high, Kathy checked her credit union for possible financing. The loan officer at the credit union quoted a 9.8% interest rate for a newcar loan and 10.5% for a used car. But to be eligible for the loan, Kathy has to be
a member of the union for at least six months. Since she joined the union two
months ago, she has to wait four more months to apply for the loan. Consequently, she decided to go ahead with the dealer’s financing, and four months later she
refinanced the balance through the credit union at an interest rate of 10.5%.
(a) Compute the monthly payment to the dealer.
(b) Compute the monthly payment to the union.
(c) What is the total interest payment on each loan?
4.66 A house can be purchased for $155,000, and you have $25,000 cash for a down
payment. You are considering the following two financing options:
• Option 1. Getting a new standard mortgage with a 7.5% (APR) interest and a
30-year term.
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• Option 2. Assuming the seller’s old mortgage, which has an interest rate of
5.5% (APR), a remaining term of 25 years (the original term was 30 years), a remaining balance of $97,218, and payments of $597 per month. You can obtain a
second mortgage for the remaining balance ($32,782) from your credit union at
9% (APR) with a 10-year repayment period.
(a) What is the effective interest rate of the combined mortgage?
(b) Compute the monthly payments for each option over the life of the mortgage.
(c) Compute the total interest payment for each option.
(d) What homeowner’s interest rate makes the two financing options equivalent?
Loans with Variable Interest Rates
4.67 A loan of $10,000 is to be financed over a period of 24 months. The agency quotes
a nominal rate of 8% for the first 12 months and a nominal rate of 9% for any remaining unpaid balance after 12 months, compounded monthly. Based on these
rates, what equal end-of-the-month payment for 24 months would be required to
repay the loan with interest?
4.68 Emily Wang financed her office furniture from a furniture dealer. The dealer’s
terms allowed her to defer payments (including interest) for six months and to
make 36 equal end-of-month payments thereafter. The original note was for
$15,000, with interest at 9% compounded monthly. After 26 monthly payments,
Emily found herself in a financial bind and went to a loan company for assistance.
The loan company offered to pay her debts in one lump sum if she would pay the
company $186 per month for the next 30 months.
(a) Determine the original monthly payment made to the furniture store.
(b) Determine the lump-sum payoff amount the loan company will make.
(c) What monthly rate of interest is the loan company charging on this loan?
4.69 If you borrow $120,000 with a 30-year term at a 9% (APR) variable rate and the
interest rate can be changed every five years,
(a) What is the initial monthly payment?
(b) If the lender’s interest rate is 9.75% (APR) at the end of five years, what will
the new monthly payments be?
Investment in Bonds
4.70 The Jimmy Corporation issued a new series of bonds on January 1, 1996. The
bonds were sold at par ($1,000), have a 12% coupon rate, and mature in 30 years,
on December 31, 2025. Coupon interest payments are made semiannually (on
June 30 and December 31).
(a) What was the yield to maturity (YTM) of the bond on January 1, 1996?
(b) Assuming that the level of interest rates had fallen to 9%, what was the price
of the bond on January 1, 2001, five years later?
(c) On July 1, 2001, the bonds sold for $922.38. What was the YTM at that date?
What was the current yield at that date?
4.71 A $1,000, 9.50% semiannual bond is purchased for $1,010. If the bond is sold at
the end of three years and six interest payments, what should the selling price be
to yield a 10% return on the investment?
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4.72 Mr. Gonzalez wishes to sell a bond that has a face value of $1,000. The bond bears
an interest rate of 8%, with bond interests payable semiannually. Four years ago,
$920 was paid for the bond. At least a 9% return (yield) on the investment is desired. What must be the minimum selling price?
4.73 Suppose you have the choice of investing in (1) a zero-coupon bond, which costs
$513.60 today, pays nothing during its life, and then pays $1,000 after five years,
or (2) a bond that costs $1,000 today, pays $113 in interest semiannually, and matures at the end of five years. Which bond would provide the higher yield?
4.74 Suppose you were offered a 12-year, 15% coupon, $1,000 par value bond at a
price of $1,298.68. What rate of interest (yield to maturity) would you earn if you
bought the bond and held it to maturity (at semiannual interest)?
4.75 The Diversified Products Company has two bond issues outstanding. Both bonds
pay $100 semiannual interest, plus $1,000 at maturity. Bond A has a remaining
maturity of 15 years, bond B a maturity of 1 year. What is the value of each of
these bonds now, when the going rate of interest is 9%?
4.76 The AirJet Service Company’s bonds have four years remaining to maturity. Interest is paid annually, the bonds have a $1,000 par value, and the coupon interest
rate is 8.75%.
(a) What is the yield to maturity at a current market price of $1,108?
(b) Would you pay $935 for one of these bonds if you thought that the market rate
of interest was 9.5%?
4.77 Suppose Ford sold an issue of bonds with a 15-year maturity, a $1,000 par value,
a 12% coupon rate, and semiannual interest payments.
(a) Two years after the bonds were issued, the going rate of interest on bonds such
as these fell to 9%. At what price would the bonds sell?
(b) Suppose that, two years after the bonds’ issue, the going interest rate had risen
to 13%. At what price would the bonds sell?
(c) Today, the closing price of the bond is $783.58. What is the current yield?
4.78 Suppose you purchased a corporate bond with a 10-year maturity, a $1,000 par
value, a 10% coupon rate, and semiannual interest payments. All this means
that you receive a $50 interest payment at the end of each six-month period for
10 years (20 times). Then, when the bond matures, you will receive the principal
amount (the face value) in a lump sum. Three years after the bonds were purchased, the going rate of interest on new bonds fell to 6% (or 6% compounded
semiannually). What is the current market value (P) of the bond (three years
after its purchase)?
Short Case Studies
ST4.1 Jim Norton, an engineering junior, was mailed two guaranteed line-of-credit applications from two different banks. Each bank offered a different annual fee and
finance charge.
Jim expects his average monthly balance after payment to the bank to be
$300 and plans to keep the credit card he chooses for only 24 months. (After
graduation, he will apply for a new card.) Jim’s interest rate on his savings account
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200 CHAPTER 4 Understanding Money and Its Management
is 6% compounded daily. The following table lists the terms of each bank:
Terms
Annual fee
Bank A
$20
Bank B
$30
Finance charge
1.55%
16.5%
monthly
interest
rate
annual
percentage
rate
(a) Compute the effective annual interest rate for each card.
(b) Which bank’s credit card should Jim choose?
(c) Suppose Jim decided to go with Bank B and used the card for one year. The
balance after one year is $1,500. If he makes just a minimum payment each
month (say, 5% of the unpaid balance), how long will it take to pay off the
card debt? Assume that he will not make any new purchases on the card until
he pays off the debt.
ST4.2 The following is an actual promotional pamphlet prepared by Trust Company
Bank in Atlanta, Georgia:
“Lower your monthly car payments as much as 48%.” Now you can buy the car you
want and keep the monthly payments as much as 48% lower than they would be if
you financed with a conventional auto loan. Trust Company’s Alternative Auto Loan
(AAL)SM makes the difference. It combines the lower monthly payment advantages
of leasing with tax and ownership of a conventional loan. And if you have your
monthly payment deducted automatically from your Trust Company checking account, you will save 12 % on your loan interest rate. Your monthly payments can be
spread over 24, 36 or 48 months.
Amount
Financed
$10,000
$20,000
Financing
Period
(months)
Monthly
Alternative
Auto Loan
Payment
Conventional
Auto Loan
24
$249
$477
36
211
339
48
191
270
24
498
955
36
422
678
48
382
541
The amount of the final payment will be based on the residual value of the car at the
end of the loan. Your monthly payments are kept low because you make principal payments on only a portion of the loan and not on the residual value of the car. Interest is
computed on the full amount of the loan. At the end of the loan period you may:
1. Make the final payment and keep the car.
2. Sell the car yourself, repay the note (remaining balance), and keep any profit
you make.
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Short Case Studies 201
3. Refinance the car.
4. Return the car to Trust Company in good working condition and pay only a return fee.
So, if you’ve been wanting a special car, but not the high monthly payments that
could go with it, consider the Alternative Auto Loan. For details, ask at any Trust
Company branch.
Note 1: The chart above is based on the following assumptions. Conventional auto
loan 13.4% annual percentage rate. Alternative Auto Loan 13.4% annual percentage rate.
Note 2: The residual value is assumed to be 50% of sticker price for 24 months;
45% for 36 months. The amount financed is 80% of sticker price.
Note 3: Monthly payments are based on principal payments equal to the depreciation amount on the car and interest in the amount of the loan.
Note 4: The residual value of the automobile is determined by a published residual
value guide in effect at the time your Trust Company’s Alternative Auto Loan is
originated.
Note 5: The minimum loan amount is $10,000 (Trust Company will lend up
to 80% of the sticker price). Annual household income requirement is $50,000.
Note 6: Trust Company reserves the right of final approval based on customer’s
credit history. Offer may vary at other Trust Company banks in Georgia.
(a) Show how the monthly payments were computed for the Alternative Auto
Loan by the bank.
(b) Suppose that you have decided to finance a new car for 36 months from Trust
Company. Suppose also that you are interested in owning the car (not leasing
it). If you decided to go with the Alternative Auto Loan, you would make the
final payment and keep the car at the end of 36 months. Assume that your
opportunity cost rate (personal interest rate) is an interest rate of 8% compounded monthly. (You may view this opportunity cost rate as an interest rate
at which you can invest your money in some financial instrument, such as
a savings account.) Compare Trust Company’s alternative option with the
conventional option and make a choice between them.
ST4.3 In 1988, the Michigan legislature enacted the nation’s first state-run program, the
Pay-Now, Learn-Later Plan, to guarantee college tuition for students whose families
invested in a special tax-free trust fund. The minimum deposit now is $1,689 for each
year of tuition that sponsors of a newborn want to prepay. The yearly amount to buy
into the plan increases with the age of the child: Parents of infants pay the least, and
parents of high school seniors pay the most—$8,800 this year. This is because high
school seniors will go to college sooner. Michigan State Treasurer Robert A. Bowman contends that the educational trust is a better deal than putting money into a certificate of deposit (CD) or a tuition prepayment plan at a bank, because the state
promises to stand behind the investment. “Regardless of how high tuition goes, you
know it’s paid for,” he said. “The disadvantage of a CD or a savings account is you
have to hope and cross your fingers that tuition won’t outpace the amount you save.”
At the newborns’ rate, $6,756 will prepay four years of college, which is 25% less
than the statewide average public-college cost of $9,000 for four years in 1988. In
2006, when a child born in 1988 will be old enough for college, four years of college
could cost $94,360 at a private institution and $36,560 at a state school if costs continue to rise the expected average of at least 7% a year. The Internal Revenue Service
issued its opinion, ruling that the person who sets aside the money would not be taxed
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202 CHAPTER 4 Understanding Money and Its Management
on the amount paid into the fund. The agency said that the student would be subject
to federal tax on the difference between the amount paid in and the amount paid out.
Assuming that you are interested in the program for a newborn, would you join it?
ST4.4 Suppose you are going to buy a home worth $110,000 and you make a down payment in the amount of $50,000. The balance will be borrowed from the Capital
Savings and Loan Bank. The loan officer offers the following two financing plans
for the property:
• Option 1. A conventional fixed loan at an interest rate of 13% over 30 years,
with 360 equal monthly payments.
• Option 2. A graduated payment schedule (FHA 235 plan) at 11.5% interest,
with the following monthly payment schedule:
Year
(n)
Monthly
Payment
Monthly
Mortgage
Insurance
1
$497.76
$25.19
2
522.65
25.56
3
548.78
25.84
4
576.22
26.01
5
605.03
26.06
6–30
635.28
25.96
For the FHA 235 plan, mortgage insurance is a must.
(a) Compute the monthly payment under Option 1.
(b) What is the effective annual interest rate you are paying under Option 2?
(c) Compute the outstanding balance at the end of five years under each option.
(d) Compute the total interest payment under each option.
(e) Assuming that your only investment alternative is a savings account that earns
an interest rate of 6% compounded monthly, which option is a better deal?
ST4.5 Consider the following advertisement seeking to sell a beachfront condominium
at SunDestin, Florida:
95% Financing 8 18% interest!!
5% Down Payment. Own a Gulf-Front Condominium for only $100,000 with a
30-year variable-rate mortgage. We’re providing incredible terms: $95,000 mortgage (30 years), year 1 at 8.125%, year 2 at 10.125%, year 3 at 12.125%, and
years 4 through 30 at 13.125%.
(a) Compute the monthly payments for each year.
(b) Calculate the total interest paid over the life of the loan.
(c) Determine the equivalent single-effect annual interest rate for the loan.
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P A R T
2
Evaluation
of Business
and Engineering
Assets
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FIVE
Present-Worth Analysis
Parking Meters Get Smarter1 —Wireless Technology
Turns Old-Fashioned Coin-Operated Device into a
Sophisticated Tool for Catching Scofflaws and Raising
Cash Technology is taking much of the fun out of finding a place to
park the car:
• In Pacific Grove, California, parking meters “know” when a car pulls
out of the spot and quickly reset to zero—eliminating drivers’ little
joy of parking for free on someone else’s quarters.
• In Montreal, when cars stay past their time limit, meters send realtime alerts to an enforcement officer’s handheld device, reducing
the number of people needed to monitor parking spaces—not to
mention drivers’ chances of getting away with violations.
• In Aspen, Colorado, wireless “in-car” meters may eliminate the
need for curbside parking meters altogether: They dangle from the
rearview mirror inside the car, ticking off a prepaid time.
Now, in cities from New York to Seattle, the door is open to a host
of wireless technologies seeking to improve the parking meter even
further. Chicago and Sacramento, California, among others, are
equipping enforcement vehicles with infrared cameras capable of
scanning license plates even at 30 miles an hour. Using a global
positioning system, the cameras can tell which individual cars have
parked too long in a two-hour parking zone. At a cost of $75,000 a
camera, the system is an expensive upgrade of the old method of
chalking tires and then coming back two hours later to see if the car
has moved.
The camera system, supplied by Canada’s Autovu Technologies, also
helps identify scofflaws and stolen vehicles, by linking to a database of
unpaid tickets and auto thefts. Sacramento bought three cameras in
August, and since then its practice of “booting,” or immobilizing, cars
1
204
Christopher Conkey, staff reporter of The Wall Street Journal, June 30, 2005, p. B1.
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with a lot of unpaid tickets has increased sharply. Revenue is soaring, too.
According to Howard Chan, Sacramento’s parking director, Sacramento
booted 189 cars and took in parking revenue of $169,000 for the fiscal
year ended in June 2004; for fiscal 2005, the city expects to boot 805 cars
and take in more than $475,000.
In downtown Montreal, more than 400 “pay-by-space” meters, each
covering 10 to 15 spaces, are a twist on regular multispace meters. Motorists
park, then go to the meter to type in the parking-space number, and pay by
card or coin. These meters, which cost about $9,000 each, identify violators in
real time for enforcement officers carrying handheld devices: A likeness of the
block emerges on the screen, and cars parked illegally show up in red.
Montreal, QC
Multispace meters,
handheld alerts
Each meter governs 10 to 15
spaces. After parking,
drivers type in the space
number and pay with a credit
card or cash. Meters send
real-time, block-by-block
information to enforcement
officers’ handheld devices.
Coral Gables, FL
Pay with cellphone
Pacific Grove, CA
Smart meters
Drivers register their
cellphone, credit card and
license plate numbers
online. After they park, they
dial a number and enter a
lot and space number to
begin their parking session.
Sensors embedded in the
concrete under a parking
space can tell when a car pulls
out, resetting the meter to
zero.
Fort Lauderdale, FL
In-car meters
Drivers can load up to $100
onto a prepaid meter that
dangles from the rearview
mirror, above; the meter
counts down remaining
parking minutes.
Handheld
device
Cars parked
legally are
displayed as
green squares,
while those that
have exceeded
their time limit
turn red.
Sacramento, CA
Infrared license plate scanners
Enforcement vehicles traveling as fast as 30 mph use cameras
to scan license plates. Using a global positioning system lets
officers check whether a car has outlasted its time on the
meter. The system also can match license plates against
databases of unpaid parking tickets and stolen vehicles.
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Parking czars in municipalities across the country are starting to realize parking meters’
original goals: generating revenue and creating a continuous turnover of parking spaces
on city streets. Clearly, their main question is “Would there be enough new revenues from
installing the expensive parking monitoring devices?” or “How many devices could be
installed to maximize the revenue streams?” From the device manufacturer’s point of
view, the question is “Would there be enough demand for their products to justify the investment required in new facilities and marketing?” If the manufacturer decides to go
ahead and market the products, but the actual demand is far less than its forecast or the
adoption of the technology is too slow, what would be the potential financial risk?
In Chapters 3 and 4, we presented the concept of the time value of money and developed techniques for establishing cash flow equivalence with compound-interest factors.
That background provides a foundation for accepting or rejecting a capital investment: the
economic evaluation of a project’s desirability. The forthcoming coverage of investment
worth in this chapter will allow us to go a step beyond merely accepting or rejecting an investment to making comparisons of alternative investments. We will learn how to compare
alternatives on an equal basis and select the wisest alternative from an economic standpoint.
The three common measures based on cash flow equivalence are (1) equivalent present
worth (PW), (2) equivalent future worth (FW), and (3) equivalent annual worth (AE).
Present worth represents a measure of future cash flow relative to the time point “now,”
with provisions that account for earning opportunities. Future worth is a measure of cash
flow at some future planning horizon and offers a consideration of the earning opportunities
of intermediate cash flows. Annual worth is a measure of cash flow in terms of equivalent
equal payments made on an annual basis.
Our treatment of measures of investment worth is divided into three chapters. Chapter 5
begins with a consideration of the payback period, a project screening tool that was the first
formal method used to evaluate investment projects. Then it introduces two measures based
on fundamental cash flow equivalence techniques: present-worth and future-worth analysis.
Because the annual-worth approach has many useful engineering applications related to
estimating the unit cost, Chapter 6 is devoted to annual cash flow analysis. Chapter 7 presents
measures of investment worth based on yield—measures known as rate-of-return analysis.
We must also recognize that one of the most important parts of the capital budgeting
process is the estimation of relevant cash flows. For all examples in this chapter, and those in
Chapters 6 and 7, however, net cash flows can be viewed as before-tax values or after-tax values for which tax effects have been recalculated. Since some organizations (e.g., governments
and nonprofit organizations) are not subject to tax, the before-tax situation provides a valid
base for this type of economic evaluation. Taking this after-tax view will allow us to focus on
our main area of concern: the economic evaluation of investment projects. The procedures for
determining after-tax net cash flows in taxable situations are developed in Chapter 10.
CHAPTER LEARNING OBJECTIVES
After completing this chapter, you should understand the following concepts:
How firms screen potential investment opportunities.
How firms evaluate the profitability of an investment project by
considering the time value of money.
How firms compare mutually exclusive investment opportunities.
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Section 5.1 Describing Project Cash Flows 207
5.1 Describing Project Cash Flows
n Chapter 1, we described many engineering economic decision problems, but we did
not provide suggestions on how to solve them. What do all engineering economic decision problems have in common? The answer is that they all involve two dissimilar
types of amounts. First, there is the investment, which is usually made in a lump sum
at the beginning of the project. Although not literally made “today,” the investment is
made at a specific point in time that, for analytical purposes, is called today, or time 0.
Second, there is a stream of cash benefits that are expected to result from the investment
over some years in the future.
I
5.1.1 Loan versus Project Cash Flows
An investment made in a fixed asset is similar to an investment made by a bank when it
lends money. The essential characteristic of both transactions is that funds are committed
today in the expectation of their earning a return in the future. In the case of the bank loan,
the future return takes the form of interest plus repayment of the principal and is known as
the loan cash flow. In the case of the fixed asset, the future return takes the form of cash
generated by productive use of the asset. As shown in Figure 5.1, the representation of
these future earnings, along with the capital expenditures and annual expenses (such as
wages, raw materials, operating costs, maintenance costs, and income taxes), is the
project cash flow. This similarity between the loan cash flow and the project cash flow
brings us to an important conclusion: We can use the same equivalence techniques developed in Chapter 3 to measure economic worth. Example 5.1 illustrates a typical procedure
for obtaining a project’s cash flows.
Loan cash flow
Loan
Bank
Customer
Repayment
Investment
Company
Project
Return
Project cash flow
Figure 5.1
A bank loan versus an investment project.
EXAMPLE 5.1 Identifying Project Cash Flows
XL Chemicals is thinking of installing a computer process control system in one of
its process plants. The plant is used about 40% of the time, or 3,500 operating hours
per year, to produce a proprietary demulsification chemical. During the remaining
60% of the time, it is used to produce other specialty chemicals. Annual production
of the demulsification chemical amounts to 30,000 kilograms, and it sells for $15 per
kilogram. The proposed computer process control system will cost $650,000 and is
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208 CHAPTER 5 Present-Worth Analysis
expected to provide the following specific benefits in the production of the demulsification chemical:
• First, the selling price of the product could be increased by $2 per kilogram because
the product will be of higher purity, which translates into better demulsification.
• Second, production volumes will increase by 4,000 kilograms per year as a result
of higher reaction yields, without any increase in the quantities of raw material or
in production time.
• Finally, the number of process operators can be reduced by one per shift, which represents a savings of $25 per hour. The new control system would result in additional
maintenance costs of $53,000 per year and has an expected useful life of eight years.
Although the system is likely to provide similar benefits in the production of the
other specialty chemicals manufactured in the process plant, these benefits have not
yet been quantified.
SOLUTION
Given: The preceding cost and benefit information.
Find: Net cash flow in each year over the life of the new system.
Although we could assume that similar benefits are derivable from the production of
the other specialty chemicals, let’s restrict our consideration to the demulsification
chemical and allocate the full initial cost of the control system and the annual maintenance costs to this chemical. (Note that you could logically argue that only 40% of
these costs belong to this production activity.) The gross benefits are the additional
revenues realized from the increased selling price and the extra production, as well as
the cost savings resulting from having one fewer operator:
• Revenues from the price increases are 30,000 kilograms per year * $2/kilogram,
or $60,000 per year. The added production volume at the new pricing adds revenues of 4,000 kilograms per year * $17 per kilogram, or $68,000 per year.
• The elimination of one operator results in an annual savings of 3,500 operating
hours per year * $25 per hour, or $87,500 per year.
• The net benefits in each of the eight years that make up the useful lifetime of the new
system are the gross benefits less the maintenance costs: 1$60,000 + $68,000 +
$87,500 - $53,0002 = $162,500 per year.
Now we are ready to summarize a cash flow table as follows:
Year
(n)
Cash Inflows
(Benefits)
Cash Outflows
(Costs)
Net
Cash Flows
0
0
$650,000
- $650,000
1
215,500
53,000
162,500
2
215,500
53,000
162,500
o
o
8
215,500
o
53,000
o
162,500
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Section 5.1 Describing Project Cash Flows 209
COMMENTS: If the company purchases the computer process control system for
$650,000 now, it can expect an annual savings of $162,500 for eight years. (Note that
these savings occur in discrete lumps at the ends of years.) We also considered only
the benefits associated with the production of the demulsification chemical. We
could also have quantified some benefits attributable to the production of the other
chemicals from this plant. Suppose that the demulsification chemical benefits alone
justify the acquisition of the new system. Then it is obvious that, had we considered
the benefits deriving from the other chemicals as well, the acquisition of the system
would have been even more clearly justified.
We draw a cash flow diagram of this situation in Figure 5.2. Assuming that these
cost savings and cash flow estimates are correct, should management give the goahead for installation of the system? If management decides not to purchase the computer control system, what should it do with the $650,000 (assuming that it has this
amount in the first place)? The company could buy $650,000 of Treasury bonds, or it
could invest the amount in other cost-saving projects. How would the company compare cash flows that differ both in timing and amount for the alternatives it is considering? This is an extremely important question, because virtually every engineering
investment decision involves a comparison of alternatives. Indeed, these are the types
of questions this chapter is designed to help you answer.
A = $162,500
0
1
2
3
4
Years
5
6
7
8
$650,000
Figure 5.2 Cash flow diagram for the computer process control project
described in Example 5.1.
5.1.2 Independent versus Mutually Exclusive Investment Projects
Most firms have a number of unrelated investment opportunities available. For example,
in the case of XL Chemicals, other projects being considered in addition to the computer
process control project in Example 5.1 are a new waste heat recovery boiler, a CAD system for the engineering department, and a new warehouse. The economic attractiveness
of each of these projects can be measured, and a decision to accept or reject the project
can be made without reference to any of the other projects. In other words, the decision
regarding any one project has no effect on the decision to accept or reject another project.
Such projects are said to be independent.
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In Section 5.5, we will see that in many engineering situations we are faced with
selecting the most economically attractive project from a number of alternative projects, all
of which solve the same problem or meet the same need. It is unnecessary to choose more
than one project in this situation, and the acceptance of one automatically entails the
rejection of all of the others. Such projects are said to be mutually exclusive.
As long as the total cost of all the independent projects found to be economically
attractive is less than the investment funds available to a firm, all of these projects could
proceed. However, this is rarely the case. The selection of those projects which should
proceed when investment funds are limited is the subject of capital budgeting. Apart from
Chapter 15, which deals with capital budgeting, the availability of funds will not be a
consideration in accepting or rejecting projects dealt with in this book.
5.2 Initial Project Screening Method
Payback period:
The length of
time required to
recover the cost
of an investment.
Let’s suppose that you are in the market for a new punch press for your company’s machine shop, and you visit an equipment dealer. As you take a serious look at one of the
punch press models in the display room, an observant equipment salesperson approaches
you and says, “That press you are looking at is the state of the art in its category. If you
buy that top-of-the-line model, it will cost a little bit more, but it will pay for itself in less
than two years.” Before studying the four measures of investment attractiveness, we will
review a simple, but nonrigorous, method commonly used to screen capital investments.
One of the primary concerns of most businesspeople is whether and when the money invested in a project can be recovered. The payback method screens projects on the basis
of how long it takes for net receipts to equal investment outlays. This calculation can take
one of two forms: either ignore time-value-of-money considerations or include them. The
former case is usually designated the conventional payback method, the latter case the
discounted payback method.
A common standard used to determine whether to pursue a project is that the project
does not merit consideration unless its payback period is shorter than some specified period.
(This time limit is determined largely by management policy. For example, a high-tech firm,
such as a computer chip manufacturer, would set a short time limit for any new investment,
because high-tech products rapidly become obsolete.) If the payback period is within the
acceptable range, a formal project evaluation (such as a present-worth analysis) may begin.
It is important to remember that payback screening is not an end in itself, but rather a
method of screening out certain obviously unacceptable investment alternatives before
progressing to an analysis of potentially acceptable ones.
5.2.1 Payback Period: The Time It Takes to Pay Back
Determining the relative worth of new production machinery by calculating the time it
will take to pay back what it cost is the single most popular method of project screening.
If a company makes investment decisions solely on the basis of the payback period, it
considers only those projects with a payback period shorter than the maximum acceptable
payback period. (However, because of shortcomings of the payback screening method,
which we will discuss later, it is rarely used as the only decision criterion.)
What does the payback period tell us? One consequence of insisting that each proposed investment have a short payback period is that investors can assure themselves of
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Section 5.2 Initial Project Screening Method 211
being restored to their initial position within a short span of time. By restoring their initial
position, investors can take advantage of additional, perhaps better, investment possibilities
that may come along.
EXAMPLE 5.2 Conventional Payback Period for the
Computer Process Control System Project
Consider the cash flows given in Example 5.1. Determine the payback period for this
computer process control system project.
SOLUTION
Given: Initial cost = $650,000 and annual net benefits = $162,500.
Find: Conventional payback period.
Given a uniform stream of receipts, we can easily calculate the payback period by dividing the initial cash outlay by the annual receipts:
Payback period =
$650,000
Initial cost
=
Uniform annual benefit
$162,500
= 4 years.
If the company’s policy is to consider only projects with a payback period of five
years or less, this computer process control system project passes the initial screening.
In Example 5.2, dividing the initial payment by annual receipts to determine the payback
period is a simplification we can make because the annual receipts are uniform. Whenever the expected cash flows vary from year to year, however, the payback period must be
determined by adding the expected cash flows for each year until the sum is equal to or
greater than zero. The significance of this procedure is easily explained. The cumulative
cash flow equals zero at the point where cash inflows exactly match, or pay back, the
cash outflows; thus, the project has reached the payback point. Similarly, if the cumulative cash flows are greater than zero, then the cash inflows exceed the cash outflows, and
the project has begun to generate a profit, thus surpassing its payback point. To illustrate,
consider Example 5.3.
EXAMPLE 5.3 Conventional Payback Period
with Salvage Value
Autonumerics Company has just bought a new spindle machine at a cost of $105,000
to replace one that had a salvage value of $20,000. The projected annual after-tax savings via improved efficiency, which will exceed the investment cost, are as follows:
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212 CHAPTER 5 Present-Worth Analysis
Cumulative
Cash Flow
Period
Cash Flow
0
- $105,000 + $20,000
- $85,000
1
15,000
-70,000
2
25,000
-45,000
3
35,000
-10,000
4
45,000
35,000
5
45,000
80,000
6
35,000
115,000
SOLUTION
Given: Cash flow series as shown in Figure 5.3(a).
Find: Conventional payback period.
$45,000
$45,000
$35,000
$35,000
Annual cash flow
$25,000
$15,000
0
1
2
3
Years (n)
4
5
6
$85,000
(a)
Cumulative cash flow ($)
150,000
3.2 years
Payback period
100,000
50,000
0
⫺50,000
⫺100,000
0
1
2
3
Years (n)
4
5
(b)
Figure 5.3
Illustration of conventional payback period (Example 5.3).
6
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Section 5.2 Initial Project Screening Method 213
The salvage value of retired equipment becomes a major consideration in most justification analysis. (In this example, the salvage value of the old machine should be
taken into account, as the company already had decided to replace the old machine.)
When used, the salvage value of the retired equipment is subtracted from the purchase price of new equipment, revealing a closer true cost of the investment. As we
see from the cumulative cash flow in Figure 5.3(b), the total investment is recovered
during year 4. If the firm’s stated maximum payback period is three years, the project will not pass the initial screening stage.
COMMENTS: In Example 5.2, we assumed that cash flows occur only in discrete
lumps at the ends of years. If instead cash flows occur continuously throughout the
year, the payback period calculation needs adjustment. A negative balance of
$10,000 remains at the start of year 4. If $45,000 is expected to be received as a more
or less continuous flow during year 4, the total investment will be recovered twotenths ($10,000/$45,000) of the way through the fourth year. Thus, in this situation,
the payback period is 3.2 years.
5.2.2 Benefits and Flaws of Payback Screening
The simplicity of the payback method is one of its most appealing qualities. Initial
project screening by the method reduces the information search by focusing on that
time at which the firm expects to recover the initial investment. The method may also
eliminate some alternatives, thus reducing the firm’s time spent analyzing. But the
much-used payback method of equipment screening has a number of serious drawbacks. The principal objection to the method is that it fails to measure profitability (i.e.,
no “profit” is made during the payback period). Simply measuring how long it will take
to recover the initial investment outlay contributes little to gauging the earning power
of a project. (In other words, you already know that the money you borrowed for the
drill press is costing you 12% per year; the payback method can’t tell you how much
your invested money is contributing toward the interest expense.) Also, because payback period analysis ignores differences in the timing of cash flows, it fails to recognize the difference between the present and future value of money. For example,
although the payback on two investments can be the same in terms of numbers of years,
a front-loaded investment is better because money available today is worth more than
that to be gained later. Finally, because payback screening ignores all proceeds after the
payback period, it does not allow for the possible advantages of a project with a longer
economic life.
By way of illustration, consider the two investment projects listed in Table 5.1. Each
requires an initial investment outlay of $90,000. Project 1, with expected annual cash
proceeds of $30,000 for the first 3 years, has a payback period of 3 years. Project 2 is expected to generate annual cash proceeds of $25,000 for 6 years; hence, its payback period is 3.6 years. If the company’s maximum payback period is set to 3 years, then project
1 would pass the initial project screening, whereas project 2 would fail even though it is
clearly the more profitable investment.
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TABLE 5.1
Investment Cash Flows for
Two Competing Projects
n
Project 1
Project 2
0
- $90,000
- $90,000
1
30,000
25,000
2
30,000
25,000
3
30,000
25,000
4
1,000
25,000
5
1,000
25,000
6
1,000
25,000
$ 3,000
$60,000
5.2.3 Discounted Payback Period
Discounted
payback period:
The length of
time required
to recover the
cost of an
investment based
on discounted
cash flows.
To remedy one of the shortcomings of the conventional payback period, we may modify
the procedure so that it takes into account the time value of money—that is, the cost of funds
(interest) used to support a project. This modified payback period is often referred to as the
discounted payback period. In other words, we may define the discounted payback period
as the number of years required to recover the investment from discounted cash flows.
For the project in Example 5.3, suppose the company requires a rate of return of
15%. To determine the period necessary to recover both the capital investment and the
cost of funds required to support the investment, we may construct Table 5.2, showing
cash flows and costs of funds to be recovered over the life of the project.
To illustrate, let’s consider the cost of funds during the first year: With $85,000 committed at the beginning of the year, the interest in year 1 would be $12,750 1$85,000 * 0.152.
Therefore, the total commitment grows to $97,750, but the $15,000 cash flow in year 1
TABLE 5.2
Payback Period Calculation Taking into Account
the Cost of Funds (Example 5.3)
Period
Cash Flow
Cost of Funds
(15%)*
Cumulative
Cash Flow
0
–$85,000
0
–$85,000
1
15,000
- $85,00010.152 = - $12,750
-82,750
2
25,000
- $82,75010.152 =
-12,413
-70,163
3
35,000
- $70,16310.152 =
-10,524
-45,687
4
45,000
- $45,68710.152 =
-6,853
-7,540
5
45,000
- $7,54010.152 =
-1,131
36,329
6
35,000
$36,32910.152 =
5,449
76,778
*Cost of funds = Unrecovered beginning balance * interest rate.
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Section 5.3 Discounted Cash Flow Analysis 215
100,000
76,778
Cumulative cash flow ($)
75,000
Discounted
payback period
50,000
36,329
25,000
0
⫺7,540
⫺25,000
⫺50,000
⫺45,687
⫺75,000
⫺100,000
0
Figure 5.4
⫺70,163
⫺82,750
⫺85,000
1
2
3
Years (n)
4
5
6
Illustration of discounted payback period.
leaves a net commitment of $82,750. The cost of funds during the second year would be
$12,413 1$82,750 * 0.152, but with the $25,000 receipt from the project, the net commitment drops to $70,163. When this process repeats for the remaining years of the project’s
life, we find that the net commitment to the project ends during year 5. Depending on which
cash flow assumption we adopt, the project must remain in use about 4.2 years (continuous
cash flows) or 5 years (year-end cash flows) in order for the company to cover its cost of
capital and also recover the funds it has invested. Figure 5.4 illustrates this relationship.
The inclusion of effects stemming from time value of money has increased the payback period calculated for this example by a year. Certainly, this modified measure is an
improved one, but it does not show the complete picture of the project’s profitability either.
5.2.4 Where Do We Go from Here?
Should we abandon the payback method? Certainly not, but if you use payback screening
exclusively to analyze capital investments, look again. You may be missing something
that another method can help you spot. Therefore, it is illogical to claim that payback is
either a good or bad method of justification. Clearly, it is not a measure of profitability.
But when it is used to supplement other methods of analysis, it can provide useful information. For example, payback can be useful when a company needs a measure of the
speed of cash recovery, when the company has a cash flow problem, when a product is
built to last only for a short time, and when the machine the company is contemplating
buying itself is known to have a short market life.
5.3 Discounted Cash Flow Analysis
Until the 1950s, the payback method was widely used as a means of making investment decisions. As flaws in this method were recognized, however, businesspeople began to search
for methods to improve project evaluations. The result was the development of discounted
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Discounted
cash flow
analysis (DCF):
A method of
evaluating an
investment by
estimating future
cash flows and
taking into
consideration the
time value of
money.
Net present
worth: The
difference
between the
present value of
cash inflows and
the present value
of cash outflows.
Investment pool
operates like a
mutual fund to
earn a targeted
return by
investing the
firm’s money in
various
investment
assets.
cash flow techniques (DCFs), which take into account the time value of money. One of the
DCFs is the net-present-worth, or net-present-value, method. A capital investment problem
is essentially a problem of determining whether the anticipated cash inflows from a proposed
project are sufficient to attract investors to invest funds in the project. In developing the NPW
criterion, we will use the concept of cash flow equivalence discussed in Chapter 3.
As we observed, the most convenient point at which to calculate the equivalent values
is often at time 0. Under the NPW criterion, the present worth of all cash inflows is compared
against the present worth of all cash outflows associated with an investment project. The
difference between the present worth of these cash flows, referred to as the net present
worth (NPW), net present value (NPV) determines whether the project is an acceptable
investment. When two or more projects are under consideration, NPW analysis further allows us to select the best project by comparing their NPW figures.
5.3.1 Net-Present-Worth Criterion
We will first summarize the basic procedure for applying the net-present-worth criterion
to a typical investment project:
• Determine the interest rate that the firm wishes to earn on its investments. The interest rate you determine represents the rate at which the firm can always invest the
money in its investment pool. This interest rate is often referred to as either a
required rate of return or a minimum attractive rate of return (MARR). Usually,
selection of the MARR is a policy decision made by top management. It is possible
for the MARR to change over the life of a project, as we saw in Section 4.4, but for
now we will use a single rate of interest in calculating the NPW.
• Estimate the service life of the project.
• Estimate the cash inflow for each period over the service life.
• Estimate the cash outflow over each service period.
• Determine the net cash flows 1net cash flow = cash inflow - cash outflow2.
• Find the present worth of each net cash flow at the MARR. Add up all the presentworth figures; their sum is defined as the project’s NPW, given by
PW1i2 = NPW calculated at i =
A0
11 + i2
0
+
+
AN
A1
11 + i2
1
+
A2
11 + i22
+ Á
11 + i2N
N
An
= a
n
n = 0 11 + i2
N
= a A n1P/F, i, n2,
(5.1)
n=0
where
A n = Net cash flow at end of period n,
i = MARR 1or cost of capital2,
N = Service life of the project.
A n will be positive if the corresponding period has a net cash inflow and negative if
there is a net cash outflow.
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Section 5.3 Discounted Cash Flow Analysis 217
• Single Project Evaluation. In this context, a positive NPW means that the equivalent
worth of the inflows is greater than the equivalent worth of outflows, so the project
makes a profit. Therefore, if the PW(i) is positive for a single project, the project should
be accepted; if the PW(i) is negative, the project should be rejected.2 The decision rule is
If PW1i2 7 0, accept the investment.
If PW1i2 = 0, remain indifferent.
If PW1i2 6 0, reject the investment.
• Comparing Multiple Alternatives. Compute the PW(i) for each alternative and select
the one with the largest PW(i). As you will learn in Section 5.5, when you compare mutually exclusive alternatives with the same revenues, they are compared on a cost-only
basis. In this situation (because you are minimizing costs, rather than maximizing profits), you should accept the project that results in the smallest, or least negative, NPW.
EXAMPLE 5.4 Net Present Worth: Uniform Flows
Consider the investment cash flows associated with the computer process control
project discussed in Example 5.1. If the firm’s MARR is 15%, compute the NPW of
this project. Is the project acceptable?
SOLUTION
Given: Cash flows in Figure 5.2 and MARR = 15% per year.
Find: NPW.
Since the computer process control project requires an initial investment of $650,000
at n = 0, followed by the eight equal annual savings of $162,000, we can easily determine the NPW as follows:
PW115%2Outflow = $650,000;
PW115%2Inflow = $162,5001P>A, 15%, 82
= $729,190.
Then the NPW of the project is
PW115%2 = PW115%2Inflow - PW115%2Outflow
= $729,190 - $650,000
= $79,190,
or, from Eq. (5.1),
PW115%2 = - $650,000 + $162,5001P>A, 15%, 82
= $79,190.
Since PW115%2 7 0, the project is acceptable.
2
Some projects (e.g., the installation of pollution control equipment) cannot be avoided. In a case such as this,
the project would be accepted even though its NPW … 0. This type of project will be discussed in Chapter 12.
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218 CHAPTER 5 Present-Worth Analysis
Now let’s consider an example in which the investment cash flows are not uniform
over the service life of the project.
EXAMPLE 5.5 Net Present Worth: Uneven Flows
Tiger Machine Tool Company is considering acquiring a new metal-cutting machine.
The required initial investment of $75,000 and the projected cash benefits3 over the
project’s three-year life are as follows:
End of Year
Net Cash Flow
0
- $75,000
1
24,400
2
27,340
3
55,760
You have been asked by the president of the company to evaluate the economic
merit of the acquisition. The firm’s MARR is known to be 15%.
SOLUTION
Given: Cash flows as tabulated and MARR = 15% per year.
Find: NPW.
If we bring each flow to its equivalent at time zero, we find that
PW115%2 = - $75,000 + $24,0001P>F, 15%, 12 + $27,3401P>F, 15%, 22
+ $55,7601P>F, 15%, 32
= $3,553.
Since the project results in a surplus of $3,553, the project is acceptable.
In Example 5.5, we computed the NPW of a project at a fixed interest rate of 15%. If
we compute the NPW at varying interest rates, we obtain the data in Table 5.3. Plotting
the NPW as a function of interest rate gives Figure 5.5, the present-worth profile. (You
may use a spreadsheet program such as Excel to generate Table 5.3 or Figure 5.5.)
Figure 5.5 indicates that the investment project has a positive NPW if the interest rate
is below 17.45% and a negative NPW if the interest rate is above 17.45%. As we will see
in Chapter 7, this break-even interest rate is known as the internal rate of return. If the
3
As we stated at the beginning of this chapter, we treat net cash flows as before-tax values or as having their
tax effects precalculated. Explaining the process of obtaining cash flows requires an understanding of income
taxes and the role of depreciation, which are discussed in Chapter 9.
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Section 5.3 Discounted Cash Flow Analysis 219
TABLE 5.3
i(%)
Present-Worth Amounts at Varying
Interest Rates (Example 5.5)
PW(i)
i(%)
PW(i)
0
$32,500
20
- $3,412
2
27,743
22
-5,924
4
23,309
24
-8,296
6
19,169
26
-10,539
8
15,296
28
-12,662
10
11,670
30
-14,673
12
8,270
32
-16,580
14
5,077
34
-18,360
16
2,076
36
-20,110
0
38
-21,745
-750
40
-23,302
17.45*
18
*Break-even interest rate (also known as the rate of return).
40
Accept
Reject
30
20
PW (i ) ($ thousands)
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Break-even interest rate
(or rate of return).
10
$3,553
17.45%
0
⫺10
⫺20
⫺30
0
5
10
15
20
25
30
Years
i = MARR(%)
Figure 5.5
Present-worth profile described in Example 5.5.
35
40
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220 CHAPTER 5 Present-Worth Analysis
firm’s MARR is 15%, then the project has an NPW of $3,553 and so may be accepted.
The $3,553 figure measures the equivalent immediate gain in present worth to the firm
following acceptance of the project. By contrast, at i = 20%, PW120%2 = - $3,412,
and the firm should reject the project. (Note that either accepting or rejecting an investment is influenced by the choice of a MARR, so it is crucial to estimate the MARR correctly. We will defer this important issue until Section 5.3.3. For now, we will assume that
the firm has an accurate MARR estimate available for use in investment analysis.)
5.3.2 Meaning of Net Present Worth
In present-worth analysis, we assume that all the funds in a firm’s treasury can be placed
in investments that yield a return equal to the MARR. We may view these funds as an
investment pool. Alternatively, if no funds are available for investment, we assume that
the firm can borrow them at the MARR (or cost of capital) from the capital market. In this
section, we will examine these two views as we explain the meaning of the MARR in
NPW calculations.
Investment Pool Concept
An investment pool is equivalent to a firm’s treasury. All fund transactions are administered and managed by the firm’s comptroller. The firm may withdraw funds from this investment pool for other investment purposes, but if left in the pool, these funds will earn
at the MARR. Thus, in investment analysis, net cash flows will be net cash flows relative
to an investment pool. To illustrate the investment pool concept, we consider again the
project in Example 5.5 that required an investment of $75,000.
If the firm did not invest in the project and left $75,000 in the investment pool for
three years, these funds would grow as follows:
$75,0001F>P, 15%, 32 = $114,066.
Suppose the company decided instead to invest $75,000 in the project described in Example 5.5. Then the firm would receive a stream of cash inflows during the project’s life
of three years in the following amounts:
Period (n)
Net Cash Flow (An)
1
$24,400
2
27,340
3
55,760
Since the funds that return to the investment pool earn interest at a rate of 15%, it is worthwhile to see how much the firm would benefit from its $75,000 investment. For this alternative, the returns after reinvestment are
$24,4001F/P, 15%, 22 = $32,269,
$27,3401F/P, 15%, 12 = $31,441,
$55,7601F/P, 15%, 02 = $55,760.
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Section 5.3 Discounted Cash Flow Analysis 221
These returns total $119,470. At the end of three years, the additional cash accumulation from investing in the project is
$119,470 - $114,066 = $5,404.
If we compute the equivalent present worth of this net cash surplus at time 0, we obtain
$5,4041P>F, 15%, 32 = $3,553,
which is exactly what we get when we compute the NPW of the project with Eq. (5.1).
Clearly, on the basis of its positive NPW, the alternative of purchasing a new machine
should be preferred to that of simply leaving the funds in the investment pool at the
MARR. Thus, in PW analysis, any investment is assumed to be returned at the MARR. If
a surplus exists at the end of the project, then PW1MARR2 7 0. Figure 5.6 illustrates the
reinvestment concept as it relates to the firm’s investment pool.
Borrowed-Funds Concept
Suppose that the firm does not have $75,000 at the outset. In fact, the firm doesn’t have to
have an investment pool at all. Suppose further that the firm borrows all of its capital
from a bank at an interest rate of 15%, invests in the project, and uses the proceeds from
the investment to pay off the principal and interest on the bank loan. How much is left
over for the firm at the end of the project’s life?
n=0
n=3
If $75,000 were
left in the
investment pool
$75,000 ( F/P,15%,3)
$114,066
At n = 3, the funds
would grow to
If $75,000
withdrawal from
the investment pool
were invested in
the project
Year
Amount
1
2
3
$24,400
$27,340
$55,760
Reinvestment of
project receipts
$24,400 (F/P,15%,2) = $32,269
$27,340 (F/P,15%,1) = $31,441
$55,760 (F/P,15%,0) = $55,760
Return
to company
$119,470
Return to
investment
pool
The net benefit of
investing in the
project is
Company (lender)
$119,470 – $114,066
= $5,404
Project (borrower)
Figure 5.6 The concept of an investment pool with the company as a lender and the
project as a borrower.
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222 CHAPTER 5 Present-Worth Analysis
At the end of first year, the interest on the project’s use of the bank loan would be
$75,00010.152 = $11,250. Therefore, the total loan balance grows to $75,00011.152 =
$86,250. Then, the firm receives $24,400 from the project and applies the entire amount
to repay the loan portion, leaving a balance due of
$75,00011 + 0.152 - $24,400 = $61,850.
This amount becomes the net amount the project is borrowing at the beginning of year 2,
which is also known as the project balance. At the end of year 2, the bank debt grows to
$61,85011.152 = $71,128, but with the receipt of $27,340, the project balance is reduced to
$71,128 - $27,340 = $43,788.
Similarly, at the end of year 3, the project balance becomes
$43,78811.152 = $50,356.
But with the receipt of $55,760 from the project, the firm should be able to pay off the
remaining balance and come out with a surplus in the amount of $5,404. This terminal
project balance is also known as the net future worth of the project. In other words,
the firm repays its initial bank loan and interest at the end of year 3, with a resulting
profit of $5,404. If we compute the equivalent present worth of this net profit at time 0,
we obtain
PW115%2 = $5,4041P>F, 15%, 32 = $3,553.
The result is identical to the case in which we directly computed the NPW of the project
at i = 15%, shown in Example 5.5. Figure 5.7 illustrates the project balance as a function
of time.4
5.3.3 Basis for Selecting the MARR
Cost of capital:
The required
return
necessary
to make an
investment
project
worthwhile.
The basic principle used to determine the discount rate in project evaluations is similar to
the concept of the required return on investment for financial assets discussed in Section
4.6.2. The first element to cover is the cost of capital, which is the required return necessary to make an investment project worthwhile. The cost of capital would include both
the cost of debt (the interest rate associated with borrowing) and the cost of equity (the
return that stockholders require for a company). Both the cost of debt and the cost of equity
reflect the presence of inflation in the economy. The cost of capital determines how a
company can raise money (through issuing a stock, borrowing, or a mix of the two).
Therefore, this is normally considered as the rate of return that a firm would receive if it
invested its money someplace else with a similar risk.
The second element is a consideration of any additional risk associated with the project. If the project belongs to the normal risk category, the cost of capital may already reflect
the risk premium. However, if you are dealing with a project with higher risk, the additional risk premium may be added onto the cost of capital.
4
Note that the sign of the project balance changes from negative to positive during year 3. The time at which
the project balance becomes zero is known as the discounted payback period.
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Section 5.4 Variations of Present-Worth Analysis 223
80,000
60,000
Project balance ($)
40,000
Terminal project balance
(net future worth, or
project surplus)
20,000
$5,404
0
Discounted
payback period
– 20,000
–$43,788
– 40,000
–$61,850
– 60,000
–$75,000
– 80,000
– 100,000
– 120,000
0
2
1
3
Year (n)
Figure 5.7 Project balance diagram as a function of time. (A negative project
balance indicates the amount of the loan remaining to be paid off or the amount
of investment to be recovered.)
In sum, the discount rate (MARR) to use for project evaluation would be equivalent
to the firm’s cost of capital for a project of normal risk, but could be much higher if you
are dealing with a risky project. Chapter 15 will detail the analytical process of determining this discount rate. For now, we assume that such a rate is already known to us, and we
will focus on the evaluation of the investment project.
5.4 Variations of Present-Worth Analysis
As variations of present-worth analysis, we will consider two additional measures of investment worth: future-worth analysis and capitalized equivalent-worth analysis.
(The equivalent annual worth measure is another variation of the present-worth measure,
but we present it in Chapter 6.) Future-worth analysis calculates the future worth of an investment undertaken. Capitalized equivalent-worth analysis calculates the present worth
of a project with a perpetual life span.
5.4.1 Future-Worth Analysis
Net present worth measures the surplus in an investment project at time 0. Net future worth
(NFW) measures this surplus at a time other than 0. Net-future-worth analysis is particularly useful in an investment situation in which we need to compute the equivalent worth of
a project at the end of its investment period, rather than at its beginning. For example, it may
MARR: this is
based on the
firm’s cost of
capital plus or
minus a risk
premium to reflect the project’s
specific risk
characteristics.
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224 CHAPTER 5 Present-Worth Analysis
take 7 to 10 years to build a nuclear power plant because of the complexities of engineering design and the many time-consuming regulatory procedures that must be followed to
ensure public safety. In this situation, it is more common to measure the worth of the investment at the time of the project’s commercialization (i.e., we conduct an NFW analysis at the end of the investment period).
Net future
worth: The value
of an asset or
cash at a specified date in the
future that is
equivalent in
value to a
specified sum
today.
Net-Future-Worth Criterion and Calculations
Let A n represent the cash flow at time n for n = 0, 1, 2, Á , N for a typical investment
project that extends over N periods. Then the net-future-worth (NFW) expression at the
end of period N is
FW1i2 = A 011 + i2N + A 111 + i2N - 1 + A 211 + i2N - 2 + Á + A N
N
= a A n11 + i2N - n
n=0
N
= a A n1F>P, i, N - n2.
(5.2)
n=0
As you might expect, the decision rule for the NFW criterion is the same as that for the
NPW criterion: For a single project evaluation,
If FW1i2 7 0, accept the investment.
If FW1i2 = 0, remain indifferent to the investment.
If FW1i2 6 0, reject the investment.
EXAMPLE 5.6 Net Future Worth: At the End of the Project
Consider the project cash flows in Example 5.5. Compute the NFW at the end of year
3 at i = 15%.
SOLUTION
Given: Cash flows in Example 5.5 and MARR = 15% per year.
Find: NFW.
As seen in Figure 5.8, the NFW of this project at an interest rate of 15% would be
FW115%2 = - $75,0001F>P, 15%, 32 + $24,4001F>P, 15%, 22
+ $27,3401F>P, 15%, 12 + $55,760
= $5,404.
Note that the net future worth of the project is equivalent to the terminal project balance as calculated in Section 5.3.2. Since FW115%2 7 0, the project is acceptable.
We reach the same conclusion under present-worth analysis.
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Section 5.4 Variations of Present-Worth Analysis 225
$55,760
$27,340
$24,400
0
1
2
3
Years
$75,000
– $75,000 (F/P,15%,3)
$24,400 (F/P,15%,2)
$27,340 (F/P,15%,1)
$55,760
–$114,066
$32,269
$31,441
$55,760
NFW = $5,404
Figure 5.8
Future-worth calculation at the end of year 3 (Example 5.6).
EXAMPLE 5.7 Future Equivalent: At an Intermediate Time
Higgins Corporation (HC), a Detroit-based robot-manufacturing company, has developed a new advanced-technology robot called Helpmate, which incorporates advanced technology such as vision systems, tactile sensing, and voice recognition.
These features allow the robot to roam the corridors of a hospital or office building
without following a predetermined track or bumping into objects. HC’s marketing
department plans to target sales of the robot toward major hospitals. The robots will
ease nurses’ workloads by performing low-level duties such as delivering medicines
and meals to patients.
• The firm would need a new plant to manufacture the Helpmates; this plant could
be built and made ready for production in two years. It would require a 30-acre
site, which can be purchased for $1.5 million in year 0. Building construction
would begin early in year 1 and continue throughout year 2. The building would
cost an estimated $10 million, with a $4 million payment due to the contractor at
the end of year 1, and with another $6 million payable at the end of year 2.
• The necessary manufacturing equipment would be installed late in year 2 and would
be paid for at the end of year 2. The equipment would cost $13 million, including
transportation and installation. When the project terminates, the land is expected
to have an after-tax market value of $2 million, the building an after-tax value of
$3 million, and the equipment an after-tax value of $3 million.
For capital budgeting purposes, assume that the cash flows occur at the end of each
year. Because the plant would begin operations at the beginning of year 3, the first operating cash flows would occur at the end of year 3. The Helpmate plant’s estimated
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226 CHAPTER 5 Present-Worth Analysis
economic life is six years after completion, with the following expected after-tax operating cash flows in millions:
’
Calendar Year
End of Year
06
0
’
07
1
’
08
2
’
09
3
’
10
4
’
’
12
6
’
$6
$8
$13
$18
$14
11
5
13
7
’
14
8
After-tax cash flows
A. Operating revenue
$8
B. Investment
Land
-1.5
Building
+2
-4
Equipment
Net cash flow
- $1.5
- $4
-6
+3
-13
+3
- $19
$6
$8
$13
$18
$14
$16
Compute the equivalent worth of this investment at the start of operations. Assume
that HC’s MARR is 15%.
SOLUTION
Given: Preceding cash flows and MARR = 15% per year.
Find: Equivalent worth of project at the end of calendar year 2.
One easily understood method involves calculating the present worth and then transforming it to the equivalent worth at the end of year 2. First, we can compute PW(15%)
at time 0 of the project:
PW115%2 = - $1.5 - $41P>F, 15%, 12 - $191P>F, 15%, 22
+ $61P>F, 15%, 32 + $81P>F, 15%, 42 + $131P>F, 15%, 52
+ $181P>F, 15%, 62 + $141P>F, 15%, 72 + $161P>F, 15%, 82
= $13.91 million.
Then, the equivalent project worth at the start of operation is
FW115%2 = PW115%21F>P, 15%, 22
= $18.40 million.
A second method brings all flows prior to year 2 up to that point and discounts future
flows back to year 2. The equivalent worth of the earlier investment, when the plant
begins full operation, is
- $1.51F>P, 15%, 22 - $41F>P, 15%, 12 - $19 = - $25.58 million,
which produces an equivalent flow as shown in Figure 5.9. If we discount the future
flows to the start of operation, we obtain
FW115%2 = - $25.58 + $61P>F, 15%, 12 + $81P>F, 15%, 22 + Á
+ $161F>P, 15%, 62
= $18.40 million.
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Section 5.4 Variations of Present-Worth Analysis 227
$18
$16
$14
$13
$8
$6
0
1
2
3
$1.5
4
Years
5
6
7
$18
$4
$16
$14
$13
$19
8
$8
$6
0
1
2
3
Equivalent
lump sum
investment
at n = 2
4
Years
5
6
7
8
(Unit: Million dollars)
$25.58
Figure 5.9 Cash flow diagram for the Helpmate project
(Example 5.7).
COMMENTS: If another company is willing to purchase the plant and the right to manufacture the robots immediately after completion of the plant (year 2), HC would set
the price of the plant at $43.98 million 1$18.40 + $25.582 at a minimum.
5.4.2 Capitalized Equivalent Method
Another special case of the PW criterion is useful when the life of a proposed project is
perpetual or the planning horizon is extremely long (say, 40 years or more). Many
public projects, such as bridges, waterway structures, irrigation systems, and hydroelectric dams, are expected to generate benefits over an extended period (or forever). In
this section, we will examine the capitalized equivalent (CE(i)) method for evaluating
such projects.
Perpetual Service Life
Consider the cash flow series shown in Figure 5.10. How do we determine the PW for an infinite (or almost infinite) uniform series of cash flows or a repeated cycle of cash flows? The
process of computing the PW cost for this infinite series is referred to as the capitalization
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228 CHAPTER 5 Present-Worth Analysis
A
A
A
A
A
A
A
A
1
2
3
4
Years
5
6
N
⬁
0
P
Figure 5.10 Equivalent present worth of an infinite cash flow series.
Capitalized cost
related to car
leasing, means
the amount that
is being financed.
of the project cost. The cost, known as the capitalized cost, represents the amount of money
that must be invested today to yield a certain return A at the end of each and every period
forever, assuming an interest rate of i. Observe the limit of the uniform series present-worth
factor as N approaches infinity:
lim 1P>A, i, N2 = lim c
N: q
N: q
11 + i2N - 1
N
i11 + i2
d =
1
.
i
Thus,
PW1i2 = A1P>A, i, N : q 2 =
A
.
i
(5.3)
Another way of looking at this problem is to ask what constant income stream could
be generated by PW(i) dollars today in perpetuity. Clearly, the answer is A = iPW1i2. If
withdrawals were greater than A, you would be eating into the principal, which would
eventually reduce it to 0.
EXAMPLE 5.8 Capitalized Equivalent Cost
An engineering school has just completed a new engineering complex worth $50 million.
A campaign targeting alumni is planned to raise funds for future maintenance costs,
which are estimated at $2 million per year. Any unforeseen costs above $2 million
per year would be obtained by raising tuition. Assuming that the school can create a
trust fund that earns 8% interest annually, how much has to be raised now to cover
the perpetual string of $2 million in annual costs?
SOLUTION
Given: A = $2 million, i = 8% per year, and N = q .
Find: CE(8%).
The capitalized cost equation is
CE1i2 =
A
,
i
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Section 5.4 Variations of Present-Worth Analysis 229
so
CE18%2 = $2,000,000>0.08
= $25,000,000.
COMMENTS: It is easy to see that this lump-sum amount should be sufficient to pay
maintenance expenses for the school forever. Suppose the school deposited $25 million in a bank that paid 8% interest annually. Then at the end of the first year, the
$25 million would earn 8%1$25 million2 = $2 million interest. If this interest were
withdrawn, the $25 million would remain in the account. At the end of the second
year, the $25 million balance would again earn 8%1$25 million2 = $2 million. This
annual withdrawal could be continued forever, and the endowment (gift funds) would
always remain at $25 million.
Project’s Service Life Is Extremely Long
The benefits of typical civil engineering projects, such as bridge and highway construction, although not perpetual, can last for many years. In this section, we will examine the
use of the CE(i) criterion to approximate the NPW of engineering projects with long lives.
EXAMPLE 5.9 Comparison of Present Worth for Long Life
and Infinite Life
Mr. Gaynor L. Bracewell amassed a small fortune developing real estate in Florida
and Georgia over the past 30 years. He sold more than 700 acres of timber and farmland to raise $800,000, with which he built a small hydroelectric plant, known as High
Shoals Hydro. The plant was a decade in the making. The design for Mr. Bracewell’s
plant, which he developed using his Army training as a civil engineer, is relatively
simple. A 22-foot-deep canal, blasted out of solid rock just above the higher of two
dams on his property, carries water 1,000 feet along the river to a “trash rack,” where
leaves and other debris are caught. A 6-foot-wide pipeline capable of holding 3 million pounds of liquid then funnels the water into the powerhouse at 7.5 feet per second,
thereby creating 33,000 pounds of thrust against the turbines. Under a 1978 federal law
designed to encourage alternative power sources, Georgia Power Company is required
to purchase any electricity Mr. Bracewell can supply. Mr. Bracewell estimates that his
plant can generate 6 million kilowatt-hours per year.
Suppose that, after paying income taxes and operating expenses, Mr. Bracewell’s
annual income from the hydroelectric plant will be $120,000. With normal maintenance,
the plant is expected to provide service for at least 50 years. Figure 5.11 illustrates when
and in what quantities Mr. Bracewell spent his $800,000 (not taking into account the
time value of money) during the last 10 years. Was Mr. Bracewell’s $800,000 investment a wise one? How long will he have to wait to recover his initial investment, and
will he ever make a profit? Examine the situation by computing the project worth at
varying interest rates.
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230 CHAPTER 5 Present-Worth Analysis
Money spent on hydro
project for the last 10 years
(undiscounted) = $800
–10 –9
–8
–7
–6
–5
–4
–3
–2
–1
$120
0
1
$50
$50
$60
$120
$60
2
3
Years
45
46
47
48
49
50
(Unit: Thousand dollars)
$80
$100
$100
$150
Figure 5.11 Net cash flow diagram for Mr. Bracewell’s hydroelectric project (Example 5.9).
(a) If Mr. Bracewell’s interest rate is 8%, compute the NPW (at time 0 in Figure
5.11) of this project with a 50-year service life and infinite service, respectively.
(b) Repeat part (a), assuming an interest rate of 12%.
SOLUTION
Given: Cash flow in Figure 5.11 (to 50 years or q ) and i = 8% or 12%.
Find: NPW at time 0.
One of the main questions is whether Mr. Bracewell’s plant will be profitable. Now
we will compute the equivalent total investment and the equivalent worth of receiving future revenues at the start of power generation (i.e., at time 0).
(a) Let i = 8%. Then
• with a plant service life of 50 years, we can make use of single-payment
compound-amount factors in the invested cash flow to help us find the equivalent total investment at the start of power generation. Using K to indicate
thousands, we obtain
V1 = - $50K1F>P, 8%, 92 - $50K1F>P, 8%, 82
- $60K1F>P, 8%, 72 Á - $100K1F>P, 8%, 12 - $60K
= - $1,101K.
The equivalent total benefit at the start of generation is
V2 = $120K1P>A, 8%, 502 = $1,468K.
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Section 5.4 Variations of Present-Worth Analysis 231
Summing, we find the net equivalent worth at the start of power generation:
V1 + V2 = - $1,101K + $1,468K
= $367K.
• With an infinite service life, the net equivalent worth is called the capitalized
equivalent worth. The investment portion prior to time 0 is identical, so the capitalized equivalent worth is
CE18%2 = - $1,101K + $120K>10.082
= $399K.
Note that the difference between the infinite situation and the planning horizon
of 50 years is only $32,000.
(b) Let i = 12%. Then
• With a service life of 50 years, proceeding as we did in part (a), we find that
the equivalent total investment at the start of power generation is
V1 = - $50K1F>P, 12%, 92 - $50K1F>P, 12%, 82
- $60K1F>P, 12%, 72 Á - $100K1F>P, 12%, 12 - 60K
= - $1,299K.
Equivalent total benefits at the start of power generation are
V2 = $120K1P>A, 12%, 502 = $997K.
The net equivalent worth at the start of power generation is
V1 + V2 = - $1,299K + $997K
= - $302K.
• With infinite cash flows, the capitalized equivalent worth at the current time is
CE112%2 = - $1,299K + $120K/10.122
= - $299K.
Note that the difference between the infinite situation and a planning horizon of 50
years is merely $3,000, which demonstrates that we may approximate the present
worth of long cash flows (i.e., 50 years or more) by using the capitalized equivalent
value. The accuracy of the approximation improves as the interest rate increases (or
the number of years is greater).
COMMENTS: At i = 12%, Mr. Bracewell’s investment is not a profitable one, but at
8% it is. This outcome indicates the importance of using the appropriate i in investment analysis. The issue of selecting an appropriate i will be presented again in
Chapter 15.
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232 CHAPTER 5 Present-Worth Analysis
5.5 Comparing Mutually Exclusive Alternatives
Until now, we have considered situations involving either a single project alone or projects
that were independent of each other. In both cases, we made the decision to reject or accept
each project individually according to whether it met the MARR requirements, evaluated
with either the PW or FW criterion.
In the real world of engineering practice, however, it is typical for us to have two or
more choices of projects that are not independent of one another in seeking to accomplish a
business objective. (As we shall see, even when it appears that we have only one project to
consider, the implicit “do-nothing” alternative must be factored into the decision-making
process.) In this section, we extend our evaluation techniques to multiple projects that are
mutually exclusive. Other dependencies between projects will be considered in Chapter 15.
Often, various projects or investments under consideration do not have the same duration or do not match the desired study period. Adjustments must then be made to account
for the differences. In this section, we explain the concept of an analysis period and the
process of accommodating for different lifetimes, two important considerations that
apply in selecting among several alternatives. Up to now in this chapter, all available
options in a decision problem were assumed to have equal lifetimes. In the current section,
this restriction is also relaxed.
5.5.1 Meaning of Mutually Exclusive and “Do Nothing”
As we briefly mentioned in Section 5.1.2, several alternatives are mutually exclusive
when any one of them will fulfill the same need and the selection of one of them implies
that the others will be excluded. Take, for example, buying versus leasing an automobile
for business use; when one alternative is accepted, the other is excluded. We use the terms
alternative and project interchangeably to mean “decision option.”
“Do Nothing” Is a Decision Option
When considering an investment, we are in one of two situations: Either the project is
aimed at replacing an existing asset or system, or it is a new endeavor. In either case, a donothing alternative may exist. On the one hand, if a process or system already in place to
accomplish our business objectives is still adequate, then we must determine which, if
any, new proposals are economical replacements. If none are feasible, then we do nothing. On the other hand, if the existing system has failed, then the choice among proposed
alternatives is mandatory (i.e., do nothing is not an option).
New endeavors occur as alternatives to the “green fields” do-nothing situation, which
has zero revenues and zero costs (i.e., nothing currently exists). For most new endeavors, do
nothing is generally an alternative, as we won’t proceed unless at least one of the proposed
alternatives is economically sound. In fact, undertaking even a single project entails making a
decision between two alternatives, because the do-nothing alternative is implicitly included.
Occasionally, a new initiative must be undertaken, cost notwithstanding, and in this case the
goal is to choose the most economical alternative, since “do nothing” is not an option.
When the option of retaining an existing asset or system is available, there are two
ways to incorporate it into the evaluation of the new proposals. One way is to treat the donothing option as a distinct alternative; we cover this approach primarily in Chapter 14,
where methodologies specific to replacement analysis are presented. The second approach,
used mostly in this chapter, is to generate the cash flows of the new proposals relative to
that of the do-nothing alternative. That is, for each new alternative, the incremental costs
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Section 5.5 Comparing Mutually Exclusive Alternatives 233
(and incremental savings or revenues if applicable) relative to “do nothing” are used in
the economic evaluation. For a replacement-type problem, these costs are calculated by
subtracting the do-nothing cash flows from those of each new alternative. For new endeavors, the incremental cash flows are the same as the absolute amounts associated with
each alternative, since the do-nothing values are all zero.
Because the main purpose of this chapter is to illustrate how to choose among mutually exclusive alternatives, most of the problems are structured so that one of the options
presented must be selected. Therefore, unless otherwise stated, it is assumed that “do
nothing” is not an option, and costs and revenues can be viewed as incremental to “do
nothing.”
Service Projects versus Revenue Projects
When comparing mutually exclusive alternatives, we need to classify investment projects
into either service or revenue projects. Service projects are projects whose revenues do
not depend on the choice of project; rather, such projects must produce the same amount
of output (revenue). In this situation, we certainly want to choose an alternative with the
least input (or cost). For example, suppose an electric utility is considering building a
new power plant to meet the peak-load demand during either hot summer or cold winter
days. Two alternative service projects could meet this demand: a combustion turbine
plant and a fuel-cell power plant. No matter which type of plant is selected, the firm will
collect the same amount of revenue from its customers. The only difference is how much
it will cost to generate electricity from each plant. If we were to compare these service
projects, we would be interested in knowing which plant could provide the cheaper power
(lower production cost). Further, if we were to use the NPW criterion to compare alternatives so as to minimize expenditures, we would choose the alternative with the lower
present-value production cost over the service life of the plant.
Revenue projects, by contrast, are projects whose revenues depend on the choice
of alternative. With revenue projects, we are not limiting the amount of input going into
the project or the amount of output that the project would generate. Then our decision
is to select the alternative with the largest net gains (output – input). For example, a TV
manufacturer is considering marketing two types of high-resolution monitors. With its
present production capacity, the firm can market only one of them. Distinct production
processes for the two models could incur very different manufacturing costs, so the
revenues from each model would be expected to differ due to divergent market prices
and potentially different sales volumes. In this situation, if we were to use the NPW criterion, we would select the model that promises to bring in the higher net present
worth.
Total-Investment Approach
Applying an evaluation criterion to each mutually exclusive alternative individually and then
comparing the results to make a decision is referred to as the total-investment approach.
We compute the PW for each individual alternative, as we did in Section 5.3, and select the
one with the highest PW. Note that this approach guarantees valid results only when PW,
FW, and AE criteria are used. (As you will see in Chapters 7 and 16, the total-investment approach does not work for any decision criterion based on either a percentage (rate of return)
or a ratio (e.g., a benefit–cost ratio). With percentages or ratios, you need to use the incremental investment approach, which also works with any decision criterion, including PW,
FW, and AE.) The incremental investment approach will be discussed in detail in Chapter 7.
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234 CHAPTER 5 Present-Worth Analysis
Scale of Investment
Frequently, mutually exclusive investment projects may require different levels of investments. At first, it seems unfair to compare a project requiring a smaller investment with
one requiring a larger investment. However, the disparity in scale of investment should
not be of concern in comparing mutually exclusive alternatives, as long as you understand the basic assumption: Funds not invested in the project will continue to earn interest at the MARR. We will look at mutually exclusive alternatives that require different
levels of investments for both service projects and revenue projects:
• Service projects. Typically, what you are asking yourself to do here is to decide
whether the higher initial investment can be justified by additional savings that will
occur in the future. More efficient machines are usually more expensive to acquire initially, but they will reduce future operating costs, thereby generating more savings.
• Revenue projects. If two mutually exclusive revenue projects require different levels
of investments with varying future revenue streams, then your question is what to do
with the difference in investment funds if you decide to go with the project that requires
the smaller investment. To illustrate, consider the two mutually exclusive revenue projects illustrated in Figure 5.12. Our objective is to compare these two projects at a
MARR of 10%.
Suppose you have exactly $4,000 to invest. If you choose Project B, you do not have
any leftover funds. However, if you go with Project A, you will have $3,000 in unused
$2,075
$2,110
$1,400
$600
$500
$450
Project A
Project B
$1,000
$3,993
$4,000
$600
$450
$500
Modified
Project A
$1,000
$3,000
Figure 5.12 Comparing mutually exclusive revenue projects
requiring different levels of investment.
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Section 5.5 Comparing Mutually Exclusive Alternatives 235
funds. Our assumption is that these unused funds will continue to earn an interest rate that
is the MARR. Therefore, the unused funds will grow at 10%, or $3,933, at the end of the
project term, or three years from now. Consequently, selecting Project A is equivalent to
having a modified project cash flow as shown in Figure 5.12.
Let’s calculate the net present worth for each option at 10%:
• Project A:
PW110%2A = - $1,000 + $4501P>F, 10%, 12 + $6001P>F, 10%, 22
+ $5001P>F, 10%, 32
= $283.
• Project B:
PW110%2B = - $4,000 + $1,4001P>A, 10%, 12
+ $2,0751P>F, 10%, 22 + $2,1101P>F, 10%, 32
= $579.
Clearly, Project B is the better choice. But how about the modified Project A? If we calculate the present worth for the modified Project A, we have the following:
• Modified Project A:
PW110%2A = - $4,000 + $4501P>A, 10%, 12 + $6001P>F, 10%, 22
+ $4,4931P>F, 10%, 32
= $283.
This is exactly the same as the net present worth without including the investment consequence of the unused funds. It is not a surprising result, as the return on investment in the
unused funds will be exactly 10%. If we also discount the funds at 10%, there will be no
surplus. So, what is the conclusion? It is this: If there is any disparity in investment scale
for mutually exclusive revenue projects, go ahead and calculate the net present worth for
each option without worrying about the investment differentials.
5.5.2 Analysis Period
The analysis period is the time span over which the economic effects of an investment
will be evaluated. The analysis period may also be called the study period or planning
horizon. The length of the analysis period may be determined in several ways: It may be
a predetermined amount of time set by company policy, or it may be either implied or explicit in the need the company is trying to fulfill. (For example, a diaper manufacturer
sees the need to dramatically increase production over a 10-year period in response to an
anticipated “baby boom.”) In either of these situations, we consider the analysis period to
be the required service period.
When the required service period is not stated at the outset, the analyst must choose
an appropriate analysis period over which to study the alternative investment projects. In
such a case, one convenient choice of analysis period is the period of the useful life of the
investment project.
When the useful life of an investment project does not match the analysis or required
service period, we must make adjustments in our analysis. A further complication in a consideration of two or more mutually exclusive projects is that the investments themselves
may have different useful lives. Accordingly, we must compare projects with different useful lives over an equal time span, which may require further adjustments in our analysis.
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236 CHAPTER 5 Present-Worth Analysis
Case 1
Case 2
Analysis Required
=
period
service
period
Finite
Required service
period
Case 3
Case 4
Analysis period
equals
project lives
Ex 5.10
Analysis period
is shorter than
project lives
Ex 5.11
Analysis period
is longer than
project lives
Ex 5.12
Analysis period is longest
Ex 5.13
of project life in
the group
Project
repeatability
likely
Analysis period is lowest
common multiple
of project lives
Project
repeatability
unlikely
Analysis period
equals one of
the project lives
Ex 5.14
Infinite
Figure 5.13
Analysis period implied in comparing mutually exclusive alternatives.
(Figure 5.13 is a flowchart showing the possible combinations of the analysis period and
the useful life of an investment.) In the sections that follow, we will explore in more detail how to handle situations in which project lives differ from the analysis period and
from each other. But we begin with the most straightforward situation: when the project
lives and the analysis period coincide.
5.5.3 Analysis Period Equals Project Lives
When the project lives equal the analysis period, we compute the NPW for each project
and select the one with the highest NPW. Example 5.10 illustrates this point.
EXAMPLE 5.10 Present-Worth Comparison (Revenue
Projects with Equal Lives): Three Alternatives
Bullard Company (BC) is considering expanding its range of industrial machinery
products by manufacturing machine tables, saddles, machine bases, and other similar
parts. Several combinations of new equipment and personnel could serve to fulfill
this new function:
• Method 1 (M1): new machining center with three operators.
• Method 2 (M2): new machining center with an automatic pallet changer and
three operators.
• Method 3 (M3): new machining center with an automatic pallet changer and two
task-sharing operators.
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Section 5.5 Comparing Mutually Exclusive Alternatives 237
Each of these arrangements incurs different costs and revenues. The time taken to
load and unload parts is reduced in the pallet-changer cases. Certainly, it costs more to
acquire, install, and tool-fit a pallet changer, but because the device is more efficient and
versatile, it can generate larger annual revenues. Although saving on labor costs, tasksharing operators take longer to train and are more inefficient initially. As the operators
become more experienced at their tasks and get used to collaborating with each other, it
is expected that the annual benefits will increase by 13% per year over the five-year
study period. BC has estimated the investment costs and additional revenues as follows:
Machining Center Methods
M1
M2
M3
Investment:
Machine tool purchase
$121,000
Automatic pallet changer
$121,000
$121,000
$ 66,600
$ 66,600
Installation
$ 30,000
$ 42,000
$ 42,000
Tooling expense
$ 58,000
$ 65,000
$ 65,000
Total investment
$209,000
$294,600
$294,600
$ 55,000
$ 69,300
Annual benefits: Year 1
Additional revenues
$ 36,000
Direct labor savings
$ 17,300
Setup savings
$
Year 1: Net revenues
$ 55,000
4,700
$
4,700
$ 74,000
$ 58,000
Years 2–5: Net revenues
constant
constant
g = 13%/year
Salvage value in year 5
$ 80,000
$120,000
$120,000
All cash flows include all tax effects. “Do nothing” is obviously an option, since
BC will not undertake this expansion if none of the proposed methods is economically viable. If a method is chosen, BC expects to operate the machining center over
the next five years. On the basis of the use of the PW measure at i = 12%, which option would be selected?
SOLUTION
Given: Cash flows for three revenue projects and i = 12% per year.
Find: NPW for each project and which project to select.
For these revenue projects, the net-present-worth figures at i = 12% would be as
follows:
• For Option M1,
PW112%2M1 = - $209,000 + $55,0001P>A, 12%, 52
+ $80,0001P>F, 12%, 52
= $34,657.
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238 CHAPTER 5 Present-Worth Analysis
• For Option M2,
PW112%2M2 = - $294,600 + $74,0001P>A, 12%, 52
+ $120,0001P>F, 12%, 52
= $40,245.
• For Option M3,
PW112%2M3 = - $294,600 + $58,0001P>A 1, 13%, 12%, 52
+ $120,0001P>F, 12%, 52
= $37,085.
Clearly, Option M2 is the most profitable. Given the nature of BC parts and shop orders, management decides that the best way to expand would be with an automatic
pallet changer, but without task sharing.
5.5.4 Analysis Period Differs from Project Lives
In Example 5.10, we assumed the simplest scenario possible when analyzing mutually
exclusive projects: The projects had useful lives equal to each other and to the required
service period. In practice, this is seldom the case. Often, project lives do not match the
required analysis period or do not match each other (or both). For example, two machines
may perform exactly the same function, but one lasts longer than the other, and both of
them last longer than the analysis period over which they are being considered. In the
sections and examples that follow, we will develop some techniques for dealing with
these complications.
Salvage value:
The estimated
value that an
asset will realize
upon its sale at
the end of its
useful life.
Project’s Life Is Longer than Analysis Period
Project lives rarely conveniently coincide with a firm’s predetermined required analysis
period; they are often too long or too short. The case of project lives that are too long is
the easier one to address.
Consider the case of a firm that undertakes a five-year production project when all of
the alternative equipment choices have useful lives of seven years. In such a case, we analyze each project for only as long as the required service period (in this case, five years).
We are then left with some unused portion of the equipment (in this case, two years’
worth), which we include as salvage value in our analysis. Salvage value is the amount
of money for which the equipment could be sold after its service to the project has been
rendered. Alternatively, salvage value is the dollar measure of the remaining usefulness
of the equipment.
A common instance of project lives that are longer than the analysis period occurs in
the construction industry: A building project may have a relatively short completion time,
but the equipment that is purchased has a much longer useful life.
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Section 5.5 Comparing Mutually Exclusive Alternatives 239
EXAMPLE 5.11 Present-Worth Comparison: Project Lives
Longer than the Analysis Period
Waste Management Company (WMC) has won a contract that requires the firm to remove radioactive material from government-owned property and transport it to a designated dumping site. This task requires a specially made ripper–bulldozer to dig and
load the material onto a transportation vehicle. Approximately 400,000 tons of waste
must be moved in a period of two years.
• Model A costs $150,000 and has a life of 6,000 hours before it will require any
major overhaul. Two units of model A would be required to remove the material
within two years, and the operating cost for each unit would run to $40,000/year
for 2,000 hours of operation. At this operational rate, the model would be operable
for three years, at the end of which time it is estimated that the salvage value will
be $25,000 for each machine.
• A more efficient model B costs $240,000 each, has a life of 12,000 hours without
any major overhaul, and costs $22,500 to operate for 2,000 hours per year to complete the job within two years. The estimated salvage value of model B at the end
of six years is $30,000. Once again, two units of model B would be required to remove the material within two years.
Since the lifetime of either model exceeds the required service period of two
years (Figure 5.14), WMC has to assume some things about the used equipment at
the end of that time. Therefore, the engineers at WMC estimate that, after two years,
the model A units could be sold for $45,000 each and the model B units for $125,000
each. After considering all tax effects, WMC summarized the resulting cash flows (in
thousand of dollars) for each project as follows:
Period
Model A
Model B
0
- $300
- $480
1
-80
-45
2
-80
3
-80
+90
+50
-45
+250
-45
4
-45
5
-45
6
-45
+60
Here, the figures in the boxes represent the estimated salvage values at the end of
the analysis period (the end of year 2). Assuming that the firm’s MARR is 15%,
which option would be acceptable?
SOLUTION
Given: Cash flows for two alternatives as shown in Figure 5.14 and i = 15% per year.
Find: NPW for each alternative and which alternative is preferred.
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240 CHAPTER 5 Present-Worth Analysis
$60,000
$50,000
Years
0
2
1
Years
3
0
2
1
3
4
5
6
$45,000
$80,000
$300,000
(a)
(b)
$480,000
Estimated salvage value
at the end of required
service period
$250,000
$90,000
$50,000
Model A
$60,000
Model B
Years
0
1
$80,000
$300,000
2
Years
0
3
1
2
3
$45,000
$80,000
Required
service period
4
5
6
$45,000
Required
service period
$480,000
(c)
Figure 5.14 (a) Cash flow for model A; (b) cash flow for model B; (c) comparison
of service projects with unequal lives when the required service period is shorter than
the individual project life (Example 5.11).
First, note that these are service projects, so we can assume the same revenues for both
configurations. Since the firm explicitly estimated the market values of the assets at the
end of the analysis period (two years), we can compare the two models directly. Because the benefits (removal of the waste) are equal, we can concentrate on the costs:
PW115%2A =
=
PW115%2B =
=
- $300 - $801P>A, 15%, 22 + $901P>F, 15%, 22
- $362;
- $480 - $451P>A, 15%, 22 + $2501P>F, 15%, 22
- $364.
Model A has the least negative PW costs and thus would be preferred.
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Section 5.5 Comparing Mutually Exclusive Alternatives 241
Project’s Life Is Shorter than Analysis Period
When project lives are shorter than the required service period, we must consider how, at
the end of the project lives, we will satisfy the rest of the required service period. Replacement projects—additional projects to be implemented when the initial project has
reached the limits of its useful life—are needed in such a case. A sufficient number of replacement projects that match or exceed the required service period must be analyzed.
To simplify our analysis, we could assume that the replacement project will be exactly
the same as the initial project, with the same costs and benefits. However, this assumption
is not necessary. For example, depending on our forecasting skills, we may decide that a
different kind of technology—in the form of equipment, materials, or processes—is a
preferable replacement. Whether we select exactly the same alternative or a new technology
as the replacement project, we are ultimately likely to have some unused portion of the
equipment to consider as salvage value, just as in the case when the project lives are longer
than the analysis period. Of course, we may instead decide to lease the necessary equipment or subcontract the remaining work for the duration of the analysis period. In this
case, we can probably match our analysis period and not worry about salvage values.
In any event, at the outset of the analysis period, we must make some initial guess
concerning the method of completing the analysis. Later, when the initial project life is
closer to its expiration, we may revise our analysis with a different replacement project.
This is only reasonable, since economic analysis is an ongoing activity in the life of a
company and an investment project, and we should always use the most reliable, up-todate data we can reasonably acquire.
EXAMPLE 5.12 Present-Worth Comparison: Project Lives
Shorter than the Analysis Period
The Smith Novelty Company, a mail-order firm, wants to install an automatic mailing
system to handle product announcements and invoices. The firm has a choice between
two different types of machines. The two machines are designed differently, but have
identical capacities and do exactly the same job. The $12,500 semiautomatic model A
will last three years, while the fully automatic model B will cost $15,000 and last four
years. The expected cash flows for the two machines, including maintenance, salvage
value, and tax effects, are as follows:
n
Model A
0
- $12,500
- $15,000
1
-5,000
-4,000
2
-5,000
-4,000
3
-5,000 + 2,000
-4,000
-4,000 + 1,500
4
Model B
5
As business grows to a certain level, neither of the models may be able to handle the
expanded volume at the end of year 5. If that happens, a fully computerized mail-order
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242 CHAPTER 5 Present-Worth Analysis
system will need to be installed to handle the increased business volume. In the scenario just presented, which model should the firm select at MARR = 15%?
SOLUTION
Given: Cash flows for two alternatives as shown in Figure 5.15, analysis period of
five years, and i = 15%.
Find: NPW of each alternative and which alternative to select.
Since both models have a shorter life than the required service period of 5 years, we
need to make an explicit assumption of how the service requirement is to be met.
Suppose that the company considers leasing equipment comparable to model A at an
annual payment of $6,000 (after taxes) and with an annual operating cost of $5,000
$2,000
Years
Model A
0
1
2
3
4
5
$5,000
$12,500
Remaining
service
requirement
met by
leasing an
asset
$11,000
$1,500
Years
Model B
0
1
2
3
4
5
$5,000
$4,000
$11,000
$15,000
Required
service
period
Figure 5.15 Comparison for service projects with unequal lives
when the required service period is longer than the individual project life (Example 5.12).
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Section 5.5 Comparing Mutually Exclusive Alternatives 243
for the remaining required service period. In this case, the cash flow would look like
that shown in Figure 5.15:
n
Model A
Model B
0
- $12,500
- $15,000
1
-5,000
-4,000
2
-5,000
-4,000
3
-5,000 +
4
-5,000 -
5
-5,000 -
2,000
6,000
6,000
-4,000
-4,000 + 1,500
-5,000 -
6,000
Here, the boxed figures represent the annual lease payments. (It costs $6,000 to lease
the equipment and $5,000 to operate it annually. Other maintenance costs will be
paid by the leasing company.) Note that both alternatives now have the same required
service period of five years. Therefore, we can use NPW analysis:
PW115%2A = - $12,500 - $5,0001P>A, 15%, 22 - $3,0001P>F, 15%, 32
- $11,0001P>A, 15%, 221P>F, 15%, 32
= - $34,359.
PW115%2B = - $15,000 - $4,0001P>A, 15%, 32 - $2,5001P>F, 15%, 42
- $11,0001P>F, 15%, 52
= - $31,031.
Since these are service projects, model B is the better choice.
Analysis Period Coincides with Longest Project Life
As seen in the preceding pages, equal future periods are generally necessary to achieve
comparability of alternatives. In some situations, however, revenue projects with different
lives can be compared if they require only a one-time investment because the task or need
within the firm is a one-time task or need. An example of this situation is the extraction of
a fixed quantity of a natural resource such as oil or coal.
Consider two mutually exclusive processes: One requires 10 years to recover some
coal, and the other can accomplish the task in only 8 years. There is no need to continue the
project if the short-lived process is used and all the coal has been retrieved. In this example,
the two processes can be compared over an analysis period of 10 years (the longest project
life of the two being considered), assuming that no cash flows after 8 years for the shorter
lived project. Because of the time value of money, the revenues must be included in the
analysis even if the price of coal is constant. Even if the total (undiscounted) revenue is
equal for either process, that for the faster process has a larger present worth. Therefore, the
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244 CHAPTER 5 Present-Worth Analysis
two projects could be compared by using the NPW of each over its own life. Note that in
this case the analysis period is determined by, and coincides with, the longest project life in
the group. (Here we are still, in effect, assuming an analysis period of 10 years.)
EXAMPLE 5.13 Present-Worth Comparison: A Case where
the Analysis Period Coincides with the
Project with the Longest Life in the
Mutually Exclusive Group
The family-operated Foothills Ranching Company (FRC) owns the mineral rights to
land used for growing grain and grazing cattle. Recently, oil was discovered on this
property. The family has decided to extract the oil, sell the land, and retire. The company can lease the necessary equipment and extract and sell the oil itself, or it can
lease the land to an oil-drilling company:
• Drill option. If the company chooses to drill, it will require $300,000 leasing
expenses up front, but the net annual cash flow after taxes from drilling operations will be $600,000 at the end of each year for the next five years. The company can sell the land for a net cash flow of $1,000,000 in five years, when the
oil is depleted.
• Lease option. If the company chooses to lease, the drilling company can extract
all the oil in only three years, and FRC can sell the land for a net cash flow of
$800,000 at that time. (The difference in resale value of the land is due to the increasing rate of land appreciation anticipated for this property.) The net cash flow
from the lease payments to FRC will be $630,000 at the beginning of each of the
next three years.
All benefits and costs associated with the two alternatives have been accounted for in
the figures listed. Which option should the firm select at i = 15%?
SOLUTION
Given: Cash flows shown in Figure 5.16 and i = 15% per year.
Find: NPW of each alternative and which alternative to select.
As illustrated in Figure 5.16, the cash flows associated with each option look like this:
n
Drill
Lease
0
- $300,000
$630,000
1
600,000
630,000
2
600,000
630,000
3
600,000
800,000
4
600,000
5
1,600,000
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Section 5.5 Comparing Mutually Exclusive Alternatives 245
$1,600
Drill option
$600
$600
$600
$600
1
2
3
4
0
5
Years
Analysis
period of
5 years
$300
Lease option
$800
$630
$630
$630
0
1
2
3
4
5
Years
Figure 5.16 Comparison of revenue projects with unequal
lives when the analysis period coincides with the project
with the longest life in the mutually exclusive group (Example 5.13). In our example, the analysis period is five years,
assuming no cash flow in years 4 and 5 for the lease option.
After depletion of the oil, the project will terminate. The present worth of each of the
two options is as follows:
PW115%2Drill = - $300,000 + $600,0001P>A, 15%, 42
+ $1,600,0001P>F, 15%, 52
= $2,208,470.
PW115%2Lease = $630,000 + $630,0001P>A, 15%, 22
+ $800,0001P>F, 15%, 32
= $2,180,210.
Note that these are revenue projects; therefore, the drill option appears to be the marginally better option.
COMMENTS: The relatively small difference between the two NPW amounts
($28,260) suggests that the actual decision between drilling and leasing might be
based on noneconomic issues. Even if the drilling option were slightly better, the
company might prefer to forgo the small amount of additional income and select the
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246 CHAPTER 5 Present-Worth Analysis
lease option, rather than undertake an entirely new business venture and do its own
drilling. A variable that might also have a critical effect on this decision is the sales
value of the land in each alternative. The value of land is often difficult to forecast
over any long period, and the firm may feel some uncertainty about the accuracy of
its guesses. In Chapter 12, we will discuss sensitivity analysis, a method by which
we can factor uncertainty about the accuracy of project cash flows into our analysis.
5.5.5 Analysis Period Is Not Specified
Our coverage so far has focused on situations in which an analysis period is known.
When an analysis period is not specified, either explicitly by company policy or practice or implicitly by the projected life of the investment project, it is up to the analyst
to choose an appropriate one. In such a case, the most convenient procedure is to
choose an analysis on the basis of the useful lives of the alternatives. When the alternatives have equal lives, this is an easy selection. When the lives of at least some of
the alternatives differ, we must select an analysis period that allows us to compare
projects with different lifetimes on an equal time basis—that is, a common service
period.
Least common
multiple of two
numbers is the
lowest number
that can be
divided by both.
Lowest Common Multiple of Project Lives
A required service period of infinity may be assumed if we anticipate that an investment
project will be proceeding at roughly the same level of production for some indefinite period. It is certainly possible to make this assumption mathematically, although the analysis is likely to be complicated and tedious. Therefore, in the case of an indefinitely
ongoing investment project, we typically select a finite analysis period by using the
lowest common multiple of project lives. For example, if alternative A has a 3-year useful life and alternative B has a 4-year life, we may select 12 years as the analysis or common service period. We would consider alternative A through four life cycles and
alternative B through three life cycles; in each case, we would use the alternatives completely. We then accept the finite model’s results as a good prediction of what will be the
economically wisest course of action for the foreseeable future. The next example is a
case in which we conveniently use the lowest common multiple of project lives as our
analysis period.
EXAMPLE 5.14 Present-Worth Comparison: Unequal Lives,
Lowest-Common-Multiple Method
Consider Example 5.12. Suppose that models A and B can each handle the increased
future volume and that the system is not going to be phased out at the end of five years.
Instead, the current mode of operation is expected to continue indefinitely. Suppose
also that the two models will be available in the future without significant changes in
price and operating costs. At MARR = 15%, which model should the firm select?
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Section 5.5 Comparing Mutually Exclusive Alternatives 247
SOLUTION
Given: Cash flows for two alternatives as shown in Figure 5.17, i = 15% per year,
and an indefinite period of need.
Find: NPW of each alternative and which alternative to select.
Recall that the two mutually exclusive alternatives have different lives, but provide
identical annual benefits. In such a case, we ignore the common benefits and can
make the decision solely on the basis of costs, as long as a common analysis period
is used for both alternatives.
To make the two projects comparable, let’s assume that, after either the 3- or 4year period, the system would be reinstalled repeatedly, using the same model, and
that the same costs would apply. The lowest common multiple of 3 and 4 is 12, so we
will use 12 years as the common analysis period. Note that any cash flow difference
between the alternatives will be revealed during the first 12 years. After that, the same
Model A
$2,000
$2,000
$2,000
$2,000
Years
0
1
2
3
4
$5,000
5
6
7
$5,000
8
9
10
$5,000
11
12
$5,000
$12,500
First
cycle
Second
cycle
Third
cycle
Fourth
cycle
Model B
$1,500
$1,500
$1,500
Years
0
1
2
3
4
5
$4,000
$15,000
First
cycle
6
7
8
9
$4,000
Second
cycle
10
11
12
$4,000
Third
cycle
Figure 5.17 Comparison of projects with unequal lives when the required
service period is infinite and the project is likely to be repeatable with the same
investment and operations and maintenance costs in all future replacement
cycles (Example 5.14).
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248 CHAPTER 5 Present-Worth Analysis
cash flow pattern repeats every 12 years for an indefinite period. The replacement cycles
and cash flows are shown in Figure 5.17. Here is our analysis:
• Model A. Four replacements occur in a 12-year period. The PW for the first
investment cycle is
PW115%2 = - $12,500 - $5,0001P/A, 15%, 32
+ $2,0001P/F, 15%, 32
= - $22,601.
With four replacement cycles, the total PW is
PW115%2 = - $22,601[1 + 1P>F, 15%, 32
+ 1P>F, 15%, 62 + 1P>F, 15%, 92]
= - $53,657.
• Model B. Three replacements occur in a 12-year period. The PW for the first
investment cycle is
PW115%2 = - $15,000 - $4,0001P>A, 15%, 42
+ $1,5001P>F, 15%, 42
= - $25,562.
With three replacement cycles in 12 years, the total PW is
PW115%2 = - $25,562[1 + 1P/F, 15%, 42 + 1P/F, 15%, 82]
= - $48,534.
Clearly, model B is a better choice, as before.
COMMENTS: In Example 5.14, an analysis period of 12 years seems reasonable. The
number of actual reinvestment cycles needed with each type of system will depend on
the technology of the future system, so we may or may not actually need the four reinvestment cycles (model A) or three (model B) we used in our analysis. The validity of
the analysis also depends on the costs of the system and labor remaining constant. If we
assume constant-dollar prices (see Chapter 11), this analysis would provide us with a
reasonable result. (As you will see in Example 6.3, the annual-worth approach makes it
mathematically easier to solve this type of comparison.) If we cannot assume constantdollar prices in future replacements, we need to estimate the costs for each replacement
over the analysis period. This will certainly complicate the problem significantly.
Other Common Analysis Periods
In some cases, the lowest common multiple of project lives is an unwieldy analysis period to
consider. Suppose, for example, that you were considering two alternatives with lives of 7
and 12 years, respectively. Besides making for tedious calculations, an 84-year analysis period may lead to inaccuracies, since, over such a long time, we can be less and less confident
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Problems 249
about the ability to install identical replacement projects with identical costs and benefits. In
a case like this, it would be reasonable to use the useful life of one of the alternatives by either factoring in a replacement project or salvaging the remaining useful life, as the case may
be. The important idea is that we must compare both projects on the same time basis.
SUMMARY
In this chapter, we presented the concept of present-worth analysis, based on cash flow
equivalence along with the payback period. We observed the following important results:
Present worth is an equivalence method of analysis in which a project’s cash flows are
discounted to a single present value. It is perhaps the most efficient analysis method
we can use in determining the acceptability of a project on an economic basis. Other
analysis methods, which we will study in Chapters 6 and 7, are built on a sound understanding of present worth.
The minimum attractive rate of return (MARR) is the interest rate at which a firm can
always earn or borrow money under a normal operating environment. It is generally
dictated by management and is the rate at which NPW analysis should be conducted.
Revenue projects are those for which the income generated depends on the choice of
project. Service projects are those for which the income remains the same, regardless
of which project is selected.
Several alternatives that meet the same need are mutually exclusive if, whenever one
of them is selected, the others will be rejected.
When not specified by management or company policy, the analysis period to use in a
comparison of mutually exclusive projects may be chosen by an individual analyst.
Several efficiencies can be applied when an analysis period is selected. In general, the
analysis period should be chosen to cover the required service period, as highlighted in
Figure 5.13.
PROBLEMS
Note: Unless otherwise stated, all cash flows are after-tax cash flows. The interest rate
(MARR) is also given on an after-tax basis.
Identifying Cash Inflows and Outflows
5.1 Camptown Togs, Inc., a children’s clothing manufacturer, has always found payroll processing to be costly because it must be done by a clerk so that the number
of piece-goods coupons received by each employee can be collected and the types
of tasks performed by each employee can be calculated. Not long ago, an industrial engineer designed a system that partially automates the process by means of
a scanner that reads the piece-goods coupons. Management is enthusiastic about
this system because it utilizes some personal computer systems that were purchased recently. It is expected that this new automated system will save $45,000
per year in labor. The new system will cost about $30,000 to build and test prior to
operation. It is expected that operating costs, including income taxes, will be
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250 CHAPTER 5 Present-Worth Analysis
about $5,000 per year. The system will have a five-year useful life. The expected
net salvage value of the system is estimated to be $3,000.
(a) Identify the cash inflows over the life of the project.
(b) Identify the cash outflows over the life of the project.
(c) Determine the net cash flows over the life of the project.
Payback Period
5.2 Refer to Problem 5.1, and answer the following questions:
(a) How long does it take to recover the investment?
(b) If the firm’s interest rate is 15% after taxes, what would be the discounted payback period for this project?
5.3 Consider the following cash flows:
Project’s Cash Flow ($)
B
C
n
A
D
0
- $2,500
- $3,000
- $5,500
- $4,000
1
300
2,000
2,000
5,000
2
300
1,500
2,000
-3,000
3
300
1,500
2,000
-2,500
4
300
500
5,000
1,000
5
300
500
5,000
1,000
6
300
1,500
7
300
8
300
2,000
3,000
(a) Calculate the payback period for each project.
(b) Determine whether it is meaningful to calculate a payback period for project D.
(c) Assuming that i = 10%, calculate the discounted payback period for each
project.
NPW Criterion
5.4 Consider the following sets of investment projects, all of which have a three-year
investment life:
Project’s Cash Flow ($)
B
C
n
A
D
0
- $1,500
- $1,200
- $1,600
- $3,100
1
0
600
-1,800
800
2
0
800
800
1,900
3
3,000
1,500
2,500
2,300
(a) Compute the net present worth of each project at i = 10%.
(b) Plot the present worth as a function of the interest rate (from 0% to 30%) for
project B.
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Problems 251
5.5 You need to know whether the building of a new warehouse is justified under the
following conditions:
The proposal is for a warehouse costing $200,000. The warehouse has an expected
useful life of 35 years and a net salvage value (net proceeds from sale after tax adjustments) of $35,000. Annual receipts of $37,000 are expected, annual maintenance
and administrative costs will be $8,000/year, and annual income taxes are $5,000.
Given the foregoing data, which of the following statements are correct?
(a) The proposal is justified for a MARR of 9%.
(b) The proposal has a net present worth of $152,512 when 6% is used as the
interest rate.
(c) The proposal is acceptable, as long as MARR … 11.81%.
(d) All of the preceding are correct.
5.6 Your firm is considering purchasing an old office building with an estimated remaining service life of 25 years. Recently, the tenants signed long-term leases,
which leads you to believe that the current rental income of $150,000 per year will
remain constant for the first 5 years. Then the rental income will increase by 10%
for every 5-year interval over the remaining life of the asset. For example, the annual rental income would be $165,000 for years 6 through 10, $181,500 for years
11 through 15, $199,650 for years 16 through 20, and $219,615 for years 21
through 25. You estimate that operating expenses, including income taxes, will be
$45,000 for the first year and that they will increase by $3,000 each year thereafter. You also estimate that razing the building and selling the lot on which it
stands will realize a net amount of $50,000 at the end of the 25-year period. If you
had the opportunity to invest your money elsewhere and thereby earn interest at
the rate of 12% per annum, what would be the maximum amount you would be
willing to pay for the building and lot at the present time?
5.7 Consider the following investment project:
n
An
i
0
- $42,000
10%
1
32,400
11
2
33,400
13
3
32,500
15
4
32,500
12
5
33,000
10
Suppose the company’s reinvestment opportunities change over the life of the
project as shown in the preceding table (i.e., the firm’s MARR changes over the
life of the project). For example, the company can invest funds available now at
10% for the first year, 11% for the second year, and so forth. Calculate the net
present worth of this investment and determine the acceptability of the investment.
5.8 Cable television companies and their equipment suppliers are on the verge of installing new technology that will pack many more channels into cable networks,
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252 CHAPTER 5 Present-Worth Analysis
thereby creating a potential programming revolution with implications for broadcasters, telephone companies, and the consumer electronics industry.
Digital compression uses computer techniques to squeeze 3 to 10 programs into
a single channel. A cable system fully using digital compression technology
would be able to offer well over 100 channels, compared with about 35 for the
average cable television system now used. If the new technology is combined
with the increased use of optical fibers, it might be possible to offer as many as
300 channels.
A cable company is considering installing this new technology to increase subscription sales and save on satellite time. The company estimates that the installation will take place over 2 years. The system is expected to have an 8-year service
life and produce the following savings and expenditures:
Digital Compression
Investment
Now
$500,000
First year
$3,200,000
Second year
$4,000,000
Annual savings in satellite time
$2,000,000
Incremental annual revenues due
to new subscriptions
$4,000,000
Incremental annual expenses
$1,500,000
Incremental annual income taxes
$1,300,000
Economic service life
Net salvage value
8 years
$1,200,000
Note that the project has a 2-year investment period, followed by an 8-year service life (a total 10-year life for the project). This implies that the first annual savings will occur at the end of year 3 and the last will occur at the end of year 10. If
the firm’s MARR is 15%, use the NPW method to justify the economic worth of
the project.
5.9 A large food-processing corporation is considering using laser technology to
speed up and eliminate waste in the potato-peeling process. To implement the system, the company anticipates needing $3.5 million to purchase the industrialstrength lasers. The system will save $1,550,000 per year in labor and materials.
However, it will require an additional operating and maintenance cost of
$350,000. Annual income taxes will also increase by $150,000. The system is expected to have a 10-year service life and will have a salvage value of about
$200,000. If the company’s MARR is 18%, use the NPW method to justify the
economics of the project.
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Problems 253
Future Worth and Project Balance
5.10 Consider the following sets of investment projects, all of which have a three-year
investment life:
Period
(n)
Project’s Cash Flow
B
C
A
D
0
- $12,500
-11,000
12,500
-13,000
1
5,400
-3,000
-7,000
5,500
2
14,400
21,000
-2,000
5,500
3
7,200
13,000
4,000
8,500
(a) Compute the net present worth of each project at i = 15%.
(b) Compute the net future worth of each project at i = 15%.
Which project or projects are acceptable?
5.11 Consider the following project balances for a typical investment project with a
service life of four years:
n
An
Project Balance
0
- $1,000
- $1,000
1
()
-1,100
2
()
-800
3
460
-500
4
()
0
(a) Construct the original cash flows of the project.
(b) Determine the interest rate used in computing the project balance.
(c) Would the project be acceptable at i = 15%?
5.12 Your R&D group has developed and tested a computer software package that
assists engineers to control the proper chemical mix for the various processmanufacturing industries. If you decide to market the software, your first-year
operating net cash flow is estimated to be $1,000,000. Because of market competition, product life will be about 4 years, and the product’s market share will
decrease by 25% each year over the previous year’s share. You are approached
by a big software house which wants to purchase the right to manufacture and
distribute the product. Assuming that your interest rate is 15%, for what minimum
price would you be willing to sell the software?
5.13 Consider the accompanying project balance diagram for a typical investment project with a service life of five years. The numbers in the figure indicate the beginning project balances.
(a) From the project balance diagram, construct the project’s original cash flows.
(b) What is the project’s conventional payback period (without interest)?
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254 CHAPTER 5 Present-Worth Analysis
n
An
PB(i)n
PB (i) n
0
0 –$10,000
1
()
2
()
3 $8,000
4
()
5
()
–$10,000
–$11,000
–$8,200
–$1,840
$7,550
$3,792
$3,792
$7,550
–$1,840
–$8,200
–$10,000
–$11,000
0
1
2
3
4
5
6
Year (n)
5.14 Consider the following cash flows and present-worth profile:
Net Cash Flows ($)
Project 1
Project 2
Year
0
- $1,000
- $1,000
1
400
300
2
800
Y
3
X
800
(a) Determine the values of X and Y.
(b) Calculate the terminal project balance of project 1 at MARR = 24%.
(c) Find the values of a, b, and c in the NPW plot.
a
PW (i)
b
24%
ic = ?
0
10
i (%)
Project 1
c
23% Project 2
5.15 Consider the project balances for a typical investment project with a service life of
five years, as shown in Table P5.15.
(a) Construct the original cash flows of the project and the terminal balance, and
fill in the blanks in the preceding table.
(b) Determine the interest rate used in the project balance calculation, and compute the present worth of this project at the computed interest rate.
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Problems 255
TABLE P5.15
Investment Project Balances
n
An
Project Balance
0
1
2
3
4
5
- $3,000
- $3,000
-2,700
-1,500
0
-300
1,470
600
5.16 Refer to Problem 5.3, and answer the following questions:
(a) Graph the project balances (at i = 10%) of each project as a function of n.
(b) By examining the graphical results in part (a), determine which project appears
to be the safest to undertake if there is some possibility of premature termination
of the projects at the end of year 2.
5.17 Consider the following investment projects:
Project’s Cash Flow
C
D
n
A
B
E
0
- $1,800
- $5,200
- $3,800
- $4,000
- $6,500
1
-500
2,500
0
500
1,000
2
900
-4,000
0
2,000
3,600
3
1,300
5,000
4,000
3,000
2,400
4
2,200
6,000
7,000
4,000
5
-700
3,000
12,000
1,250
(a) Compute the future worth at the end of life for each project at i = 12%.
(b) Determine the acceptability of each project.
5.18 Refer to Problem 5.17, and answer the following questions:
(a) Plot the future worth for each project as a function of the interest rate (0%–50%).
(b) Compute the project balance of each project at i = 12%.
(c) Compare the terminal project balances calculated in (b) with the results obtained in Problem 5.17(a). Without using the interest factor tables, compute the
future worth on the basis of the project balance concept.
5.19 Covington Corporation purchased a vibratory finishing machine for $20,000 in
year 0. The useful life of the machine is 10 years, at the end of which the machine
is estimated to have a salvage value of zero. The machine generates net annual revenues of $6,000. The annual operating and maintenance expenses are estimated to
be $1,000. If Covington’s MARR is 15%, how many years will it take before this
machine becomes profitable?
5.20 Gene Research, Inc., just finished a 4-year program of R&D and clinical trial. It expects a quick approval from the Food and Drug Administration. If Gene markets
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256 CHAPTER 5 Present-Worth Analysis
the product its own, the company will require $30 million immediately 1n = 02 to
build a new manufacturing facility, and it is expected to have a 10-year product life.
The R&D expenditure in the previous years and the anticipated revenues that the
company can generate over the next 10 years are summarized as follows:
Period (n)
Cash Flow
(Unit: $ million)
-4
–$10
-3
–10
-2
–10
-1
–10
0
-10 - 30
1–10
100
Merck, a large drug company is interested, in purchasing the R&D project and the
right to commercialize the product from Gene Research, Inc.; it wants to do so
immediately 1n = 02. What would be a starting negotiating price for the project
from Merck? Assume that Gene’s MARR = 20%.
5.21 Consider the following independent investment projects:
Project Cash Flows
B
C
n
A
0
- $400
- $300
$100
1
150
140
-40
2
150
140
-40
3
350
140
-40
4
-200
110
5
400
110
6
300
Assume that MARR = 10%, and answer the following questions:
(a) Compute the net present worth for each project, and determine the acceptability
of each.
(b) Compute the net future worth of each project at the end of each project period,
and determine the acceptability of each project.
(c) Compute the project worth of each project at the end of six years with variable
MARRs as follows: 10% for n = 0 to n = 3 and 15% for n = 4 to n = 6.
5.22 Consider the project balance profiles shown in Table P5.22 for proposed investment
projects. Project balance figures are rounded to nearest dollars.
(a) Compute the net present worth of projects A and C.
(b) Determine the cash flows for project A.
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(c) Identify the net future worth of project C.
(d) What interest rate would be used in the project balance calculations for project B?
TABLE P5.22
Profiles for Proposed Investment Projects
n
A
Project Balances
B
0
- $1,000
- $1,000
- $1,000
1
-1,000
-650
-1,200
2
-900
-348
-1,440
3
-690
-100
-1,328
4
-359
85
-1,194
5
105
198
-1,000
Interest
rate used
NPW
10%
?
?
$79.57
C
20%
?
5.23 Consider the following project balance profiles for proposed investment projects:
Project Balances
B
n
A
0
- $1,000
- $1,000
- $1,000
1
-800
-680
-530
2
-600
-302
X
3
-400
-57
-211
4
-200
233
-89
5
0
575
0
Interest
rate used
10%
18%
C
12%
Project balance figures are rounded to the nearest dollar.
(a) Compute the net present worth of each investment.
(b) Determine the project balance for project C at the end of period 2 if A 2 = $500.
(c) Determine the cash flows for each project.
(d) Identify the net future worth of each project.
Capitalized Equivalent Worth
5.24 Maintenance money for a new building has been sought. Mr. Kendall would like
to make a donation to cover all future expected maintenance costs for the building.
These maintenance costs are expected to be $50,000 each year for the first 5 years,
$70,000 each year for years 6 through 10, and $90,000 each year after that. (The
building has an indefinite service life.)
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(a) If the money is placed in an account that will pay 13% interest compounded
annually, how large should the gift be?
(b) What is the equivalent annual maintenance cost over the infinite service life of
the building?
5.25 Consider an investment project, the cash flow pattern of which repeats itself every
five years forever as shown in the accompanying diagram. At an interest rate of
14%, compute the capitalized equivalent amount for this project.
$100
$100
$40
$100
$40
$20
$20
0 1 2 3 4 5 6 7 8 9 10 11 12
Years
5.26 A group of concerned citizens has established a trust fund that pays 6% interest,
compounded monthly, to preserve a historical building by providing annual maintenance funds of $30,000 forever. Compute the capitalized equivalent amount for
these building maintenance expenses.
5.27 A newly constructed bridge costs $5,000,000. The same bridge is estimated to
need renovation every 15 years at a cost of $1,000,000. Annual repairs and maintenance are estimated to be $100,000 per year.
(a) If the interest rate is 5%, determine the capitalized cost of the bridge.
(b) Suppose that, in (a), the bridge must be renovated every 20 years, not every 15
years. What is the capitalized cost of the bridge?
(c) Repeat (a) and (b) with an interest rate of 10%. What have you to say about the
effect of interest on the results?
5.28 To decrease the costs of operating a lock in a large river, a new system of operation is proposed. The system will cost $650,000 to design and build. It is estimated that it will have to be reworked every 10 years at a cost of $100,000. In
addition, an expenditure of $50,000 will have to be made at the end of the fifth
year for a new type of gear that will not be available until then. Annual operating
costs are expected to be $30,000 for the first 15 years and $35,000 a year thereafter. Compute the capitalized cost of perpetual service at i = 8%.
Mutually Exclusive Alternatives
5.29 Consider the following cash flow data for two competing investment projects:
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n
Cash Flow Data
(Unit: $ thousand)
Project A
Project B
0
- $800
- $2,635
1
-1,500
-565
2
-435
820
3
775
820
4
775
1,080
5
1,275
1,880
6
1,275
1,500
7
975
980
8
675
580
9
375
380
10
660
840
At i = 12%, which of the two projects would be a better choice?
5.30 Consider the cash flows for the following investment projects.
Project’s Cash Flow
B
C
D
n
A
E
0
- $1,500
- $1,500
- $3,000
$1,500
- $1,800
1
1,350
1,000
1,000
-450
600
2
800
800
X
-450
600
3
200
800
1,500
-450
600
4
100
150
X
-450
600
(a) Suppose projects A and B are mutually exclusive. On the basis of the NPW
criterion, which project would be selected? Assume that MARR = 15%.
(b) Repeat (a), using the NFW criterion.
(c) Find the minimum value of X that makes project C acceptable.
(d) Would you accept project D at i = 18%?
(e) Assume that projects D and E are mutually exclusive. On the basis of the NPW
criterion, which project would you select?
5.31 Consider two mutually exclusive investment projects, each with MARR = 12%,
as shown in Table 5.31.
(a) On the basis of the NPW criterion, which alternative would be selected?
(b) On the basis of the NFW criterion, which alternative would be selected?
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TABLE P5.31
Two Mutually Exclusive
Investment Projects
Project’s Cash Flow
A
B
n
0
- $14,500
- $12,900
1
12,610
11,210
2
12,930
11,720
3
12,300
11,500
5.32 Consider the following two mutually exclusive investment projects, each with
MARR = 15%:
n
Project’s Cash Flow
A
B
0
- $6,000
- $8,000
1
800
11,500
2
14,000
400
(a) On the basis of the NPW criterion, which project would be selected?
(b) Sketch the PW(i) function for each alternative on the same chart between 0%
and 50%. For what range of i would you prefer project B?
5.33 Two methods of carrying away surface runoff water from a new subdivision are
being evaluated:
Method A. Dig a ditch. The first cost would be $60,000, and $25,000 of redigging
and shaping would be required at five-year intervals forever.
Method B. Lay concrete pipe. The first cost would be $150,000, and a replacement would be required at 50-year intervals at a net cost of $180,000 forever.
At i = 12%, which method is the better one?
5.34 A local car dealer is advertising a standard 24-month lease of $1,150 per month
for its new XT 3000 series sports car. The standard lease requires a down payment
of $4,500, plus a $1,000 refundable initial deposit now. The first lease payment is
due at the end of month 1. In addition, the company offers a 24-month lease plan
that has a single up-front payment of $30,500, plus a refundable initial deposit of
$1,000. Under both options, the initial deposit will be refunded at the end of
month 24. Assume an interest rate of 6% compounded monthly. With the presentworth criterion, which option is preferred?
5.35 Two machines are being considered for a manufacturing process. Machine A has a
first cost of $75,200, and its salvage value at the end of six years of estimated
service life is $21,000. The operating costs of this machine are estimated to be
$6,800 per year. Extra income taxes are estimated at $2,400 per year. Machine B
has a first cost of $44,000, and its salvage value at the end of six years’ service is
estimated to be negligible. The annual operating costs will be $11,500. Compare
these two mutually exclusive alternatives by the present-worth method at i = 13%.
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5.36 An electric motor is rated at 10 horsepower (HP) and costs $800. Its full load efficiency is specified to be 85%. A newly designed high-efficiency motor of the
same size has an efficiency of 90%, but costs $1,200. It is estimated that the
motors will operate at a rated 10 HP output for 1,500 hours a year, and the cost of
energy will be $0.07 per kilowatt-hour. Each motor is expected to have a 15-year
life. At the end of 15 years, the first motor will have a salvage value of $50 and the
second motor will have a salvage value of $100. Consider the MARR to be 8%.
(Note: 1 HP = 0.7457 kW.)
(a) Use the NPW criterion to determine which motor should be installed.
(b) In (a), what if the motors operated 2,500 hours a year instead of 1,500 hours a
year? Would the motor you chose in (a) still be the choice?
5.37 Consider the following two mutually exclusive investment projects:
n
Project’s Cash Flow
A
B
0
- $20,000
- $25,000
1
17,500
25,500
2
17,000
18,000
3
15,000
On the basis of the NPW criterion, which project would be selected if you use an
infinite planning horizon with project repeatability (the same costs and benefits)
likely? Assume that i = 12%.
5.38 Consider the following two mutually exclusive investment projects, which have
unequal service lives:
n
Project’s Cash Flow
A1
A2
0
- $900
- $1,800
1
-400
-300
2
-400
-300
3
-400 + 200
-300
4
-300
5
-300
6
-300
7
-300
8
-300 + 500
(a) What assumption(s) do you need in order to compare a set of mutually exclusive investments with unequal service lives?
(b) With the assumption(s) defined in (a) and using i = 10%, determine which
project should be selected.
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(c) If your analysis period (study period) is just three years, what should be the
salvage value of project A2 at the end of year 3 to make the two alternatives
economically indifferent?
5.39 Consider the following two mutually exclusive projects:
B1
B2
n
Cash
Flow
Salvage
Value
Cash
Flow
Salvage
Value
0
- $18,000
1
-2,000
6,000
-2,100
6,000
2
-2,000
4,000
-2,100
3,000
3
-2,000
3,000
-2,100
1,000
4
-2,000
2,000
5
-2,000
2,000
- $15,000
Salvage values represent the net proceeds (after tax) from disposal of the assets if
they are sold at the end of each year. Both B1 and B2 will be available (or can be
repeated) with the same costs and salvage values for an indefinite period.
(a) Assuming an infinite planning horizon, which project is a better choice at
MARR = 12%?
(b) With a 10-year planning horizon, which project is a better choice at
MARR = 12%?
5.40 Consider the following cash flows for two types of models:
n
Project’s Cash Flow
Model A
Model B
0
- $6,000
- $15,000
1
3,500
10,000
2
3,500
10,000
3
3,500
Both models will have no salvage value upon their disposal (at the end of their respective service lives). The firm’s MARR is known to be 15%.
(a) Notice that the models have different service lives. However, model A will be
available in the future with the same cash flows. Model B is available at one
time only. If you select model B now, you will have to replace it with model A
at the end of year 2. If your firm uses the present worth as a decision criterion,
which model should be selected, assuming that the firm will need either model
for an indefinite period?
(b) Suppose that your firm will need either model for only two years. Determine the
salvage value of model A at the end of year 2 that makes both models indifferent
(equally likely).
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5.41 An electric utility is taking bids on the purchase, installation, and operation of microwave towers. Following are some details associated with the two bids that were
received:
Cost per Tower
Bid A
Bid B
Equipment cost
$112,000
$98,000
Installation cost
$25,000
$30,000
$2,000
$2,500
Annual maintenance
and inspection fee
Annual extra income taxes
$800
Life
40 years
35 years
$0
$0
Salvage value
Which is the most economical bid if the interest rate is considered to be 11%? Either tower will have no salvage value after 20 years of use.
Use the NPW method to compare these two mutually exclusive plans.
5.42 Consider the following two investment alternatives:
n
Project’s Cash Flow
A1
A2
0
- $15,000
- $25,000
1
9,500
0
2
12,500
X
3
7,500
X
PW(15%)
?
9,300
The firm’s MARR is known to be 15%.
(a) Compute PW(15%) for A1.
(b) Compute the unknown cash flow X in years 2 and 3 for A2.
(c) Compute the project balance (at 15%) of A1 at the end of period 3.
(d) If these two projects are mutually exclusive alternatives, which one would you
select?
5.43 Consider each of the after-tax cash flows shown in Table P5.43.
(a) Compute the project balances for projects A and D as a function of the project
year at i = 10%.
(b) Compute the net future-worth values for projects A and D at i = 10%.
(c) Suppose that projects B and C are mutually exclusive. Suppose also that the
required service period is eight years and that the company is considering leasing comparable equipment with an annual lease expense of $3,000 for the remaining years of the required service period. Which project is a better choice?
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TABLE P5.43
After-Tax Cash Flows
Project’s Cash Flow
B
C
n
A
D
0
- $2,500
- $7,000
- $5,000
- $5,000
1
650
-2,500
-2,000
-500
2
650
-2,000
-2,000
-500
3
650
-1,500
-2,000
4,000
4
600
-1,500
-2,000
3,000
5
600
-1,500
-2,000
3,000
6
600
-1,500
-2,000
2,000
7
300
-2,000
3,000
8
300
5.44 A mall with two levels is under construction. The plan is to install only 9 escalators at the start, although the ultimate design calls for 16. The question arises as to
whether to provide necessary facilities (stair supports, wiring conduits, motor
foundations, etc.) that would permit the installation of the additional escalators at
the mere cost of their purchase and installation or to defer investment in these facilities until the escalators need to be installed.
• Option 1. Provide these facilities now for all 7 future escalators at $200,000.
• Option 2. Defer the investment in the facility as needed. Install 2 more escalators in two years, 3 more in five years, and the last 2 in eight years. The installation of these facilities at the time they are required is estimated to cost $100,000
in year 2, $160,000 in year 5, and $140,000 in year 8.
Additional annual expenses are estimated at $3,000 for each escalator facility
installed. At an interest rate of 12%, compare the net present worth of each option over eight years.
5.45 An electrical utility is experiencing a sharp power demand that continues to grow
at a high rate in a certain local area.
Two alternatives are under consideration. Each is designed to provide enough capacity during the next 25 years, and both will consume the same amount of fuel,
so fuel cost is not considered in the analysis.
• Alternative A. Increase the generating capacity now so that the ultimate demand
can be met with additional expenditures later. An initial investment of $30 million
would be required, and it is estimated that this plant facility would be in service
for 25 years and have a salvage value of $0.85 million. The annual operating and
maintenance costs (including income taxes) would be $0.4 million.
• Alternative B. Spend $10 million now and follow this expenditure with future
additions during the 10th year and the 15th year. These additions would cost
$18 million and $12 million, respectively. The facility would be sold 25 years
from now, with a salvage value of $1.5 million. The annual operating and maintenance costs (including income taxes) initially will be $250,000 and will increase
to $0.35 million after the second addition (from the 11th year to the 15th year)
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and to $0.45 million during the final 10 years. (Assume that these costs begin 1
year subsequent to the actual addition.)
On the basis of the present-worth criterion, if the firm uses 15% as a MARR,
which alternative should be undertaken?
5.46 A large refinery–petrochemical complex is to manufacture caustic soda, which
will use feedwater of 10,000 gallons per day. Two types of feedwater storage installation are being considered over the 40 years of their useful life.
Option 1. Build a 20,000-gallon tank on a tower. The cost of installing the tank and
tower is estimated to be $164,000. The salvage value is estimated to be negligible.
Option 2. Place a tank of 20,000-gallon capacity on a hill, which is 150 yards
away from the refinery. The cost of installing the tank on the hill, including the
extra length of service lines, is estimated to be $120,000, with negligible salvage
value. Because of the tank’s location on the hill, an additional investment of
$12,000 in pumping equipment is required. The pumping equipment is expected
to have a service life of 20 years, with a salvage value of $1,000 at the end of that
time. The annual operating and maintenance cost (including any income tax effects) for the pumping operation is estimated at $1,000.
If the firm’s MARR is known to be 12%, which option is better, on the basis of the
present-worth criterion?
Short Case Studies
ST5.1 Apex Corporation requires a chemical finishing process for a product under contract for a period of six years. Three options are available. Neither Option 1 nor
Option 2 can be repeated after its process life. However, Option 3 will always be
available from H&H Chemical Corporation at the same cost during the period that
the contract is operative. Here are the options:
• Option 1. Process device A, which costs $100,000, has annual operating and
labor costs of $60,000 and a useful service life of four years with an estimated
salvage value of $10,000.
• Option 2. Process device B, which costs $150,000, has annual operating and
labor costs of $50,000 and a useful service life of six years with an estimated
salvage value of $30,000.
• Option 3. Subcontract out the process at a cost of $100,000 per year.
According to the present-worth criterion, which option would you recommend at
i = 12%?
ST5.2 Tampa Electric Company, an investor-owned electric utility serving approximately
2,000 square miles in west central Florida, was faced with providing electricity to a
newly developed industrial park complex. The distribution engineering department
needs to develop guidelines for the design of the distribution circuit. The “main
feeder,” which is the backbone of each 13-kV distribution circuit, represents a substantial investment by the company.5
5
Example provided by Andrew Hanson from Tampa Electric Company.
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Tampa Electric has four approved main feeder construction configurations—
(1) crossarm, (2) vertical (horizontal line post), (3) vertical (standoff pin), and
(4) triangular, as illustrated in the accompanying figure. The width of the easement sought depends on the planned construction configuration. If crossarm
construction is planned, a 15-foot easement is sought. A 10-foot wide easement
is sought for vertical and triangular configurations.
Crossarm
Vertical
(Standoff pin)
Triangular
Vertical
(Horizontal line post)
Once the required easements are obtained, the line clearance department clears
any foliage that would impede the construction of the line. The clearance cost is dictated by the typical tree densities along road rights-of-way. The average cost to trim
1 tree is estimated at $20, and the average tree density in the service area is estimated to be 75 trees per mile. The costs of each type of construction are as follows:
Design Configurations
Triangular Horizontal Line
Factors
Crossarm
Easements
$487,000
$388,000
$388,000
$388,000
$613
$1,188
$1,188
$1,188
$7,630
$7,625
$12,828
$8,812
Line clearance
Line construction
Standoff
Additional factors to consider in selecting the best main feeder configuration are
as follows: In certain sections of Tampa Electric’s service territory, ospreys often
nest on transmission and distribution poles. The nests reduce the structural and
electrical integrity of the poles. Crossarm construction is most vulnerable to osprey nesting, since the crossarm and braces provide a secure area for construction
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of the nest. Vertical and triangular construction do not provide such spaces. Furthermore, in areas where ospreys are known to nest, vertical and triangular configuration have added advantages. The insulation strength of a construction
configuration may favorably or adversely affect the reliability of the line for
which the configuration is used. A common measure of line insulation strength is
the critical flashover (CFO) voltage. The greater the CFO, the less susceptible the
line is to nuisance flashovers from lightning and other electrical phenomena.
The utility’s existing inventory of crossarms is used primarily for main feeder
construction and maintenance. The use of another configuration for main feeder
construction would result in a substantial reduction in the inventory of crossarms.
The line crews complain that line spacing on vertical and triangular construction
is too restrictive for safe live line work. Each accident would cost $65,000 in lost
work and other medical expenses. The average cost of each flashover repair would
be $3,000. The following table lists the values of the factors involved in the four
design configurations:
Factors
Crossarm
Nesting
Severe
Insulation
strength, CFO (kV)
None
None
Standoff
None
387
474
476
462
2
1
1
1
$4,521
$4,521
$4,521
Problem
Problem
Problem
Annual flashover
occurrence (n)
Annual inventory
savings
Safety
Design Configurations
Triangular
Horizontal Line
OK
All configurations would last about 20 years, with no salvage value. It appears that
noncrossarm designs are better, but engineers need to consider other design factors,
such as safety, rather than just monetary factors when implementing the project. It
is true that the line spacing on triangular construction is restrictive. However, with
a better clearance design between phases for vertical construction, the hazard
would be minimized. In the utility industry, the typical opposition to new types of
construction is caused by the confidence acquired from constructing lines in the
crossarm configuration for many years. As more vertical and triangular lines are
built, opposition to these configurations should decrease. Which of the four designs described in the preceding table would you recommend to the management?
Assume Tampa Electric’s interest rate to be 12%.
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SIX
CHAPTER
Annual Equivalent-Worth
Analysis
Thermally Activated Technologies: Absorption Chillers
for Buildings1 Absorption chillers provide cooling to buildings by
using heat.This seemingly paradoxical, but highly efficient, technology is
most cost effective in large facilities with significant heat loads. Not
only do absorption chillers use less energy than conventional equipment
does, but they also cool buildings without the use of ozone-depleting
chlorofluorocarbons (CFCs). Unlike conventional electric chillers,
which use mechanical energy in a vapor compression process to provide refrigeration, absorption chillers primarily use heat energy, with
limited mechanical energy for pumping. Absorption chillers can be
powered by natural gas, steam, or waste heat.
• The most promising markets for absorption chillers are in commercial buildings, government facilities, college campuses, hospital
complexes, industrial parks, and municipalities.
• Absorption chillers generally become economically attractive when
there is a source of inexpensive thermal energy at temperatures
between 212°F and 392°F.
An absorption chiller transfers thermal energy from the heat
source to the heat sink through an absorbent fluid and a refrigerant.
The absorption chiller accomplishes its refrigerative effect by absorbing and then releasing water vapor into and out of a lithium
bromide solution. Absorption chiller systems are classified by single-,
double-, or triple-stage effects, which indicate the number of
generators in the given system.The greater the number of stages,
the higher is the overall efficiency of the system. Double-effect absorption chillers typically have a higher first cost, but a significantly
lower energy cost, than single-effect chillers, resulting in a lower net
present worth.
268
1
Tech Brief, Office of Power Technology, U.S. Department of Energy, Washington, DC.
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Single-Effect Absorption Chiller
Refrigerant vapor
Separator
Condenser
Orifice
Evaporator
Lift pipe
Chilled water
Heat medium
Cooling water
Generator
Heat
Exchanger
Liquid refrigerant
Concentrated lithium
bromide solution
Dilute lithium bromide/
refrigerant solution
Absorber
Chilled water
Cooling water
Heat medium
(solar or gas)
Some of the known economic benefits of the absorption chiller over the
conventional mechanical chiller are as follows: 2
• In a plant where low-pressure steam is currently being vented to the atmosphere, a mechanical chiller with a Coefficient of Performance (COP) of 4.0 is
used 4,000 hours a year to produce an average of 300 tons of refrigeration.
• The plant’s cost of electricity is $0.05 a kilowatt-hour. An absorption
unit requiring 5,400 lb/hr of 15-psig steam could replace the mechanical
chiller, providing the following annual electrical cost savings:
Annual Savings = 300 tons * (12,000 Btu>ton>4.0) * 4,000 hrs>yr
* $0.05>kWh * kWh>3,413 Btu = $52,740.
2
Source: EcoGeneration Solutions™, LLC, Companies, 12615 Jones Rd., Suite 209, Houston, Texas 77070
(http://www.cogeneration.net/Absorption_Chillers.htm).
269
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Suppose you plan to install the chiller and expect to operate it continuously for 10
years. How would you calculate the operating cost per hour? Suppose you are considering buying a new car. If you expect to drive 12,000 miles per year, can you figure out how
much the car costs per mile? You would have good reason to want to know the cost if you
were being reimbursed by your employer on a per mile basis for the business use of your
car. Or consider a real-estate developer who is planning to build a 500,000-square-foot
shopping center. What would be the minimum annual rental fee per square foot required
to recover the initial investment?
Annual equivalence analysis is the method by which these and other unit costs (or
profits) are calculated. Along with present-worth analysis, annual equivalence analysis is
the second major equivalence technique for putting alternatives on a common basis of
comparison. In this chapter, we develop the annual equivalent-worth criterion and
demonstrate a number of situations in which annual equivalence analysis is preferable to
other methods of comparison.
CHAPTER LEARNING OBJECTIVES
After completing this chapter, you should understand the following concepts:
How to determine the equivalent annual worth (cost) for a given project.
Why the annual equivalent approach facilitates the comparison of unequal
service life problems.
How to determine the capital cost (or ownership cost) when you purchase
an asset.
How to determine the unit cost or unit profit.
How to conduct a life-cycle cost analysis.
How to optimize design parameters in engineering design.
6.1 Annual Equivalent-Worth Criterion
n this section, we set forth a fundamental decision rule based on annual equivalent
worth by considering both revenue and cost streams of a project. If revenue streams
are irrelevant, then we make a decision solely on the basis of cost. This leads to a
popular decision tool known as “life-cycle cost analysis,” which we will discuss in
Section 6.4.
I
6.1.1 Fundamental Decision Rule
The annual equivalent worth (AE) criterion provides a basis for measuring the worth of
an investment by determining equal payments on an annual basis. Knowing that any
lump-sum cash amount can be converted into a series of equal annual payments, we may
first find the net present worth (NPW) of the original series and then multiply this amount
by the capital recovery factor:
AE1i2 = PW1i21A>P, i, N2.
(6.1)
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Section 6.1 Annual Equivalent-Worth Criterion 271
• Single-project evaluation: The accept–reject selection rule for a single revenue
project is as follows:
If AE1i2 7 0, accept the investment.
If AE1i2 = 0, remain indifferent to the investment.
If AE1i2 6 0, reject the investment.
Notice that the factor (A/P, i, N) in Eq. (6.1) is positive for -1 6 i 6 q , which indicates that the value of AE(i) will be positive if, and only if, PW(i) is positive. In
other words, accepting a project that has a positive AE(i) is equivalent to accepting a
project that has a positive PW(i). Therefore, the AE criterion for evaluating a project
is consistent with the NPW criterion.
• Comparing mutually exclusive alternatives: As with present-worth analysis,
when you compare mutually exclusive service projects whose revenues are the
same, you may compare them on a cost-only basis. In this situation, the alternative
with the minimum annual equivalent cost (or least negative annual equivalent
worth) is selected.
Example 6.1 illustrates how to find the equivalent annual worth for a proposed energysavings project. As you will see, you first calculate the net present worth of the project and
then convert this present worth into an equivalent annual basis.
EXAMPLE 6.1 Annual Equivalent Worth: A Single-Project
Evaluation
A utility company is considering adding a second feedwater heater to its existing system unit to increase the efficiency of the system and thereby reduce fuel costs. The
150-MW unit will cost $1,650,000 and has a service life of 25 years. The expected
salvage value of the unit is considered negligible. With the second unit installed, the
efficiency of the system will improve from 55% to 56%. The fuel cost to run the
feedwater is estimated at $0.05 kWh. The system unit will have a load factor of 85%,
meaning that the system will run 85% of the year.
(a) Determine the equivalent annual worth of adding the second unit with an interest
rate of 12%.
(b) If the fuel cost increases at the annual rate of 4% after first year, what is the
equivalent annual worth of having the second feedwater unit at i = 12%?
DISCUSSION: Whenever we compare machines with different efficiency ratings, we
need to determine the input powers required to operate the machines. Since the percent
efficiency is equal to the ratio of the output power to the input power, we can determine
the input power by dividing the output power by the motor’s percent efficiency:
Input power =
output power
.
percent efficiency
A feedwater
heater is a
power-plant
component used
to preheat water
delivered to a
boiler. Preheating
the feedwater
reduces the
amount of
energy needed
to make steam
and thus reduces
plant operation
costs.
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272 CHAPTER 6 Annual Equivalent-Worth Analysis
For example, a 30-HP motor with 90% efficiency will require an input power of
130 HP * 0.746 kW>HP2
0.90
= 24.87 kW.
Input power =
Therefore, energy consumption with and without the second unit can be calculated as
follows:
150,000 kW
= 272,727 kW
0.55
150,000 kW
= 267,857 kW
• After adding the second unit,
0.56
• Before adding the second unit,
So the reduction in energy consumption is 4,871 kW.
Since the system unit will operate only 85% of the year, the total annual operating
hours are calculated as follows:
Annual operating hours = 13652124210.852 = 7,446 hours/year.
SOLUTION
Given: I = $1,650,000, N = 25 years, S = 0, annual fuel savings, and i = 12%.
Find: AE of fuel savings due to improved efficiency.
(a) With the assumption of constant fuel cost over the service life of the second
heater,
A fuel savings = 1reduction in fuel requirement2 * 1fuel cost2
* 1operating hours per year2
= a
150,000 kW
150,000 kW
b * 1$0.05/kWh2
0.55
0.56
* 118,760210.852 hours/year2
= 14,871 kW2 * 1$0.05/kWh2 * 17,446 hours/year2
= $1,813,473/year;
PW112%2 = - $1,650,000 + $1,813,4731P/A, 12%, 252
= $12,573,321;
AE112%2 = $12,573,3211A/P, 12%, 252
= $1,603,098.
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Section 6.1 Annual Equivalent-Worth Criterion 273
(b) With the assumption of escalating energy cost at the annual rate of 4%, since the
first year’s fuel savings is already calculated in (a), we use it as A 1 in the geometric-gradient-series present-worth factor (P/A1 , g, i, N):
A 1 = $1,813,473
PW112%2 = - $1,650,000 + $1,813,4731P/A 1 , 4%, 12%, 252
= $17,463,697
AE112%2 = $17,463,6971A/P, 12%, 252
= $2,226,621
Clearly, either situation generates enough fuel savings to justify adding the
second unit of the feedwater. Figure 6.1 illustrates the cash flow series associated with the required investment and fuel savings over the heater’s service life
of 25 years.
A $1,813,473
0
(a) Constant fuel price
1 2 3 4 5
24 25
$1,650,000
g 4%
A1 $1,813,473
0
(b) Escalating fuel price
1 2 3 4 5
24 25
$1,650,000
Figure 6.1
Cash flow diagram (Example 6.1).
6.1.2 Annual-Worth Calculation with Repeating
Cash Flow Cycles
In some situations, a cyclic cash flow pattern may be observed over the life of the
project. Unlike the situation in Example 6.1, where we first computed the NPW of
the entire cash flow and then calculated the AE, we can compute the AE by examining the first cash flow cycle. Then we can calculate the NPW for the first cash flow
cycle and derive the AE over that cycle. This shortcut method gives the same solution
when the NPW of the entire project is calculated, and then the AE can be computed
from this NPW.
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274 CHAPTER 6 Annual Equivalent-Worth Analysis
EXAMPLE 6.2 Annual Equivalent Worth:
Repeating Cash Flow Cycles
SOLEX Company is producing electricity directly from a solar source by using a
large array of solar cells and selling the power to the local utility company. SOLEX
decided to use amorphous silicon cells because of their low initial cost, but these
cells degrade over time, thereby resulting in lower conversion efficiency and power
output. The cells must be replaced every four years, which results in a particular cash
flow pattern that repeats itself as shown in Figure 6.2. Determine the annual equivalent cash flows at i = 12%.
Revenue declines as cells
degrade
$800
$700
$600
First replacement
$500
$800
0
1
2
3
$700
4
$600
Second replacement
$500
$800
$1,000
Initial cost
4
5
6
7
$700
$600
8
$1,000
Cost to
replace cells —
same as
initial cost
8
(Unit: Thousand dollars)
9
10
11
$500
12
12
$1,000
$1,000
0
1
2
3
4
5
6
7
8
9
10
11
12
Figure 6.2 Conversion of repeating cash flow cycles into an equivalent annual payment
(Example 6.2).
SOLUTION
Given: Cash flows in Figure 6.2 and i = 12%.
Find: Annual equivalent benefit.
To calculate the AE, we need only consider one cycle over the four-year replacement
period of the cells. For i = 12%, we first obtain the NPW for the first cycle as
follows:
PW112%2 = - $1,000,000
+ [1$800,000 - $100,0001A>G, 12%, 42]1P>A, 12%, 42
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Section 6.1 Annual Equivalent-Worth Criterion 275
= - $1,000,000 + $2,017,150.
= $1,017,150.
Then we calculate the AE over the four-year life cycle:
AE112%2 = $1,017,1501A>P, 12%, 42
= $334,880.
We can now say that the two cash flow series are equivalent:
Original Cash Flows
n
An
Annual Equivalent Flows
n
An
0
- $1,000,000
0
0
1
800,000
1
$334,880
2
700,000
2
334,880
3
600,000
3
334,880
4
500,000
4
334,880
We can extend this equivalency over the remaining cycles of the cash flow. The
reasoning is that each similar set of five values (one disbursement and four receipts)
is equivalent to four annual receipts of $334,880 each. In other words, the $1 million
investment in the solar project will recover the entire investment and generate equivalent annual savings of $334,880 over a four-year life cycle.
6.1.3 Comparing Mutually Exclusive Alternatives
In this section, we consider a situation in which two or more mutually exclusive alternatives need to be compared on the basis of annual equivalent worth. In Section 5.5, we discussed the general principles that should be applied when mutually exclusive alternatives
with unequal service lives were compared. The same general principles should be applied
in comparing mutually exclusive alternatives on the basis of annual equivalent worth:
Mutually exclusive alternatives in equal time spans must be compared. Therefore, we
must give careful consideration to the period covered by the analysis: the analysis period. We will consider two situations: (1) The analysis period equals project lives and
(2) the analysis period differs from project lives.
With situation (1), we compute AE for each project and select the project that has the
least negative AE for service projects (or the largest AE for revenue projects). With situation (2), we need to consider the issue of unequal project lives. As we saw in Chapter 5,
comparing projects with unequal service lives is complicated by the need to determine
the lowest common multiple life. For the special situation of an indefinite service period
and replacement with identical projects, we can avoid this complication by the use of AE
analysis, provided that the following criteria are met:
1. The service of the selected alternative is required on a continuous basis.
2. Each alternative will be replaced by an identical asset that has the same costs and
performance.
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276 CHAPTER 6 Annual Equivalent-Worth Analysis
When these two criteria are met, we may solve for the AE of each project on the basis
of its initial life span, rather than on that of the lowest common multiple of the projects’
lives. Example 6.3 illustrates the process of comparing unequal service projects.
EXAMPLE 6.3 Annual Equivalent Cost Comparison:
Unequal Project Lives
Consider again Example 5.14, in which we compared two types of equipment with
unequal service lives. Apply the annual equivalent approach to select the most economical equipment.
SOLUTION
Given: Cost cash flows shown in Figure 6.3 and i = 15% per year.
Find: AE cost and which alternative is the preferred one.
Model A
Years
0
1
2
3
4
5
$3,000
$5,000
6
7
$3,000
$5,000
9
8
10 11 12
$3,000
$5,000
0 1 2 3 4 5 6 7 8 9 10 11 12
$3,000
$5,000
A = $9,899
$12,500
$12,500
$12,500
$12,500
Model B
Years
0
1
2
3
4
5
6
$2,500
$4,000
7
8
9
10 11 12
$2,500
$4,000
0 1 2 3 4 5 6 7 8 9 10 11 12
$2,500
$4,000
A = $8,954
$15,000
$15,000
$15,000
Figure 6.3 Comparison of projects with unequal lives and an indefinite analysis period
using the annual equivalent-worth criterion (Example 6.3).
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Section 6.2 Capital Costs versus Operating Costs
277
An alternative procedure for solving Example 5.14 is to compute the annual equivalent cost of an outlay of $12,500 for model A every 3 years and the annual equivalent
cost of an outlay of $15,000 for model B every 4 years. Notice that the AE of each
12-year cash flow is the same as that of the corresponding 3- or 4-year cash flow
(Figure 6.3). From Example 5.14, we calculate
• Model A:
For a 3-year life,
PW115%2 = $22,601
AE115%2 = 22,6011A>P, 15%, 32
= $9,899.
For a 12-year period (the entire analysis period),
PW115%2 = $53,657
AE115%2 = 53,6571A>P, 15%, 122
= $9,899.
• Model B: For a 4-year life,
PW115%2 = $25,562
AE115%2 = $25,5621A>P, 15%, 42
= $8,954.
For a 12-year period (the entire analysis period),
PW115%2 = $48,534
AE115%2 = $48,5341A>P, 15%, 122
= $8,954.
Notice that the annual equivalent values that were calculated on the basis of the common service period are the same as those which were obtained over their initial life
spans. Thus, for alternatives with unequal lives, we will obtain the same selection by
comparing the NPW over a common service period using repeated projects or by
comparing the AE for initial lives.
6.2 Capital Costs versus Operating Costs
When only costs are involved, the AE method is sometimes called the annual equivalent
cost (AEC) method. In this case, revenues must cover two kinds of costs: operating
costs and capital costs. Operating costs are incurred through the operation of physical
plant or equipment needed to provide service; examples include items such as labor and
Capital cost:
the amount of
net investment.
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278 CHAPTER 6 Annual Equivalent-Worth Analysis
Capital recovery
cost: The annual
payment that will
repay the cost of
a fixed asset
over the useful
life of the asset
and will provide
an economic
rate of return on
the investment.
raw materials. Capital costs are incurred by purchasing assets to be used in production
and service. Normally, capital costs are nonrecurring (i.e., one-time) costs, whereas operating costs recur for as long as an asset is owned.
Because operating costs recur over the life of a project, they tend to be estimated on
an annual basis anyway, so, for the purposes of annual equivalent cost analysis, no special
calculation is required. However, because capital costs tend to be one-time costs, in conducting an annual equivalent cost analysis we must translate this one-time cost into its annual equivalent over the life of the project. The annual equivalent of a capital cost is given
a special name: capital recovery cost, designated CR(i).
Two general monetary transactions are associated with the purchase and eventual retirement of a capital asset: its initial cost (I) and its salvage value (S). Taking into account
these sums, we calculate the capital recovery factor as follows:
CR1i2 = I1A>P, i, N2 - S1A>F, i, N2.
(6.2)
Now, recall algebraic relationships between factors in Table 3.4, and notice that the factor
(A/F, i, N) can be expressed as
1A>F, i, N2 = 1A>P, i, N2 - i.
Then we may rewrite CR(i) as
CR1i2 = I1A>P, i, N2 - S[1A>P, i, N2 - i]
= 1I - S21A>P, i, N2 + iS.
(6.3)
Since we are calculating the equivalent annual costs, we treat cost items with a positive
sign. Then the salvage value is treated as having a negative sign in Eq. (6.3). We may interpret this situation thus: To obtain the machine, one borrows a total of I dollars, S dollars
of which are returned at the end of the Nth year. The first term, 1I - S2(A/P, i, N), implies
that the balance 1I - S2 will be paid back in equal installments over the N-year period at
a rate of i. The second term, iS, implies that simple interest in the amount iS is paid on S
until it is repaid (Figure 6.4). Thus, the amount to be financed is I - S (P/F, i, N), and the
installments of this loan over the N-year period are
AE1i2 = [I - S1P>F, i, N2]1A>P, i, N2
= I1A>P, i, N2 - S1P>F, i, N21A>P, i, N2
= [I1A>P, i, N2 - S1A>F, i, N2]
= CR1i2.
(6.4)
Therefore, CR(i) tells us what the bank would charge each year. Many auto leases are
based on this arrangement, in that most require a guarantee of S dollars in salvage. From
an industry viewpoint, CR(i) is the annual cost to the firm of owning the asset.
With this information, the amount of annual savings required to recover the capital and operating costs associated with a project can be determined, as Example 6.4
illustrates.
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Section 6.2 Capital Costs versus Operating Costs
S
0
N
I
CR(i) = (I – S)(A/P,i,N) + iS
0
1
2
3
4
N–1
N
CR(i)
Figure 6.4
Capital recovery (ownership) cost calculation.
EXAMPLE 6.4 Capital Recovery Cost
Consider a machine that costs $20,000 and has a five-year useful life. At the end of the
five years, it can be sold for $4,000 after tax adjustment. The annual operating and
maintenance (O&M) costs are about $500. If the firm could earn an after-tax revenue
of $5,000 per year with this machine, should it be purchased at an interest rate of 10%?
(All benefits and costs associated with the machine are accounted for in these figures.)
SOLUTION
Given: I = $20,000, S = $4,000, O&M = $500, A = $5,000, N = 5 years, and
i = 10% per year.
Find: AE, and determine whether to purchase the machine.
The first task is to separate cash flows associated with acquisition and disposal of the
asset from the normal operating cash flows. Since the operating cash flows—the
$5,000 yearly revenue—are already given in equivalent annual flows, we need to convert only the cash flows associated with acquisition and disposal of the asset into
equivalent annual flows (Figure 6.5). Using Eq. (6.3), we obtain
CR1i2 = 1I - S21A>P, i, N2 + iS
= 1$20,000 - $4,00021A>P, 10%, 52 + 10.102 $4,000
= $4,620.76,
O&M1i2 = $500,
279
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280 CHAPTER 6 Annual Equivalent-Worth Analysis
$4,000
$5,000
0
1
2
3
4
Operating revenue $5,000
5
$500
1
2
3
4
5
Ownership cost
0
1
2
3
4
0
1
2
3
4
5
$4,000
$20,000
0
0
$20,000
1
2
3
4
5
5
Capital cost O&M cost $5,120.76
O&M $500
Figure 6.5 Separating ownership cost (capital cost) and
operating cost from operating revenue, which must exceed the
annual equivalent cost to make the project acceptable.
AEC110%2 = CR110%2 + O&M110%2
= $4,620.76 + $500
= $5,120.76,
AE110%2 = $5,000 - $5,120.76
= - $120.76.
This negative AE value indicates that the machine does not generate sufficient revenue to recover the original investment, so we must reject the project. In fact, there
will be an equivalent loss of $120.76 per year over the life of the machine.
COMMENTS: We may interpret the value found for the annual equivalent cost as asserting that the annual operating revenues must be at least $5,120.76 in order to recover the cost of owning and operating the asset. However, the annual operating
revenues actually amount to only $5,000, resulting in a loss of $120.76 per year.
Therefore, the project is not worth undertaking.
6.3 Applying Annual-Worth Analysis
In general, most engineering economic analysis problems can be solved by the presentworth methods that were introduced in Chapter 5. However, some economic analysis
problems can be solved more efficiently by annual-worth analysis. In this section, we introduce several applications that call for such an analysis.
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Section 6.3 Applying Annual-Worth Analysis 281
6.3.1 Benefits of AE Analysis
Example 6.1 should look familiar to you: It is exactly the situation we encountered in
Chapter 4 when we converted a mixed cash flow into a single present value and then into
a series of equivalent cash flows. In the case of Example 6.1, you may wonder why we
bother to convert NPW to AE at all, since we already know that the project is acceptable
from NPW analysis. In fact, the example was mainly an exercise to familiarize you with
the AE calculation.
However, in the real world, a number of situations can occur in which AE analysis
is preferred, or even demanded, over NPW analysis. For example, corporations issue
annual reports and develop yearly budgets. For these purposes, a company may find it
more useful to present the annual cost or benefit of an ongoing project, rather than its
overall cost or benefit. Following are some additional situations in which AE analysis
is preferred:
1. Consistency of report formats. Financial managers commonly work with annual
rather than overall costs in any number of internal and external reports. Engineering managers may be required to submit project analyses on an annual basis for
consistency and ease of use by other members of the corporation and stockholders.
2. Need for unit costs or profits. In many situations, projects must be broken into
unit costs (or profits) for ease of comparison with alternatives. Make-or-buy and
reimbursement analyses are key examples, and these will be discussed in the
chapter.
3. Life-cycle cost analysis. When there is no need for estimating the revenue stream
for a proposed project, we can consider only the cost streams of the project. In that
case, it is common to convert this life-cycle cost (LCC) into an equivalent annual
cost for purposes of comparison. Industry has used the LCC to help determine
which project will cost less over the life of a product. LCC analysis has had a long
tradition in the Department of Defense, having been applied to virtually every new
weapon system proposed or under development.
6.3.2 Unit Profit or Cost Calculation
In many situations, we need to know the unit profit (or cost) of operating an asset. To obtain this quantity, we may proceed as follows:
• Determine the number of units to be produced (or serviced) each year over the life of
the asset.
• Identify the cash flow series associated with production or service over the life of the
asset.
• Calculate the net present worth of the project cash flow series at a given interest rate,
and then determine the equivalent annual worth.
• Divide the equivalent annual worth by the number of units to be produced or serviced
during each year. When the number of units varies each year, you may need to convert them into equivalent annual units.
To illustrate the procedure, Example 6.5 uses the annual equivalent concept in estimating the savings per machine hour for the proposed acquisition of a machine.
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EXAMPLE 6.5 Unit Profit per Machine Hour When Annual
Operating Hours Remain Constant
Consider the investment in the metal-cutting machine of Example 5.5. Recall that
this three-year investment was expected to generate an NPW of $3,553. Suppose that
the machine will be operated for 2,000 hours per year. Compute the equivalent savings per machine hour at i = 15%.
SOLUTION
Given: NPW = $3,553, N = 3 years, i = 15% per year, and 2,000 machine hours
per year.
Find: Equivalent savings per machine hour.
We first compute the annual equivalent savings from the use of the machine. Since
we already know the NPW of the project, we obtain the AE by the formula
AE115%2 = $3,5531A>P, 15%, 32 = $1,556.
With an annual usage of 2,000 hours, the equivalent savings per machine hour
would be
Savings per machine hour = $1,556>2,000 hours = $0.78>hour.
COMMENTS: Note that we cannot simply divide the NPW ($3,553) by the total number of machine hours over the three-year period (6,000 hours) to obtain $0.59/hour.
This $0.59 figure represents the instant savings in present worth for each hourly use
of the equipment, but does not consider the time over which the savings occur. Once
we have the annual equivalent worth, we can divide by the desired time unit if the
compounding period is one year. If the compounding period is shorter, then the
equivalent worth should be calculated for the compounding period.
EXAMPLE 6.6 Unit Profit per Machine Hour When Annual
Operating Hours Fluctuate
Consider again Example 6.5, and suppose that the metal-cutting machine will be operated according to varying hours: 1,500 hours the first year, 2,500 hours the second
year, and 2,000 hours the third year. The total operating hours still remain at 6,000
over three years. Compute the equivalent savings per machine hour at i = 15%.
SOLUTION
Given: NPW = $3,553, N = 3 years, i = 15% per year, and operating hours of
1,500 the first year, 2,500 the second year, and 2,000 the third year.
Find: Equivalent savings per machine hour.
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Section 6.3 Applying Annual-Worth Analysis 283
As calculated in Example 6.5, the annual equivalent savings is $1,556. Let C denote
the equivalent annual savings per machine hour, which we need to determine. Now,
with varying annual usages of the machine, we can set up the equivalent annual savings as a function of C:
Equivalent annual savings = [C11,50021P>F, 15%, 12
+ C12,50021P>F, 15%, 22
+ C12,00021P>F, 15%, 32]1A>P, 15%, 32
= 1,975.16C.
We can equate this amount to the $1,556 we calculated in Example 6.5 and solve for
C. This gives us
C = $1,556>1,975.16 = $0.79>hour,
which is about a penny more than the $0.78 we found in Example 6.5.
6.3.3 Make-or-Buy Decision
Make-or-buy problems are among the most common of business decisions. At any
given time, a firm may have the option of either buying an item or producing it. Unlike
the make-or-buy situation we will consider in Chapter 8, if either the “make” or the
“buy” alternative requires the acquisition of machinery or equipment, then it becomes
an investment decision. Since the cost of an outside service (the “buy” alternative) is
usually quoted in terms of dollars per unit, it is easier to compare the two alternatives if
the differential costs of the “make” alternative are also given in dollars per unit. This
unit cost comparison requires the use of annual-worth analysis. The specific procedure
is as follows:
Step 1. Determine the time span (planning horizon) for which the part (or product) will
be needed.
Step 2. Determine the annual quantity of the part (or product).
Step 3. Obtain the unit cost of purchasing the part (or product) from the outside
firm.
Step 4. Determine the equipment, manpower, and all other resources required to make
the part (or product) in-house.
Step 5. Estimate the net cash flows associated with the “make” option over the planning horizon.
Step 6. Compute the annual equivalent cost of producing the part (or product).
Step 7. Compute the unit cost of making the part (or product) by dividing the annual
equivalent cost by the required annual volume.
Step 8. Choose the option with the minimum unit cost.
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EXAMPLE 6.7 Equivalent Worth: Outsourcing the Manufacture
of Cassettes and Tapes
Ampex Corporation currently produces both videocassette cases and metal particle
magnetic tape for commercial use. An increased demand for metal particle tapes is
projected, and Ampex is deciding between increasing the internal production of
empty cassette cases and magnetic tape or purchasing empty cassette cases from an
outside vendor. If Ampex purchases the cases from a vendor, the company must also
buy specialized equipment to load the magnetic tapes, since its current loading machine is not compatible with the cassette cases produced by the vendor under consideration. The projected production rate of cassettes is 79,815 units per week for 48
weeks of operation per year. The planning horizon is seven years. After considering
the effects of income taxes, the accounting department has itemized the annual costs
associated with each option as follows:
• Make option (annual costs):
Labor
$1,445,633
Materials
$2,048,511
Incremental overhead
$1,088,110
Total annual cost
$4,582,254
• Buy option:
Capital expenditure
Acquisition of a new loading machine
$ 405,000
Salvage value at end of seven years
$
45,000
Annual Operating Costs
Labor
$ 251,956
Purchasing empty cassette ($0.85/unit)
$3,256,452
Incremental overhead
$ 822,719
Total annual operating costs
$4,331,127
(Note the conventional assumption that cash flows occur in discrete lumps at the
ends of years, as shown in Figure 6.6.) Assuming that Ampex’s MARR is 14%, calculate the unit cost under each option.
SOLUTION
Given: Cash flows for two options and i = 14%.
Find: Unit cost for each option and which option is preferred.
The required annual production volume is
79,815 units>week * 48 weeks = 3,831,120 units per year.
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Section 6.3 Applying Annual-Worth Analysis 285
Make option
0
1
2
3
4
5
6
7 Years
5
6
7 Years
$4,582,254
Buy option
0
1
2
3
4
$45,000
$405,000
$4,331,127
Figure 6.6
Make-or-buy analysis.
We now need to calculate the annual equivalent cost under each option:
• Make option. Since the “make option” is already given on an annual basis, the
equivalent annual cost will be
AEC114%2Make = $4,582,254.
• Buy option. The two cost components are capital cost and operating cost.
Capital cost:
CR114%2 = 1$405,000 - $45,00021A>P, 14%, 72
+ 10.1421$45,0002
= $90,249
AEC114%21 = CR114%2 = $90,249.
Operating cost:
AEC114%22 = $4,331,127.
Total annual equivalent cost:
AEC114%2Buy = AEC114%21 + AEC114%22 = $4,421,376.
Obviously, this annual equivalent calculation indicates that Ampex would be better off
buying cassette cases from the outside vendor. However, Ampex wants to know the unit
costs in order to set a price for the product. In such a situation, we need to calculate the
unit cost of producing the cassette tapes under each option. We do this by dividing the
magnitude of the annual equivalent cost for each option by the annual quantity required:
• Make option:
Unit cost = $4,582,254>3,831,120 = $1.20>unit.
• Buy option:
Unit cost = $4,421,376>3,831,120 = $1.15>unit.
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286 CHAPTER 6 Annual Equivalent-Worth Analysis
Buying the empty cassette cases from the outside vendor and loading the tape inhouse will save Ampex 5 cents per cassette before any consideration of taxes.
COMMENTS: Two important noneconomic factors should also be considered. The
first is the question of whether the quality of the supplier’s component is better than,
equal to, or worse than the component the firm is presently manufacturing. The second is the reliability of the supplier in terms of providing the needed quantities of the
cassette cases on a timely basis. A reduction in quality or reliability should virtually
always rule out buying.
6.3.4 Pricing the Use of an Asset
Companies often need to calculate the cost of equipment that corresponds to a unit of use of
that equipment. For example, if you own an asset such as a building, you would be interested
in knowing the cost per square foot of owning and operating the asset. This information will
be the basis for determining the rental fee for the asset. A familiar example is an employer’s
reimbursement of costs for the use of an employee’s personal car for business purposes. If an
employee’s job depends on obtaining and using a personal vehicle on the employer’s behalf,
reimbursement on the basis of the employee’s overall costs per mile seems fair.
EXAMPLE 6.8 Pricing an Apartment Rental Fee
Sunbelt Corporation, an investment company, is considering building a 50-unit apartment complex in a growing area near Tucson, Arizona. Since the long-term growth
potential of the town is excellent, it is believed that the company could average 85%
full occupancy for the complex each year. If the following financial data are reasonably accurate estimates, determine the minimum monthly rent that should be charged
if a 15% rate of return is desired:
•
•
•
•
•
•
Land investment cost = $1,000,000
Building investment cost = $2,500,000
Annual upkeep cost = $150,000
Property taxes and insurance = 5% of total initial investment
Study period = 25 years
Salvage value = Only land cost can be recovered in full.
SOLUTION
Given: Preceding financial data, study period = 25 years, and i = 15%.
Find: Minimum monthly rental charge.
First we need to determine the capital cost associated with ownership of the
property:
Total investment required = land cost + building cost = $3,500,000,
Salvage value = $1,000,000 at the end of 25 years,
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Section 6.4 Life-Cycle Cost Analysis 287
CR115%2 = 1$3,500,000 - $1,000,00021A/P, 15%, 252
+ 1$1,000,000210.152
= $536,749.
Second, the annual operating cost has two elements: (1) property taxes and insurance
and (2) annual upkeep cost. Thus,
O&M cost = 10.0521$3,500,0002 + $150,000
= $325,000.
So the total annual equivalent cost is
AEC115%2 = $536,749 + $325,000
= $861,749,
which is the minimum annual rental required to achieve a 15% rate of return. Therefore, with annual compounding, the monthly rental amount is
$861,749
112 * 50210.852
= $1,690.
Required monthly charge =
COMMENTS: The rental charge that is exactly equal to the cost of owning and operating
the building is known as the break-even point.
6.4 Life-Cycle Cost Analysis
Because 80% of the total life-cycle cost of a system occurs after the system has entered
service, the best long-term system acquisition and support decisions are based on a full understanding of the total cost of acquiring, operating, and supporting the system (Figure 6.7).
Life-cycle cost analysis (LCCA) enables the analyst to make sure that the selection of a
design alternative is not based solely on the lowest initial costs, but also takes into account all the future costs over the project’s usable life. Some of the unique features of
LCCA are as follows:
• LCCA is used appropriately only to select from among design alternatives that would
yield the same level of performance or benefits to the project’s users during normal
operations. If benefits vary among the design alternatives, then the alternatives cannot
be compared solely on the basis of cost. Rather, the analyst would need to employ
present-worth analysis or benefit–cost analysis (BCA), which measures the monetary
value of life-cycle benefits as well as costs. BCA is discussed in Chapter 16.
• LCCA is a way to predict the most cost-effective solution; it does not guarantee a
particular result, but allows the plant designer or manager to make a reasonable comparison among alternative solutions within the limits of the available data.
• To make a fair comparison, the plant designer or manager might need to consider the
measure used. For example, the same process output volume should be considered,
and if the two items being examined cannot give the same output volume, it may be
appropriate to express the figures in cost per unit of output (e.g., $/ton, or euro/kg),
which requires a calculation of annual equivalent dollars generated. This calculation
is based on the annual output.
Life-cycle cost
analysis is
useful when
project
alternatives that
fulfill the same
performance
requirements,
but differ with
respect to initial
costs and
operating costs.
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Phase 5: Phasing out
operation
Phase 4: Operating and
support
Phase 3: Production and
development
Annual cost
Phase 1: Concept refinement and
technology development
Phase 2: System development and
demonstration
288 CHAPTER 6 Annual Equivalent-Worth Analysis
Operating & Maintenance
Dis
al
pos
Investment
in physical assets
R&D
0
Life Cycle
Useful life (N)
Figure 6.7 Stages of life-cycle cost. These include the concept refinement and technology
development phase, the system development and demonstration phase, the production and
deployment phase, the operating phase, and the disposal phase.
In many situations, we need to compare a set of different design alternatives, each
of which would produce the same number of units (constant revenues), but would require different amounts of investment and operating costs (because of different degrees
of mechanization). Example 6.9 illustrates the use of the annual equivalent-cost concept to compare the cost of operating an existing pumping system with that of an improved system.
EXAMPLE 6.9 Pumping System with a Problem Valve3
Consider a single-pump circuit that transports a process fluid containing some
solids from a storage tank to a pressurized tank. A heat exchanger heats the fluid,
and a control valve regulates the rate of flow into the pressurized tank to 80 cubic
meters per hour 1m3/h2, or 350 gallons per minute (gpm). The process is depicted
in Figure 6.8.
The plant engineer is experiencing problems with a fluid control valve (FCV)
that fails due to erosion caused by cavitations. The valve fails every 10 to 12 months
at a cost of $4,000 per repair. A change in the control valve is being considered: Replace the existing valve with one that can resist cavitations. Before the control valve
3
Pump Life Cycle Costs: A Guide to LCC Analysis for Pumping Systems. DOE/GO-102001-1190. December
2000, Office of Industrial Technologies, U.S. Department of Energy and Hydraulic Institute.
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Section 6.4 Life-Cycle Cost Analysis 289
Pressure tank
2.0 bar
Storage tank
Pump
FCV@15%
Heat exchanger
Figure 6.8 Sketch of a pumping system in which the control valve fails.
is repaired again, the project engineer wants to look at other options and perform an
LCCA on alternative solutions.
Engineering Solution Alternatives
The first step is to determine how the system is currently operating and why the control valve fails. Then the engineer can see what can be done to correct the problem.
The control valve currently operates between 15 and 20% open and with considerable cavitation noise from the valve. It appears that the valve was not sized properly for the application. After reviewing the original design calculations, it was
discovered that the pump was oversized: 110 m3/h (485 gpm) instead of 80 m3/h
(350 gpm). This resulted in a larger pressure drop across the control valve than was
originally intended.
As a result of the large differential pressure at the operating rate of flow, and because the valve is showing cavitation damage at regular intervals, the engineer determines that the control valve is not suitable for this process.
The following four options are suggested:
• Option A. A new control valve can be installed to accommodate the high pressure
differential.
• Option B. The pump impeller can be trimmed so that the pump does not develop
as much head, resulting in a lower pressure drop across the current
valve.
• Option C. A variable-frequency drive (VFD) can be installed and the flow control
valve removed. The VFD can vary the pump speed and thus achieve the
desired process flow.
• Option D. The system can be left as it is, with a yearly repair of the flow control
valve to be expected.
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290 CHAPTER 6 Annual Equivalent-Worth Analysis
Cost Summary:
Pumping systems often have a life span of 15 to 20 years. Some cost elements will be
incurred at the outset, and others will be incurred at different times throughout the
lives of the different solutions evaluated. Therefore, you need to calculate a present
value of the LCC to accurately assess the different solutions.
Some of the major LCC elements related to a typical pumping system are summarized as follows:
LCC = Cic + Cin + Ce + Co + Cm + Cs + Cenv + Cd
LCC = life-cycle cost
Cic = initial costs, purchase price (costs of pump, system, pipe,
auxiliary services2
Cin = installation and commissioning cost 1including cost of training2
Ce = energy costs 1predicted cost for system operation, including costs of
pump driver, controls, and any auxiliary services2
Co = operation cost 1labor cost of normal system supervision2
Cm = maintenance and repair costs 1costs of routine and predicted repairs2
Cs = downtime costs 1cost of loss of production2
Cenv = environmental costs 1costs due to contamination from pumped
liquid and auxiliary equipment2
Cd = decommissioning and disposal costs (including the cost of
of restoration of the local environment and disposal
of auxiliary services2
For each option, the major cost elements identified are as follows:
• Option A. The cost of a new control valve that is properly sized is $5,000. The cost
of modifying the pump’s performance by reducing the diameter of the
impeller is $2,250. The process operates at 80 m3/h for 6,000 h/year.
The energy cost is $0.08 per kWh and the motor efficiency is 90%.
• Option B. By trimming the impeller to 375 mm, the pump’s total head is reduced
to 42.0 m (138 ft) at 80 m3/h. This drop in pressure reduces the differential pressure across the control valve to less than 10 m (33 ft), which
better matches the valve’s original design intent. The resulting annual
energy cost with the smaller impeller is $6,720 per year. It costs $2,250
to trim the impeller. This cost includes the machining cost as well as
the cost to disassemble and reassemble the pump.
• Option C. A 30-kW VFD costs $20,000 and an additional $1,500 to install. The
VFD will cost $500 to maintain each year. It is assumed that it will not
need any repairs over the project’s eight-year life.
• Option D. The option to leave the system unchanged will result in a yearly cost of
$4,000 for repairs to the cavitating flow control value.
Table 6.1 summarizes financial as well as technical data related to the various options.
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Section 6.4 Life-Cycle Cost Analysis 291
TABLE 6.1
Cost Comparison for Options A through D in the System with a
Failing Control Valve
Cost
Change Control Trim Impeller
Valve (A)
(B)
VFD
(C)
Repair Control
Valve (D)
430 mm
430 mm
Pump Cost Data
Impeller
diameter
Pump head
Pump
efficiency
430 mm
375 mm
71.7 m (235 ft) 42.0 m (138 ft) 34.5 m (113 ft) 71.7 m (235 ft)
75.1%
72.1%
77%
75.1%
80 m >h
(350 gpm)
80 m >h
(350 gpm)
80 m >h
(350 gpm)
80 m3>h
(350 gpm)
23.1 kW
14.0 kW
11.6 kW
23.1 kW
Energy cost/year
$11,088
$6,720
$ 5,568
$11,088
New valve
$ 5,000
0
0
0
Modify impeller
0
$2,250
0
0
VFD
0
0
$20,000
0
Installation of VFD
0
0
$ 1,500
Valve repair/year
0
0
0
Rate of flow
Power
consumed
3
3
3
$ 4,000
SOLUTION
Given: Financial data as summarized in Table 6.1.
Find: Which design option to choose.
Assumptions:
• The current energy price is $0.08/kWh.
• The process is operated for 6,000 hours/year.
• The company has a cost of $500 per year for routine maintenance of pumps of
this size, with a repair cost of $2,500 every second year.
• There is no decommissioning cost or environmental disposal cost associated
with this project.
• The project has an eight-year life.
• The interest rate for new capital projects is 8%, and an inflation rate of 4% is
expected.
A sample LCC calculation for Option A is shown in Table 6.2. Note that the energy
cost and other cost data are escalated at the annual rate of 4%. For example, the
current estimate of energy cost is $11,088. To find the cost at the end of year 1, we
292
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
1
A
$5,000
$5,000
$91,827
$75,129
$3,388
$15,979
$2.66
Sum costs
Present costs
Present costs, energy
Present costs, routine maintenance
Sum of present costs
Sum of energy costs
Sum of routine maintenance costs
Annual equivalent cost (8%)
Cost per operating hour
Decommissioning costs
Environmental & disposal costs
2
$0
$11,159 $13,064
$10,677 $10,282
$481
$464
$12,052 $15,238
$0
$11,532 $11,993
$520
$541
$2,704
$0
$0
$0
$0
1
D
$10,348
$9,901
$446
$13,035
$0
$0
$0
$12,472
$562
3
E
$12,114
$9,534
$430
$16,481
$0
$12,971
$585
$2,925
$0
$0
4
F
$9,595
$9,181
$414
$14,099
$0
$0
$0
$13,490
$608
5
G
$11,233
$8,841
$399
$17,826
$0
$14,030
$633
$3,163
$0
$0
6
H
$8,898
$8,514
$384
$15,249
$0
$0
$0
$14,591
$658
7
I
$10,417
$8,198
$370
$19,280
$0
$0
$0
$15,175
$684
$3,421
$0
$0
8
J
9:10 PM
Downtime costs
0
$5,000
$0
C
5/25/06
Energy cost/year (calculated)
Routine maintenance
Repair cost every 2nd year
Operating costs
Other costs
Year No
Initial investment cost:
Installation and commissioning cost:
B
LCC Calculation for Option A
Calculation Chart for LCC: Option A
TABLE 6.2
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Section 6.4 Life-Cycle Cost Analysis 293
TABLE 6.3
Comparison of LCC for Options A through D
Option A
Change
Control
Valve
Option B
Trim
Impeller
Option C
Option D
VFD and
Repair
Remove
Control
Control Valve
Valve
Initial investment cost:
$5,000
$2,250
$21,500
0
Energy price (present)
per kWh:
0.080
0.080
0.080
0.080
23.1
14.0
11.6
23.1
Average operating hours/year:
6,000
6,000
6,000
6,000
Energy cost/year (calculated)
energy price weighted
average power average
operating hours/year:
11,088
6,720
5,568
11,088
Maintenance cost (routine
maintenance/year:
500
500
1,000
500
Repair every second year:
2,500
2,500
2,500
2,500
Other yearly costs:
0
0
0
4,000
Downtime cost/year:
0
0
0
0
Environmental cost:
0
0
0
0
Decommissioning/disposal
(salvage) cost:
0
0
0
0
Lifetime in years:
8
8
8
8
Interest rate (%):
8.0%
8.0%
8.0%
8.0%
Inflation rate (%):
4.0%
4.0%
4.0%
4.0%
$91,827
$59,481
$74,313
$113,930
$2.66
$1.73
$2.16
$3.30
Input
Weighted average power of
equipment in kW:
Output
Present LCC value:
Cost per operating hour
multiply $11,088 by 11 + 0.042, yielding $11,532. For year 2, we multiply
$11,532 by 11 + 0.042 to obtain $11,993. Once we calculate the LCC in present
worth ($91,827), we find the equivalent annual value ($15,979) at 8% interest. Finally, to calculate the cost per hour, we divide $15,979 by 6,000 hours, resulting in
$2.66. We can calculate the unit costs for other options in a similar fashion. (See
Table 6.3.)
Option B, trimming the impeller, has the lowest life-cycle cost ($1.73 per hour)
and is the preferred option for this example.
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294 CHAPTER 6 Annual Equivalent-Worth Analysis
6.5 Design Economics
Engineers are frequently involved in making design decisions that provide the required
functional quality at the lowest cost. Another valuable extension of AE analysis is minimum-cost analysis, which is used to determine optimal engineering designs. The AE
analysis method is useful when two or more cost components are affected differently by
the same design element (i.e., for a single design variable, some costs may increase while
others decrease). When the equivalent annual total cost of a design variable is a function
of increasing and decreasing cost components, we can usually find the optimal value that
will minimize the cost of the design with the formula
AEC1i2 = a + bx +
c
,
x
(6.5)
where x is a common design variable and a, b, and c are constants.
To find the value of the common design variable that minimizes AE(i), we need to
take the first derivative, equate the result to zero, and solve for x:
dAEC1i2
c
= b - 2
dx
x
= 0
c
x = x* =
.
Ab
(6.6)
(It is common engineering practice to denote the optimal solution with an asterisk.)
The logic of the first-order requirement is that an activity should, if possible, be carried to a point where its marginal yield dAEC(i)/dx is zero. However, to be sure whether
we have found a maximum or a minimum when we have located a point whose marginal
yield is zero, we must examine it in terms of what are called the second-order conditions.
Second-order conditions are equivalent to the usual requirements that the second derivative be negative in the case of a maximum and positive for a minimum. In our situation,
d2AEC1i2
dx
2
=
2C
.
x3
As long as C 7 0, the second derivative will be positive, indicating that the value x* is
the minimum-cost point for the design alternative. To illustrate the optimization concept,
two examples are given, the first having to do with designing the optimal cross-sectional
area of a conductor and the second dealing with selecting an optimal size for a pipe.
EXAMPLE 6.10 Optimal Cross-Sectional Area
A constant electric current of 5,000 amps is to be transmitted a distance of 1,000 feet
from a power plant to a substation for 24 hours a day, 365 days a year. A copper conductor can be installed for $8.25 per pound. The conductor will have an estimated
life of 25 years and a salvage value of $0.75 per pound. The power loss from a conductor is inversely proportional to the cross-sectional area A of the conductor.
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Section 6.5 Design Economics 295
It is known that the resistance of the conductor sought is 0.8145 * 10-5 ohm for
1 square inch per foot of cross section. The cost of energy is $0.05 per kilowatt-hour,
the interest rate is 9%, and the density of copper is 555 lb/ft 3. For the given data, calculate the optimum cross-sectional area A of the conductor.
DISCUSSION: The resistance of the conductor is the most important cause of
power loss in a transmission line. The resistance of an electrical conductor varies
directly with its length and inversely with its cross-sectional area according to the
equation
L
R = ra b,
A
(6.7)
where R is the resistance of the conductor, L is the length of the conductor, A is
the cross-sectional area of the conductor, and r is the resistivity of the conductor
material.
Any consistent set of units may be used. In power work in the United States, L is
usually given in feet, A in circular mils (cmil), and r in ohm-circular mils per foot.
A circular mil is the area of a circle that has a diameter of 1 mil, equal to
1 * 10-3 in. The cross-sectional area of a wire in square inches equals its area in circular mils multiplied by 0.7854 * 10-6. More specifically, one unit relates to another
as follows:
1 linear mil = 0.001 in.
= 0.0254 millimeter.
1 circular mil = area of circle 1 linear mil in diameter
= 10.522p mil2
= 0.7854 * 10-6 in.2.
1 in.2 = 1/10.7854 * 10-62
= 1.27324 * 106 cmil.
In SI units (the official designation for the Système International d’Unités), L is in
meters, A in square meters, and r in ohm-meters. In terms of SI units, the copper
conductor has a r value of 1.7241 * 10-8 Æ-meter. We can easily convert this value
into units of Æ-in.2 per foot. From Eq. (6.7), solving for r yields
r = RA>L = 1.7241 * 10-8 Æ11 meter22>1 meter
= 1.7241 * 10-8 Æ139.37 in.22>3.2808 ft
= 0.8145421 * 10-5 Æ in.2>ft.
When current flows through a circuit, power is used to overcome resistance. The unit
of electrical work is the kilowatt hour (kWh), which is equal to the power in kilowatts, multiplied by the hours during which work is performed. If the current I is
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296 CHAPTER 6 Annual Equivalent-Worth Analysis
steady, then the charge passing through the wire in time T is equivalent to the power
that is converted to heat (known as energy loss) and is equal to
Power = I 2
RT
,
1,000 kWh
(6.8)
where I is the current in amperes, R is the resistance in ohms, and T is the duration, in
hours, during which work is performed.
SOLUTION
Given: Cost components as a function of cross-sectional area (A), N = 25 years, and
i = 9%.
Find: Optimal value of A.
Step 1: This classic minimum-cost example, the design of the cross-sectional area
of an electrical conductor, involves increasing and decreasing cost components. Since the resistance of a conductor is inversely proportional to the
size of the conductor, energy loss will decrease as the size of the conductor
increases. More specifically, the energy loss in kilowatt-hours (kWh) in a
conductor due to resistance is equal to
Energy loss in kilowatt-hours =
=
=
I 2R
T
1,000A
15,0002210.0081452
124 * 3652
1,000A
1,783,755
kWh.
A
Step 2: Again, since the electrical resistance is inversely proportional to the crosssectional area A of the conductor, the total energy loss in dollars per year for
a specified conductor material is
1,783,755
1f2
A
1,783,755
=
1$0.052
A
$89,188
,
=
A
Cost of energy loss =
where f is the cost of energy in dollars per kWh.
Step 3: As we increase the size of the conductor, however, it costs more to build.
First, we need to calculate the total amount of conductor material in pounds.
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Section 6.5 Design Economics 297
Since the cross-sectional area is given in square inches, we need to convert
the total length in feet to inches before finding the weight of the material:
Weight of material in pounds =
11,0002112215552A
123
= 3,8541A2
Total material cost = 3,8541A21$8.252
= $31,7971A2.
Here, we are looking for the trade-off between the cost of installation and
the cost of energy loss.
Step 4: Since, at the end of 25 years, the copper material will be salvaged at the rate
of $0.75 per pound, we can compute the capital recovery cost as
CR19%2 = [31,797A - 0.7513,854A2]1A>P, 9%, 252 + 0.7513,854A210.092
= 2,943A + 260A
= 3,203A.
Step 5: Using Eq. (6.5), we express the total annual equivalent cost as a function of
the design variable A:
Capital cost
$'%'&
AEC19%2 = 3,203A +
89,188
.
A
(')'*
Operating cost
To find the minimum annual equivalent cost, we use Eq. (6.6):
dAEC19%2
89,188
= 0,
= 3,203 dA
A2
89,188
A* =
A 3,203
= 5.276 in.2.
The minimum annual equivalent total cost is
AEC19%2 = 3,20315.2762 +
89,188
5.276
= $33,802.
Figure 6.9 illustrates the nature of this design trade-off problem.
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298 CHAPTER 6 Annual Equivalent-Worth Analysis
90,000
80,000
2.5921 in
Cross-sectional area (5.276 sq in)
Annual equivalent cost ($)
70,000
60,000
6.7163 x 106 circular mils
2.5921 in
50,000
40,000
Total cost
30,000
Capital cost
20,000
Operating cost
10,000
5.276
0
1.0
2.0
3.0
4.0
5.0
6.0
7.0
Cross-sectional area in square inches (A)
Figure 6.9 Optimal cross-sectional areas for a copper conductor. Note
that the minimum point almost coincides with the crossing point of the
capital-cost and operating-cost lines. This is not always true. Since the
cost components can have a variety of cost patterns, the minimum point
does not in general occur at the crossing point (Example 6.10).
EXAMPLE 6.11 Economical Pipe Size
As a result of the 1990 conflict in the Persian Gulf, Kuwait is studying the feasibility
of running a steel pipeline across the Arabian Peninsula to the Red Sea. The pipeline
will be designed to handle 3 million barrels of crude oil per day under optimum conditions. The length of the line will be 600 miles. Calculate the optimum pipeline
diameter that will be used for 20 years for the following data at i = 10%:
1.333Q¢P
HP
1,980,000
Q = volume flow rate, ft 3/hour
128QmL
, pressure drop, lb/ft 2
¢P =
4
gpD
L = pipe length, ft
Pumping power =
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Section 6.5 Design Economics 299
D = inside pipe diameter, ft
t = 0.01D, pipeline wall thickness, ft
m = 8,500 lb>hour ft, oil viscosity
g = 32.3 * 12,960,000 ft>hour2
Power cost, $0.015 per HP hour
Oil cost, $50 per barrel
Pipeline cost, $1.00 per pound of steel
Pump and motor costs, $195 per HP
The salvage value of the steel after 20 years is assumed to be zero because removal costs
exhaust scrap profits from steel. (See Figure 6.10 for the relationship between D and t.)
Cross-sectional view of pipeline
t = 0.01D
D
Outside
diameter
= D + 2t
Figure 6.10 Designing economical pipe size to handle 3 million
barrels of crude oil per day (Example 6.11).
DISCUSSION: In general, when a progressively larger size of pipe is used to carry a
given fluid at a given volume flow rate, the energy required to move the fluid will
progressively decrease. However, as we increase the pipe size, the cost of its construction will increase. In practice, to obtain the best pipe size for a particular situation,
you may choose a reasonable, but small, starting size. Compute the energy cost of
pumping fluid through this size of pipe and the total construction cost, and then compare the difference in energy cost with the difference in construction cost. When the
savings in energy cost exceeds the added construction cost, the process may be repeated with progressively larger pipe sizes until the added construction cost exceeds
the savings in energy cost. As soon as this happens, the best pipe size to use in the
particular application is identified. However, this search process can be simplified by
using the minimum-cost concept encompassed by Eqs. (6.5) and (6.6).
SOLUTION
Given: Cost components as a function of pipe diameter 1D2, N = 20 years, and
i = 10%.
Find: Optimal pipe size (D).
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Step 1: Several different units are introduced; however, we need to work with common units. We will assume the following conversions:
1 mile
600 miles
1 barrel
1 barrel
=
=
=
=
5,280 ft
600 * 5,280 = 3,168,000 ft
42 U.S. gallons
42 gallons * 231 in.3>gallon = 9,702 in.3
1 barrel = 9,702 in.3>123 = 5.6146 ft 3
Density of steel = 490.75 lb>ft 3
Step 2: Power required to pump oil:
It is well known that, for any given set of operating conditions involving the
flow of a noncompressible fluid, such as oil, through a pipe of constant diameter, a small-diameter pipe will have a high fluid velocity and a high fluid
pressure drop through the pipe. This will require a pump that will deliver a
high discharge pressure and a motor with large energy consumption. To determine the pumping power required to boost the pressure drop, we need to
determine both the volume flow rate and the amount of pressure drop. Then
we can calculate the cost of the power required to pump oil.
• Volume flow rate per hour:
Q = 3,000,000 barrels>day * 5.6146 ft 3>barrel
= 16,843,800 ft 3>day
= 701,825 ft 3>hour.
• Pressure drop:
128QmL
¢P =
=
gpD4
128 * 701,825 * 8,500 * 3,168,000
32.3 * 12,960,000 * 3.14159D4
1,845,153,595
lb>ft 2.
D4
• Pumping power required to boost the pressure drop:
=
Power =
1.333Q¢P
1,980,000
1,845,153,595
D4
1,980,000
1.333 * 701,825 *
=
=
871,818,975
HP.
D4
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Section 6.5 Design Economics 301
• Power cost to pump oil:
Power cost =
871,818,975
HP * $0.015>HP. hour
D4
* 24 hours>day * 365 days>year
=
$114,557,013,315
/year.
D4
Step 3: Pump and motor cost calculation:
Once we determine the required pumping power, we can find the size of the
pump and the motor costs. This is because the capacity of the pump and
motor is proportional to the required pumping power. Thus,
Pump and motor cost =
=
871,818,975
* $195>HP
D4
$170,004,700,125
.
D4
Step 4: Required amount and cost of steel:
The pumping cost will be counterbalanced by the lower costs for the smaller
pipe, valves, and fittings. If the pipe diameter is made larger, the fluid velocity drops markedly and the pumping costs become substantially lower. Conversely, the capital cost for the larger pipe, fittings, and valves becomes
greater. For a given cross-sectional area of the pipe, we can determine the
total volume of the pipe as well as the weight. Once the total weight of the
pipe is determined, we can easily convert it into the required investment
cost. The calculations are as follows:
Cross-sectional area = 3.14159[10.51D22 - 10.50D22]
= 0.032D2 ft 2,
Total volume of pipe = 0.032D2 ft 2 * 3,168,000 ft
= 101,376D2 ft 3,
Total weight of steel = 101,376D2 ft 3 * 490.75 lb>ft 3
= 49,750,272D2 lb,
Total pipeline cost = $1.00>lb * 49,750,272D 2 lb
= $49,750,272D2.
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Step 5: Annual equivalent cost calculation:
For a given total pipeline cost and its salvage value at the end of 20 years
of service life, we can find the equivalent annual capital cost by using the
formula for the capital recovery factor with return:
Capital cost = a $49,750,272D2 +
$170,004,700,125
b 1A>P, 10%, 202
D4
19,968,752,076
= 5,843,648D2 +
D4
$114,557,013,315
.
Annual power cost =
D4
Step 6: Economical pipe size:
Now that we have determined the annual pumping and motor costs and the
equivalent annual capital cost, we can express the total equivalent annual
cost as a fraction of the pipe diameter (D):
AEC110%2 = 5,843,648D2 +
19,968,752,076
114,557,013,315
+
.
4
D
D4
To find the optimal pipe size (D) that results in the minimum annual equivalent cost, we take the first derivative of AEC(10%) with respect to D, equate
the result to zero, and solve for D:
dAEC110%2
538,103,061,567
= 11,687,297D dD
D5
= 0;
11,687,297D6 = 538,103,061,567,
D6 = 46,041.70,
D* = 5.9868 ft.
Note that, ideally, the velocity in a pipe should be no more than approximately 10 ft/sec, because friction wears the pipe. To check whether the preceding answer is reasonable, we may compute
Q = velocity * pipe inner area,
701,825 ft 3>hr *
3.1415915.986822
1
hr>sec = V
,
3,600
4
V = 6.93 ft>sec,
which is less than 10 ft/sec. Therefore, the optimal answer as calculated is
practical.
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Summary 303
Step 7: Equivalent annual cost at optimal pipe size:
Capital cost = c $49,750,27215.986822 +
= 5,843,64815.986822 +
$170,004,700,125
d1A>P, 10%, 202
5.98684
19,968,752,076
5.98684
= $224,991,039;
Annual power cost =
114,557,013,315
5.98684
= $89,174,911;
Total annual cost = $224,991,039 + $89,174,911
= $314,165,950.
Step 8: Total annual oil revenue:
Annual oil revenue = $50>bbl * 3,000,000 bbl>day
= * 365 day>year
= $54,750,000,000>year.
Enough revenues are available to offset the capital cost as well as the operating cost.
COMMENTS: A number of other criteria exist for choosing the pipe size for a particular fluid transfer application. For example, low velocity may be required when erosion
or corrosion concerns must be considered. Alternatively, higher velocities may be desirable for slurries when settling is a concern. Ease of construction may also weigh
significantly in choosing a pipe size. On the one hand, a small pipe size may not accommodate the head and flow requirements efficiently; on the other, space limitations
may prohibit the selection of large pipe sizes.
SUMMARY
Annual equivalent worth analysis, or AE, is—along with present-worth analysis—one
of the two main analysis techniques based on the concept of equivalence. The equation
for AE is
AE1i2 = PW1i21A>P, i, N2.
AE analysis yields the same decision result as PW analysis.
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304 CHAPTER 6 Annual Equivalent-Worth Analysis
The capital recovery cost factor, or CR(i), is one of the most important applications of
AE analysis, in that it allows managers to calculate an annual equivalent cost of capital for ease of itemization with annual operating costs. The equation for CR(i) is
CR1i2 = 1I - S21A>P, i, N2 + iS,
where I = initial cost and S = salvage value.
AE analysis is recommended over NPW analysis in many key real-world situations for
the following reasons:
1. In many financial reports, an annual equivalent value is preferred to a present-worth
value.
2. Unit costs often must be calculated to determine reasonable pricing for items that
are on sale.
3. The cost per unit of use of an item must be calculated in order to reimburse employees for the business use of personal cars.
4. Make-or-buy decisions usually require the development of unit costs for the various
alternatives.
5. Minimum-cost analysis is easy to do when it is based on annual equivalent cost.
LCCA is a way to predict the most cost-effective solution by allowing engineers
to make a reasonable comparison among alternatives within the limit of the available data.
PROBLEMS
Note: Unless otherwise stated, all cash flows given in the problems that follow represent
after-tax cash flows.
Annual Equivalent-Worth Calculation
6.1 Consider the following cash flows and compute the equivalent annual worth at
i = 10%:
n
0
An
Investment
Revenue
- $5,000
1
$2,000
2
2,000
3
3,000
4
3,000
5
1,000
6
2,000
500
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Problems 305
6.2 Consider the accompanying cash flow diagram. Compute the equivalent annual
worth at i = 12%.
$20,000
Years
1
2
3
4
5
6
$5,000
$8,000
$11,000
$14,000
$17,000
6.3 Consider the accompanying cash flow diagram. Compute the equivalent annual
worth at i = 10%.
$5,000
$6,000
$4,000
$3,000
0
1
2
3
4
5
6
Years
$3,000
6.4 Consider the accompanying cash flow diagram. Compute the equivalent annual
worth at i = 13%.
$7,000
$6,000
$5,000
$4,000
$3,000
$2,000
0
1
2
$4,000
$8,000
3
4
5
Years
$4,000
6
$1,000
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306 CHAPTER 6 Annual Equivalent-Worth Analysis
6.5 Consider the accompanying cash flow diagram. Compute the equivalent annual
worth at i = 8%.
$2,000
$1,500
$1,000
$1,500
$2,000
0
1
2
3
4
Years
5
6
$5,000
6.6 Consider the following sets of investment projects:
Project’s Cash Flow ($)
n
A
0 - $2,500
B
C
D
- $4,500
- $8,000
- $12,000
1
400
3,000
-2,000
2,000
2
500
2,000
6,000
4,000
3
600
1,000
2,000
8,000
4
700
500
4,000
8,000
5
800
500
2,000
4,000
Compute the equivalent annual worth of each project at i = 10%, and determine
the acceptability of each project.
6.7 Sun-Devil Company is producing electricity directly from a solar source by using
a large array of solar cells and selling the power to the local utility company. Because these cells degrade over time, thereby resulting in lower conversion efficiency
and power output, the cells must be replaced every four years, which results in a
particular cash flow pattern that repeats itself as follows: n = 0, - $500,000; n = 1,
$600,000; n = 2, $400,000; n = 3, $300,000, and n = 4, $200,000. Determine
the annual equivalent cash flows at i = 12%.
6.8 Consider the following sets of investment projects:
Project’s Cash Flow
n
A
B
C
D
0
- $7,500
- $4,000
- $5,000
- $6,600
1
0
1,500
4,000
3,800
2
0
1,800
3,000
3,800
3
15,500
2,100
2,000
3,800
Compute the equivalent annual worth of each project at i = 13%, and determine
the acceptability of each project.
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Problems 307
6.9 The cash flows for a certain project are as follows:
Net Cash Flow
n
Investment
0
- $800
Operating Income
1st cycle
1
$900
2
700
3
-800
500
4
900
5
700
6
-800
500
7
900
8
700
9
500
2nd cycle
3rd cycle
Find the equivalent annual worth for this project at i = 10%, and determine the
acceptability of the project.
6.10 Beginning next year, a foundation will support an annual seminar on campus
by the earnings of a $100,000 gift it received this year. It is felt that 8% interest
will be realized for the first 10 years, but that plans should be made to anticipate an interest rate of 6% after that time. What amount should be added to the
foundation now to fund the seminar at the $10,000 level into infinity?
Capital (Recovery) Cost/Annual Equivalent Cost
6.11 The owner of a business is considering investing $55,000 in new equipment.
He estimates that the net cash flows will be $5,000 during the first year, but will
increase by $2,500 per year the next year and each year thereafter. The equipment is estimated to have a 10-year service life and a net salvage value of
$6,000 at that time. The firm’s interest rate is 12%.
(a) Determine the annual capital cost (ownership cost) for the equipment.
(b) Determine the equivalent annual savings (revenues).
(c) Determine whether this is a wise investment.
6.12 You are considering purchasing a dump truck. The truck will cost $45,000 and
have an operating and maintenance cost that starts at $15,000 the first year and
increases by $2,000 per year. Assume that the salvage value at the end of five
years is $9,000 and interest rate is 12%. What is the equivalent annual cost of
owning and operating the truck?
6.13 Emerson Electronics Company just purchased a soldering machine to be used in
its assembly cell for flexible disk drives. The soldering machine cost $250,000.
Because of the specialized function it performs, its useful life is estimated to be
five years. It is also estimated that at that time its salvage value will be $40,000.
What is the capital cost for this investment if the firm’s interest rate is 18%?
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6.14 The present price (year 0) of kerosene is $1.80 per gallon, and its cost is expected
to increase by $.15 per year. (At the end of year 1, kerosene will cost $1.95 per
gallon.) Mr. Garcia uses about 800 gallons of kerosene for space heating during a
winter season. He has an opportunity to buy a storage tank for $600, and at the end
of four years he can sell the storage tank for $100. The tank has a capacity to supply four years of Mr. Garcia’s heating needs, so he can buy four years’ worth of
kerosene at its present price ($1.80), or he can invest his money elsewhere at 8%.
Should he purchase the storage tank? Assume that kerosene purchased on a payas-you-go basis is paid for at the end of the year. (However, kerosene purchased
for the storage tank is purchased now.)
6.15 Consider the following advertisement, which appeared in a local paper.
Pools-Spas-Hot Tubs—Pure Water without Toxic Chemicals: The comparative
costs between the conventional chemical system (chlorine) and the IONETICS
systems are as follows:
Conventional
System
Item
IONETICS
System
Annual costs
Chemical
$471
IONETICS
$
Pump
($0.667/kWh)
$576
85
$ 100
Capital investment
$1,200
Note that the IONETICS system pays for itself in less than 2 years.
Assume that the IONETICS system has a 12-year service life and the interest rate
is 6%. What is the equivalent annual cost of operating the IONETICS system?
6.16 The cash flows for two investment projects are as follows:
Project’s Cash Flow
n
A
B
0
- $4,000
$5,500
1
1,000
-1,400
2
X
-1,400
3
1,000
-1,400
4
1,000
-1,400
(a) For project A, find the value of X that makes the equivalent annual receipts
equal the equivalent annual disbursement at i = 13%.
(b) Would you accept project B at i = 15%, based on an AE criterion?
6.17 An industrial firm can purchase a special machine for $50,000. A down payment
of $5,000 is required, and the unpaid balance can be paid off in five equal year-end
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Problems 309
installments at 7% interest. As an alternative, the machine can be purchased for
$46,000 in cash. If the firm’s MARR is 10%, use the annual equivalent method to
determine which alternative should be accepted.
6.18 An industrial firm is considering purchasing several programmable controllers
and automating the company’s manufacturing operations. It is estimated that the
equipment will initially cost $100,000 and the labor to install it will cost $35,000.
A service contract to maintain the equipment will cost $5,000 per year. Trained
service personnel will have to be hired at an annual salary of $30,000. Also estimated is an approximate $10,000 annual income-tax savings (cash inflow). How
much will this investment in equipment and services have to increase the annual
revenues after taxes in order to break even? The equipment is estimated to have an
operating life of 10 years, with no salvage value because of obsolescence. The
firm’s MARR is 10%.
6.19 A construction firm is considering establishing an engineering computing center.
The center will be equipped with three engineering workstations that cost $35,000
each, and each has a service life of five years. The expected salvage value of each
workstation is $2,000. The annual operating and maintenance cost would be
$15,000 for each workstation. At a MARR of 15%, determine the equivalent annual cost for operating the engineering center.
Unit-Cost Profit Calculation
6.20 You have purchased a machine costing $20,000. The machine will be used for two
years, at the end of which time its salvage value is expected to be $10,000. The
machine will be used 6,000 hours during the first year and 8,000 hours during the
second year. The expected annual net savings will be $30,000 during the first year
and $40,000 during the second year. If your interest rate is 10%, what would be
the equivalent net savings per machine hour?
6.21 The engineering department of a large firm is overly crowded. In many cases, several engineers share one office. It is evident that the distraction caused by the
crowded conditions reduces the productive capacity of the engineers considerably.
Management is considering the possibility of providing new facilities for the department, which could result in fewer engineers per office and a private office for
some. For an office presently occupied by five engineers, what minimum individual increase in effectiveness must result to warrant the assignment of only three
engineers to an office if the following data apply?
•
•
•
•
•
•
•
The office size is 16 * 20 feet.
The average annual salary of each engineer is $80,000.
The cost of the building is $110 per square foot.
The estimated life of the building is 25 years.
The estimated salvage value of the building is 10% of the initial cost.
The annual taxes, insurance, and maintenance are 6% of the initial cost.
The cost of janitorial service, heating, and illumination is $5.00 per square foot
per year.
• The interest rate is 12%.
Assume that engineers reassigned to other office space will maintain their present
productive capability as a minimum.
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6.22 Sam Tucker is a sales engineer at Buford Chemical Engineering Company. Sam
owns two vehicles, and one of them is entirely dedicated to business use. His business car is a used small pickup truck, which he purchased with $11,000 of personal savings. On the basis of his own records and with data compiled by the U.S.
Department of Transportation, Sam has estimated the costs of owning and operating his business vehicle for the first three years as follows:
First Year
Depreciation
Second Year
Third Year
$2,879
$1,776
$1,545
Scheduled maintenance
100
153
220
Insurance
635
635
635
78
57
50
$3,692
$2,621
$2,450
Nonscheduled repairs
35
85
200
Replacement tires
35
30
27
Accessories
15
13
12
688
650
522
80
100
100
135
125
110
$988
$1,003
$971
Total of all costs
$4,680
$3,624
$3,421
Expected miles driven
14,500
13,000
11,500
Registration and taxes
Total ownership cost
Gasoline and taxes
Oil
Parking and tolls
Total operating costs
If his interest rate is 6%, what should be Sam’s reimbursement rate per mile so
that he can break even?
6.23 Two 150-horsepower (HP) motors are being considered for installation at a municipal sewage-treatment plant. The first costs $4,500 and has an operating efficiency of 83%. The second costs $3,600 and has an efficiency of 80%. Both
motors are projected to have zero salvage value after a life of 10 years. If all the
annual charges, such as insurance, maintenance, etc., amount to a total of 15% of
the original cost of each motor, and if power costs are a flat 5 cents per kilowatthour, how many minimum hours of full-load operation per year are necessary to
justify purchasing the more expensive motor at i = 6%? (A conversion factor
you might find useful is 1 HP = 746 watts = 0.746 kilowatts.)
6.24 Danford Company, a manufacturer of farm equipment, currently produces 20,000
units of gas filters per year for use in its lawn-mower production. The costs, based
on the previous year’s production, are reported in Table P6.24.
It is anticipated that gas-filter production will last five years. If the company continues to produce the product in-house, annual direct material costs will increase
at the rate of 5%. (For example, annual material costs during the first production
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Problems 311
TABLE P6.24
Production costs
Item
Expense ($)
Direct materials
$ 60,000
Direct labor
180,000
Variable overhead
(power and water)
135,000
Fixed overhead
(light and heat)
Total cost
70,000
$445,000
year will be $63,000.) Direct labor will also increase at the rate of 6% per year.
However, variable overhead costs will increase at the rate of 3%, but the fixed
overhead will remain at its current level over the next five years. John Holland
Company has offered to sell Danford 20,000 units of gas filters for $25 per unit. If
Danford accepts the offer, some of the manufacturing facilities currently used to
manufacture the filter could be rented to a third party for $35,000 per year. In addition, $3.5 per unit of the fixed overheard applied to the production of gas filters
would be eliminated. The firm’s interest rate is known to be 15%. What is the unit
cost of buying the gas filter from the outside source? Should Danford accept John
Holland’s offer, and why?
6.25 Southern Environmental Consulting (SEC), Inc., designs plans and specifications
for asbestos abatement (removal) projects in public, private, and governmental
buildings. Currently, SEC must conduct an air test before allowing the reoccupancy of a building from which asbestos has been removed. SEC subcontracts air-test
samples to a laboratory for analysis by transmission electron microscopy (TEM).
To offset the cost of TEM analysis, SEC charges its clients $100 more than the
subcontractor’s fee. The only expenses in this system are the costs of shipping the
air-test samples to the subcontractor and the labor involved in shipping the samples. With the growth of the business, SEC is having to consider either continuing
to subcontract the TEM analysis to outside companies or developing its own TEM
laboratory. Because of the passage of the Asbestos Hazard Emergency Response
Act (AHERA) by the U.S. Congress, SEC expects about 1,000 air-sample testings
per year over eight years. The firm’s MARR is known to be 15%.
• Subcontract option. The client is charged $400 per sample, which is $100
above the subcontracting fee of $300. Labor expenses are $1,500 per year, and
shipping expenses are estimated to be $0.50 per sample.
• TEM purchase option. The purchase and installation cost for the TEM is
$415,000. The equipment would last for eight years, at which time it should have
no salvage value. The design and renovation cost is estimated to be $9,500. The
client is charged $300 per sample, based on the current market price. One full-time
manager and two part-time technicians are needed to operate the laboratory. Their
combined annual salaries will be $50,000. Material required to operate the lab includes carbon rods, copper grids, filter equipment, and acetone. The costs of these
materials are estimated at $6,000 per year. Utility costs, operating and maintenance
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6.26
6.27
6.28
6.29
costs, and the indirect labor needed to maintain the lab are estimated at $18,000
per year. The extra income-tax expenses would be $20,000.
(a) Determine the cost of an air-sample test by the TEM (in-house).
(b) What is the required number of air samples per year to make the two options
equivalent?
A company is currently paying its employees $0.38 per mile to drive their own
cars on company business. The company is considering supplying employees with
cars, which would involve purchasing at $25,000, with an estimated three-year
life, a net salvage value of $8,000, taxes and insurance at a cost of $900 per year,
and operating and maintenance expenses of $0.22 per mile. If the interest rate is
10% and the company anticipates an employee’s annual travel to be 22,000 miles,
what is the equivalent cost per mile (neglecting income taxes)?
An automobile that runs on electricity can be purchased for $25,000. The automobile is estimated to have a life of 12 years with annual travel of 20,000 miles.
Every 3 years, a new set of batteries will have to be purchased at a cost of $3,000.
Annual maintenance of the vehicle is estimated to cost $700. The cost of recharging the batteries is estimated at $0.015 per mile. The salvage value of the batteries
and the vehicle at the end of 12 years is estimated to be $2,000. Suppose the
MARR is 7%. What is the cost per mile to own and operate this vehicle, based on
the preceding estimates? The $3,000 cost of the batteries is a net value, with the
old batteries traded in for the new ones.
The estimated cost of a completely installed and ready-to-operate 40-kilowatt generator is $30,000. Its annual maintenance costs are estimated at $500. The energy
that can be generated annually at full load is estimated to be 100,000 kilowatt-hours.
If the value of the energy generated is $0.08 per kilowatt-hour, how long will it take
before this machine becomes profitable? Take the MARR to be 9% and the salvage
value of the machine to be $2,000 at the end of its estimated life of 15 years.
A large land-grant university that is currently facing severe parking problems on its
campus is considering constructing parking decks off campus. A shuttle service could
pick up students at the off-campus parking deck and transport them to various locations on campus. The university would charge a small fee for each shuttle ride, and the
students could be quickly and economically transported to their classes. The funds
raised by the shuttle would be used to pay for trolleys, which cost about $150,000
each. Each trolley has a 12-year service life, with an estimated salvage value of
$3,000. To operate each trolley, the following additional expenses will be incurred:
Item
Annual Expenses ($)
Driver
$50,000
Maintenance
Insurance
10,000
3,000
If students pay 10 cents for each ride, determine the annual ridership per trolley
(number of shuttle rides per year) required to justify the shuttle project, assuming
an interest rate of 6%.
6.30 Eradicator Food Prep, Inc., has invested $7 million to construct a food irradiation
plant. This technology destroys organisms that cause spoilage and disease, thus
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extending the shelf life of fresh foods and the distances over which they can be
shipped. The plant can handle about 200,000 pounds of produce in an hour, and it
will be operated for 3,600 hours a year. The net expected operating and maintenance costs (taking into account income-tax effects) would be $4 million per year.
The plant is expected to have a useful life of 15 years, with a net salvage value of
$700,000. The firm’s interest rate is 15%.
(a) If investors in the company want to recover the plant investment within 6 years
of operation (rather than 15 years), what would be the equivalent after-tax annual revenues that must be generated?
(b) To generate annual revenues determined in part (a), what minimum processing
fee per pound should the company charge to its producers?
6.31 The local government of Santa Catalina Island, off the coast of Long Beach, California, is completing plans to build a desalination plant to help ease a critical
drought on the island. The drought has combined with new construction on Catalina
to leave the island with an urgent need for a new water source. A modern desalination plant could produce fresh water from seawater for $1,000 an acre-foot (326,000
gallons), or enough to supply two households for 1 year. On Catalina, the cost of acquiring water from natural sources is about the same as that for desalting. The
$3 million plant, with a daily desalting capacity of 0.4 acre-foot, can produce
132,000 gallons of fresh water a day (enough to supply 295 households daily), more
than a quarter of the island’s total needs. The desalination plant has an estimated
service life of 20 years, with no appreciable salvage value. The annual operating and
maintenance costs would be about $250,000. Assuming an interest rate of 10%,
what should be the minimum monthly water bill for each household?
6.32 A California utility firm is considering building a 50-megawatt geothermal plant
that generates electricity from naturally occurring underground heat. The binary
geothermal system will cost $85 million to build and $6 million (including any income-tax effect) to operate per year. (Virtually no fuel costs will accrue compared
with fuel costs related to a conventional fossil-fuel plant.) The geothermal plant is
to last for 25 years. At that time, its expected salvage value will be about the same
as the cost to remove the plant. The plant will be in operation for 70% (plant utilization factor) of the year (or 70% of 8,760 hours per year). If the firm’s MARR
is 14% per year, determine the cost per kilowatt-hour of generating electricity.
6.33 A corporate executive jet with a seating capacity of 20 has the following cost factors:
Item
Initial cost
Service life
Salvage value
Crew costs per year
Cost
$12,000,000
15 years
$2,000,000
$225,000
Fuel cost per mile
$1.10
Landing fee
$250
Maintenance per year
$237,500
Insurance cost per year
$166,000
Catering per passenger trip
$75
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314 CHAPTER 6 Annual Equivalent-Worth Analysis
The company flies three round trips from Boston to London per week, a distance
of 3,280 miles one way. How many passengers must be carried on an average trip
in order to justify the use of the jet if the first-class round-trip fare is $3,400? The
firm’s MARR is 15%. (Ignore income-tax consequences.)
Comparing Mutually Exclusive Alternatives by the AE Method
6.34 The following cash flows represent the potential annual savings associated with
two different types of production processes, each of which requires an investment
of $12,000:
n
Process A
Process B
0
- $12,000
- $12,000
1
9,120
6,350
2
6,840
6,350
3
4,560
6,350
4
2,280
6,350
Assuming an interest rate of 15%,
(a) Determine the equivalent annual savings for each process.
(b) Determine the hourly savings for each process if it is in operation 2,000 hours
per year.
(c) Which process should be selected?
6.35 Birmingham Steel, Inc., is considering replacing 20 conventional 25-HP, 230-V,
60-Hz, 1800-rpm induction motors in its plant with modern premium efficiency
(PE) motors. Both types of motors have a power output of 18.650 kW per motor
125 HP * 0.746 kW/HP2. Conventional motors have a published efficiency of
89.5%, while the PE motors are 93% efficient. The initial cost of the conventional
motors is $13,000, whereas the initial cost of the proposed PE motors is $15,600.
The motors are operated 12 hours per day, five days per week, 52 weeks per year,
with a local utility cost of $0.07 per kilowatt-hour (kWh). The motors are to be operated at 75% load, and the life cycle of both the conventional and PE motor is 20
years, with no appreciable salvage value.
(a) At an interest rate of 13%, what are the savings per kWh achieved by switching from the conventional motors to the PE motors?
(b) At what operating hours are the two motors equally economical?
6.36 A certain factory building has an old lighting system, and lighting the building
costs, on average, $20,000 a year. A lighting consultant tells the factory supervisor
that the lighting bill can be reduced to $8,000 a year if $50,000 were invested in
relighting the building. If the new lighting system is installed, an incremental
maintenance cost of $3,000 per year must be taken into account. The new lighting
system has zero salvage value at the end of its life. If the old lighting system also
has zero salvage value, and the new lighting system is estimated to have a life of
20 years, what is the net annual benefit for this investment in new lighting? Take
the MARR to be 12%.
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6.37 Travis Wenzel has $2,000 to invest. Usually, he would deposit the money in his
savings account, which earns 6% interest compounded monthly. However, he is
considering three alternative investment opportunities:
• Option 1. Purchasing a bond for $2,000. The bond has a face value of $2,000 and
pays $100 every six months for three years, after which time the bond matures.
• Option 2. Buying and holding a stock that grows 11% per year for three years.
• Option 3. Making a personal loan of $2,000 to a friend and receiving $150 per
year for three years.
Determine the equivalent annual cash flows for each option, and select the best
option.
6.38 A chemical company is considering two types of incinerators to burn solid waste
generated by a chemical operation. Both incinerators have a burning capacity of
20 tons per day. The following data have been compiled for comparison:
Incinerator A
Incinerator B
Installed
cost
$1,200,000
$750,000
Annual
O&M costs
$
50,000
$ 80,000
20 years
10 years
Service life
Salvage
value
$
60,000
$ 30,000
Income
taxes
$
40,000
$ 30,000
If the firm’s MARR is known to be 13%, determine the processing cost per ton of
solid waste incurred by each incinerator. Assume that incinerator B will be available in the future at the same cost.
6.39 Consider the cash flows for the following investment projects 1MARR = 15%2:
Project’s Cash Flow
n
A
B
C
0
- $2,500
- $4,000
- $5,000
1
1,000
1,600
1,800
2
1,800
1,500
1,800
3
1,000
1,500
2,000
4
400
1,500
2,000
(a) Suppose that projects A and B are mutually exclusive. Which project would
you select, based on the AE criterion?
(b) Assume that projects B and C are mutually exclusive. Which project would
you select, based on the AE criterion?
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Life-Cycle Cost Analysis
6.40 An airline is considering two types of engine systems for use in its planes. Each
has the same life and the same maintenance and repair record.
• System A costs $100,000 and uses 40,000 gallons per 1,000 hours of operation
at the average load encountered in passenger service.
• System B costs $200,000 and uses 32,000 gallons per 1,000 hours of operation
at the same level.
Both engine systems have three-year lives before any major overhaul is required. On
the basis of the initial investment, the systems have 10% salvage values. If jet fuel
currently costs $2.10 a gallon and fuel consumption is expected to increase at the rate
of 6% per year because of degrading engine efficiency, which engine system should
the firm install? Assume 2,000 hours of operation per year and a MARR of 10%. Use
the AE criterion. What is the equivalent operating cost per hour for each engine?
6.41 Mustang Auto-Parts, Inc., is considering one of two forklift trucks for its assembly
plant. Truck A costs $15,000 and requires $3,000 annually in operating expenses.
It will have a $5,000 salvage value at the end of its three-year service life. Truck B
costs $20,000, but requires only $2,000 annually in operating expenses; its service
life is four years, at which time its expected salvage value will be $8,000. The
firm’s MARR is 12%. Assuming that the trucks are needed for 12 years and that
no significant changes are expected in the future price and functional capacity of
each truck, select the most economical truck, on the basis of AE analysis.
6.42 A small manufacturing firm is considering purchasing a new machine to modernize
one of its current production lines. Two types of machines are available on the market. The lives of machine A and machine B are four years and six years, respectively,
but the firm does not expect to need the service of either machine for more than five
years. The machines have the following expected receipts and disbursements:
Item
Machine A
Machine B
First cost
$6,500
$8,500
Service life
4 years
6 years
Estimated
salvage value
$600
$1,000
Annual O&M costs
$800
$520
Change oil filter
every other year
$100
None
$200
(every
3 years)
$280
(every
4 years)
Engine overhaul
The firm always has another option: leasing a machine at $3,000 per year, fully
maintained by the leasing company. After four years of use, the salvage value for
machine B will remain at $1,000.
(a) How many decision alternatives are there?
(b) Which decision appears to be the best at i = 10%?
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6.43 A plastic-manufacturing company owns and operates a polypropylene production
facility that converts the propylene from one of its cracking facilities to polypropylene plastics for outside sale. The polypropylene production facility is currently
forced to operate at less than capacity due to an insufficiency of propylene production capacity in its hydrocarbon cracking facility. The chemical engineers are considering alternatives for supplying additional propylene to the polypropylene
production facility. Two feasible alternatives are to build a pipeline to the nearest
outside supply source and to provide additional propylene by truck from an outside
source. The engineers also gathered the following projected cost estimates:
•
•
•
•
•
•
•
•
Future costs for purchased propylene excluding delivery: $0.215 per lb.
Cost of pipeline construction: $200,000 per pipeline mile.
Estimated length of pipeline: 180 miles.
Transportation costs by tank truck: $0.05 per lb, utilizing a common carrier.
Pipeline operating costs: $0.005 per lb, excluding capital costs.
Projected additional propylene needs: 180 million lb per year.
Projected project life: 20 years.
Estimated salvage value of the pipeline: 8% of the installed costs.
Determine the propylene cost per pound under each option if the firm’s MARR is
18%. Which option is more economical?
6.44 The City of Prattsville is comparing two plans for supplying water to a newly developed subdivision:
• Plan A will take care of requirements for the next 15 years, at the end of which
time the initial cost of $400,000 will have to be duplicated to meet the requirements of subsequent years. The facilities installed at dates 0 and 15 may be considered permanent; however, certain supporting equipment will have to be
replaced every 30 years from the installation dates, at a cost of $75,000. Operating costs are $31,000 a year for the first 15 years and $62,000 thereafter, although they are expected to increase by $1,000 a year beginning in the 21st year.
• Plan B will supply all requirements for water indefinitely into the future, although it will be operated only at half capacity for the first 15 years. Annual
costs over this period will be $35,000 and will increase to $55,000 beginning in
the 16th year. The initial cost of Plan B is $550,000; the facilities can be considered permanent, although it will be necessary to replace $150,000 worth of
equipment every 30 years after the initial installation.
The city will charge the subdivision for the use of water on the basis of the equivalent annual cost. At an interest rate of 10%, determine the equivalent annual cost
for each plan, and make a recommendation to the city.
Minimum-Cost Analysis
6.45 A continuous electric current of 2,000 amps is to be transmitted from a generator
to a transformer located 200 feet away. A copper conductor can be installed for $6
per pound, will have an estimated life of 25 years, and can be salvaged for $1 per
pound. Power loss from the conductor will be inversely proportional to the crosssectional area of the conductor and may be expressed as 6.516/A kilowatt, where
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318 CHAPTER 6 Annual Equivalent-Worth Analysis
A is in square inches. The cost of energy is $0.0825 per kilowatt-hour, the interest
rate is 11%, and the density of copper is 555 pounds per cubic foot.
(a) Calculate the optimum cross-sectional area of the conductor.
(b) Calculate the annual equivalent total cost for the value you obtained in part (a).
(c) Graph the two individual cost factors (capital cost and power-loss cost) and the
total cost as a function of the cross-sectional area A, and discuss the impact of
increasing energy cost on the optimum obtained in part (a).
Short Case Studies
ST6.1 Automotive engineers at Ford are considering the laser blank welding (LBW)
technique to produce a windshield frame rail blank. The engineers believe that,
compared with the conventional sheet metal blanks, LBW would result in a significant savings as follows:
1. Scrap reduction through more efficient blank nesting on coil.
2. Scrap reclamation (weld scrap offal into a larger usable blank).
The use of a laser welded blank provides a reduction in engineered scrap for the
production of a window frame rail blank.
On the basis of an annual volume of 3,000 blanks, Ford engineers have estimated
the following financial data:
Blanking Method
Description
Conventional
Laser Blank
Welding
Weight per
blank (lb/part)
63.764
34.870
Steel cost/part
$
14.98
$
8.19
Transportation/part
$
0.67
$
0.42
Blanking/part
$
0.50
$
0.40
Die investment
$106,480
$83,000
The LBW technique appears to achieve significant savings, so Ford’s engineers
are leaning toward adopting it. Since the engineers have had no previous experience with LBW, they are not sure whether producing the windshield frames inhouse at this time is a good strategy. For this windshield frame, it may be cheaper
to use the services of a supplier that has both the experience with, and the machinery for, laser blanking. Ford’s lack of skill in laser blanking may mean that it
will take six months to get up to the required production volume. If, however, Ford
relies on a supplier, it can only assume that supplier labor problems will not halt
the production of Ford’s parts. The make-or-buy decision depends on two factors:
the amount of new investment that is required in laser welding and whether additional machinery will be required for future products. Assuming a lifetime of 10 years
and an interest rate of 16%, recommend the best course of action. Assume also that
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72.00
Conventional
processing
the salvage value at the end of 10 years is estimated to be insignificant for either
system. If Ford considers the subcontracting option, what would be the acceptable
range of contract bid (unit cost per part)?
coil feed
46.25
94.29
coil feed
laser weld
46.25
Laser welded
blank
39.12
coil feed
40.92
ST6.2 The proliferation of computers into all aspects of business has created an everincreasing need for data capture systems that are fast, reliable, and cost effective.
One technology that has been adopted by many manufacturers, distributors, and
retailers is a bar-coding system. Hermes Electronics, a leading manufacturer of
underwater surveillance equipment, evaluated the economic benefits of installing
an automated data acquisition system into its plant. The company could use the
system on a lim-ited scale, such as for tracking parts and assemblies for inventory
management, or it could opt for a broader implementation by recording information that is useful for quality control, operator efficiency, attendance, and other
functions. All these aspects are currently monitored, but although computers are
used to manage the information, the recording is conducted primarily manually.
The advantages of an automated data collection system, which include faster and
more accurate data capture, quicker analysis of and response to production
changes, and savings due to tighter control over operations, could easily outweigh
the cost of the new system. Two alternative systems from competing suppliers are
under consideration:
• System 1 relies on handheld bar-code scanners that transmit radio frequencies. The
hub of this wireless network can then be connected to the company’s existing LAN
and integrated with its current MRP II system and other management software.
• System 2 consists primarily of specialized data terminals installed at every collection point, with connected bar-code scanners where required. This system is
configured in such a way as to facilitate phasing in the components over two
stages or installing the system all at once. The former would allow Hermes to
defer some of the capital investment, while becoming thoroughly familiar with
the functions introduced in the first stage.
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Either of these systems would satisfy Hermes’s data collection needs. They each
have some unique elements, and the company needs to compare the relative benefits of the features offered by each system. From the point of view of engineering
economics, the two systems have different capital costs, and their operating and
maintenance costs are not identical. One system may also be rated to last longer
than the other before replacement is required, particularly if the option of acquiring System 2 in phases is selected. Discuss many issues to be considered before
making the best choice of those types of technology in manufacturing.
ST6.3 A Veterans Administration (VA) hospital is to decide which type of boiler fuel system will most efficiently provide the required steam energy output for heating,
laundry, and sterilization purposes. The current boilers were installed in the early
1950s and are now obsolete. Much of the auxiliary equipment is also old and in
need of repair. Because of these general conditions, an engineering recommendation was made to replace the entire plant with a new boiler plant building that
would house modern equipment. The cost of demolishing the old boiler plant
would be almost a complete loss, as the salvage value of the scrap steel and used
brick was estimated to be only about $1,000. The VA hospital’s engineer finally
selected two alternative proposals as being worthy of more intensive analysis. The
hospital’s annual energy requirement, measured in terms of steam output, is approximately 145,000,000 pounds of steam. As a rule of thumb for analysis, 1 pound
of steam is approximately 1,000 Btu, and 1 cubic foot of natural gas is also approximately 1,000 Btu. The two alternatives are as follows:
• Proposal 1. Replace the old plant with a new coal-fired boiler plant that costs
$1,770,300. To meet the requirements for particulate emission as set by the Environmental Protection Agency, this coal-fired boiler, even if it burned low-sulfur
coal, would need an electrostatic precipitator, which would cost approximately
$100,000. The plant would last for 20 years. One pound of dry coal yields about
14,300 Btu. To convert the 145,000,000 pounds of steam energy to the common
denominator of Btu, it is necessary to multiply by 1,000. To find the Btu input
requirements, it is necessary to divide by the relative boiler efficiency for the
type of fuel. The boiler efficiency for coal is 0.75. The price of coal is estimated
to be $55.50 per ton.
• Proposal 2. Build a gas-fired boiler plant with No. 2 fuel oil, and use the new
plant as a standby. This system would cost $889,200 and have an expected service life of 20 years. Since small household or commercial gas users that are entirely dependent on gas have priority, large plants must have an oil switchover
capability. It has been estimated that 6% of 145,000,000 pounds of steam energy (or 8,700,000 pounds) would come about as a result of the switch to oil.
The boiler efficiency with each fuel would be 0.78 for gas and 0.81 for oil, respectively. The heat value of natural gas is approximately 1,000,000 Btu/MCF
(thousand cubic feet), and for No. 2 fuel oil it is 139,400 Btu/gal. The estimated
gas price is $9.50/MCF, and the price of No. 2 fuel oil is $1.35 per gallon.
(a) Calculate the annual fuel costs for each proposal.
(b) Determine the unit cost per steam pound for each proposal. Assume that
i = 10%.
(c) Which proposal is the more economical?
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ST6.4 The following is a letter that I received from a local city engineer:
Dear Professor Park:
Thank you for taking the time to assist with this problem. I’m really embarrassed
at not being able to solve it myself, since it seems rather straightforward. The situation is as follows:
A citizen of Opelika paid for concrete drainage pipe approximately 20 years ago
to be installed on his property. (We have a policy that if drainage trouble exists on
private property and the owner agrees to pay for the material, city crews will install it.) That was the case in this problem. Therefore, we are dealing with only
material costs, disregarding labor.
However, this past year, we removed the pipe purchased by the citizen, due to a
larger area drainage project. He thinks, and we agree, that he is due some refund
for salvage value of the pipe due to its remaining life.
Problem:
• Known: 80¿ of 48– pipe purchased 20 years ago. Current quoted price of
48– pipe = $52.60/foot, times 80 feet = $4,208 total cost in today’s dollars.
• Unknown: Original purchase price.
• Assumptions: 50-year life; therefore, assume 30 years of life remaining at removal after 20 years. A 4% price increase per year, average, over 20 years.
Thus, we wish to calculate the cost of the pipe 20 years ago. Then we will calculate, in today’s dollars, the present salvage value after 20 years, use with 30 years
of life remaining. Thank you again for your help. We look forward to your reply.
Charlie Thomas, P.E.
Director of Engineering
City of Opelika
After reading this letter, recommend a reasonable amount of compensation to the
citizen for the replaced drainage pipe.
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SEVEN
CHAPTER
Rate-of-Return Analysis
Will That be Cash, Credit—or Fingertip?1 Have you ever
found yourself short of cash or without a wallet when you want to
buy something? Consider the following two types of technologies
available in retail stores to speed up checkouts:
• Pay By Touch takes fingerprints when customers enroll in the
program. The image is then converted to about 40 unique points
of the finger. Those points are stored in a computer system
with “military-level encryption.” They want this to be your cash
replacement because of the time savings, and a lot of customers
who are paying cash will find it more convenient now to use these
cards.
• A contactless card allows the shopper to pay in seconds by
waving his or her contactless card in front of a reader, which lights
up and beeps to tell the shopper the transaction is done.
A contactless payment is twice as fast as a no-signature credit card
purchase and three times as fast as using cash.That’s why it’s
catching on at fast-food restaurants and convenience stores.
These stores’ profits depend, in part, on how quickly they get
customers—typically with small purchases—through the line.
These new technologies being rolled out at convenience stores,
supermarkets, and gas stations could some day make it passé to carry
bulky wallets. Without the need to dig for cash and checks at the
register, the quick stop-and-go payments promise speedier transactions for consumers—and perhaps fatter profits for retailers.
The appeal is that there’s no need to run them through a machine. A
contactless-card transaction is usually more expensive for a retailer to
process than a cash payment. But retailers that adopt contactless payments hope they’ll bring in more customers, offsetting higher costs. If
that turns out to be false, then some could turn their backs on the
new technology.
1
322
“Will that be cash, credit—or fingertip?” Kathy Chu, USA Today, Section B1, Friday, December 2, 2005.
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Getting Through the Checkout Line Faster
Contactless Payment
1. Hold the card 1 to
2 inches from
electronic reader.
1. Place your finger
on the electronic
reader.
2. Microchip in card
helps encrypt data
and sends account
number.
2. Unique points
from your finger
identify you.
3. Unique transaction number generated to deter
fraud.
3. Enter phone number and choose
payment method.
4. No signing, no
swiping. Transaction processed
normally.
4. No signing, no
swiping. Transaction processed
normally.
One retailer who just installed a Pay By Touch™ system hopes to increase its
customer traffic so that a 10% return on investment can be attained. The Pay
By Touch™ scanners cost about $50 each, the monthly service fee ranges
between $38 and $45, and each transaction fee costs 10 cents. In a society
driven by convenience, anything that speeds up the payment process attracts
consumers. But technology providers will need to convince consumers of the
safety of their information before the technologies can become a staple in the
checkout line.
What does the 10% rate of return for the retailer really represent? How
do we compute the figure from the projected additional retail revenues?
And once computed, how do we use the figure when evaluating an investment project? Our consideration of the concept of rate of return in this
chapter will answer these and other questions.
Along with the NPW and the AE criteria, the third primary measure of investment worth is rate of return. As shown in Chapter 5, the NPW measure
is easy to calculate and apply. Nevertheless, many engineers and financial managers prefer rate-of-return analysis to the NPW method because they find it
323
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324 CHAPTER 7 Rate-of-Return Analysis
intuitively more appealing to analyze investments in terms of percentage rates of return
rather than dollars of NPW. Consider the following statements regarding an investment’s
profitability:
• This project will bring in a 15% rate of return on the investment.
• This project will result in a net surplus of $10,000 in the NPW.
Neither statement describes the nature of the investment project in any complete
sense. However, the rate of return is somewhat easier to understand because many of us
are so familiar with savings-and-loan interest rates, which are in fact rates of return.
In this chapter, we will examine four aspects of rate-of-return analysis: (1) the concept of return on investment, (2) the calculation of a rate of return, (3) the development of
an internal rate-of-return criterion, and (4) the comparison of mutually exclusive alternatives based on a rate of return.
CHAPTER LEARNING OBJECTIVES
After completing this chapter, you should understand the following concepts:
The meaning of the rate of return.
The various methods to compute the rate of return.
How you make an accept and reject decision with the rate of return.
How to resolve the multiple rates of return problem.
How you conduct an incremental analysis with the rate of return.
7.1 Rate of Return
Yield: The
annual rate of
return on an
investment,
expressed as a
percentage.
any different terms are used to refer to rate of return, including yield (i.e.,
the yield to maturity, commonly used in bond valuation), internal rate of
return, and marginal efficiency of capital. First we will review three
common definitions of rate of return. Then we will use the definition of internal rate of return as a measure of profitability for a single investment project
throughout the text.
M
7.1.1 Return on Investment
There are several ways of defining the concept of a rate of return on investment. The first
is based on a typical loan transaction, the second on the mathematical expression of the
present-worth function, and the third on the project cash flow series.
Definition 1
The rate of return is the interest rate earned on the unpaid balance of an amortized loan.
Suppose that a bank lends $10,000 and is repaid $4,021 at the end of each year for
three years. How would you determine the interest rate that the bank charges on this
transaction? As we learned in Chapter 3, you would set up the equivalence equation
$10,000 = $4,0211P>A, i, 32
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Section 7.1 Rate of Return
and solve for i. It turns out that i = 10%. In this situation, the bank will earn a return of
10% on its investment of $10,000. The bank calculates the balances over the life of the
loan as follows:
Year
Unpaid Balance
at Beginning
of Year
Return on
Unpaid
Balance (10%)
Payment
Received
Unpaid
Balance at
End of Year
0
- $10,000
$0
$0
- $10,000
1
-10,000
-1,000
+4,021
-6,979
2
-6,979
-698
+4,021
-3,656
3
-3,656
-366
+4,021
0
A negative balance indicates an unpaid balance. In other words, the customer still owes
money to the bank.
Observe that, for the repayment schedule shown, the 10% interest is calculated only
on each year’s outstanding balance. In this situation, only part of the $4,021 annual payment represents interest; the remainder goes toward repaying the principal. Thus, the
three annual payments repay the loan itself and additionally provide a return of 10% on
the amount still outstanding each year.
Note that when the last payment is made, the outstanding principal is eventually reduced
to zero.2 If we calculate the NPW of the loan transaction at its rate of return (10%), we see that
PW110%2 = - $10,000 + $4,0211P>A, 10%, 32 = 0,
which indicates that the bank can break even at a 10% rate of interest. In other words, the rate
of return becomes the rate of interest that equates the present value of future cash repayments
to the amount of the loan. This observation prompts the second definition of rate of return.
Definition 2
The rate of return is the break-even interest rate i* that equates the present worth of a
project’s cash outflows to the present worth of its cash inflows, or
PW1i*2 = PWCash inflows - PWCash outflows
= 0.
Note that the expression for the NPW is equivalent to
PW1i*2 =
A0
11 + i*2
0
+
A1
11 + i*2
1
+ ... +
AN
11 + i*2N
= 0.
(7.1)
Here we know the value of A n for each period, but not the value of i*. Since it is the only
unknown, however, we can solve for i*. (Inevitably, there will be N values of i* that satisfy this equation. In most project cash flows, you would be able to find a unique positive
i* that satisfies Eq. (7.1). However, you may encounter some cash flows that cannot be
solved for a single rate of return greater than 100%. By the nature of the NPW function in
2
As we learned in Section 5.3.2, this terminal balance is equivalent to the net future worth of the investment.
If the net future worth of the investment is zero, its NPW should also be zero.
325
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326 CHAPTER 7 Rate-of-Return Analysis
Eq. (7.1), it is possible to have more than one rate of return for certain types of cash
flows. For some cash flows, we may not find a specific rate of return at all.)3
Note that the formula in Eq. (7.1) is simply the NPW formula solved for the particular interest rate 1i*2 at which PW(i) is equal to zero. By multiplying both sides of
Eq. (7.1) by 11 + i*2N, we obtain
PW1i*211 + i*2N = FW1i*2 = 0.
If we multiply both sides of Eq. (7.1) by the capital recovery factor 1A/P, i*, N2, we obtain the relationship AE1i*2 = 0. Therefore, the i* of a project may be defined as the rate
of interest that equates the present worth, future worth, and annual equivalent worth of
the entire series of cash flows to zero.
7.1.2 Return on Invested Capital
Investment projects can be viewed as analogous to bank loans. We will now introduce the
concept of rate of return based on the return on invested capital in terms of a project investment. A project’s return is referred to as the internal rate of return (IRR) or the yield
promised by an investment project over its useful life.
Definition 3
The internal rate of return is the interest rate charged on the unrecovered project balance of the investment such that, when the project terminates, the unrecovered project
balance will be zero.
Suppose a company invests $10,000 in a computer with a three-year useful life and
equivalent annual labor savings of $4,021. Here, we may view the investing firm as the
lender and the project as the borrower. The cash flow transaction between them would be
identical to the amortized loan transaction described under Definition 1:
Internal rate of
return: This is
the return that
a company
would earn if
it invested in
itself, rather
than investing
that money
elsewhere.
Return
on
Invested
Capital
n
Beginning
Project
Balance
Ending
Cash
Payment
Project
Balance
0
$0
$0
- $10,000
- $10,000
1
-10,000
-1,000
4,021
6,979
2
-6,979
-697
4,021
3,656
3
-3,656
-365
4,021
0
In our project balance calculation, we see that 10% is earned (or charged) on $10,000 during year 1, 10% is earned on $6,979 during year 2, and 10% is earned on $3,656 during year
3. This indicates that the firm earns a 10% rate of return on funds that remain internally invested in the project. Since it is a return that is internal to the project, we refer to it as the
internal rate of return, or IRR. This means that the computer project under consideration
brings in enough cash to pay for itself in three years and also to provide the firm with a return
3
You will always have N rates of return. The issue is whether they are real or imaginary. If they are real, the
question “Are they in the 1-100%, q 2 interval?” should be asked. A negative rate of return implies that you
never recover your initial investment.
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Section 7.2 Methods for Finding the Rate of Return 327
of 10% on its invested capital. Put differently, if the computer is financed with funds costing 10% annually, the cash generated by the investment will be exactly sufficient to repay
the principal and the annual interest charge on the fund in three years.
Notice that only one cash outflow occurs at time 0, and the present worth of this outflow is simply $10,000. There are three equal receipts, and the present worth of these inflows is $4,0211P/A, 10%, 32 = $10,000. Since the NPW = PWInflow - PWOutflow =
$10,000 - $10,000 = 0, 10% also satisfies Definition 2 of the rate of return. Even
though the preceding simple example implies that i* coincides with IRR, only Definitions 1 and 3 correctly describe the true meaning of the internal rate of return. As we will
see later, if the cash expenditures of an investment are not restricted to the initial period,
several break-even interest rates 1i*’s2 may exist that satisfy Eq. (7.1). However, there
may not be a rate of return that is internal to the project.
7.2 Methods for Finding the Rate of Return
We may find i* by several procedures, each of which has its advantages and disadvantages. To facilitate the process of finding the rate of return for an investment project, we
will first classify various types of investment cash flow.
7.2.1 Simple versus Nonsimple Investments
We can classify an investment project by counting the number of sign changes in its net
cash flow sequence. A change from either “+” to “-” or “-” to “+” is counted as one
sign change. (We ignore a zero cash flow.) Then,
• A simple investment is an investment in which the initial cash flows are negative
and only one sign change occurs in the remaining net cash flow series. If the initial
flows are positive and only one sign change occurs in the subsequent net cash flows,
they are referred to as simple borrowing cash flows.
• A nonsimple investment is an investment in which more than one sign change occurs in the cash flow series.
Multiple i*’s, as we will see later, occur only in nonsimple investments. Three different
types of investment possibilities are illustrated in Example 7.1.
EXAMPLE 7.1 Investment Classification
Consider the following three cash flow series and classify them into either simple or
nonsimple investments:
Period
n
Project A
Net Cash Flow
Project B
Project C
0
- $1,000
- $1,000
$1,000
1
-500
3,900
-450
2
800
-5,030
-450
3
1,500
2,145
-450
4
2,000
Simple
investment:
The project with
only one sign
change in the
net cash flow
series.
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328 CHAPTER 7 Rate-of-Return Analysis
SOLUTION
Given: Preceding cash flow sequences.
Find: Classify the investments shown into either simple and nonsimple investments.
• Project A represents many common simple investments. This situation reveals the NPW profile shown in Figure 7.1(a). The curve crosses the i-axis
only once.
• Project B represents a nonsimple investment. The NPW profile for this investment has the shape shown in Figure 7.1(b). The i-axis is crossed at 10%,
30%, and 50%.
• Project C represents neither a simple nor a nonsimple investment, even though
only one sign change occurs in the cash flow sequence. Since the first cash
flow is positive, this is a simple borrowing cash flow, not an investment flow.
Figure 7.1(c) depicts the NPW profile for this type of investment.
Project A
PW (i)
i*
i
44.24%
(a)
Project B
PW (i)
A
C
B
i*1
i*2
10%
30%
(b)
D
i*3
i
50%
Project C
PW (i)
i*
i
16.66%
(c)
Figure 7.1 Present-worth profiles:
(a) Simple investment, (b) nonsimple investment with multiple rates of return,
and (c) simple borrowing cash flows.
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Section 7.2 Methods for Finding the Rate of Return 329
COMMENTS: Not all NPW profiles for nonsimple investments have multiple crossings of the i-axis. Clearly, then, we should place a high priority on discovering this
situation early in our analysis of a project’s cash flows. The quickest way to predict
multiple i*’s is to generate an NPW profile on a computer and check whether it
crosses the horizontal axis more than once. In the next section, we illustrate when to
expect such multiple crossings by examining types of cash flows.
7.2.2 Predicting Multiple i*’s
As hinted at in Example 7.1, for certain series of project cash flows, we may uncover
the complication of multiple i* values that satisfy Eq. (7.1). By analyzing and classifying cash flows, we may anticipate this difficulty and adjust our analysis approach
later. Here we will focus on the initial problem of whether we can predict a unique i*
for a project by examining its cash flow pattern. Two useful rules allow us to focus on
sign changes (1) in net cash flows and (2) in accounting net profit (accumulated net
cash flows).
Net Cash Flow Rule of Signs
One useful method for predicting an upper limit on the number of positive i*’s of a cash
flow stream is to apply the rule of signs: The number of real i*’s that are greater than
-100% for a project with N periods is never greater than the number of sign changes in
the sequence of the A n’s. A zero cash flow is ignored.
An example is
An
Sign Change
0
- $100
1
-20
2
Period
3
0
4
+60
6
5
+50
1
-30
1
+100
1
Three sign changes occur in the cash flow sequence, so three or fewer real positive
i*’s exist.
It must be emphasized that the rule of signs provides an indication only of the possibility of multiple rates of return: The rule predicts only the maximum number of possible
i*’s. Many projects have multiple sign changes in their cash flow sequence, but still possess a unique real i* in the 1-100%, q 2 range.
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330 CHAPTER 7 Rate-of-Return Analysis
Accumulated Cash Flow Sign Test
The accumulated cash flow is the sum of the net cash flows up to and including a
given time. If the rule of cash flow signs indicates multiple i*’s, we should proceed to
the accumulated cash flow sign test to eliminate some possibility of multiple rates
of return.
If we let A n represent the net cash flow in period n and Sn represent the accumulated
cash flow (the accounting sum) up to period n, we have the following:
Period (n)
Cash Flow (A n)
Accumulated Cash Flow (Sn)
0
A0
S0 = A 0
1
A1
S1 = S0 + A 1
2
A2
S2 = S1 + A 2
o
o
N
AN
o
SN = SN - 1 + A N
We then examine the sequence of accumulated cash flows 1S0 , S1 , S2 , S3 , Á , SN2 to
determine the number of sign changes. If the series Sn starts negatively and changes
sign only once, then a unique positive i* exists. This cumulative cash flow sign rule is a
more discriminating test for identifying the uniqueness of i* than the previously described method.
EXAMPLE 7.2 Predicting the Number of i *’s
Predict the number of real positive rates of return for each of the following cash flow
series:
Period
A
B
C
D
0
- $100
- $100
$0
- $100
1
-200
+50
-50
+50
2
+200
-100
+115
0
3
+200
+60
-66
+200
4
+200
-100
-50
SOLUTION
Given: Four cash flow series and cumulative flow series.
Find: The upper limit on number of i*’s for each series.
The cash flow rule of signs indicates the following possibilities for the positive values
of i*:
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Section 7.2 Methods for Finding the Rate of Return 331
Project
Number of Sign Changes Possible Number of
in Net Cash Flows
Positive Values of i*
A
1
1 or 0
B
4
4, 3, 2, 1, or 0
C
2
2, 1, or 0
D
2
2, 1, or 0
For cash flows B, C, and D, we would like to apply the more discriminating cumulative cash flow test to see if we can specify a smaller number of possible values of i*.
Accordingly, we write
Project B
Sn
Project C
An
Sn
Project D
n
An
An
Sn
0
- $100
- $100
$0
$0
- $100
- $100
1
+50
-50
-50
-50
+50
-50
2
-100
-150
+115
+65
0
-50
3
+60
-90
-66
-1
+200
+150
4
-100
-190
-50
+100
Recall the test: If the series starts negatively and changes sign only once, a unique
positive i* exists.
• Only project D begins negatively and passes the test; therefore, we may predict a
unique i* value, rather than 2, 1, or 0 as predicted by the cash flow rule of signs.
(i1* = - 75.16% and i *2 = 35.05%)
• Project B, with no sign change in the cumulative cash flow series, has no rate
of return.
• Project C fails the test, and we cannot eliminate the possibility of multiple
i*’s. (i1* = 10% and i *2 = 20%)
7.2.3 Computational Methods
Once we identify the type of an investment cash flow, several ways to determine its rate
of return are available. Some of the most practical methods are as follows:
• Direct solution method,
• Trial-and-error method, and
• Computer solution method.
Direct Solution Method
For the special case of a project with only a two-flow transaction (an investment followed
by a single future payment) or a project with a service life of two years of return, we can
seek a direct mathematical solution for determining the rate of return. These two cases
are examined in Example 7.3.
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332 CHAPTER 7 Rate-of-Return Analysis
EXAMPLE 7.3 Finding i * by Direct Solution: Two Flows
and Two Periods
Consider two investment projects with the following cash flow transactions:
n
Project 1
Project 2
0
- $2,000
- $2,000
1
0
1,300
2
0
1,500
3
0
4
3,500
Compute the rate of return for each project.
SOLUTION
Given: Cash flows for two projects.
Find: i* for each project.
Project 1: Solving for i* in PW1i*2 = 0 is identical to solving FW1i*2 = 0, because
FW equals PW times a constant. We could do either here, but we will set FW1i*2 = 0
to demonstrate the latter. Using the single-payment future-worth relationship, we obtain
FW1i*2 = - $2,0001F>P, i*, 42 + $3,500 = 0,
$3,500 = $2,0001F>P, i*, 42 = $2,00011 + i*24,
1.75 = 11 + i*24.
Solving for i* yields
4
i* = 21.75 - 1
= 0.1502 or 15.02%.
Project 2: We may write the NPW expression for this project as
PW1i2 = - $2,000 +
$1,500
$1,300
+
= 0.
11 + i2
11 + i22
Let X = 1/11 + i2. We may then rewrite PW(i) as a function of X as follows:
PW1X2 = - $2,000 + $1,300X + $1,500X2 = 0.
This is a quadratic equation that has the following solution:4
4
The solution of the quadratic equation aX2 + bX + c = 0 is X =
-b ; 3b2 - 4ac
.
2a
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Section 7.2 Methods for Finding the Rate of Return 333
X =
=
-1,300 ; 31,3002 - 411,50021-2,0002
211,5002
-1,300 ; 3,700
3,000
= 0.8 or -1.667.
Replacing X values and solving for i gives us
0.8 =
-1.667 =
1
: i = 25%,
11 + i2
1
: i = -160%.
11 + i2
Since an interest rate less than -100% has no economic significance, we find that the
project’s i* is 25%.
COMMENTS: In both projects, one sign change occurred in the net cash flow series,
so we expected a unique i*. Also, these projects had very simple cash flows. When
cash flows are more complicated, generally we use a trial-and-error method or a
computer to find i*.
Trial-and-Error Method
The first step in the trial-and-error method is to make an estimated guess5 at the value of
i*. For a simple investment, we use the “guessed” interest rate to compute the present
worth of net cash flows and observe whether it is positive, negative, or zero. Suppose,
then, that PW(i) is negative.
Since we are aiming for a value of i that makes PW1i2 = 0, we must raise the present worth of the cash flow. To do this, we lower the interest rate and repeat the process. If
PW(i) is positive, however, we raise the interest rate in order to lower PW(i). The process
is continued until PW(i) is approximately equal to zero. Whenever we reach the point
where PW(i) is bounded by one negative and one positive value, we use linear interpolation to approximate i*. This process is somewhat tedious and inefficient. (The trialand-error method does not work for nonsimple investments in which the NPW function is
not, in general, a monotonically decreasing function of the interest rate.)
EXAMPLE 7.4 Finding i * by Trial and Error
The Imperial Chemical Company is considering purchasing a chemical analysis machine worth $13,000. Although the purchase of this machine will not produce any
5
As we shall see later in this chapter, the ultimate objective of finding i* is to compare it against the MARR.
Therefore, it is a good idea to use the MARR as the initial guess.
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334 CHAPTER 7 Rate-of-Return Analysis
increase in sales revenues, it will result in a reduction of labor costs. In order to operate the machine properly, it must be calibrated each year. The machine has an expected life of six years, after which it will have no salvage value. The following table
summarizes the annual savings in labor cost and the annual maintenance cost in calibration over six years:
Year (n)
Costs ($)
Savings ($)
Net Cash Flow ($)
0
13,000
1
2,300
6,000
3,700
2
2,300
7,000
4,700
3
2,300
9,000
6,700
4
2,300
9,000
6,700
5
2,300
9,000
6,700
6
2,300
9,000
6,700
-13,000
Find the rate of return for this project.
SOLUTION
Given: Cash flows over six years as shown in Figure 7.2.
Find: i*.
We start with a guessed interest rate of 25%. The present worth of the cash flows is
PW125%2 = - $13,000 + $3,7001P>F, 25%, 12 + $4,7001P>F, 25%, 22
+ $6,7001P/A, 25%, 421P/F, 25%, 22
= $3,095.
$6,700
$3,700
0
1
$4,700
2
3
4
5
6
$13,000
IRR(B3:B9,10%)
Figure 7.2
Cash flow diagram for a simple investment (Example 7.4).
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Section 7.2 Methods for Finding the Rate of Return 335
Since this present worth is positive, we must raise the interest rate to bring PW toward zero. When we use an interest rate of 35%, we find that
PW135%2 = - $13,000 + $3,7001P>F, 35%, 12 + $4,7001P>F, 35%, 22
+ $6,7001P/A, 35%, 421P/F, 35%, 22
= - $339.
We have now bracketed the solution: PW(i) will be zero at i somewhere between
25% and 35%. Using straight-line interpolation, we approximate
i* 25% + 135% - 25%2c
3,095 - 0
d
3,095 - 1-3392
= 25% + 10%10.90132
= 34.01%.
Now we will check to see how close this value is to the precise value of i*. If we
compute the present worth at this interpolated value, we obtain
PW134%2 = - $13,000 + $3,7001P>F, 34%, 12 + $4,7001P>F, 34%, 22
+ $6,7001P/A, 34%, 421P/F, 34%, 22
= - $50.58.
As this is not zero, we may recompute i* at a lower interest rate, say, 33%:
PW133%2 = - $13,000 + $3,7001P>F, 33%, 12 + $4,7001P>F, 33%, 22
+ $6,7001P/A, 33%, 421P/F, 33%, 22
= $248.56.
With another round of linear interpolation, we approximate
i* 33% + 134% - 33%2c
248.56 - 0
d
248.56 - 1-50.582
= 33% + 1%10.83092
= 33.83%.
At this interest rate,
PW133.83%2 = - $13,000 + $3,7001P>F, 33.83%, 12
+ $4,7001P>F, 33.83%, 22
+ $6,7001P/A, 33.83%, 421P/F, 33.83%, 22
= - $0.49,
which is practically zero, so we may stop here. In fact, there is no need to be more
precise about these interpolations, because the final result can be no more accurate
than the basic data, which ordinarily are only rough estimates.
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336 CHAPTER 7 Rate-of-Return Analysis
COMMENT: With Excel, you can evaluate the IRR for the project as = IRR
1range,guess2, where you specify the cell range for the cash flow (e.g., B3:B9) and
the initial guess, such as 25%. Computing i* for this problem in Excel, incidentally,
gives us 33.8283%. Instead of using the factor notations, you may attempt use a tabular approach as follows:
Internal Rate of Return: What It Looks Like
Discount Rate: 25%
Discount Rate: 35%
Year
Cash Flow
Factor
Amount
Factor
Amount
0
- $13,000
1.0000
- $13,000
1.0000
- $13,000
1
3,700
0.8000
2,960
0.7407
2,741
2
4,700
0.6400
3,008
0.5487
2,579
3
6,700
0.5120
3,430
0.4064
2,723
4
6,700
0.4096
2,744
0.3011
2,017
5
6,700
0.3277
2,196
0.2230
1,494
6,700
0.2621
1,756
0.1652
1,107
6
Total
+ $22,200
NPW $3,095
NPW $339
IRR close to 34%
Graphical Method
We don’t need to do laborious manual calculations to find i*. Many financial calculators
have built-in functions for calculating i*. It is worth noting that many online financial calculators or spreadsheet packages have i* functions, which solve Eq. (7.1) very rapidly,6
usually with the user entering the cash flows via a computer keyboard or by reading a
cash flow data file. (For example, Microsoft Excel has an IRR financial function that analyzes investment cash flows, as illustrated in Example 7.4.)
The most easily generated and understandable graphic method of solving for i* is to
create the NPW profile on a computer. On the graph, the horizontal axis indicates the interest rate and the vertical axis indicates the NPW. For a given project’s cash flows, the
NPW is calculated at an interest rate of zero (which gives the vertical-axis intercept) and
several other interest rates. Points are plotted and a curve is sketched. Since i* is defined
as the interest rate at which PW1i*2 = 0, the point at which the curve crosses the horizontal axis closely approximates i*. The graphical approach works for both simple and
nonsimple investments.
6
An alternative method of solving for i* is to use a computer-aided economic analysis program. Cash Flow
Analyzer (CFA) finds i* visually by specifying the lower and upper bounds of the interest search limit and
generates NPW profiles when given a cash flow series. In addition to the savings in calculation time, the advantage of computer-generated profiles is their precision. CFA can be found from the book’s website.
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Section 7.2 Methods for Finding the Rate of Return 337
EXAMPLE 7.5 Graphical Approach to Estimate i *
Consider the cash flow series shown in Figure 7.3(a). Estimate the rate of return by
generating the NPW profile on a computer.
SOLUTION
Given: Cash flow series in Figure 7.3.
Find: (a) i* by plotting the NPW profile and (b) i* by using Excel.
Figure 7.3 Graphical solution to rate-of-return problem for a typical nonsimple investment
(Example 7.5).
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338 CHAPTER 7 Rate-of-Return Analysis
(a) The present-worth function for the project cash flow series is
PW1i2 = - $10,000 + $20,0001P>A, i, 22 - $25,0001P>F, i, 32.
First we use i = 0 in this equation to obtain NPW = $5,000, which is the
vertical-axis intercept. Then we substitute several other interest rates—10%,
20%, Á , 140%—and plot these values of PW(i) as well. The result is Figure
7.3, which shows the curve crossing the horizontal axis at roughly 140%. This
value can be verified by other methods if we desire to do so. Note that, in addition to establishing the interest rate that makes NPW = 0, the NPW profile indicates where positive and negative NPW values fall, thus giving us a broad
picture of those interest rates for which the project is acceptable or unacceptable. (Note also that a trial-and-error method would lead to some confusion: As
you increase the interest rate from 0% to 20%, the NPW value also keeps increasing, instead of decreasing.) Even though the project is a nonsimple investment, the curve crosses the horizontal axis only once. As mentioned in the
previous section, however, most nonsimple projects have more than one value
of i* that makes NPW = 0 (i.e., more than one i* per project). In such a case,
the NPW profile would cross the horizontal axis more than once.7
(b) With Excel, you can evaluate the IRR for the project with the function
= IRR1range,guess2
in which you specify the cell range for the cash flow and the initial guess, such
as 10%.
7.3 Internal-Rate-of-Return Criterion
Now that we have classified investment projects and learned methods for determining the
i* value for a given project’s cash flows, our objective is to develop an accept–reject decision rule that gives results consistent with those obtained from NPW analysis.
7.3.1 Relationship to PW Analysis
As we already observed in Chapter 5, NPW analysis depends on the rate of interest used
for the computation of NPW. A different rate may change a project from being considered
acceptable to being unacceptable, or it may change the ranking of several projects:
• Consider again the NPW profile as drawn for the simple project in Figure 7.1(a). For
interest rates below i*, this project should be accepted because NPW 7 0; for interest rates above i*, it should be rejected.
• By contrast, for certain nonsimple projects, the NPW may look like the one shown in
Figure 7.1(b). NPW analysis would lead you to accept the projects in regions A and C,
but reject those in regions B and D. Of course, this result goes against intuition: A higher
interest rate would change an unacceptable project into an acceptable one. The situation
graphed in Figure 7.1(b) is one of the cases of multiple i*’s mentioned in Definition 2.
7
In Section 7.2.2, we discuss methods of predicting the number of i* values by looking at cash flows. However, generating an NPW profile to discover multiple i*’s is as practical and informative as any other method.
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Section 7.3 Internal-Rate-of-Return Criterion 339
Therefore, for the simple investment situation in Figure 7.1(a), i* can serve as an appropriate index for either accepting or rejecting the investment. However, for the nonsimple
investment of Figure 7.1(b), it is not clear which i* to use to make an accept–reject decision. Therefore, the i* value fails to provide an appropriate measure of profitability for an
investment project with multiple rates of return.
7.3.2 Net-Investment Test: Pure versus Mixed Investments
To develop a consistent accept–reject decision rule with the NPW, we need to further
classify a project into either a pure or a mixed investment:
• A project is said to be a net investment when the project balances computed at the
project’s i* values, PB1i*2n , are either less than or equal to zero throughout the life
of the investment, with the first cash flow being negative 1A 0 6 02. The investment
is net in the sense that the firm does not overdraw on its return at any point and hence
investment is not indebted to the project. This type of project is called a pure investment. In contrast, pure borrowing is defined as the situation in which PB1i*2n values
are positive or zero throughout the life of the loan, with A 0 7 0. Simple investments
will always be pure investments.
• If any of the project balances calculated at the project’s i* is positive, the project
is not a pure investment. A positive project balance indicates that, at some time
during the project life, the firm acts as a borrower [PB1i*2n 7 0] rather than an
investor in the project [PB1i*2n 6 0]. This type of investment is called a mixed
investment.
Net investment
test: A process
to determine
whether or not
a firm borrows
money from a
project during
the investment
period.
Pure investment: An
investment in
which a firm
never borrows
money from the
project.
EXAMPLE 7.6 Pure versus Mixed Investments
Consider the following four investment projects with known i* values:
Project Cash Flows
n
A
B
C
0
- $1,000
- $1,000
- $1,000
- $1,000
1
1,000
1,600
500
3,900
2
2,000
-300
-500
-5,030
3
1,500
-200
2,000
2,145
i*
33.64%
21.95%
29.95%
(10%, 30%, 50%)
Determine which projects are pure investments.
SOLUTION
Given: Four projects with cash flows and i*’s as shown.
Find: Which projects are pure investments?
D
Mixed investment: An
investment in
which a firm
borrows money
from the project
during the
investment
period.
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340 CHAPTER 7 Rate-of-Return Analysis
We will first compute the project balances at the projects’ respective i*’s. If multiple
rates of return exist, we may use the largest value of i* greater than zero.8
Project A:
PB133.64%20 = - $1,000,
PB133.64%21 = - $1,00011 + 0.33642 + 1- $1,0002 = - $2,336.40,
PB133.64%22 = - $2,336.4011 + 0.33642 + $2,000 = - $1,122.36,
PB133.64%23 = - $1,122.3611 + 0.33642 + $1,500 = 0.
1-, -, -, 02: passes the net-investment test (pure investment).
Project B:
PB121.95%20 = - $1,000,
PB121.95%21 = - $1,00011 + 0.21952 + $1,600 = $380.50,
PB121.95%22 = + $380.5011 + 0.21952 - $300 = $164.02,
PB121.95%23 = + $164.0211 + 0.21952 - $200 = 0.
1-, +, +, 02: fails the net-investment test (mixed investment).
Project C:
PB129.95%20 = - $1,000,
PB129.95%21 = - $1,00011 + 0.29952 + $500 = - $799.50,
PB129.95%22 = - $799.5011 + 0.29952 - $500 = - $1,538.95,
PB129.95%23 = - $1,538.9511 + 0.29952 + $2,000 = 0.
1-, -, -, 02: passes the net-investment test (pure investment).
Project D: There are three rates of return. We can use any of them for the net investment test. Thus,
PB150%20 = - $1,000,
PB150%21 = - $1,00011 + 0.502 + $3,900 = $2,400,
PB150%22 = + $2,40011 + 0.502 - $5,030 = - $1,430,
PB150%23 = - $1,43011 + 0.502 + $2,145 = 0.
1-, +, -, 02: fails the net-investment test (mixed investment).
COMMENTS: As shown in Figure 7.4, projects A and C are the only pure investments.
Project B demonstrates that the existence of a unique i* is a necessary but not sufficient condition for a pure investment.
8
In fact, it does not matter which rate we use in applying the net-investment test. If one value passes the test,
they will all pass. If one value fails, they will all fail.
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Section 7.3 Internal-Rate-of-Return Criterion 341
Project B
Mixed investment
Project A
Pure investment
3,000
3,000
2,000
2,000
1,000
0
0
1
2
3
–1,000
0
2,000
2,000
4
N
2,400
1,000
0
1
2
–1,000
3
4
N
PB
1,000
PB
3
Project D
Mixed investment
3,000
0
0
–1,538
1
2
–1,000
–799
–2,000
3
4
N
–1,430
–3,000
–3,000
Figure 7.4
2
0
–3,000
3,000
–2,000
1
164
–2,000
–2,236
Project C
Pure investment
0
380
–1,000
–1,122
–2,000
–3,000
4
N
PB
PB
1,000
Net-investment test (Example 7.6).
7.3.3 Decision Rule for Pure Investments
Suppose we have a pure investment. (Recall that all simple investments are pure investments as well.) Why are we interested in finding the particular interest rate that equates a
project’s cost with the present worth of its receipts? Again, we may easily answer this
question by examining Figure 7.1(a). In this figure, we notice two important characteristics of the NPW profile. First, as we compute the project’s PW(i) at a varying interest rate
i, we see that the NPW is positive for i 6 i*, indicating that the project would be acceptable under PW analysis for those values of i. Second, the NPW is negative for i 7 i*, indicating that the project is unacceptable for those values of i. Therefore, i* serves as a
benchmark interest rate, knowledge of which will enable us to make an accept–reject
decision consistent with NPW analysis.
Note that, for a pure investment, i* is indeed the IRR of the investment. (See Section
7.1.2.) Merely knowing i*, however, is not enough to apply this method. Because firms
typically wish to do better than break even (recall that at NPW = 0 we were indifferent
to the project), a minimum acceptable rate of return (MARR) is indicated by company
policy, management, or the project decision maker. If the IRR exceeds this MARR, we
are assured that the company will more than break even. Thus, the IRR becomes a useful
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342 CHAPTER 7 Rate-of-Return Analysis
gauge against which to judge a project’s acceptability, and the decision rule for a pure
project is as follows:
If IRR 7 MARR, accept the project.
If IRR = MARR, remain indifferent.
If IRR 6 MARR, reject the project.
Note that this decision rule is designed to be applied for a single project evaluation.
When we have to compare mutually exclusive investment projects, we need to apply the
incremental analysis approach, as we shall see in Section 7.4.2.
EXAMPLE 7.7 Investment Decision for a Pure Investment
Merco, Inc., a machinery builder in Louisville, Kentucky, is considering investing
$1,250,000 in a complete structural beam-fabrication system. The increased productivity
resulting from the installation of the drilling system is central to the project’s justification. Merco estimates the following figures as a basis for calculating productivity:
•
•
•
•
•
•
•
Increased fabricated steel production: 2,000 tons/year.
Average sales price/ton fabricated steel: $2,566.50/ton.
Labor rate: $10.50/hour.
Tons of steel produced in a year: 15,000 tons.
Cost of steel per ton (2,205 lb): $1,950/ton.
Number of workers on layout, hole making, sawing, and material handling: 17.
Additional maintenance cost: $128,500/year.
With the cost of steel at $1,950 per ton and the direct labor cost of fabricating 1 lb at
10 cents, the cost of producing a ton of fabricated steel is about $2,170.50. With a
selling price of $2,566.50 per ton, the resulting contribution to overhead and profit
becomes $396 per ton. Assuming that Merco will be able to sustain an increased production of 2,000 tons per year by purchasing the system, engineers have estimated
the projected additional contribution to be 2,000 tons * $396 = $792,000.
Since the drilling system has the capacity to fabricate the full range of structural
steel, two workers can run the system, one on the saw and the other on the drill. A third
operator is required to operate a crane for loading and unloading materials. Merco estimates that, to do the equivalent work of these three workers with conventional
manufacturing techniques would require, on the average, an additional 14 people for
center punching, hole making with a radial or magnetic drill, and material handling.
This translates into a labor savings in the amount of $294,000 per year 114 * $10.50 *
40 hours/week * 50 weeks/year2. The system can last for 15 years, with an estimated
after-tax salvage value of $80,000. However, after an annual deduction of $226,000 in
corporate income taxes, the net investment costs, as well as savings, are as follows:
• Project investment cost: $1,250,000.
• Projected annual net savings:
1$792,000 + $294,0002 - $128,500 - $226,000 = $731,500.
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Section 7.3 Internal-Rate-of-Return Criterion 343
• Projected after-tax salvage value at the end of year 15: $80,000.
(a) What is the projected IRR on this fabrication investment?
(b) If Merco’s MARR is known to be 18%, is this investment justifiable?
SOLUTION
Given: Projected cash flows as shown in Figure 7.5 and MARR = 18%.
Find: (a) The IRR and (b) whether to accept or reject the investment.
$80,000
$731,500
0
1
2
3
4
5
6
7 8 9 10 11 12 13 14 15
Years
$1,250,000
Figure 7.5
Cash flow diagram (Example 7.7).
(a) Since only one sign change occurs in the net cash flow series, the fabrication project is a simple investment. This indicates that there will be a unique rate of return
that is internal to the project:
PW1i2 = - $1,250,000 + $731,5001P>A, i, 152
+ $80,0001P/F, i, 152
= 0
i* = 58.71%.
With Excel, you will also find that the IRR is about 58.71% for the net investment of $1,250,000.
(b) The IRR figure far exceeds Merco’s MARR, indicating that the fabrication
system project is an economically attractive one. Merco’s management believes that, over a broad base of structural products, there is no doubt that the
installation of the fabricating system would result in a significant savings,
even after considering some potential deviations from the estimates used in
the analysis.
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344 CHAPTER 7 Rate-of-Return Analysis
7.3.4 Decision Rule for Mixed Investments
Applied to pure projects, i* provides an unambiguous criterion for measuring profitability. However, when multiple rates of return occur, none of them is an accurate portrayal of a project’s acceptability or profitability. However, there is a correct method,
which uses an external interest rate, for refining our analysis when we do discover
multiple i*’s. An external rate of return allows us to calculate a single accurate rate of
return; if you choose to avoid these more complicated applications of rate-of-return techniques, you must be able to predict multiple i*’s via the NPW profile and, when they
occur, select an alternative method such as NPW or AE analysis for determining the
project’s acceptability.
Project balance:
The amount
of money
committed to
a project at a
specific period.
Need for an External Interest Rate for Mixed Investments
In the case of a mixed investment, we can extend the economic interpretation of the IRR
to the return-on-invested-capital measure if we are willing to make an assumption about
what happens to the extra cash that the investor gets from the project during the intermediate years.
First, the project balance (PB), or investment balance, can also be interpreted from
the viewpoint of a financial institution that borrows money from an investor and then
pays interest on the PB. Thus, a negative PB means that the investor has money in a bank
account; a positive PB means that the investor has borrowed money from the bank. Negative PBs represent interest paid by the bank to the investor; positive PBs represent interest paid by the investor to the bank.
Now, can we assume that the interest paid by the bank and the interest received from
the investor are the same for the same amount of balance? In our banking experience, we
know that is not the case. Normally, the borrowing rate (interest paid by the investor) is
higher than the interest rate on your deposit (interest paid by the bank).
However, when we calculate the project balance at an i* for mixed investments, we
notice an important point: Cash borrowed (released) from the project is assumed to earn
the same interest rate through external investment as money that remains internally invested. In other words, in solving a cash flow for an unknown interest rate, it is assumed
that money released from a project can be reinvested to yield a rate of return equal to that
received from the project. In fact, we have been making this assumption regardless of
whether a cash flow does or does not produce a unique positive i*. Note that money is
borrowed only when PB1i*2 7 0, and the magnitude of the borrowed amount is the project balance. When PB1i*2 6 0, no money is borrowed, even though the cash flow may
be positive at that time.
In reality, it is not always possible for cash borrowed (released) from a project to be
reinvested to yield a rate of return equal to that received from the project. Instead, it is
likely that the rate of return available on a capital investment in the business is much
different—usually higher—from the rate of return available on other external investments. Thus, it may be necessary to compute the project balances for a project’s cash
flow at two rates of interest—one on the internal investment and one on the external investments. As we will see later, by separating the interest rates, we can measure the true
rate of return of any internal portion of an investment project.
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Section 7.3 Internal-Rate-of-Return Criterion 345
Calculation of Return on Invested Capital for Mixed Investments
For a mixed investment, we must calculate a rate of return on the portion of capital that
remains invested internally. This rate is defined as the true IRR for the mixed investment
and is commonly known as the return on invested capital (RIC). Then, what interest
rate should we assume for the portion of external investment? Insofar as a project is not a
net investment, one or more periods when the project has a net outflow of money (a positive project balance) must later be returned to the project. This money can be put into the
firm’s investment pool until such time as it is needed in the project. The interest rate of
this investment pool is the interest rate at which the money can in fact be invested outside
the project.
Recall that the NPW method assumed that the interest rate charged to any funds
withdrawn from a firm’s investment pool would be equal to the MARR. In this book, we
will use the MARR as an established external interest rate (i.e., the rate earned by money
invested outside of the project). We can then compute the RIC as a function of the MARR
by finding the value of the RIC that will make the terminal project balance equal to zero.
(This implies that the firm wants to fully recover any investment made in the project and
pays off any borrowed funds at the end of the project life.) This way of computing the rate
of return is an accurate measure of the profitability of the project as represented by the
cash flow. The following procedure outlines the steps for determining the IRR for a mixed
investment:
Step 1. Identify the MARR (or external interest rate).
Step 2. Calculate PB1i, MARR2n (or simply PBn) according to the rule
PB1i, MARR20 = A 0 .
PB1i, MARR21 = e
PB011 + i2 + A 1 , if PB0 6 0
,
PB011 + MARR2 + A 1 , if PB0 7 0
o
PB1i, MARR2n = e
PBn - 111 + i2 + A n , if PBn - 1 6 0
.
PBn - 111 + MARR2 + A n , if PBn - 1 7 0
(As defined in the text, A n stands for the net cash flow at the end of period n. Note that
the terminal project balance must be zero.)
Step 3. Determine the value of i by solving the terminal project balance equation
PB1i, MARR2N = 0.
The interest rate i is the RIC (or IRR) for the mixed investment.
Using the MARR as an external interest rate, we may accept a project if the IRR exceeds the MARR, and we should reject the project otherwise. Figure 7.6 summarizes the
IRR computation for a mixed investment.
Return on
invested capital
(RIC): The
amount that a
company earns
on the total
investment it
has made in its
project.
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346 CHAPTER 7 Rate-of-Return Analysis
Internal investment (negative balance)
External investment (positive balance)
+
Project balance
Terminal
balance = 0
MARR
0
1
IRR
2
3
End of year
0
4
IRR
5
N
–
Figure 7.6
Computational logic for IRR (mixed investment).
EXAMPLE 7.8 IRR for a Mixed Investment
By outbidding its competitors, Trane Image Processing (TIP), a defense contractor,
received a contract worth $7,300,000 to build navy flight simulators for U.S. Navy
pilot training over two years. With some defense contracts, the U.S. government
makes an advance payment when the contract is signed, but in this case the government will make two progressive payments: $4,300,000 at the end of the first year and
the $3,000,000 balance at the end of the second year. The expected cash outflows required to produce the simulators are estimated to be $1,000,000 now, $2,000,000
during the first year, and $4,320,000 during the second year. The expected net cash
flows from this project are summarized as follows:
Year
Cash Inflow
Cash Outflow
Net Cash Flow
$1,000,000
- $1,000,000
1
$4,300,000
2,000,000
2,300,000
2
3,000,000
4,320,000
-1,320,000
0
In normal situations, TIP would not even consider a marginal project such as this
one. However, hoping that the company can establish itself as a technology leader in
the field, management felt that it was worth outbidding its competitors. Financially,
what is the economic worth of outbidding the competitors for this project? That is,
(a) Compute the values of i*’s for this project.
(b) Make an accept–reject decision based on the results in part (a). Assume that the
contractor’s MARR is 15%.
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Section 7.3 Internal-Rate-of-Return Criterion 347
SOLUTION
Given: Cash flow shown and MARR = 15%.
Find: (a) Compute the NPW, (b) i*, and (c) RIC at MARR = 15%, and determine
whether to accept the project.
(a)
PW115%2 = - $1,000,000 + $2,300,0001P>F, 15%, 12
= - $1,320,0001P>F, 15%, 22
= $1,890 > 0.
(b) Since the project has a two-year life, we may solve the net-present-worth equation directly via the quadratic formula:
- $1,000,000 + $2,300,000>11 + i*2 - $1,320,000>11 + i*22 = 0.
If we let X = 1/11 + i*2, we can rewrite the preceding expression as
-1,000,000 + 2,300,000X - 1,320,000X2 = 0.
Solving for X gives X = 10/11 and 10/12, or i* = 10% and 20%. As shown in
Figure 7.7, the NPW profile intersects the horizontal axis twice, once at 10% and
again at 20%. The investment is obviously not a simple one; thus, neither 10%
nor 20% represents the true internal rate of return of this government project.
2,000
Net present worth ($)
1,000
PW(i) > 0
i*1 = 10%
0
i*2 = 20%
–1,000
–2,000
5
10
15
20
Interest rate (%)
Figure 7.7 NPW plot for a nonsimple investment with multiple
rates of return (Example 7.8).
25
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348 CHAPTER 7 Rate-of-Return Analysis
(c) As calculated in (b), the project has multiple rates of return. This is obviously
not a net investment, as the following table shows:
Net Investment
Test
N
Beginning
balance
Return on
investment
Payment
Ending balance
0
Using i* 10%
1
2
Using i* 20%
1
0
2
$0
- $1,000
$1,200
$0
- $1,000
$1,100
0
-100
120
0
-200
220
-1,000
2,300
-1,320
-1,000
2,300
-1,320
-$1,000
$1,200
0
- $1,000
$1,100
0
(Unit: $1,000)
At n = 0, there is a net investment to the firm, so the project balance expression
becomes
PB1i, 15%20 = - $1,000,000.
The net investment of $1,000,000 that remains invested internally grows at the interest
rate i for the next period. With the receipt of $2,300,000 in year 1, the project balance
becomes
PB1i, 15%21 = - $1,000,00011 + i2 + $2,300,000
= $1,300,000 - $1,000,000i
= $1,000,00011.3 - i2.
At this point, we do not know whether PB1i, 15%21 is positive or negative; we want
to know this in order to test for net investment and the presence of a unique i*. It depends on the value of i, which we want to determine. Therefore, we need to consider
two situations: (1) i 6 1.3 and (2) i 7 1.3.
• Case 1: i 6 1.3 : PB1i, 15%21 7 0.
Since this indicates a positive balance, the cash released from the project
would be returned to the firm’s investment pool to grow at the MARR until it
is required back in the project. By the end of year 2, the cash placed in the
investment pool would have grown at the rate of 15% [to $1,000,00011.3 - i2
11 + 0.152] and must equal the investment into the project of $1,320,000
required at that time. Then the terminal balance must be
PB1i, 15%22 = $1,000,00011.3 - i211 + 0.152 - $1,320,000
= $175,000 - $1,150,000i
= 0.
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Section 7.3 Internal-Rate-of-Return Criterion 349
1,500
1,148
1,000
15%
PB
500
0
0
1
–500
2
3
15.22%
–1,000
–1,152
–1,500
Figure 7.8 Calculation of the IRR for a mixed investment (Example 7.8).
Solving for i yields
RIC = IRR = 0.1522, or 15.22% 7 15%.
The computational process is shown graphically in Figure 7.8.
• Case 2: i 7 1.3 : PB1i, 15%21 6 0.
The firm is still in an investment mode. Therefore, the balance at the end of year 1
that remains invested will grow at the rate i for the next period. With the investment of $1,320,000 required in year 2 and the fact that the net investment must
be zero at the end of the project life, the balance at the end of year 2 should be
PB1i, 15%22 = $1,000,00011.3 - i211 + i2 - $1,320,000
= - $20,000 + $300,000i - $1,000,000i2
= 0.
Solving for i gives
IRR = 0.1 or 0.2 6 1.3,
which violates the initial assumption that i 7 1.3. Therefore, Case 1 is the only
correct situation. Since it indicates that IRR 7 MARR, the project is acceptable,
resulting in the same decision as obtained in (a) by applying the NPW criterion.
COMMENTS: In this example, we could have seen by inspection that Case 1 was correct.
Since the project required an investment as the final cash flow, the project balance at
the end of the previous period (year 1) had to be positive in order for the final balance
to equal zero. Inspection does not generally work with more complicated cash flows.
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350 CHAPTER 7 Rate-of-Return Analysis
Trial-and-Error Method for Computing IRR for Mixed Investments
The trial-and-error approach to finding the IRR (RIC) for a mixed investment is similar to
the trial-and-error approach to finding i*. We begin with a given MARR and a guess for
IRR and solve for the project balance. (A value of IRR close to the MARR is a good starting point for most problems.) Since we desire the project balance to approach zero, we can
adjust the value of IRR as needed after seeing the result of the initial guess. For example, for
a given pair of interest rates (IRRguess, MARR), if the terminal project balance is positive,
the IRRguess value is too low, so we raise it and recalculate. We can continue adjusting our
IRR guesses in this way until we obtain a project balance equal or close to zero.
EXAMPLE 7.9 IRR for a Mixed Investment by Trial and Error
Consider project D in Example 7.6. The project has the following cash flow:
n
An
0
- $1,000
1
3,900
2
-5,030
3
2,145
We know from an earlier calculation that this is a mixed investment. Compute the
IRR for this project. Assume that MARR = 6%.
SOLUTION
Given: Cash flow as stated for mixed investment and MARR = 6%.
Find: IRR.
For MARR = 6%, we must compute i by trial and error. Suppose we guess i = 8%:
PB18%, 6%20 = - $1,000,
PB18%, 6%21 = - $1,00011 + 0.082 + $3,900 = $2,820.
PB18%, 6%22 = + $2,82011 + 0.062 - $5,030 = - $2,040.80,
PB18%, 6%23 = - $2,040.8011 + 0.082 + $2,145 = - $59.06.
The net investment is negative at the end of the project, indicating that our trial
i = 8% is in error. After several trials, we conclude that, for MARR = 6%, the IRR
is approximately 6.13%. To verify the results, we write
PB16.13%, 6%20 = - $1,000,
PB16.13%, 6%21 = - $1,000.0011 + 0.06132 + $3,900 = $2,838.66.
PB16.13%, 6%22 = + $2,838.6611 + 0.06002 - $5,030 = - $2,021.02,
PB16.13%, 6%23 = - $2,021.0211 + 0.06132 + $2,145 = 0.
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Section 7.3 Internal-Rate-of-Return Criterion 351
The positive balance at the end of year 1 indicates the need to borrow from the project
during year 2. However, note that the net investment becomes zero at the end of the
project life, confirming that 6.13% is the IRR for the cash flow. Since IRR 7 MARR,
the investment is acceptable.
COMMENTS: On the basis of the NPW criterion, the investment would be acceptable
if the MARR was between zero and 10% or between 30% and 50%. The rejection region is 10% 6 i 6 30% and i 7 50%. This can be verified in Figure 7.1(b). Note
that the project also would be marginally accepted under the NPW analysis at
MARR = i = 6%:
PW16%2 = - $1,000 + 3,9001P>F, 6%, 12
= - $5,0301P/F, 6%, 22 + 2,1451P/F, 6%, 32
= $3.55 > 0.
The flowchart in Figure 7.9 summarizes how you should proceed to apply the net cash
flow sign test, accumulated cash flow sign test, and net-investment test to calculate an
IRR and make an accept–reject decision for a single project. Given the complications
Cash flow
sign rule
Simple
investment
Calculate i* by one
of the methods in
Section 7.2.3
Nonsimple
investment
Apply net
investment test
IRR = i*
Select one
possible
value of i*
Estimate i*s by
generating NPW
profile and locate
at least one i*
Fail
Mixed
investment
Pass
Pure
investment
Calculate IRR using
external rate
(MARR)
Accept project if
IRR > MARR
Figure 7.9 Summary of IRR criterion: A flowchart that summarizes how you may
proceed to apply the net cash flow sign rule and net-investment test to calculate IRR for
a pure as well as a mixed investment.
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352 CHAPTER 7 Rate-of-Return Analysis
involved in using IRR analysis to compare alternative projects, it is usually more desirable to use one of the other equivalence techniques for this purpose. As an engineering
manager, you should keep in mind the intuitive appeal of the rate-of-return measure.
Once you have selected a project on the basis of NPW or AE analysis, you may also wish
to express its worth as a rate of return, for the benefit of your associates.
7.4 Mutually Exclusive Alternatives
In this section, we present the decision procedures that should be used in comparing two
or more mutually exclusive projects on the basis of the rate-of-return measure. We will
consider two situations: (1) alternatives that have the same economic service life and (2)
alternatives that have unequal service lives.
7.4.1 Flaws in Project Ranking by IRR
Under NPW or AE analysis, the mutually exclusive project with the highest worth was
preferred. (This is known as the “total investment approach.”) Unfortunately, the analogy
does not carry over to IRR analysis: The project with the highest IRR may not be the preferred alternative. To illustrate the flaws inherent in comparing IRRs in order to choose
from mutually exclusive projects, suppose you have two mutually exclusive alternatives,
each with a 1-year service life: One requires an investment of $1,000 with a return of
$2,000, and the other requires $5,000 with a return of $7,000. You already obtained the
IRRs and NPWs at MARR = 10% as follows:
n
A1
A2
0
- $1,000
- $5,000
1
2,000
7,000
IRR
100%
40%
$818
$1,364
PW(10%)
Assuming that you have enough money in your investment pool to select either alternative, would you prefer the first project simply because you expect a higher rate of return?
On the one hand, we can see that A2 is preferred over A1 by the NPW measure. On
the other hand, the IRR measure gives a numerically higher rating for A1. This inconsistency in ranking occurs because the NPW, NFW, and AE are absolute (dollar) measures of
investment worth, whereas the IRR is a relative (percentage) measure and cannot be
applied in the same way. That is, the IRR measure ignores the scale of the investment.
Therefore, the answer to our question in the previous paragraph is no; instead, you
would prefer the second project, with the lower rate of return, but higher NPW. Either
the NPW or the AE measure would lead to that choice, but a comparison of IRRs would
rank the smaller project higher. Another approach, referred to as incremental analysis,
is needed.
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Section 7.4 Mutually Exclusive Alternatives 353
7.4.2 Incremental Investment Analysis
In the previous example, the more costly option requires an incremental investment of
$4,000 at an incremental return of $5,000. Let’s assume that you have exactly $5,000 in
your investment pool.
• If you decide to invest in option A1, you will need to withdraw only $1,000 from
your investment pool. The remaining $4,000 will continue to earn 10% interest. One
year later, you will have $2,000 from the outside investment and $4,400 from the investment pool. With an investment of $5,000, in one year you will have $6,400.
The equivalent present worth of this change in wealth is PW110%2 = - $5,000 +
$6,4001P/F, 10%, 12 = $818.
• If you decide to invest in option A2, you will need to withdraw $5,000 from your investment pool, leaving no money in the pool, but you will have $7,000 from your
outside investment. Your total wealth changes from $5,000 to $7,000 in a year.
The equivalent present worth of this change in wealth is PW110%2 = - $5,000 +
$7,0001P/F, 10%, 12 = $1,364.
In other words, if you decide to take the more costly option, certainly you would be interested in knowing that this additional investment can be justified at the MARR. The 10%of-MARR value implies that you can always earn that rate from other investment sources
(i.e., $4,400 at the end of 1 year for a $4,000 investment). However, in the second option, by
investing the additional $4,000, you would make an additional $5,000, which is equivalent
to earning at the rate of 25%. Therefore, the incremental investment can be justified.
Now we can generalize the decision rule for comparing mutually exclusive projects.
For a pair of mutually exclusive projects (A and B, with B defined as the more costly option), we may rewrite B as
B = A + 1B - A2.
In other words, B has two cash flow components: (1) the same cash flow as A and (2) the
incremental component 1B - A2. Therefore, the only situation in which B is preferred
to A is when the rate of return on the incremental component 1B - A2 exceeds the
MARR. Therefore, for two mutually exclusive projects, rate-of-return analysis is done
by computing the internal rate of return on the incremental investment 1IRR¢2 between
the projects. Since we want to consider increments of investment, we compute the cash
flow for the difference between the projects by subtracting the cash flow for the lower
investment-cost project (A) from that of the higher investment-cost project (B). Then the
decision rule is
If IRR B - A 7 MARR, select B,
If IRR B - A = MARR, select either project,
If IRR B - A 6 MARR, select A,
where B - A is an investment increment (negative cash flow). If a “do-nothing” alternative is allowed, the smaller cost option must be profitable (its IRR must be greater than
the MARR) at first. This means that you compute the rate of return for each alternative
in the mutually exclusive group and then eliminate the alternatives whose IRRs are less
than the MARR before applying the incremental analysis.
It may seem odd to you how this simple rule allows us to select the right project.
Example 7.10 illustrates the incremental investment decision rule.
Incremental
IRR: IRR on the
incremental
investment from
choosing a large
project instead
of a smaller
project.
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354 CHAPTER 7 Rate-of-Return Analysis
EXAMPLE 7.10 IRR on Incremental Investment:
Two Alternatives
John Covington, a college student, wants to start a small-scale painting business during his off-school hours. To economize the start-up business, he decides to purchase
some used painting equipment. He has two mutually exclusive options: Do most of
the painting by himself by limiting his business to only residential painting jobs (B1)
or purchase more painting equipment and hire some helpers to do both residential
and commercial painting jobs that he expects will have a higher equipment cost, but
provide higher revenues as well (B2). In either case, John expects to fold up the business in three years, when he graduates from college.
The cash flows for the two mutually exclusive alternatives are as follows:
n
B1
B2
B2 B1
0
- $3,000
- $12,000
- $9,000
1
1,350
4,200
2,850
2
1,800
6,225
4,425
3
1,500
6,330
4,830
IRR
25%
17.43%
Knowing that both alternatives are revenue projects, which project would John
select at MARR = 10%? (Note that both projects are also profitable at 10%.)
SOLUTION
Given: Incremental cash flow between two alternatives and MARR = 10%.
Find: (a) IRR on the increment and (b) which alternative is preferable.
(a) To choose the best project, we compute the incremental cash flow B2 - B1. Then
we compute the IRR on this increment of investment by solving the equation
- $9,000 + $2,8501P>F, i, 12 + $4,4251P>F, i, 22
+ $4,8301P>F, i, 32 = 0.
(b) We obtain i*B2 - B1 = 15%, as plotted in Figure 7.10. By inspection of the incremental cash flow, we know that it is a simple investment, so IRR B2 - B1 =
i*B2 - B1 . Since IRR B2 - B1 7 MARR, we select B2, which is consistent with
the NPW analysis. Note that, at MARR 7 25%, neither project would be
acceptable.
COMMENTS: Why did we choose to look at the increment B2 - B1 instead of
B1 - B2? Because we want the first flow of the incremental cash flow series to be
negative (an investment flow), so that we can calculate an IRR. By subtracting the
lower initial investment project from the higher, we guarantee that the first increment
will be an investment flow. If we ignore the investment ranking, we might end up
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Section 7.4 Mutually Exclusive Alternatives 355
Select B2
Select B1
Do nothing
6,000
5,000
Net present worth ($)
4,000
3,000
i*B2–B1 = 15%
2,000
1,000
0
B1
–1,000
PW(i)B2 > PW(i)B1
–2,000
–3,000
B2
0
Figure 7.10
5
10
15
20
25
30
Interest rate (%)
35
40
45
NPW profiles for B1 and B2 (Example 7. 10).
with an increment that involves borrowing cash flow and has no internal rate of return.
This is indeed the case for B1 - B2. (i*B1 - B2 is also 15%, not -15%, but it has a different meaning: it is a borrowing rate, not a rate of return on your investment.) If, erroneously, we had compared this i* with the MARR, we might have accepted project B1
over B2. This undoubtedly would have damaged our credibility with management!
The next example indicates that the inconsistency in ranking between NPW and IRR can
also occur when differences in the timing of a project’s future cash flows exist, even if
their initial investments are the same.
EXAMPLE 7.11 IRR on Incremental Investment When Initial
Flows Are Equal
Consider the following two mutually exclusive investment projects that require the
same amount of investment:
Which project would you select on the basis of the rate of return on incremental
investment, assuming that MARR = 12%? (Once again, both projects are profitable
at 12%.)
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356 CHAPTER 7 Rate-of-Return Analysis
n
C1
C2
0
- $9,000
- $9,000
1
480
5,800
2
3,700
3,250
3
6,550
2,000
4
3,780
1,561
IRR
18%
20%
SOLUTION
Given: Cash flows for two mutually exclusive alternatives as shown and
MARR = 12%.
Find: (a) IRR on incremental investment and (b) which alternative is preferable.
(a) When the initial investments are equal, we progress through the cash flows
until we find the first difference and then set up the increment so that this first
nonzero flow is negative (i.e., an investment). Thus, we set up the incremental
investment by taking 1C1 - C22:
n
C1 C2
0
$0
1
-5,320
2
450
3
4,550
4
2,219
We next set the PW equation equal to zero:
- $5,320 + $4501P>F, i, 12 + $4,5501P>F, i, 22
+ $2,2191P>F, i, 32 = 0.
(b) Solving for i yields i* = 14.71%, which is also an IRR, since the increment is a
simple investment. Since IRR C1 - C2 = 14.71% 7 MARR, we would select
C1. If we used NPW analysis, we would obtain PW112%2C1 = $1,443 and
PW112%2C2 = $1,185, indicating that C1 is preferred over C2.
When you have more than two mutually exclusive alternatives, they can be compared in
pairs by successive examination. Example 7.12 illustrates how to compare three mutually
exclusive alternatives. (In Chapter 15, we will examine some multiple-alternative problems in the context of capital budgeting.)
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Section 7.4 Mutually Exclusive Alternatives 357
EXAMPLE 7.12 IRR on Incremental Investment: Three
Alternatives
Consider the following three sets of mutually exclusive alternatives:
n
D1
D2
D3
0
- $2,000
- $1,000
- $3,000
1
1,500
800
1,500
2
1,000
500
2,000
3
800
500
1,000
IRR
34.37%
40.76%
24.81%
Which project would you select on the basis of the rate of return on incremental investment, assuming that MARR = 15%?
SOLUTION
Given: Preceding cash flows and MARR = 15%.
Find: IRR on incremental investment and which alternative is preferable.
Step 1: Examine the IRR of each alternative. At this point, we can eliminate any alternative that fails to meet the MARR. In this example, all three alternatives
exceed the MARR.
Step 2: Compare D1 and D2 in pairs.9 Because D2 has a lower initial cost, compute
the rate of return on the increment (D1D2), which represents an increment
of investment.
n
D1 D2
0
- $1,000
1
700
2
3
500
300
The incremental cash flow represents a simple investment. To find the
incremental rate of return, we write
- $1,000 + $7001P>F, i, 12 + $5001P>F, i, 22 + $3001P>F, i, 32 = 0.
Solving for i*D1 - D2 yields 27.61%, which exceeds the MARR; therefore,
D1 is preferred over D2. Now you eliminate D2 from further consideration.
Step 3: Compare D1 and D3. Once again, D1 has a lower initial cost. Examine the
increment 1D3 - D12:
9
When faced with many alternatives, you may arrange them in order of increasing initial cost. This is not a required step, but it makes the comparison more tractable.
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358 CHAPTER 7 Rate-of-Return Analysis
n
D3 D1
0
- $1,000
1
0
2
1,000
3
200
Here, the incremental cash flow represents another simple investment. The increment
1D3 - D12 has an unsatisfactory 8.8% rate of return; therefore, D1 is preferred over
D3. Accordingly, we conclude that D1 is the best alternative.
EXAMPLE 7.13 Incremental Analysis for Cost-Only Projects
Falk Corporation is considering two types of manufacturing systems to produce its
shaft couplings over six years: (1) a cellular manufacturing system (CMS) and (2) a
flexible manufacturing system (FMS). The average number of pieces to be produced
with either system would be 544,000 per year. The operating cost, initial investment,
and salvage value for each alternative are estimated as follows:
Items
CMS Option
FMS Option
Annual O&M costs:
Annual labor cost
Annual material cost
$1,169,600
$707,200
832,320
598,400
3,150,000
1,950,000
Annual tooling cost
470,000
300,000
Annual inventory cost
141,000
31,500
Annual overhead cost
1,650,000
1,917,000
Total annual costs
$7,412,920
$5,504,100
Investment
$4,500,000
$12,500,000
$500,000
$1,000,000
Annual income taxes
Net salvage value
Figure 7.11 illustrates the cash flows associated with each alternative. The firm’s
MARR is 15%. On the basis of the IRR criterion, which alternative would be a better
choice?
DISCUSSION: Since we can assume that both manufacturing systems would provide
the same level of revenues over the analysis period, we can compare the two alternatives on the basis of cost only. (These are service projects.) Although we cannot
compute the IRR for each option without knowing the revenue figures, we can still
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Section 7.4 Mutually Exclusive Alternatives 359
$500,000
CMS Option
Years
0
1
2
3
4
5
6
$4,500,000
$7,412,920
$1,000,000
FMS Option
Years
0
1
2
3
4
5
6
$5,504,100
$12,500,000
$2,408,820
$1,908,820
0
1
2
3
4
5
Years
Incremental cash flow
(FMS option – CMS option)
$8,000,000
6
Figure 7.11 Comparison of mutually exclusive alternatives
with equal revenues (cost only) (Example 7.13).
calculate the IRR on incremental cash flows. Since the FMS option requires a higher
initial investment than that for the CMS, the incremental cash flow is the difference
1FMS - CMS2.
Incremental
1FMS CMS2
n
CMS Option
FMS Option
0
- $4,500,000
- $12,500,000
- $8,000,000
1
-7,412,920
-5,504,100
1,908,820
2
-7,412,920
-5,504,100
1,908,820
3
-7,412,920
-5,504,100
1,908,820
4
-7,412,920
-5,504,100
1,908,820
5
-7,412,920
-5,504,100
1,908,820
6
-7,412,920
-5,504,100
+ $500,000
+ $1,000,000
$2,408,820
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360 CHAPTER 7 Rate-of-Return Analysis
SOLUTION
Given: Cash flows shown in Figure 7.11 and i = 15% per year.
Find: Incremental cash flows, and select the better alternative on the basis of the IRR
criterion.
First, we have
PW1i2FMS - CMS = - $8,000,000 + $1,908,8201P>A, i, 52
= + $2,408,8201P>F, i, 62
= 0.
Solving for i by trial and error yields 12.43%. Since IRR FMS - CMS = 12.43% 6 15%,
we would select CMS. Although the FMS would provide an incremental annual savings of $1,908,820 in operating costs, the savings are not large enough to justify the
incremental investment of $8,000,000.
COMMENTS: Note that the CMS option is marginally preferred to the FMS option.
However, there are dangers in relying solely on the easily quantified savings in
input factors—such as labor, energy, and materials—from the FMS and in not
considering gains from improved manufacturing performance that are more difficult and subjective to quantify. Factors such as improved product quality, increased
manufacturing flexibility (rapid response to customer demand), reduced inventory
levels, and increased capacity for product innovation are frequently ignored because we have inadequate means for quantifying their benefits. If these intangible
benefits were considered, the FMS option could come out better than the CMS
option.
7.4.3 Handling Unequal Service Lives
In Chapters 5 and 6, we discussed the use of the NPW and AE criteria as bases for comparing projects with unequal lives. The IRR measure can also be used to compare
projects with unequal lives, as long as we can establish a common analysis period.
The decision procedure is then exactly the same as for projects with equal lives. It is
likely, however, that we will have a multiple-root problem, which creates a substantial
computational burden. For example, suppose we apply the IRR measure to a case in
which one project has a 5-year life and the other project has an 8-year life, resulting in
a least common multiple of 40 years. Then when we determine the incremental cash
flows over the analysis period, we are bound to observe many sign changes. This leads
to the possibility of having many i*’s. Example 7.14 uses i* to compare mutually exclusive projects, in which the incremental cash flows reveal several sign changes. (Our
purpose is not to encourage you to use the IRR approach to compare projects with unequal lives; rather, it is to show the correct way to compare them if the IRR approach
must be used.)
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Section 7.4 Mutually Exclusive Alternatives 361
EXAMPLE 7.14 IRR Analysis for Projects with Different
Lives in Which the Increment is a Nonsimple
Investment
Consider Example 5.14, in which a mail-order firm wants to install an automatic mailing system to handle product announcements and invoices. The firm had a choice between two different types of machines. Using the IRR as a decision criterion, select
the best machine. Assume a MARR of 15%, as before.
SOLUTION
Given: Cash flows for two projects with unequal lives, as shown in Figure 5.11, and
MARR = 15%.
Find: The alternative that is preferable.
Since the analysis period is equal to the least common multiple of 12 years, we
may compute the incremental cash flow over this 12-year period. As shown in
Figure 7.12, we subtract the cash flows of model A from those of model B to form
the increment of investment. (Recall that we want the first cash flow difference to
be a negative value.) We can then compute the IRR on this incremental cash flow.
Five sign changes occur in the incremental cash flows, indicating a nonsimple
incremental investment:
n
0
Model A
- $12,500
Model B
- $15,000
Model B Model A
- $2,500
1
-5,000
-4,000
1,000
2
-5,000
-4,000
1,000
3
-12,500
-4,000
11,500
-2,500
-12,500
-5,000
-4,000
1,000
-3,000
-4,000
11,500
-3,000
4
-5,000
5
6
-12,500
7
-5,000
8
-5,000
9
10
-12,500
-15,000
-4,000
1,000
-2,500
-12,500
-3,000
-4,000
11,500
-5,000
-4,000
1,000
-15,000
11
-5,000
-4,000
1,000
12
-3,000
-2,500
500
Four replacement
cycles
Three replacement
cycles
Incremental cash
flows
Even though there are five sign changes in the cash flow, there is only one positive i* for this problem: 63.12%. Unfortunately, however, the investment is not a pure
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362 CHAPTER 7 Rate-of-Return Analysis
Years
0
1
2
3
4
5
6
7
8
9
10
11
12
Model A
$3,000
$3,000
$3,000
$12,500
$12,500
$12,500
$3,000
$5,000
$12,500
Years
0
1
2
3
4
5
6
7
8
9
10
11
12
Model B
$2,500
$2,500
$15,000
$15,000
$2,500
$4,000
$15,000
Model B – Model A (Incremental cash flows)
$11,500
$11,500
$1,000
$1,000
$11,500
$1,000
$1,000
$500
0
4
Model B –
Model A
1
2
3
8
5
6
7
Years
9
10
11
12
$2,500
$12,500
Figure 7.12
$12,500
Comparison of projects with unequal lives (Example 7.14).
investment. We need to employ an external rate to compute the IRR in order to make
a proper accept–reject decision. Assuming that the firm’s MARR is 15%, we will use
a trial-and-error approach. Try i = 20%:
PB120%, 15%20 = - $2,500,
PB120%, 15%21 = - $2,50011.202 + $1,000 = - $2,000.
PB120%, 15%22 = - $2,00011.202 + $1,000 = - $1,400,
PB120%, 15%23 = - $1,40011.202 + $11,500 = $9,820.
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Summary 363
PB120%, 15%24 = $9,82011.152 - $12,500 = - $1,207,
PB120%, 15%25 = - $1,20711.202 + $1,000 = - $448.40.
PB120%, 15%26 = - $448.4011.202 + $11,500 = $10,961.92,
PB120%, 15%27 = $10,961.9211.152 + $1,000 = $13,606.21.
PB120%, 15%28 = $13,606.2111.152 - $12,500 = $3,147.14,
PB120%, 15%29 = $3,147.1411.152 + $11,500 = $15,119.21.
PB120%, 15%210 = $15,119.2111.152 + $1,000 = $18,387.09,
PB120%, 15%211 = $18,387.0911.152 + $1,000 = $22,145.16,
PB120%, 15%212 = $22,145.1611.152 + $500 = $25,966.93.
Since PB120%, 15%212 7 0, the guessed 20% is not the RIC. We may increase the
value of i and repeat the calculations. After several trials, we find that the RIC is
50.68%.10 Since IRR B - A 7 MARR, model B would be selected, which is consistent
with the NPW analysis. In other words, the additional investment over the years to obtain model B ( - $2,500 at n = 0, - $12,500 at n = 4, and - $12,500 at n = 8)
yields a satisfactory rate of return.
COMMENTS: Given the complications inherent in IRR analysis in comparing alternative projects, it is usually more desirable to employ one of the other equivalence
techniques for this purpose. As an engineering manager, you should keep in mind
the intuitive appeal of the rate-of-return measure. Once you have selected a project
on the basis of NPW or AE analysis, you may also wish to express its worth as a
rate of return, for the benefit of your associates.
SUMMARY
The rate of return (ROR) is the interest rate earned on unrecovered project bal-
ances such that an investment’s cash receipts make the terminal project balance
equal to zero. The rate of return is an intuitively familiar and understandable measure of project profitability that many managers prefer to NPW or other equivalence
measures.
Mathematically, we can determine the rate of return for a given project cash flow series by locating an interest rate that equates the net present worth of the project’s cash
flows to zero. This break-even interest rate is denoted by the symbol i*.
The internal rate of return (IRR) is another term for ROR which stresses the fact
that we are concerned with the interest earned on the portion of the project that is
internally invested, not those portions released by (borrowed from) the project.
10
It is tedious to solve this type of problem by a trial-and-error method on your calculator. The problem can be
solved quickly by using the Cash Flow Analyzer, which can be found from the book’s website.
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364 CHAPTER 7 Rate-of-Return Analysis
To apply rate-of-return analysis correctly, we need to classify an investment as ei-
ther simple or nonsimple. A simple investment is defined as an investment in
which the initial cash flows are negative and only one sign change in the net cash
flow occurs, whereas a nonsimple investment is an investment for which more
than one sign change in the cash flow series occurs. Multiple i*’s occur only in
nonsimple investments. However, not all nonsimple investments will have multiple
i*’s. In this regard,
1. The possible presence of multiple i*’s (rates of return) can be predicted by
• The net cash flow sign test.
• The accumulated cash flow sign test.
When multiple rates of return cannot be ruled out by the two methods, it is useful to
generate an NPW profile to approximate the value of i*.
2. All i* values should be exposed to the net investment test. Passing this test indicates that i* is an internal rate of return and is therefore a suitable measure of project profitability. Failing to pass the test indicates project borrowing, a situation that
requires further analysis with the use of an external interest rate.
3. Return-on-invested-capital analysis uses one rate (the firm’s MARR) on externally
invested balances and solves for another rate 1i*2 on internally invested balances.
For a pure investment, i* is the rate of return that is internal to the project. For a mixed
investment, the RIC calculated with the use of the external interest rate (or MARR) is
the true IRR; so the decision rule is as follows:
If IRR 7 MARR, accept the project.
If IRR = MARR, remain indifferent.
If IRR 6 MARR, reject the project.
IRR analysis yields results consistent with NPW and other equivalence methods.
In properly selecting among alternative projects by IRR analysis, incremental invest-
ment must be used. In creating an incremental investment, we always subtract the lower
cost investment from the higher cost one. Basically, you want to know that the extra investment required can be justified on the basis of the extra benefits generated in the future.
PROBLEMS
Note: The symbol i* represents the interest rate that makes the net present value of the
project in question equal to zero. The symbol IRR represents the internal rate of return of
the investment. For a simple investment, IRR = i*. For a nonsimple investment, i* is
generally not equal to IRR.
Concept of Rate of Return
7.1 You are going to buy a new car worth $14,500. The dealer computes your monthly
payment to be $267 for 72 months’ financing. What is the dealer’s rate of return
on this loan transaction?
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Problems 365
7.2 You wish to sell a bond that has a face value of $1,000. The bond bears an interest
rate of 6%, payable semiannually. Four years ago, the bond was purchased at
$900. At least an 8% annual return on the investment is desired. What must be the
minimum selling price of the bond now in order to make the desired return on the
investment?
7.3 In 1970, Wal-Mart offered 300,000 shares of its common stock to the public at a
price of $16.50 per share. Since that time, Wal-Mart has had 11 two-for-one stock
splits. On a purchase of 100 shares at $16.50 per share on the company’s first offering, the number of shares has grown to 204,800 shares worth $10,649,600 on
January 2006. What is the return on investment for investors who purchased the
stock in 1970 (say, over a 36-year ownership period)? Assume that the dividends
received during that period were not reinvested.
7.4 Johnson Controls spent more than $2.5 million retrofitting a government complex
and installing a computerized energy-management system for the State of Massachusetts. As a result, the state’s energy bill dropped from an average of $6 million
a year to $3.5 million. Moreover, both parties will benefit from the 10-year life of
the contract. Johnson recovers half the money it saved in reduced utility costs
(about $1.2 million a year over 10 years); Massachusetts has its half to spend on
other things. What is the rate of return realized by Johnson Controls in this energycontrol system?
7.5 Pablo Picasso’s 1905 portrait Boy with a Pipe sold for $104.2 million in an
auction at Sotheby’s Holdings, Inc., on June 24, 2004, shattering the existing
record for art and ushering in a new era in pricing for 20th-century paintings.
The Picasso, sold by the philanthropic Greentree Foundation, cost Mr. Whitney
about $30,000 in 1950. Determine the annual rate of appreciation of the artwork over 54 years.
Investment Classification and Calculation of i*
7.6 Consider four investments with the following sequences of cash flows:
Net Cash Flow
n
Project A
Project B
Project C
Project D
0
- $18,000
- $30,000
$34,578
- $56,500
1
30,000
32,000
-18,000
2,500
2
20,000
32,000
-18,000
6,459
3
10,000
-22,000
-18,000
-78,345
(a) Identify all the simple investments.
(b) Identify all the nonsimple investments.
(c) Compute i* for each investment.
(d) Which project has no rate of return?
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7.7 Consider the following infinite cash flow series with repeated cash flow patterns:
n
An
0
- $1,000
1
400
2
800
3
500
4
500
5
400
6
800
7
500
8
500
o
o
Determine i* for this infinite cash flow series.
7.8 Consider the following investment projects:
Project Cash Flow
n
A
B
C
D
E
0
- $100
- $100
- $200
- $50
- $50
1
60
70
$20
120
-100
2
150
70
10
40
-50
3
40
5
40
0
4
40
-180
-20
150
5
60
40
150
6
50
30
100
7
400
100
(a) Classify each project as either simple or nonsimple.
(b) Use the quadratic equation to compute i* for project A.
(c) Obtain the rate(s) of return for each project by plotting the NPW as a function
of the interest rate.
7.9 Consider the projects in Table P7.9.
(a) Classify each project as either simple or nonsimple.
(b) Identify all positive i*’s for each project.
(c) For each project, plot the present worth as a function of the interest rate (i).
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Problems 367
TABLE P7.9
Net Cash Flow for Four Projects
Net Cash Flow
n
A
B
C
D
0
- $2,000
- $1,500
- $1,800
- $1,500
1
500
800
5,600
-360
2
100
600
4,900
4,675
3
100
500
-3,500
2,288
4
2,000
700
7,000
5
-1,400
6
2,100
7
900
7.10 Consider the following financial data for a project:
Initial investment
$50,000
Project life
8 years
Salvage value
Annual revenue
$10,000
$25,000
Annual expenses
(including income taxes)
$ 9,000
(a) What is i* for this project?
(b) If the annual expense increases at a 7% rate over the previous year’s expenses,
but the annual income is unchanged, what is the new i*?
(c) In part (b), at what annual rate will the annual income have to increase to maintain the same i* obtained in part (a)?
7.11 Consider two investments, A and B, with the following sequences of cash flows:
Net Cash Flow
n
Project A
Project B
0
- $25,000
- $25,000
1
2,000
10,000
2
6,000
10,000
3
12,000
10,000
4
24,000
10,000
5
28,000
5,000
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368 CHAPTER 7 Rate-of-Return Analysis
(a) Compute i* for each investment.
(b) Plot the present-worth curve for each project on the same chart, and find the
interest rate that makes the two projects equivalent.
7.12 Consider the following investment projects:
Project Cash Flows
N
A
B
C
D
E
F
0
- $100
- $100
- $100
- $100
- $100
- $100
1
200
470
-200
0
300
300
2
300
720
200
0
250
100
3
400
360
250
500
-40
400
(a) For each project, apply the sign rule to predict the number of possible i*’s.
(b) For each project, plot the NPW profile as a function of i between 0 and 200%.
(c) For each project, compute the value(s) of i*.
7.13 Consider an investment project with the following cash flows:
n
Net Cash Flow
0
- $120,000
1
94,000
2
144,000
3
72,000
(a) Find the IRR for this investment.
(b) Plot the present worth of the cash flow as a function of i.
(c) On the basis of the IRR criterion, should the project be accepted at
MARR = 15%?
Mixed Investments
7.14 Consider the following investment projects:
Net Cash Flow
n
Project 1
Project 2
Project 3
0
- $1,000
- $2,000
- $1,000
1
500
1,560
1,400
2
840
944
-100
IRR
?
?
?
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Problems 369
Assume that MARR = 12% in the following questions:
(a) Compute i* for each investment. If the problem has more than one i*, identify
all of them.
(b) Compute IRR(true) for each project.
(c) Determine the acceptability of each investment.
7.15 Consider the following investment projects:
Project Cash Flow
n
A
B
C
D
E
0
- $100
- $100
- $5
- $100
$200
1
100
30
10
30
100
2
24
30
30
30
-500
3
70
-40
30
-500
4
70
30
200
30
600
5
(a) Use the quadratic equation to compute i* for A.
(b) Classify each project as either simple or nonsimple.
(c) Apply the cash flow sign rules to each project, and determine the number of
possible positive i*’s. Identify all projects having a unique i*.
(d) Compute the IRRs for projects B through E.
(e) Apply the net-investment test to each project.
7.16 Consider the following investment projects:
Net Cash Flow
n
Project 1
Project 2
Project 3
0
- $1,600
- $5,000
- $1,000
1
10,000
10,000
4,000
2
10,000
30,000
-4,000
3
-40,000
Assume that MARR = 12% in the following questions:
(a) Identify the i*1’s2 for each investment. If the project has more than one i*,
identify all of them.
(b) Which project(s) is (are) a mixed investment?
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370 CHAPTER 7 Rate-of-Return Analysis
(c) Compute the IRR for each project.
(d) Determine the acceptability of each project.
7.17 Consider the following investment projects:
Net Cash Flow
n
Project A
Project B
Project C
0
- $100
- $150
- $100
1
30
50
410
2
50
50
-558
3
80
50
252
4
100
IRR
(23.24%)
(21.11%)
(20%, 40%, 50%)
Assume that MARR = 12% for the following questions:
(a) Identify the pure investment(s).
(b) Identify the mixed investment(s).
(c) Determine the IRR for each investment.
(d) Which project would be acceptable?
7.18 The Boeing Company has received a NASA contract worth $460 million to
build rocket boosters for future space missions. NASA will pay $50 million
when the contract is signed, another $360 million at the end of the first year, and
the $50 million balance at the end of second year. The expected cash outflows
required to produce these rocket boosters are estimated to be $150 million now,
$100 million during the first year, and $218 million during the second year. The
firm’s MARR is 12%. The cash flow is as follows:
n
Outflow
Inflow
Net Cash
Flow
0
$150
$50
- $100
1
100
360
260
2
218
50
-168
(a) Show whether this project is or is not a mixed investment.
(b) Compute the IRR for this investment.
(c) Should Boeing accept the project?
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Problems 371
7.19 Consider the following investment projects:
Net Cash Flow
n
Project A
0
- $100
1
216
-150
50
2
-116
100
-50
3
50
200
4
40
i*
Project B
Project C
- $100
?
15.51%
29.95%
(a) Compute i* for project A. If there is more than one i*, identify all of them.
(b) Identify the mixed investment(s).
(c) Assuming that MARR = 10%, determine the acceptability of each project on
the basis of the IRR criterion.
7.20 Consider the following investment projects:
Net Cash Flow
n
A
B
C
D
E
0
- $1,000
- $5,000
- $2,000
- $2,000
- $1,000
1
3,100
20,000
1,560
2,800
3,600
2
-2,200
12,000
944
-200
-5,700
3
i*
3,600
-3,000
?
?
18%
32.45%
35.39%
Assume that MARR = 12% in the following questions:
(a) Compute i* for projects A and B. If the project has more than one i*, identify
all of them.
(b) Classify each project as either a pure or a mixed investment.
(c) Compute the IRR for each investment.
(d) Determine the acceptability of each project.
7.21 Consider an investment project whose cash flows are as follows:
n
Net Cash Flow
0
- $5,000
1
10,000
2
30,000
3
-40,000
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372 CHAPTER 7 Rate-of-Return Analysis
(a) Plot the present-worth curve by varying i from 0% to 250%.
(b) Is this a mixed investment?
(c) Should the investment be accepted at MARR = 18%?
7.22 Consider the following two mutually exclusive investment projects:
Net Cash Flow
n
Project A
Project B
0
- $300
- $800
1
0
1,150
2
690
40
51.66%
46.31%
i*
Assume that MARR = 15%.
(a) According to the IRR criterion, which project would be selected?
(b) Sketch the PW(i) function on the incremental investment 1B - A2.
7.23 Consider the following cash flows of a certain project:
n
Net Cash Flow
0
- $100,000
1
310,000
2
-220,000
The project’s i*’s are computed as 10% and 100%, respectively. The firm’s MARR
is 8%.
(a) Show why this investment project fails the net-investment test.
(b) Compute the IRR, and determine the acceptability of this project.
7.24 Consider the following investment projects:
Net Cash Flow
n
Project 1
Project 2
Project 3
0
- $1,000
- $1,000
- $1,000
1
-1,000
1,600
1,500
2
2,000
-300
-500
3
3,000
-200
2,000
Which of the following statements is correct?
(a) All projects are nonsimple investments.
(b) Project 3 should have three real rates of return.
(c) All projects will have a unique positive real rate of return.
(d) None of the above.
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Problems 373
IRR Analysis
7.25 Agdist Corporation distributes agricultural equipment. The board of directors is
considering a proposal to establish a facility to manufacture an electronically controlled “intelligent” crop sprayer invented by a professor at a local university. This
crop sprayer project would require an investment of $10 million in assets and
would produce an annual after-tax net benefit of $1.8 million over a service life of
eight years. All costs and benefits are included in these figures. When the project
terminates, the net proceeds from the sale of the assets will be $1 million. Compute the rate of return of this project. Is this a good project at MARR = 10%?
7.26 Consider an investment project with the following cash flows:
n
Cash Flow
0
- $5,000
1
0
2
4,840
3
1,331
Compute the IRR for this investment. Is the project acceptable at MARR = 10%?
7.27 Consider the following cash flow of a certain project:
n
Net Cash Flow
0
- $2,000
1
800
2
900
3
X
If the project’s IRR is 10%,
(a) Find the value of X.
(b) Is this project acceptable at MARR = 8%?
7.28 You are considering a luxury apartment building project that requires an investment of $12,500,000. The building has 50 units. You expect the maintenance cost
for the apartment building to be $250,000 the first year and $300,000 the second
year. The maintenance cost will continue to increase by $50,000 in subsequent
years. The cost to hire a manager for the building is estimated to be $80,000
per year. After five years of operation, the apartment building can be sold for
$14,000,000. What is the annual rent per apartment unit that will provide a return
on investment of 15%? Assume that the building will remain fully occupied during
its five years of operation.
7.29 A machine costing $25,000 to buy and $3,000 per year to operate will save mainly
labor expenses in packaging over six years. The anticipated salvage value of the
machine at the end of the six years is $5,000. To receive a 10% return on investment
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374 CHAPTER 7 Rate-of-Return Analysis
(rate of return), what is the minimum required annual savings in labor from this
machine?
7.30 Champion Chemical Corporation is planning to expand one of its propylenemanufacturing facilities. At n = 0, a piece of property costing $1.5 million must
be purchased to build a plant. The building, which needs to be expanded during
the first year, costs $3 million. At the end of the first year, the company needs to
spend about $4 million on equipment and other start-up costs. Once the building
becomes operational, it will generate revenue in the amount of $3.5 million during
the first operating year. This will increase at the annual rate of 5% over the previous year’s revenue for the next 9 years. After 10 years, the sales revenue will stay
constant for another 3 years before the operation is phased out. (It will have a
project life of 13 years after construction.) The expected salvage value of the
land at the end of the project’s life would be about $2 million, the building about
$1.4 million, and the equipment about $500,000. The annual operating and maintenance costs are estimated to be approximately 40% of the sales revenue each
year. What is the IRR for this investment? If the company’s MARR is 15%, determine whether the investment is a good one. (Assume that all figures represent the
effect of the income tax.)
7.31 Recent technology has made possible a computerized vending machine that can
grind coffee beans and brew fresh coffee on demand. The computer also makes
possible such complicated functions as changing $5 and $10 bills, tracking the age
of an item, and moving the oldest stock to the front of the line, thus cutting down
on spoilage. With a price tag of $4,500 for each unit, Easy Snack has estimated the
cash flows in millions of dollars over the product’s six-year useful life, including
the initial investment, as follows:
n
Net Cash Flow
0
- $20
1
8
2
17
3
19
4
18
5
10
6
3
(a) On the basis of the IRR criterion, if the firm’s MARR is 18%, is this product
worth marketing?
(b) If the required investment remains unchanged, but the future cash flows are
expected to be 10% higher than the original estimates, how much of an increase in IRR do you expect?
(c) If the required investment has increased from $20 million to $22 million, but
the expected future cash flows are projected to be 10% smaller than the original estimates, how much of a decrease in IRR do you expect?
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Problems 375
Comparing Alternatives
7.32 Consider two investments A and B with the following sequences of cash flows:
Net Cash Flow
Project A
Project B
n
0
- $120,000
- $100,000
1
20,000
15,000
2
20,000
15,000
3
120,000
130,000
(a) Compute the IRR for each investment.
(b) At MARR = 15%, determine the acceptability of each project.
(c) If A and B are mutually exclusive projects, which project would you select,
based on the rate of return on incremental investment?
7.33 With $10,000 available, you have two investment options. The first is to buy a certificate of deposit from a bank at an interest rate of 10% annually for five years.
The second choice is to purchase a bond for $10,000 and invest the bond’s interest
in the bank at an interest rate of 9%. The bond pays 10% interest annually and will
mature to its face value of $10,000 in five years. Which option is better? Assume
that your MARR is 9% per year.
7.34 A manufacturing firm is considering the following mutually exclusive alternatives:
Net Cash Flow
n
Project A1
Project A2
0
- $2,000
- $3,000
1
1,400
2,400
2
1,640
2,000
Determine which project is a better choice at a MARR = 15%, based on the IRR
criterion.
7.35 Consider the following two mutually exclusive alternatives:
Net Cash Flow
n
Project A1
Project A2
0
- $10,000
- $12,000
1
5,000
6,100
2
5,000
6,100
3
5,000
6,100
(a) Determine the IRR on the incremental investment in the amount of $2,000.
(b) If the firm’s MARR is 10%, which alternative is the better choice?
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7.36 Consider the following two mutually exclusive investment alternatives:
n
Net Cash Flow
Project A1
Project A2
0
- $15,000
- $20,000
1
7,500
8,000
2
7,500
15,000
3
7,500
5,000
IRR
23.5%
20%
(a) Determine the IRR on the incremental investment in the amount of $5,000.
(Assume that MARR = 10%.)
(b) If the firm’s MARR is 10%, which alternative is the better choice?
7.37 You are considering two types of automobiles. Model A costs $18,000 and model B
costs $15,624. Although the two models are essentially the same, after four years of
use model A can be sold for $9,000, while model B can be sold for $6,500. Model A
commands a better resale value because its styling is popular among young college
students. Determine the rate of return on the incremental investment of $2,376. For
what range of values of your MARR is model A preferable?
7.38 A plant engineer is considering two types of solar water heating system:
Model A
Initial cost
Model B
$7,000
$10,000
Annual savings
$700
$1,000
Annual maintenance
$100
$50
Expected life
20 years
20 years
Salvage value
$400
$500
The firm’s MARR is 12%. On the basis of the IRR criterion, which system is the
better choice?
7.39 Consider the following investment projects:
Net Cash Flow
n
A
B
C
D
E
F
0
- $100
- $200
- $4,000
- $2,000
- $2,000
- $3,000
1
60
120
2,410
1,400
3,700
2,500
2
50
150
2,930
1,720
1,640
1,500
3
50
21.65%
21.86%
31.10%
121.95%
23.74%
i* 28.89%
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Problems 377
Assume that MARR = 15%.
(a) Projects A and B are mutually exclusive. Assuming that both projects can be
repeated for an indefinite period, which one would you select on the basis of
the IRR criterion?
(b) Suppose projects C and D are mutually exclusive. According to the IRR criterion, which project would be selected?
(c) Suppose projects E and F are mutually exclusive. Which project is better according to the IRR criterion?
7.40 Fulton National Hospital is reviewing ways of cutting the cost of stocking medical
supplies. Two new stockless systems are being considered, to lower the hospital’s
holding and handling costs. The hospital’s industrial engineer has compiled the relevant financial data for each system as follows (dollar values are in millions):
Current
Practice
Just-inTime
System
Stockless
Supply
System
Start-up
cost
$0
$2.5
$5
Annual stock
holding cost
$3
$1.4
$0.2
Annual
operating cost
$2
$1.5
$1.2
8 years
8 years
8 years
System life
The system life of eight years represents the period that the contract with the medical suppliers is in force. If the hospital’s MARR is 10%, which system is more
economical?
7.41 Consider the cash flows for the following investment projects:
Project Cash Flow
n
A
B
C
D
E
0
- $1,000
- $1,000
- $2,000
$1,000
- $1,200
1
900
600
900
-300
400
2
500
500
900
-300
400
3
100
500
900
-300
400
4
50
100
900
-300
400
Assume that the MARR = 12%.
(a) Suppose A, B, and C are mutually exclusive projects. Which project would be
selected on the basis of the IRR criterion?
(b) What is the borrowing rate of return (BRR) for project D?
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378 CHAPTER 7 Rate-of-Return Analysis
(c) Would you accept project D at MARR = 20%?
(d) Assume that projects C and E are mutually exclusive. Using the IRR criterion,
which project would you select?
7.42 Consider the following investment projects:
Net Cash Flow
n
Project 1
Project 2
Project 3
0
- $1,000
- $5,000
- $2,000
1
500
7,500
1,500
2
2,500
600
2,000
Assume that MARR = 15%.
(a) Compute the IRR for each project.
(b) On the basis of the IRR criterion, if the three projects are mutually exclusive
investments, which project should be selected?
7.43 Consider the following two investment alternatives:
Net Cash Flow
N
Project A
0
- $10,000
- $20,000
1
5,500
0
2
5,500
0
3
5,500
40,000
30%
?
?
6300
IRR
PW(15%)
Project B
The firm’s MARR is known to be 15%.
(a) Compute the IRR of project B.
(b) Compute the NPW of project A.
(c) Suppose that projects A and B are mutually exclusive. Using the IRR, which
project would you select?
7.44 The E. F. Fedele Company is considering acquiring an automatic screwing machine for its assembly operation of a personal computer. Three different models
with varying automatic features are under consideration. The required investments are $360,000 for model A, $380,000 for model B, and $405,000 for model C.
All three models are expected to have the same service life of eight years. The following financial information, in which model 1B - A2 represents the incremental
cash flow determined by subtracting model A’s cash flow from model B’s, is
available:
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Problems 379
Model
IRR (%)
A
30%
B
15
C
25
Model
Incremental IRR (%)
1B - A2
5%
1C - B2
40
1C - A2
15
If the firm’s MARR is known to be 12%, which model should be selected?
7.45 The GeoStar Company, a leading manufacturer of wireless communication devices, is considering three cost-reduction proposals in its batch job-shop manufacturing operations. The company has already calculated rates of return for the three
projects, along with some incremental rates of return:
Incremental
Investment
Incremental
Rate of Return (%)
A1 - A0
18%
A2 - A0
20
A3 - A0
25
A2 - A1
10
A3 - A1
18
A3 - A2
23
A 0 denotes the do-nothing alternative. The required investments are $420,000 for
A 1 , $550,000 for A 2 , and $720,000 for A 3 . If the MARR is 15%, what system
should be selected?
7.46 A manufacturer of electronic circuit boards is considering six mutually exclusive
cost-reduction projects for its PC-board manufacturing plant. All have lives of 10
years and zero salvage values. The required investment and the estimated after-tax
reduction in annual disbursements for each alternative are as follows, along with
computed rates of return on incremental investments:
Proposal
Required After-Tax
Aj
Investment
Savings
Rate of
Return (%)
A1
$60,000
$22,000
35.0%
A2
100,000
28,200
25.2
A3
110,000
32,600
27.0
A4
120,000
33,600
25.0
A5
140,000
38,400
24.0
A6
150,000
42,200
25.1
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380 CHAPTER 7 Rate-of-Return Analysis
Incremental
Investment
Incremental
Rate of Return (%)
A2 - A1
9.0%
A3 - A2
42.8
A4 - A3
0.0
A5 - A4
20.2
A6 - A5
36.3
If the MARR is 15%, which project would you select, based on the rate of return
on incremental investment?
7.47 Baby Doll Shop manufactures wooden parts for dollhouses. The worker is paid $8.10
an hour and, using a handsaw, can produce a year’s required production (1,600 parts)
in just eight 40-hour weeks. That is, the worker averages five parts per hour when
working by hand. The shop is considering purchasing of a power band saw with associated fixtures, to improve the productivity of this operation. Three models of power
saw could be purchased: Model A (the economy version), model B (the high-powered
version), and model C (the deluxe high-end version). The major operating difference
between these models is their speed of operation. The investment costs, including the
required fixtures and other operating characteristics, are summarized as follows:
Category
Production rate
(parts/hour)
By
Hand
Model
A
Model
B
Model
C
5
10
15
20
320
160
107
80
2,592
1,296
867
648
400
420
480
Initial
investment ($)
4,000
6,000
7,000
Salvage value ($)
400
600
700
20
20
20
Labor hours required
(hours/year)
Annual labor cost
(@ $8.10/hour)
Annual power
cost ($)
Service life (years)
Assume that MARR = 10%. Are there enough savings to purchase any of the
power band saws? Which model is most economical, based on the rate-of-return
principle? (Assume that any effect of income tax has been already considered in
the dollar estimates.) (Source: This problem is adapted with the permission of Professor Peter Jackson of Cornell University.)
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Short Case Studies 381
Unequal Service Lives
7.48 Consider the following two mutually exclusive investment projects for which
MARR = 15%:
Net Cash Flow
Project A
Project B
n
0
- $100
- $200
1
60
120
2
50
150
3
50
IRR
28.89%
21.65%
On the basis of the IRR criterion, which project would be selected under an infinite planning horizon with project repeatability likely?
7.49 Consider the following two mutually exclusive investment projects:
Net Cash Flow
n
Project A1
Project A2
0
- $10,000
- $15,000
1
5,000
20,000
2
5,000
3
5,000
(a) To use the IRR criterion, what assumption must be made in comparing a set of
mutually exclusive investments with unequal service lives?
(b) With the assumption defined in part (a), determine the range of MARRs which
will indicate that project A1 should be selected.
Short Case Studies
ST7.1 Critics have charged that, in carrying out an economic analysis, the commercial
nuclear power industry does not consider the cost of decommissioning, or “mothballing,” a nuclear power plant and that the analysis is therefore unduly optimistic.
As an example, consider the Tennessee Valley Authority’s Bellefont twin nuclear
generating facility under construction at Scottsboro, in northern Alabama: The initial cost is $1.5 billion (present worth at the start of operations), the estimated life
is 40 years, the annual operating and maintenance costs the first year are assumed
to be 4.6% of the initial cost and are expected to increase at the fixed rate of 0.05%
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382 CHAPTER 7 Rate-of-Return Analysis
of the initial cost each year, and annual revenues are estimated to be three times
the annual operating and maintenance costs throughout the life of the plant.
(a) The criticism that the economic analysis is overoptimistic because it omits
“mothballing” costs is not justified, since the addition of a cost of 50% of the
initial cost to “mothball” the plant decreases the 10% rate of return only to approximately 9.9%.
(b) If the estimated life of the plants is more realistically taken to be 25 years instead of 40 years, then the criticism is justified. By reducing the life to 25
years, the rate of return of approximately 9% without a “mothballing” cost
drops to approximately 7.7% when a cost to “mothball” the plant equal to 50%
of the initial cost is added to the analysis.
Comment on these statements.
ST7.2 The B&E Cooling Technology Company, a maker of automobile air-conditioners,
faces an uncertain, but impending, deadline to phase out the traditional chilling
technique, which uses chlorofluorocarbons (CFCs), a family of refrigerant chemicals believed to attack the earth’s protective ozone layer. B&E has been pursuing
other means of cooling and refrigeration. As a near-term solution, the engineers
recommend a cold technology known as absorption chiller, which uses plain water
as a refrigerant and semiconductors that cool down when charged with electricity.
B&E is considering two options:
• Option 1. Retrofit the plant now to adapt the absorption chiller and continue to
be a market leader in cooling technology. Because of untested technology on a
large scale, it may cost more to operate the new facility while personnel are
learning the new system.
• Option 2. Defer the retrofitting until the federal deadline, which is three years
away. With expected improvement in cooling technology and technical knowhow, the retrofitting cost will be cheaper, but there will be tough market competition, and the revenue would be less than that of Option 1.
The financial data for the two options are as follows:
Option 1
Now
Option 2
3 years
from now
Initial
investment
$6 million
$5 million
System life
8 years
8 years
Salvage value
$1 million
$2 million
Annual revenue
$15 million
$11million
Annual O&M
costs
$6 million
$7 million
Investment
timing
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Short Case Studies 383
(a) What assumptions must be made to compare these two options?
(b) If B&E’s MARR is 15%, which option is the better choice, based on the IRR
criterion?
ST7.3 An oil company is considering changing the size of a small pump that is currently
operational in wells in an oil field. If this pump is kept, it will extract 50% of the
known crude-oil reserve in the first year of its operation and the remaining 50% in
the second year. A pump larger than the current pump will cost $1.6 million, but it
will extract 100% of the known reserve in the first year. The total oil revenues over
the two years are the same for both pumps, namely, $20 million. The advantage of
the large pump is that it allows 50% of the revenues to be realized a year earlier
than with the small pump.
Current
Pump
Larger
Pump
Investment, year 0
0
$1.6 million
Revenue, year 1
$10 million
$20 million
Revenue, year 2
$10 million
0
If the firm’s MARR is known to be 20%, what do you recommend, based on the
IRR criterion?
ST7.4 You have been asked by the president of the company you work for to evaluate
the proposed acquisition of a new injection molding machine for the firm’s manufacturing plant. Two types of injection molding machines have been identified,
with the following estimated cash flows:
Net Cash Flow
n
Project 1
Project 2
0
- $30,000
- $40,000
1
20,000
43,000
2
18,200
5,000
IRR
18.1%
18.1%
You return to your office, quickly retrieve your old engineering economics text,
and then begin to smile: Aha—this is a classic rate-of-return problem! Now,
using a calculator, you find out that both projects have about the same rate of
return: 18.1%. This figure seems to be high enough to justify accepting the project, but you recall that the ultimate justification should be done with reference to
the firm’s MARR. You call the accounting department to find out the current
MARR the firm should use in justifying a project. “Oh boy, I wish I could tell
you, but my boss will be back next week, and he can tell you what to use,” says
the accounting clerk.
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384 CHAPTER 7 Rate-of-Return Analysis
A fellow engineer approaches you and says, “I couldn’t help overhearing you
talking to the clerk. I think I can help you. You see, both projects have the same
IRR, and on top of that, project 1 requires less investment, but returns more cash
flows ( - $30,000 + $20,000 + $18,200 = $8,200, and - $40,000 + $43,000 +
$5,000 = $8,000); thus, project 1 dominates project 2. For this type of decision
problem, you don’t need to know a MARR!”
(a) Comment on your fellow engineer’s statement.
(b) At what range of MARRs would you recommend the selection of project 2?
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P A R T
3
Analysis of Project
Cash Flows
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EIGHT
CHAPTER
Cost Concepts Relevant
to Decision Making
High Hopes for Beer Bottles1 Three hundred billion beer
bottles a year worldwide is a mighty tempting target for the plastics
industry. Brewers generally say they need a bottle that provides shelf
life of over 120 days with less than 15% loss of CO2 and admittance
of no more than 1 ppm of oxygen. Internal or external coatings, and
three- or five-layer polyethylene terephthalate (PET) structures using
barrier materials are being evaluated to reach that performance.
What is the least expensive way to make a 0.5L PET barrier bottle?
Summit International LLC, Smyrna, GA, which specializes in preform
and container development and market research, compared the manufacturing costs of five different barrier technologies against a standard
monolayer PET bottle. Summit looked at three-layer and five-layer and
at internally and externally coated containers, all of them reheat stretch
blow molded. It found the bottle with an external coating to be least
costly, while the internally coated bottle was the most expensive.
The firm compared a five-layer structure with an oxygen-scavenger
material, a three-layer bottle with a $2.50/lb barrier material, and a
second three-layer structure with a $6/lb barrier. Also compared
were a bottle coated inside using Sidel’s new Actis plasma technology
and a bottle coated on the outside.
Capital investment (preform and bottle machines, utilities, downstream equipment, quality control, spare parts, and installation) for
producing 20,000 bottles/hr is $10.8 million for the five-layer bottle,
$9.9 million for both three-layer structures, $9.2 million for internal
coating, $7.5 for external coating, and $6.8 million with no barrier.
1
“Blow Molding Close-Up: Prospects Brighten for PET Beer Bottles,” Mikell Knights, Plastic
Technology Online, © copyright 2005, Gardner Publications, Inc.
386
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Coat
Page 387
Evaporate
Cure
Spray coating of external PET bottles
Direct manufacturing cost per 1,000 (materials, energy, labor, maintenance, and scrap) amounts to $66.57 for three layers with the expensive
barrier, $59.35 for five layers, $55.34 for the external coating, $54.63 for
three layers with the low-cost barrier, $46.90 for internal coating, and
$44.63 without barrier.
You may be curious how all of those cost data were estimated in the
chapter opening story. Before we study the different kinds of engineering
economic decision problems, we need to understand the concept of various costs. At the level of plant operations, engineers must make decisions
involving materials, plant facilities, and the in-house capabilities of company
personnel. Consider, for example, the manufacture of food processors.
In terms of selecting materials, several of the parts could be made of
plastic, whereas others must be made of metal. Once materials have been
chosen, engineers must consider the production methods, the shipping
weight, and the method of packaging necessary to protect the different types
of materials. In terms of actual production, parts may be made in-house
or purchased from an outside vendor, depending on the availability of
machinery and labor.
387
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388 CHAPTER 8 Cost Concepts Relevant to Decision Making
Present
economic
studies: Various
economic
analyses for
short operating
decisions.
All these operational decisions (commonly known as present economic studies in
traditional engineering economic texts) require estimating the costs associated with various production or manufacturing activities. Because these costs also provide the basis for
developing successful business strategies and planning future operations, it is important to
understand how various costs respond to changes in levels of business activity. In this
chapter, we discuss many of the possible uses of cost data. We also discuss how to define,
classify, and estimate costs for each use. Our ultimate task in doing so is to explain how
costs are classified in making various engineering economic decisions.
CHAPTER LEARNING OBJECTIVES
After completing this chapter, you should understand the following concepts:
Various cost terminologies that are common in cost accounting and
engineering economic studies.
How a cost item reacts or responds to changes in the level of production
or business activities.
The types of cost data that management needs in making choice between
alternative courses of action.
The types of present economic studies frequently performed by engineers
in manufacturing and business environment.
How to develop a production budget related to operating activities.
8.1 General Cost Terms
n engineering economics, the term cost is used in many different ways. Because there
are many types of costs, each is classified differently according to the immediate
needs of management. For example, engineers may want cost data to prepare external
reports, to prepare planning budgets, or to make decisions. Also, each different use of
cost data demands a different classification and definition of cost. For example, the
preparation of external financial reports requires the use of historical cost data, whereas
decision making may require current cost data or estimated future cost data.
Our initial focus in this chapter is on manufacturing companies, because their basic activities (acquiring raw materials, producing finished goods, marketing, etc.) are commonly
found in most other businesses. Therefore, the understanding of costs in a manufacturing
company can be helpful in understanding costs in other types of business organizations.
I
8.1.1 Manufacturing Costs
Several types of manufacturing costs incurred by a typical manufacturer are illustrated in
Figure 8.1. In converting raw materials into finished goods, a manufacturer incurs various
costs associated with operating a factory. Most manufacturing companies divide manufacturing costs into three broad categories: direct raw material costs, direct labor costs,
and manufacturing overhead.
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Section 8.1 General Cost Terms
Contains
inventory of
raw materials
Purchase of
raw materials
Raw materials
warehouse
Raw materials
purchased
389
Contains
inventory
of work
in process
Manufacturing
costs
Factory
Raw materials
used
Direct labor
Finished
Manufacturing
goods
overhead
Cost of
goods sold
Consumers
Cost of
finished goods
manufactured
Finished goods
warehouse
Goods sold
Contains
inventory
of finished
goods
Figure 8.1
Various types of manufacturing costs incurred by a manufacturer.
Direct Raw Materials
Direct raw materials are any materials that are used in the final product and that can be
easily traced to it. Some examples are wood in furniture, steel in bridge construction,
paper in printing firms, and fabric for clothing manufacturers. It is important to note that
the finished product of one company can become the raw materials of another company.
For example, the computer chips produced by Intel are a raw material used by Dell in its
personal computers.
Direct Labor
Like direct raw materials, direct labor incurs costs that go into the production of a product.
The labor costs of assembly-line workers, for example, would be direct labor costs, as
would the labor costs of welders in metal-fabricating industries, carpenters or bricklayers
in home building, and machine operators in various manufacturing operations.
Manufacturing Overhead
Manufacturing overhead, the third element of manufacturing cost, includes all costs of
manufacturing except the costs of direct materials and direct labor. In particular, manufacturing overhead includes such items as the costs of indirect materials; indirect labor;
maintenance and repairs on production equipment; heat and light, property taxes, depreciation, and insurance on manufacturing facilities; and overtime premiums. The most
Direct cost: A
cost that can be
directly traced
to producing
specific goods
or services.
Overhead: A
reference in
accounting to all
costs not
including or
related to direct
labor, materials,
or administration
costs.
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390 CHAPTER 8 Cost Concepts Relevant to Decision Making
important thing to note about manufacturing overhead is the fact that, unlike direct materials and direct labor, it is not easily traceable to specific units of output. In addition,
many manufacturing overhead costs do not change as output changes, as long as the production volume stays within the capacity of the plant. For example, depreciation of factory
buildings is unaffected by the amount of production during any particular period. If, however, a new building is required to meet any increased production, manufacturing overhead
will certainly increase.
• Sometimes it may not be worth the effort to trace the costs of materials that are relatively insignificant in the finished products. Such minor items include the solder used
to make electrical connections in a computer circuit board and the glue used to bind
this textbook. Materials such as solder and glue are called indirect materials and are
included as part of manufacturing overhead.
• Sometimes we may not be able to trace some of the labor costs to the creation of a
product. We treat this type of labor cost as a part of manufacturing overhead, along
with indirect materials. Indirect labor includes the wages of janitors, supervisors,
material handlers, and night security guards. Although the efforts of these workers
are essential to production, it would be either impractical or impossible to trace their
costs to specific units of product. Therefore, we treat such labor costs as indirect
labor costs.
8.1.2 Nonmanufacturing Costs
Matching
concept: The
accounting
principle that
requires the
recognition of all
costs that are
associated with
the generation
of the revenue
reported in the
income
statement.
Two additional costs incurred in supporting any manufacturing operation are (1) marketing or selling costs and (2) administrative costs. Marketing or selling costs include all
costs necessary to secure customer orders and get the finished product or service into the
hands of the customer. Breakdowns of these types of costs provide data for control over
selling and administrative functions in the same way that manufacturing cost breakdowns
provide data for control over manufacturing functions. For example, a company incurs
costs for
• Overhead. Heat and light, property taxes, and depreciation or similar items associated with the company’s selling and administrative functions.
• Marketing. Advertising, shipping, sales travel, sales commissions, and sales salaries.
Marketing costs include all executive, organizational, and clerical costs associated
with sales activities.
• Administrative functions. Executive compensation, general accounting, public
relations, and secretarial support, associated with the general management of an
organization.
8.2 Classifying Costs for Financial Statements
For purposes of preparing financial statements, we often classify costs as either period costs
or product costs. To understand the difference between them, we must introduce the matching concept essential to any accounting studies. In financial accounting, the matching principle states that the costs incurred in generating a certain amount of revenue should be
recognized as expenses in the same period that the revenue is recognized. This matching
principle is the key to distinguishing between period costs and product costs. Some costs
are matched against periods and become expenses immediately. Other costs are matched
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Section 8.2 Classifying Costs for Financial Statements 391
against products and do not become expenses until the products are sold, which may be in
the next accounting period.
8.2.1 Period Costs
Period costs are costs charged to expenses in the period in which they are incurred. The
underlying assumption is that the associated benefits are received in the same period the
cost is incurred. Some specific examples are all general and administrative expenses, selling expenses, and insurance and income tax expenses. Therefore, advertising costs, executives’ salaries, sales commissions, public-relations costs, and other nonmanufacturing
costs discussed earlier would all be period costs. Such costs are not related to the production and flow of manufactured goods, but are deducted from revenue in the income
statement. In other words, period costs will appear on the income statement as expenses
during the time in which they occur.
8.2.2 Product Costs
Some costs are better matched against products than they are against periods. Costs of
this type—called product costs—consist of the costs involved in the purchase or manufacture of goods. In the case of manufactured goods, product costs are the costs of direct
materials, direct labor costs, and manufacturing overhead. Product costs are not viewed
as expenses; rather, they are the cost of creating inventory. Thus, product costs are considered an asset until the associated goods are sold. At the time they are sold, the costs are
released from inventory as expenses (typically called cost of goods sold) and matched
against sales revenue. Since product costs are assigned to inventories, they are also
known as inventory costs. In theory, product costs include all manufacturing costs—that
is, all costs relating to the manufacturing process. As shown in Figure 8.2, product costs
appear on financial statements when the inventory, or final goods, is sold, not when the
product is manufactured.
To understand product costs more fully, let us look briefly at the flow of costs in a
manufacturing company. By doing so, we will be able to see how product costs move
Income Statement
as
incurred
Period costs
Product costs
as
incurred
Revenue
Cost of goods sold
Gross profit
Expenses
Net income
when goods
are sold
Balance Sheet
Current assets:
Inventory
Figure 8.2 How the period costs and product costs flow
through financial statements from the manufacturing floor to
sales.
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Costs
Period costs
Product costs
Direct materials
Balance Sheet
Raw materials inventory
Direct materials
used in production
Direct labor
Work in process inventory
Manufacturing
overhead
Goods completed
(cost of goods
manufactured)
Finished goods inventory
Selling and
administrative
Goods
get
sold
Income
Statement
Cost of Revenue
Selling and
administrative expenses
Figure 8.3
Cost flows and classifications in a manufacturing company.
through the various accounts and affect the balance sheet and the income statement in the
course of the manufacture and sale of goods. The flows of period costs and product costs
through the financial statements are illustrated in Figure 8.3. All product costs filter
through the balance-sheet statement as “inventory cost.” If the product gets sold, the inventory costs in the balance-sheet statement are transferred to the income statement
under the head “cost of goods sold.”
• Raw-materials inventory. This account in the balance-sheet statement represents
the unused portion of the raw materials on hand at the end of the fiscal year.
• Work-in-process inventory. This balance-sheet entry consists of the partially
completed goods on hand in the factory at year-end. When raw materials are used
in production, their costs are transferred to the work-in-process inventory account
as direct materials. Note that direct labor cost and manufacturing overhead cost are
also added directly to work-in-process, which can be viewed as the assembly line in
a manufacturing plant, where workers are stationed and where products slowly take
shape as they move from one end of the line to the other.
• Finished-goods inventory. This account shows the cost of finished goods that are
on hand and awaiting sale to customers at year-end. As the goods are completed, accountants transfer the cost from the work-in-process account into the finished-goods
account. At this stage, the goods await sale to a customer. As the goods are sold, their
cost is transferred from the finished-goods account into the cost-of-goods-sold (or
cost-of-revenue) account. At this point, we finally treat the various material, labor,
and overhead costs that were involved in the manufacture of the units being sold as
expenses in the income statement.
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Section 8.2 Classifying Costs for Financial Statements 393
Example 8.1 serves to explain the classification scheme for financial statements.
EXAMPLE 8.1 Classifying Costs for Uptown Ice Cream Shop
Here is a look at why it costs $2.50 for a single-dip ice cream cone at a typical store
in Washington, DC. The annual sales volume (the number of ice cream cones sold)
averages around 185,000 cones, bringing in revenue of $462,500. This is equivalent
to selling more than 500 cones a day, assuming a seven-day operation. The following
table shows the unit price of an ice cream cone and the costs that go into producing
the product:
Total Cost
Unit Price*
% of Price
$120,250
$0.65
26%
Cone
9,250
0.05
2
Rent
112,850
0.61
24
46,250
0.25
10
9,250
0.05
2
Items
Ice cream (cream, sugar, milk, and
milk solids)
Wages
Payroll taxes
Sales taxes
42,550
0.23
9
Business taxes
14,800
0.08
3
Debt service
42,550
0.23
9
Supplies
16,650
0.09
4
Utilities
14,800
0.08
3
9,250
0.05
2
Profit
24,050
0.13
5
Total
$462,500
$2.50
100
Other expenses (insurance,
advertising, fees, and heating
and lighting for shop)
*Based on an annual volume of 185,000 cones.
If you were to classify the operating costs into either product costs or period costs,
how would you do it?
SOLUTION:
Given: Financial data just described.
Find: Classify the cost elements into product costs and period costs.
The following is a breakdown of the two kinds of costs:
• Product costs: Costs incurred in preparing 185,000 ice cream cones per year.
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Raw materials:
Ice cream @ $0.65
$120,250
Cone @ $0.05
9,250
Labor:
Wages @ $0.25
46,250
Overhead:
Supplies @ $0.09
16,650
Utilities @ $0.08
14,800
Total product cost
$207,200
• Period costs: Costs incurred in running the shop regardless of sales volume.
Business taxes:
Payroll taxes @ $0.05
$
9,250
Sales taxes @ $0.23
42,550
Business taxes @ $0.08
14,800
Operating expenses:
Rent @ $0.61
Debt service @ $0.23
112,850
42,550
Other @ $0.05
9,250
Total period cost
$231,250
8.3 Cost Classification for Predicting Cost Behavior
In engineering economic analysis, we need to predict how a certain cost will behave in response to a change in activity. For example, a manager will want to estimate the impact a
5% increase in production will have on the company’s total wages before he or she decides whether to alter production. Cost behavior describes how a cost item will react or
respond to changes in the level of business activity.
8.3.1 Volume Index
In general, the operating costs of any company are likely to respond in some way to
changes in the company’s operating volume. In studying cost behavior, we need to determine
some measurable volume or activity that has a strong influence on the amount of cost incurred. The unit of measure used to define volume is called a volume index. A volume
index may be based on production inputs, such as tons of coal processed, direct labor
hours used, or machine-hours worked; or it may be based on production outputs, such as
the number of kilowatt-hours generated. For a vehicle, the number of miles driven per
year may be used as a volume index. Once we identify a volume index, we try to find out
how costs change in response to changes in the index.
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8.3.2 Cost Behaviors
Accounting systems typically record the cost of resources acquired and track their subsequent usage. Fixed costs and variable costs are the two most common cost behavior patterns.
An additional category known as “mixed costs” contains two parts, the first of which is
fixed and the other of which varies with the volume of output.
Fixed Costs
The costs of providing a company’s basic operating capacity are known as the company’s
fixed cost or capacity cost. For a cost item to be classified as fixed, it must have a relatively
wide span of output over which costs are expected to remain constant. This span is called the
relevant range. In other words, fixed costs do not change within a given period, although
volume may change. In the case of an automobile, for example, the annual insurance premium, property tax, and license fee are fixed costs, since they are independent of the number
of miles driven per year. Some other typical examples are building rents; depreciation of
buildings, machinery, and equipment; and salaries of administrative and production personnel. In our Uptown Scoop Ice Cream Store example, we may classify expenses such as rent,
business taxes, debt service, and other (insurance, advertising, professional fees) as fixed
costs (costs that are fixed in total for a given period of time and for given production levels).
Fixed cost: A
cost that remains
constant,
regardless of any
change in a
company’s
activity.
Variable Costs
In contrast to fixed operating costs, variable operating costs have a close relationship to
the level of volume of a business. If, for example, volume increases 10%, a total variable
cost will also increase by approximately 10%. Gasoline is a good example of a variable
automobile cost, because fuel consumption is directly related to miles driven. Similarly,
the cost of replacing tires will increase as a vehicle is driven more.
In a typical manufacturing environment, direct labor and material costs are major
variable costs. In our Uptown Scoop example, the variable costs would include the cost of
the ice cream and cone (direct materials), wages, payroll taxes, sales taxes, and supplies.
Both payroll and sales taxes are related to sales volume. In other words, if the store becomes busy, more servers are needed, which will increase the payroll as well as taxes.
Variable cost: A
cost that changes
in proportion
to a change in
a company’s
activity or
business.
Mixed Costs
Some costs do not fall precisely into either the fixed or the variable category, but contain
elements of both. We refer to these as mixed costs (or semivariable costs). In our automobile example, depreciation (loss of value) is a mixed cost. On the one hand, some
depreciation occurs simply from the passage of time, regardless of how many miles a car
is driven, and this represents the fixed portion of depreciation. On the other hand, the more
miles an automobile is driven a year, the faster it loses its market value, and this represents
the variable portion of depreciation. A typical example of a mixed cost in manufacturing is
the cost of electric power. Some components of power consumption, such as lighting,
are independent of the operating volume, while other components (e.g., the number of
machine-hours equipment is operated) may vary directly with volume. In our Uptown Scoop
example, the utility cost can be a mixed cost item: Some lighting and heating requirements
might stay the same, but the use of power mixers will be in proportion to sales volume.
Mixed cost:
Costs are fixed
for a set level
of production
or consumption,
becoming
variable after
the level is
exceeded.
Average Unit Cost
The foregoing description of fixed, variable, and mixed costs was expressed in terms
of total volume over a given period. We often use the term average cost to express
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activity cost on a per unit basis. In terms of unit costs, the description of cost is quite
different:
• The variable cost per unit of volume is a constant.
• Fixed cost per unit varies with changes in volume: As the volume increases, the fixed
cost per unit decreases.
• The mixed cost per unit also changes as volume changes, but the amount of change is
smaller than that for fixed costs.
To explain the behavior of the fixed, variable, mixed, and average costs in relation to volume, let’s consider a medium-size car, say, the 2005 Ford Taurus SEL Deluxe six-cylinder
(3.0-liter) four-door sedan. In calculating the vehicle’s operating costs, we may use the procedure outlined by the American Automobile Association (AAA) as shown in Table 8.1.
On the basis of the assumptions in Table 8.1, the operating and ownership costs of
our Ford vehicle are estimated as follows:
Operating costs:
• Gas and oil
8.5 cents
• Maintenance
5.8 cents
• Tires
0.7 cent
Cost per mile
TABLE 8.1
15.0 cents
Assumptions Used in Calculating the Average Cost of Owning
and Operating a New Vehicle
What’s Covered
Costs Base
Fuel
U.S. price of unleaded gasoline from AAA’s Fuel Gauge
Report, weighted 60% city and 40% highway driving.
Maintenance
Costs of retail parts and labor for normal, routine maintenance,
as specified by the vehicle manufacturer.
Tires
Costs are based on the price of one set of replacement tires
of the same quality, size, and ratings as those which came
with the vehicle.
Insurance
A full-coverage policy for a married 47-year-old male with
a good driving record living a small city and commuting
3 to 10 miles daily to work.
License, Registration,
and Taxes
All government taxes and fees payable at time of purchase, as
well as fees due each year to keep the vehicle licensed and
registered.
Depreciation
Based on the difference between the new-vehicle purchase price
and the estimated trade-in value at the end of five years.
Finance
Based on a five-year loan at 6% interest with a 10% down
payment.
Source: American Automobile Association (AAA).
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Section 8.3 Cost Classification for Predicting Cost Behavior 397
Ownership costs:
• Comprehensive insurance ($250 deductible)
• License, registration, taxes
• Depreciation (15,000 miles annually)
• Finance charge (20% down, loan @8.5%/4 years)
Cost per year
Cost per day
Added depreciation costs (per 1,000 miles over
15,000 miles annually)
Total cost per mile:
• Cost per mile * 15,000 miles
• Cost per day * 365 days per year
Total cost per year
Average cost per mile ($8,580/15,000)
$1,195
$390
$4,005
$740
$6,330
$17.34
$185
$2,250
$6,330
$8,580
57.2 cents
Now, if you drive the same vehicle for 20,000 miles instead of 15,000 miles, you may
be interested in knowing how the average cost per mile would change. Example 8.2 illustrates how you determine the average cost as a function of mileage.
EXAMPLE 8.2 Calculating Average Cost per Mile as a
Function of Mileage
Table 8.2 itemizes the operating and ownership costs associated with driving a passenger car by fixed, variable, and mixed classes. Note that the only change from the preceding list is in the depreciation amount. Using the given data, develop a cost–volume
chart and calculate the average cost per mile as a function of the annual mileage.
DISCUSSION: First we may examine the effect of driving an additional 1,000 miles
over the 15,000 allotted miles. Since the loss in the car’s value due to driving an additional 1,000 miles over 15,000 miles is estimated to be $185, the total cost per year
and the average cost per mile, based on 16,000 miles, can be recalculated as follows:
•
•
•
•
Added depreciation cost: $185.
Added operating cost: 1,000 miles * 15 cents = $150.
Total cost per year: $8,580 + $185 + $150 = $8,915.
Average cost per mile ($8,915/16,000 miles): 55.72 cents.
Note that the average cost comes down as you drive more, as the ownership cost per
mile is further reduced.
SOLUTION
Given: Financial data.
Find: The average cost per mile at an annual operating volume between 15,000 and
20,000 miles.
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TABLE 8.2
Cost Classification of Owning
and Operating a Passenger Car
Cost Classification
Reference
Cost
Variable costs:
Standard miles per gallon
20 miles/gallon
Average fuel price per gallon
$1.939/gallon
Fuel and oil per mile
$0.085
Maintenance per mile
$0.058
Tires per mile
$0.007
Annual fixed costs:
Insurance (comprehensive)
$1,195
License, registration, taxes
$390
Finance charge
$740
Mixed costs: Depreciation
Fixed portion per year
(15,000 miles)
$4,005
Variable portion per mile
(above 15,000 miles)
$0.185
In Table 8.3, we summarize the costs of owning and operating the automobile at
various annual operating volumes from 15,000 to 20,000 miles. Once the total
cost figures are available at specific volumes, we can calculate the effect of volume on unit (per mile) costs by converting the total cost figures to average unit
costs.
TABLE 8.3
Operating Costs as a Function of Mileage Driven
Volume Index (miles)
15,000
Variable costs
(15¢ per mile)
$ 2,250 $ 2,400 $ 2,550 $ 2,700 $ 2,850 $ 3,000
Fixed costs
16,000
17,000
18,000
19,000
20,000
2,325
2,325
2,325
2,325
2,325
2,325
0
185
370
555
740
925
Fixed portion
4,005
4,005
4,005
4,005
4,005
4,005
Total variable cost
2,250
2,585
2,920
3,255
3,590
3,925
6,330
6,330
6,330
6,330
6,330
6,330
Mixed costs (depreciation):
Variable portion
Total fixed cost
Total costs
$ 8,580 $ 8,915 $ 9,250 $ 9,585 $ 9,920 $10,255
Cost per mile
$0.5720 $0.5572 $0.5441 $0.5325 $0.5221 $0.5128
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Section 8.3 Cost Classification for Predicting Cost Behavior 399
$12,000
5M
0.33
Total annual cost
$9,000
0.15M
$6,330
$6,000
$3,000
0
$6,000
3,000
6,000
9,000
12,000
15,000
Miles driven per year (M)
(a) Total cost
$6,000
18,000
21,000
$6,000
85M
$4,005
$4,000
$4,000
0.1
$4,000
$2,325
$2,000
5M
$2,000
0
10,000
20,000
M
(b) Fixed cost
0.1
0
10,000
20,000
M
(c) Variable cost
$2,000
0
10,000
20,000
M
(d) Mixed cost
Figure 8.4 Cost–volume relationships pertaining to annual automobile costs
(Example 8.2).
To estimate annual costs for any assumed mileage, we construct a cost–volume diagram
as shown in Figure 8.4(a). Further, we can show the relation between volume (miles
driven per year) and the three cost classes separately, as in Figure 8.4(b) through (d).
We can use these cost–volume graphs to estimate both the separate and combined
costs of operating the car at other possible volumes. For example, an owner who expects to drive 16,500 miles in a given year may estimate the total cost at
$9,082.50, or 55.05 cents per mile. In Figure 8.4, all costs more than those necessary to operate at the zero level are known as variable costs. Since the fixed cost is
$6,330 a year, the remaining $2,752.50 is variable cost. By combining all the
fixed and variable elements of cost, we can state simply that the cost of owning
and operating an automobile is $6,330 per year, plus 16.68 cents per mile driven
during the year.
Figure 8.5 illustrates graphically the average unit cost of operating the automobile. The average fixed unit cost, represented by the height of the middle curve in
the figure, declines steadily as the volume increases. The average unit cost is high
when volume is low because the total fixed cost is spread over a relatively few units
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400 CHAPTER 8 Cost Concepts Relevant to Decision Making
1.75
1.50
$1.4160
Cost per mile ($)
1.25
$1.266
1.00
Average total unit cost
$0.7830
0.75
$0.5720
Average fixed unit cost
0.50
$0.5128
$0.633
$0.422
$0.3165
Average variable unit cost
0.25
$0.1963
$0.15
0
5,000
$0.15
$0.15
10,000
15,000
Miles driven per year (M)
20,000
Figure 8.5 Average cost per mile of owning and operating a car (Example 8.2).
of volume. In other words, the total fixed costs remain the same regardless of the
number of miles driven, but the average fixed cost decreases on a per mile basis as
the number of miles driven increases.
8.4 Future Costs for Business Decisions
In the previous sections, our focus has been on classifying cost data that serve management’s need to control and evaluate the operations of a firm. However, these data are historical in nature; they may not be suitable for management’s planning future business
operations. We are not saying that historical cost data are of no use in making decisions
about the future. In fact, they serve primarily as the first step in predicting the uncertain
future. However, the types of cost data that management needs in making choices between alternative courses of action are different from historical cost data.
8.4.1 Differential Cost and Revenue
As we have seen throughout the text, decisions involve choosing among alternatives. In
business decisions, each alternative has certain costs and benefits that must be compared
with the costs and benefits of the other available alternatives. A difference in cost between
any two alternatives is known as a differential cost. Similarly, a difference in revenue
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Section 8.4 Future Costs for Business Decisions 401
between any two alternatives is known as differential revenue. A differential cost is also
known as an incremental cost, although, technically, an incremental cost should refer only
to an increase in cost from one alternative to another.
Cost–volume relationships based on differential costs find many engineering applications. In particular, they are useful in making a variety of short-term operational decisions.
Many short-run problems have the following characteristics:
• The base case is the status quo (the current operation or existing method), and we
propose an alternative to the base case. If we find the alternative to have lower costs
than the base case, we accept the alternative, assuming that nonquantitative factors do
not offset the cost advantage. The differential (incremental) cost is the difference in
total cost that results from selecting one alternative instead of another. If several alternatives are possible, we select the one with the maximum savings from the base.
Problems of this type are often called trade-off problems, because one type of cost is
traded off for another.
• New investments in physical assets are not required.
• The planning horizon is relatively short (a week or a month—certainly less than
a year).
• Relatively few cost items are subject to change by management decision.
Some common examples of short-run problems are method changes, operations planning, and make-or-buy decisions.
Method Changes
Often, we may derive the best information about future costs from an analysis of historical
costs. Suppose the proposed alternative is to consider some new method of performing an
activity. Then, as Example 8.3 shows, if the differential costs of the proposed method are
significantly lower than the current method, we adopt the new method.
EXAMPLE 8.3 Differential Cost Associated with Adopting
a New Production Method
The engineering department at an auto-parts manufacturer recommends that the current dies (the base case) be replaced with higher quality dies (the alternative), which
would result in substantial savings in manufacturing one of the company’s products.
The higher cost of materials would be more than offset by the savings in machining
time and electricity. If estimated monthly costs of the two alternatives are as shown
in the accompanying table, what is the differential cost for going with better dies?
SOLUTION
Given: Financial data.
Find: Which production method is preferred.
In this problem, the differential cost is - $5,000 a month. The differential cost’s
being negative indicates a saving, rather than an addition to total cost. An important
point to remember is that differential costs usually include variable costs, but fixed
costs are affected only if the decision involves going outside of the relevant range.
Although the production volume remains unchanged in our example, the slight increase
Incremental
cost is the
overall change
that a company
experiences by
producing one
additional unit
of good.
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Current Dies
Better Dies
Differential Cost
$150,000
$170,000
+ $20,000
Machining labor
85,000
64,000
-21,000
Electricity
73,000
66,000
-7,000
Supervision
25,000
25,000
0
Taxes
16,000
16,000
0
Depreciation
40,000
43,000
+3,000
$392,000
$387,000
- $5,000
Variable costs:
Materials
Fixed costs:
Total
in depreciation expense is due to the acquisition of new machine tools (dies). All other
items of fixed cost remained within their relevant ranges.
Operations Planning
In a typical manufacturing environment, when demand is high, managers are interested in
whether to use a one-shift-plus-overtime operation or to add a second shift. When demand is low, it is equally possible to explore whether to operate temporarily at very low
volume or to shut down until operations at normal volume become economical. In a
chemical plant, several routes exist for scheduling products through the plant. The problem is which route provides the lowest cost. Example 8.4 illustrates how engineers may
use cost–volume relationships in a typical operational analysis.
Break-even
analysis: An
analysis of the
level of sales at
which a project
would make
zero profit.
EXAMPLE 8.4 Break-Even Volume Analysis
Sandstone Corporation has one of its manufacturing plants operating on a singleshift five-day week. The plant is operating at its full capacity (24,000 units of output
per week) without the use of overtime or extra shifts. Fixed costs for single-shift operation amount to $90,000 per week. The average variable cost is a constant $30 per
unit, at all output rates, up to 24,000 units per week. The company has received an
order to produce an extra 4,000 units per week beyond the current single-shift maximum capacity. Two options are being considered to fill the new order:
• Option 1. Increase the plant’s output to 36,000 units a week by adding overtime, by adding Saturday operations, or both. No increase in fixed costs is entailed, but the variable cost is $36 per unit for any output in excess of 24,000
units per week, up to a 36,000-unit capacity.
• Option 2. Operate a second shift. The maximum capacity of the second shift is
21,000 units per week. The variable cost of the second shift is $31.50 per unit, and
the operation of a second shift entails additional fixed costs of $13,500 per week.
Determine the range of operating volume that will make Option 2 profitable.
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Section 8.4 Future Costs for Business Decisions 403
SOLUTION
Given: Financial data.
Find: The break-even volume that will make both options indifferent.
In this example, the operating costs related to the first-shift operation will remain unchanged if one alternative is chosen instead of another. Therefore, those costs are irrelevant to the current decision and
can safely be left out of the analysis. Consequently, we need to examine only the increased total cost
due to the additional operating volume under each option. (This kind of study is known as incremental
analysis.) Let Q denote the additional operating volume. Then we have
• Option 1. Overtime and Saturday operation: $36Q.
• Option 2. Second-shift operation: $13,500 + $31.50Q.
We can find the break-even volume 1Qb2 by equating the incremental cost functions and solving for Q:
36Q = 13,500 + 31.5Q,
4.5Q = 13,500,
Qb = 3,000 units.
Relevant production
range for Option 2
$700,000
Second-shift
operation
$650,000
$600,000
Relevant production
range for Option 1
.5Q
Overtime and
Saturday
operation
$400,000
$1
$450,000
3,5
00
+
31
$500,000
$350,000
Q
$300,000
36
Increased operating cost ($)
$550,000
$250,000
$200,000
$108,000
$150,000
$100,000
Break-even volume
$50,000
3,000
0
5,000
10,000
15,000
20,000
Additional weekly operating volume (units)
Figure 8.6 Cost–volume relationships of operating overtime and a
Saturday operation versus second-shift operation beyond 24,000 units
(Example 8.4).
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If the additional volume exceeds 3,000 units, the second-shift operation becomes
more efficient than overtime or Saturday operation. A break-even (or cost–volume)
graph based on the foregoing data is shown in Figure 8.6. The horizontal scale represents additional volume per week. The upper limit of the relevant volume range for
Option 1 is 12,000 units, whereas the upper limit for Option 2 is 21,000 units. The
vertical scale is in dollars of cost. The operating savings expected at any volume may
be read from the cost–volume graph. For example, the break-even point (zero savings)
is 3,000 units per week. Option 2 is a better choice, since the additional weekly volume from the new order exceeds 3,000 units.
Make-or-Buy (Outsourcing) Decision
In business, the make-or-buy decision arises on a fairly frequent basis. Many firms perform certain activities using their own resources and pay outside firms to perform certain
other activities. It is a good policy to constantly seek to improve the balance between
these two types of activities by asking whether we should outsource some function that
we are now performing ourselves or vice versa. Make-or-buy decisions are often grounded
in the concept of opportunity cost.
Opportunity
cost: The
benefits you
could have
received by
taking an
alternative
action.
8.4.2 Opportunity Cost
Opportunity cost may be defined as the potential benefit that is given up as you seek an
alternative course of action. In fact, virtually every alternative has some opportunity cost
associated with it. For example, suppose you have a part-time job while attending college
that pays you $200 per week. You would like to spend a week at the beach during spring
break, and your employer has agreed to give you the week off. What would be the opportunity cost of taking the time off to be at the beach? The $200 in lost wages would be an
opportunity cost.
In an economic sense, opportunity cost could mean the contribution to income that
is forgone by not using a limited resource in the best way possible. Or we may view opportunity costs as cash flows that could be generated from an asset the firm already owns,
provided that such flows are not used for the alternative in question. In general, accountants
do not post opportunity cost in the accounting records of an organization. However, this
cost must be explicitly considered in every decision. In sum,
• An opportunity cost arises when a project uses a resource that may already have been
paid for by the firm.
• When a resource that is already owned by a firm is being considered for use in a project,
that resource has to be priced on its next-best alternative use, which may be
1. A sale of the asset, in which case the opportunity cost is the expected proceeds
from the sale, net of any taxes on gains.
2. Renting or leasing the asset out, in which case the opportunity cost is the expected
present value of the after-tax revenue from the rental or lease.
3. Some use elsewhere in the business, in which case the opportunity cost is the cost
of replacing the resource.
4. That the asset has been abandoned or is of no use. Then the opportunity cost is zero.
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EXAMPLE 8.5 Opportunity Cost: Lost Rental Income
(Opportunity Cost)
Benson Company is a farm equipment manufacturer that currently produces 20,000
units of gas filters annually for use in its lawn-mower production. The expected annual
production cost of the gas filters is summarized as follows:
Variable costs:
Direct materials
Direct labor
Power and water
$100,000
190,000
35,000
Fixed costs:
Heating and light
Depreciation
Total cost
20,000
100,000
$445,000
Tompkins Company has offered to sell Benson 20,000 units of gas filters for $17.00 per
unit. If Benson accepts the offer, some of the manufacturing facilities currently used to
manufacture the filters could be rented to a third party at an annual rent of $35,000.
Should Benson accept Tompkins’s offer, and why?
SOLUTION
Given: Financial data; production volume = 20,000 units.
Find: Whether Benson should outsource the gas filter operation.
Make Option
Buy Option
Differential Cost
(Make-Buy)
Variable costs:
Direct materials
Direct labor
Power and water
$100,000
$100,000
190,000
190,000
35,000
35,000
340,000
340,000
20,000
20,000
0
100,000
100,000
0
Gas filters
Fixed costs:
Heating and light
Depreciation
Rental income lost
35,000
35,000
Total cost
$480,000
$460,000
$20,000
Unit cost
$24.00
$23.00
$1.00
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This problem is unusual in the sense that the buy option would generate a rental fee of
$35,000. In other words, Benson could rent out the current manufacturing facilities if it
were to purchase the gas filters from Tompkins. To compare the two options, we need
to examine the cost of each option.
The buy option has a lower unit cost and saves $1 for each use of a gas filter. If
the lost rental income (opportunity cost) were not considered, however, the decision
would favor the make option.
Sunk cost: A
cost that has
been incurred
and cannot be
reversed
8.4.3 Sunk Costs
A sunk cost is a cost that has already been incurred by past actions. Sunk costs are not relevant to decisions, because they cannot be changed regardless of what decision is made
now or in the future. The only costs relevant to a decision are costs that vary among the
alternative courses of action being considered. To illustrate a sunk cost, suppose you have
a very old car that requires frequent repairs. You want to sell the car, and you figure that
the current market value would be about $1,200 at best. While you are in the process of
advertising the car, you find that the car’s water pump is leaking. You decided to have the
pump repaired, which cost you $200. A friend of yours is interested in buying your car
and has offered $1,300 for it. Would you take the offer, or would you decline it simply because you cannot recoup the repair cost with that offer? In this example, the $200 repair
cost is a sunk cost. You cannot change this repair cost, regardless of whether you keep or
sell the car. Since your friend’s offer is $100 more than the best market value, it would be
better to accept the offer.
8.4.4 Marginal Cost
We make decisions every day, as entrepreneurs, professionals, executives, investors, and
consumers, with little thought as to where our motivations come from or how our assumptions of logic fit a particular academic regimen. As you have seen, the engineering
economic decisions that we describe throughout this text owe much to English economist
Alfred Marshall (1842–1924) and his concepts of marginalism. In our daily quest for
material gain, rarely do we recall the precepts of microeconomics or marginal utility, yet
the ideas articulated by Marshall remain some of the most useful core principles guiding
economic decision making.
Marginal cost:
The cost
associated with
one additional
unit of
production.
Definition
Another cost term useful in cost–volume analysis is marginal cost. We define marginal
cost as the added cost that would result from increasing the rate of output by a single unit.
The accountant’s differential-cost concept can be compared to the economist’s marginalcost concept. In speaking of changes in cost and revenue, the economist employs the
terms marginal cost and marginal revenue. The revenue that can be obtained from selling
one more unit of product is called marginal revenue. The cost involved in producing one
more unit of product is called marginal cost.
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EXAMPLE 8.6 Marginal Costs versus Average Costs
Consider a company that has an available electric load of 37 horsepower and that
purchases its electricity at the following rates:
kWh/Month
First 1,500
@$/kWh
Average Cost ($/kWh)
$0.050
Next 1,250
Next 3,000
All over 5,750
0.035
0.020
0.010
$0.050
$75 + 0.03501X - 1,5002
X
$118.75 + 0.0201X - 2,7502
X
$178.25 + 0.0101X - 5,7502
X
According to this rate schedule, the unit variable cost in each rate class represents the
marginal cost per kilowatt-hours (kWh). Alternatively, we may determine the average costs in the third column by finding the cumulative total cost and dividing it
by the total number of kWh (X). Suppose that the current monthly consumption of
electric power averages 3,200 kWh. On the basis of this rate schedule, determine
the marginal cost of adding one more kWh and, for a given operating volume
(3,200 kWh), the average cost per kWh.
SOLUTION
Given: Marginal cost schedule for electricity; operating volume = 3,200 kWh.
Find: Marginal and average cost per kWh.
The marginal cost of adding one more kWh is $0.020. The average variable cost per
kWh is calculated as follows:
kWh
Rate ($/kWh)
Cost
First 1,500
0.050
$75.00
Next 1,250
0.035
43.75
Remaining 450
0.020
Total
9.00
$127.75
The average variable cost per kWh is $127.75/3,200 kWh = $0.0399 kWh. Or we
can find the value by using the formulas in the third column of the rate schedule:
$118.75 + 0.02013,200 - 2,7502
$127.75
=
= $0.0399 kWh.
3,200
3,200 kWh
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Changes in the average variable cost per unit are the result of changes in the marginal cost. As shown in Figure 8.7, the average variable cost continues to fall because
the marginal cost is lower than the average variable cost over the entire volume.
0.06
$0.0500
0.05
Cost per kWh (dollars)
$0.0432
$0.0399
0.04
Average unit cost
$0.0311
0.03
0.02
0.01
Marginal cost
3,200
0
Figure 8.7
Marginal
analysis: A
technique used in
microeconomics
by which very
small changes in
specific variables
are studied in
terms of the
effect on related
variables and the
system as a
whole.
1,000
2,000
3,000
4,000
Electricity use per month (kWh)
6,000
Marginal versus average cost per kWh (Example 8.6).
Marginal Analysis
The fundamental concept of marginal thinking is that you are where you are, and what is
past is irrelevant. (This means that we should ignore the sunk cost.) The point is whether
you move forward, and you will do so if the benefits outweigh the costs, even if they do
so by a smaller margin than before. You will continue to produce a product or service
until the cost of doing so equals the revenue derived. Microeconomics suggests that businesspersons and consumers measure progress not in great leaps, but in small incremental
steps. Rational persons will reevaluate strategies and take new actions if benefits exceed
costs on a marginal basis. If there is more than one alternative to choose, do so always
with the alternative with greatest marginal benefit.
As we have mentioned, the economist’s marginal-cost concept is the same as the accountant’s differential-cost concept. For a business problem on maximization of profit,
economists place a great deal of emphasis on the importance of marginal analysis: Rational individuals and institutions should perform the most cost-effective activities first.
Specifically,
• When you examine marginal costs, you want to use the least expensive methods first
and the most expensive last (if at all).
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Section 8.4 Future Costs for Business Decisions 409
• Similarly, when you examine marginal revenue (or demand) the most revenue-enhancing
methods should be used first, the least revenue enhancing last.
• If we need to consider the revenue along with the cost, the volume at which the total
revenue and total cost are the same is known as the break-even point. If the cost and
revenue function are assumed to be linear, this point may also be calculated, by
means of the following formula:
Break-even volume =
=
Fixed costs
Sales price per unit - Variable cost per unit
Fixed costs
Marginal contribution per unit
(8.1)
The difference between the unit sales price and the unit variable cost is the producer’s
marginal contribution, also known as marginal income. This means that each unit sold
contributes toward absorbing the company’s fixed costs.
EXAMPLE 8.7 Profit-Maximization Problem: Marginal Analysis2
Suppose you are a chief executive officer (CEO) of a small pharmaceutical company
that manufactures generic aspirin. You want the company to maximize its profits. You
can sell as many aspirins as you make at the prevailing market price. You have only
one manufacturing plant, which is the constraint. You have the plant working at full
capacity Monday through Saturday, but you close the plant on Sunday because on
Sundays you have to pay workers overtime rates, and it is not worth it. The marginal
costs of production are constant Monday through Saturday. Marginal costs are higher
on Sundays, only because labor costs are higher.
Now you obtain a long-term contract to manufacture a brand-name aspirin. The
costs of making the generic aspirin or the brand-name aspirin are identical. In fact,
there is no cost or time involved in switching from the manufacture of one to the other.
You will make much larger profits from the brand-name aspirin, but the demand is
limited. One day of manufacturing each week will permit you to fulfill the contract.
You can manufacture both the brand-name and the generic aspirin. Compared with the
situation before you obtained the contract, your profits will be much higher if you now
begin to manufacture on Sundays—even if you have to pay overtime wages.
• Generic aspirin. Each day, you can make 1,000 cases of generic aspirin. You can
sell as many as you make, for the market price of $10 per case. Every week you
have fixed costs of $5,000 (land tax and insurance). No matter how many cases you
manufacture, the cost of materials and supplies is $2 per case; the cost of labor is
$5 per case, except on Sundays, when it is $10 per case.
• Brand-name aspirin. Your order for the brand-name aspirin requires that you manufacture 1,000 cases per week, which you sell for $30 per case. The cost for the
brand-name aspirin is identical to the cost of the generic aspirin.
What do you do?
2
Source: “Profit Maximization Problem,” by David Hemenway and Elon Kohlberg, Economic Inquiry, October 1, 1997, Page 862, Copyright 1997 Western Economic Association International.
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SOLUTION
Given: Sales price, $10 per case for generic aspirin, $30 per case for band aspirin;
fixed cost, $5,000; variable cost, $7 per case during weekdays, $12 per case on Sunday operation; weekly production, 6,000 cases of generic aspirin, 1,000 cases of
brand-name aspirin.
Find: (a) Optimal production mix and (b) break-even volume.
(a) Optimal production mix: The marginal costs of manufacturing brand-name aspirin are constant Monday through Saturday; they rise substantially on Sunday
and are above the marginal revenue from manufacturing generic aspirin. In
other words, your company should manufacture the brand-name aspirin first.
Your marginal revenue is the highest the “first” day, when you manufacture the
brand-name aspirin. It then falls and remains constant for the rest of the week.
On the seventh day (Sunday), the marginal revenue from manufacturing the
generic aspirin is still below the marginal cost. You should manufacture brandname aspirin one day a week and generic aspirin five days a week. On Sundays,
the plant should close.
• The marginal revenue from manufacturing on Sunday is $10,000 (1,000 cases
times $10 per case).
• The marginal cost from manufacturing on Sunday is $12,000 (1,000 cases
times $12 per case— $10 labor + $2 materials).
• Profits will be $2,000 lower than revenue if the plant operates on Sunday.
(b) Break-even volume: The total revenue and cost functions can be represented as
follows:
Total revenue function: e
Total cost function: e
30Q
30,000 + 10Q
5,000 + 7Q
47,000 + 12Q
for 0 … Q … 1,000
for 1,000 6 Q … 6,000,
for 0 … Q … 6,000
for 6,000 … Q … 7,000.
Table 8.4 shows the various factors involved in the production of the brandname and the generic aspirin.
• If you produce the brand-name aspirin first, the break-even volume is
30Q - 7Q - 5,000 = 0,
Qb = 217.39.
• If you produce the generic aspirin first, the break-even volume is
10Q - 7Q - 5,000 = 0,
Qb = 1,666.67.
Clearly, scheduling the production of the brand-name aspirin first is the better
strategy, as you can recover the fixed cost ($5,000) much faster by selling just
217.39 cases of brand-name aspirin. Also, Sunday operation is not economical, as
the marginal cost exceeds the marginal revenue by $2,000, as shown in Figure 8.8.
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Section 8.5 Estimating Profit from Production 411
TABLE 8.4
Net Profit Calculation as a Function of Production Volume
Production Product
Total
Variable
Volume (Q)
Mix
Revenue Revenue
Cost
0
0
0
0
Fixed
Cost
Total
Cost
Net
Profit
$5,000 $5,000 - $5,000
Mon
1,000
Brand
name
$30,000
$30,000
$7,000
0 12,000
18,000
Tue
2,000
Generic
10,000
40,000
7,000
0 19,000
21,000
Wed
3,000
Generic
10,000
50,000
7,000
0 26,000
24,000
Thu
4,000
Generic
10,000
60,000
7,000
0 33,000
27,000
Fri
5,000
Generic
10,000
70,000
7,000
0 40,000
30,000
Sat
6,000
Generic
10,000
80,000
7,000
0 47,000
33,000
Sun
7,000
Generic
10,000
90,000
12,000
0 59,000
31,000
75,000
Dollars ($)
Generic aspirin
Brand-name aspirin
90,000
60,000
45,000
e
enu
rev
l
a
t
To
al
Tot
30,000
cost
Unprofitable
operation
15,000
Total profit
0
Mon
Tue
Wed
Thu
Time
Fri
Sat
Sun
Figure 8.8 Weekly profits as a function of time. Sunday operation becomes unprofitable, because the marginal revenue stays at
$10 per case whereas the marginal cost increases to $12 per case
(Example 8.7).
8.5 Estimating Profit from Production
Up to this point, we have defined various cost elements and their behaviors in a manufacturing environment. Our ultimate objective is to develop a project’s cash flows; we do so in
Chapter 10, where the first step is to formulate the budget associated with the sales and
production of the product. As we will explain in that chapter, the income statement developed in the current chapter will be the focal point for laying out the project’s cash flows.
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8.5.1 Calculation of Operating Income
Accountants measure the net income of a specified operating period by subtracting expenses from revenues for that period. These terms can be defined as follows:
1. The project revenue is the income earned3 by a business as a result of providing
products or services to customers. Revenue comes from sales of merchandise to
customers and from fees earned by services performed for clients or others.
2. The project expenses that are incurred4 are the cost of doing business to generate
the revenues of the specified operating period. Some common expenses are the cost
of the goods sold (labor, material, inventory, and supplies), depreciation, the cost of
employees’ salaries, the business’s operating costs (such as the cost of renting
buildings and the cost of insurance coverage), and income taxes.
The business expenses just listed are accounted for in a straightforward fashion on a
company’s income statement and balance sheet: The amount paid by the organization
for each item would translate, dollar for dollar, into expenses in financial reports for the
period. One additional category of expenses—the purchase of new assets—is treated by
depreciating the total cost gradually over time. Because capital goods are given this
unique accounting treatment, depreciation is accounted for as a separate expense in financial reports. Because of its significance in engineering economic analysis, we will
treat depreciation accounting in a separate chapter.
Figure 8.9 will be used as a road map to construct the income statement related to the
proposed manufacturing activities.
8.5.2 Sales Budget for a Manufacturing Business
The first step in laying out a sales budget for a manufacturing business is to predict the dollar
sales, which is the key to the entire process. The estimates are typically expressed in both dollars and units of the product. Using the expected selling price, we can easily extend unit sales
Production Budget
Direct Materials
Budget
Product
Cost
Sales
Budget
Period
Cost
Direct Labor
Budget
Cost of Goods Sold
Budget
Manufacturing
Overhead Budget
+
Selling Expenses
Budget
Non-manufacturing
Cost Budget
Administrative
Expenses Budget
Budgeted
Income Statement
Figure 8.9
3
4
Process of creating a master production budget.
Note that the cash may be received in a different accounting period.
Note that the cash may be paid in a different accounting period.
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to revenues. Often, the process is initiated by individual salespersons or managers predicting
sales in their own area for the budget period. Of course, an aggregate sales budget is influenced by economic conditions, pricing decisions, competition, marketing programs, etc. To illustrate, suppose that you want to create a sales budget schedule on a quarterly basis.
Projected sales units are 1,000 units during the first quarter, 1,200 units during the second
quarter, 1,300 units during the third quarter, and 1,500 units during the fourth quarter. The projected sales price is $15 per unit. Then an estimated sales schedule looks like the following:
Sales Budget Schedule (Year 2006)—Product X
1Q
2Q
3Q
4Q
1,000
1,200
1,300
1,500
5,000
$ 15
$ 15
$ 15
$ 15
$ 15
$ 15,000
$ 18,000
$ 19,500
$ 22,500
$ 75,000
Budgeted units
Sales price
Estimated sales
Annual
Total
8.5.3 Preparing the Production Budget
Once the sales budget is known, we can prepare the production activities. Using the sales
budget for the number of project units needed, we prepare an estimate of the number of
units to be produced in the budget period. Note that the production budget is the basis for
projecting the cost-of-goods-manufactured budget, which we will discuss subsequently.
To illustrate, we may use the following steps:
1. From the sales budget, record the projected number of units to be sold.
2. Determine the desired number of units to carry in the ending inventory; usually, that
number is a percentage of next quarter’s needs. In our example, we will assume that
20% of the budgeted units will be the desirable ending inventory position in each
quarter.
3. Add to determine the total number of units needed each period.
4. Determine the projected number of units in the beginning inventory. Note that
the beginning inventory is last quarter’s ending inventory. In our example, we
will assume that the first quarter’s beginning inventory is 100 units.
5. Subtract to determine the projected production budget.
Then a typical production budget may look like the following:
Production Budget (Year 2006)—Product X
1Q
Budgeted units to be produced
Desired ending inventory
Total units needed
Less beginning inventory
Units to produce
2Q
3Q
4Q
Annual
Total
1,000
1,200
1,300
1,500
5,000
200
240
260
300
1,000
1,200
1,440
1,560
1,800
6,000
100
200
240
260
800
1,100
1,240
1,320
1,540
5,200
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Materials Budgets
Once we know how many units we need to produce, we are ready to develop direct materials budgets. We may use the following steps:
1. From the production budget, copy the projected number of units to be produced.
2. Multiply by the amount of raw materials needed per unit to calculate the amount of
materials needed. In our example, we will assume that each production unit consumes $4 of materials.
3. Calculate the desired ending inventory (the number of units required in the ending
inventory * $4).
4. Add to calculate the total amount of materials needed.
5. Subtract the beginning inventory, which is last quarter’s ending inventory, to calculate the amount of raw materials needed to be purchased.
6. Calculate the net cost of raw materials.
Then a typical direct materials budget may look like the following:
Direct Materials Budget (Year 2006)—Product X
Annual
Total
1Q
2Q
3Q
4Q
1,100
1,240
1,320
1,540
$4
$4
$4
$4
Cost of materials for
units to be produced
$ 4,400
$ 4,960
$ 5,280
$ 6,160
$ 20,800
Plus cost of materials
in ending inventory
$
$
960
$ 1,040
$ 1,200
$ 4,000
Total cost of materials
needed
$ 5,200
$ 5,920
$ 6,320
$ 7,360
$ 24,800
Less cost of materials in
beginning inventory
$
$
$
960
$ 1,040
$ 3,200
Cost of materials
to purchase
$ 4,800
$ 5,360
$ 6,320
$ 21,600
Units to produce
Unit cost of materials
800
400
800
$ 5,120
5,200
Direct Labor Budget for a Manufacturing Business
As with the materials budgets, once the production budget has been completed, we can
easily prepare the direct labor budget. This budget allows the firm to estimate labor requirements—both labor hours and dollars—in advance. To illustrate, use the following
steps:
1. From the production budget, copy the projected number of units to be produced.
2. Multiply by the direct labor cost per unit to calculate the total direct labor cost. In
our example, we will assume that the direct labor cost is $3 per unit.
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Then a typical direct labor budget may look like the following:
Direct Labor Budget (Year 2006)—Product X
Annual
Total
1Q
2Q
3Q
4Q
Units to produce
× Direct labor cost per unit
1,100
$ 3.00
1,240
$ 3.00
1,320
$ 3.00
1,540
$ 3.00
5,200
Total direct labor cost ($)
$ 3,300
$ 3,720
$ 3,960
$ 4,620
$ 15,600
Overhead Budget for a Manufacturing Business
The overhead budget should provide a schedule of all costs of production other than direct materials and direct labor. Typically, the overhead budget is expressed in dollars,
based on a predetermined overhead rate. In preparing a manufacturing overhead budget,
we may take the following steps:
1. From the production budget, copy the projected number of units to be produced.
2. Multiply by the variable overhead rate to calculate the budgeted variable overhead.
In our example, we will assume the variable overhead rate to be $1.50 per unit.
There are several ways to determine this overhead rate. One common approach
(known as traditional standard costing) is to divide the expected total overhead
cost by the budgeted number of direct labor hours (or units). Another approach is
to adopt an activity-based costing concept, allocating indirect costs against the
activities that caused them. We will not review this accounting method here, but it
can more accurately reflect indirect cost improvement than traditional standard
costing can.
3. Add any budgeted fixed overhead to calculate the total budgeted overhead. In our
example, we will assume the fixed overhead to be $230 each quarter.
Then a typical manufacturing overhead budget may look like the following:
Manufacturing Overhead Budget (Year 2006)—Product X
Units to produce
1Q
2Q
3Q
4Q
1,100
1,240
1,320
1,540
Annual
Total
5,200
Variable mfg overhead
rate per unit ($1.50)
$ 1,650
$ 1,860
$ 1,980
$ 2,310
$ 7,800
Fixed mfg overhead
$
$
$
$
$
Total overhead
$ 1,880
230
230
$ 2,090
230
$ 2,210
230
$ 2,540
920
$ 8,720
8.5.4 Preparing the Cost-of-Goods-Sold Budget
The production budget developed in the previous section shows how much it would cost
to produce the required production volume. Note that the number of units to be produced
includes both the anticipated number of sales units and the number of units in inventory.
Activity-based
costing (ABC)
identifies
opportunities
to improve business process
effectiveness and
efficiency by
determining the
“true” cost of a
product or
service.
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The cost-of-goods-sold budget is different from the production budget, because we do
not count the costs incurred to carry the inventory. Therefore, we need to prepare a budget
that details the costs related to the sales, not the inventory. Typical steps in preparing a
cost-of-goods-sold budget are as follows:
1. From the sales budget, and not the production budget, copy the budgeted number of
sales units.
2. Multiply by the direct material cost per unit to estimate the amount of direct materials.
3. Multiply by the direct labor cost per unit to estimate the direct labor.
4. Multiply by the manufacturing overhead per unit to estimate the overhead.
Then a typical cost-of-goods-sold budget may look like the following:
Cost of Goods Sold (Year 2006)—Product X
Cost of goods
sold: A figure
reflecting the
cost of the
product or good
that a company
sells to generate
revenue.
Annual
Total
1Q
2Q
3Q
4Q
1,000
1,200
1,300
1,500
5,000
Direct materials ($4/unit)
$ 4,000
$ 4,800
$ 5,200
$ 6,000
$ 20,000
Direct labor ($3/unit)
$ 3,000
$ 3,600
$ 3,900
$ 4,500
$ 15,000
Variable ($1.50 per unit)
$ 1,500
$ 1,800
$ 1,950
$ 2,250
$ 7,500
Fixed
$
$
$
$
$
Budgeted sales units
Mfg overhead:
Cost of goods sold
230
$ 7,000
230
$ 8,400
230
$ 9,100
230
$ 10,500
920
$ 35,000
8.5.5 Preparing the Nonmanufacturing Cost Budget
To complete the entire production budget, we need to add two more items related to production: the selling expenses and the administrative expenses.
Selling Expenses Budget for a Manufacturing Business
Since we know the sales volume, we can develop a selling expenses budget by considering the budgets of various individuals involved in marketing the products. The budget includes both variable cost items (shipping, handling, and sales commission) and fixed
items, such as advertising and salaries for marketing personnel. To prepare the selling expenses budget, we may take the following steps:
1. From the sales budget, copy the projected number of unit sales.
2. List the variable selling expenses, such as the sales commission. In our example, we
will assume that the sales commission is calculated at 5% of unit sales.
3. List the fixed selling expenses, typically rent, depreciation expenses, advertising,
and other office expenses. In our example, we will assume the following: rent, $500
per quarter; advertising, $300 per quarter; office expense, $200 per quarter.
4. Add to calculate the total budgeted selling expenses.
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Then a typical selling expenses budget may look like the following:
Selling Expenses (Year 2006)—Product X
1Q
2Q
3Q
4Q
Annual
Total
$ 15,000
$ 18,000
$ 19,500
$ 22,500
$ 75,000
$
750
$
900
$
975
$ 1,125
$ 3,750
Rent
$
500
$
500
$
500
$
500
$ 2,000
Advertising
$
300
$
300
$
300
$
300
$ 1,200
Office expenses
$
200
$
200
$
200
$
200
$
Budgeted unit sales ($)
Variable expenses:
Commission
Fixed expenses:
Total selling expenses
$ 1,750
$ 1,900
$ 1,975
$ 2,125
800
$ 7,750
Administrative Expenses Budget for a Manufacturing Business
Another nonmanufacturing expense category to consider is administrative expenses, a
category that contains mostly fixed-cost items such as executive salaries and the depreciation of administrative buildings and office furniture. To prepare an administrative expense budget, we may take the following steps:
1. List the variable administrative expenses, if any.
2. List the fixed administrative expenses, which include salaries, insurance, office
supplies, and utilities. We will assume the following quarterly expenses: salaries,
$1,400; insurance, $135; office supplies, $300; other office expenses, $150.
3. Add to calculate the total administrative expenses.
Then a typical administrative expenses budget may look like the following:
Administrative Expenses (Year 2006)—Product X
1Q
2Q
3Q
4Q
Annual
Total
Salaries
$ 1,400
$ 1,400
$ 1,400
$ 1,400
$ 5,600
Insurance
$
135
$
135
$
135
$
135
$
Office supplies
$
300
$
300
$
300
$
300
$ 1,200
Utilities (phone, power, water, etc.)
$
500
$
500
$ 500
$
500
$ 2,000
Other office expenses
$
150
$
150
$
$
150
$
Variable expenses:
Fixed expenses:
Total administrative expenses
$ 2,485
$ 2,485
150
$ 2,485
$ 2,485
540
600
$ 9,940
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8.5.6 Putting It All Together: The Budgeted Income Statement
Now we are ready to develop a consolidated budget that details a company’s projected
sales, costs, expenses, and net income. To determine the net income, we need to include any
other operating expenses, such as interest expenses and renting or leasing expenses. Once
we calculate the taxable income, we can determine the federal taxes to pay. The procedure
to follow is
1.
2.
3.
4.
5.
6.
7.
Copy the net sales from the sales budget.
Copy the cost of goods sold.
Copy both the selling and administrative expenses.
Gross income = Sales - Cost of goods sold.
Operating income = Sales - 1Cost of goods sold + Operating expenses2.
Determine any interest expense incurred during the budget period.
Calculate the federal income tax at a percentage of taxable income. In our example,
let’s assume a 35% tax rate.
8. Net income = Income from operations - Federal income tax, which summarizes
the projected profit from the budgeted sales units.
Then the budgeted income statement would look like the format shown in Table 8.5.
Once you have developed a budgeted income statement, you may be able to examine
the profitability of the manufacturing operation. Three margin figures are commonly
used to quickly assess the profitability of the operation: gross margin, operating margin,
and net profit margin.
TABLE 8.5
Budgeted Income Statement
Budgeted Income Statement (Year 2006)—Product X
1Q
2Q
3Q
4Q
Annual
Total
Sales
$ 15,000
$ 18,000
$ 19,500
$ 22,500
$ 75,000
Cost of goods sold
$ 7,000
$ 8,400
$ 9,100
$ 10,500
$ 35,000
Gross income
$ 8,000
$ 9,600
$ 10,400
$ 12,000
$ 40,000
Selling expenses
$ 1,750
$ 1,900
$ 1,975
$ 2,125
$ 7,750
Administrative expenses
$ 2,485
$ 2,485
$ 2,485
$ 2,485
$ 9,940
$ 3,765
$ 5,215
$ 5,940
$ 7,390
$ 22,310
$
$
$
$
$
Operating expenses:
Operating income
Interest expenses
—
—
—
—
—
Net income before taxes
$ 3,765
$ 5,215
$ 5,940
$ 7,390
$ 22,310
Income taxes (35%)
$ 1,318
$ 1,825
$ 2,079
$ 2,587
$ 7,809
Net income
$ 2,447
$ 3,390
$ 3,861
$ 4,804
$ 14,502
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Section 8.5 Estimating Profit from Production 419
Gross Margin
The gross margin reveals how much a company earns, taking into consideration the costs
that it incurs for producing its products or services. In other words, gross margin is equal
to gross income divided by net sales and is expressed as a percentage:
Gross margin =
Gross income
.
Net sales
(8.2)
In our example, gross margin = $40,000/$75,000 = 53%. Gross margin indicates
how profitable a company is at the most fundamental level. Companies with higher gross
margins will have more money left over to spend on other business operations, such as research and development or marketing.
Operating Margin
The operating margin is defined as the operating profit for a certain period, divided by
revenues for that period:
Operating income
(8.3)
Operating margin =
.
Net sales
In our example, operating margin = $22,310/$75,000 = 30%. Operating profit
margin indicates how effective a company is at controlling the costs and expenses associated with its normal business operations.
Net Profit Margin
As we defined the term in Chapter 2, the net profit margin is obtained by dividing net
profit by net revenues, often expressed as a percentage:
Net profit margin =
Net income
.
Net sales
(8.4)
This number is an indication of how effective a company is at cost control. The
higher the net profit margin, the more effective the company is at converting revenue
into actual profit. In our example, the net profit margin = $14,502/$75,000 = 19%.
The net profit margin is a good way of comparing companies in the same industry, since
such companies are generally subject to similar business conditions.
8.5.7 Looking Ahead
The operational decision problem described in this chapter tends to have a relatively
short-term horizon (weekly or monthly). That is, such decisions do not commit a firm to
a certain course of action over a relatively long period. If operational decision problems
significantly affect the amount of funds that must be invested in a firm, fixed costs will
have to increase. In Example 8.7, if the daily aspirin production increases to 10,000 cases
per day, exceeding the current production capacity, the firm must make new investments
in plant and equipment. Since the benefits of the expansion will continue to occur over an
extended period, we need to consider the economic effects of the fixed costs over the life
of the assets. Also, we have not discussed how we actually calculate the depreciation
amount related to production or the amount of income taxes to be paid from profits generated from the operation. Doing so requires an understanding the concepts of capital investment, depreciation, and income taxes, which we will discuss in the next chapter.
Operating
margin gives
analysts an idea
of how much a
company makes
(before interest
and taxes) on
each dollar of
sales.
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420 CHAPTER 8 Cost Concepts Relevant to Decision Making
SUMMARY
In this chapter, we examined several ways in which managers classify costs. How the
costs will be used dictates how they will be classified:
Most manufacturing companies divide manufacturing costs into three broad cate-
gories: direct materials, direct labor, and manufacturing overhead. Nonmanufacturing
costs are classified into two categories: marketing or selling costs and administrative
costs.
For the purpose of valuing inventories and determining expenses for the balance sheet
and income statement, costs are classified as either product costs or period costs.
For the purpose of predicting cost behavior—how costs will react to changes in ac-
tivity—managers commonly classify costs into two categories: variable costs and
fixed costs.
An understanding of the following cost–volume relationships is essential to develop-
ing successful business strategies and to planning future operations:
Fixed operating costs: Costs that do not vary with production volume.
Variable operating costs: Costs that vary with the level of production or sales.
Average costs: Costs expressed in terms of units obtained by dividing total costs by
total volumes.
Differential (incremental) costs: Costs that represent differences in total costs, which
result from selecting one alternative instead of another.
Opportunity costs: Benefits that could have been obtained by taking an alternative
action.
Sunk costs: Past costs not relevant to decisions because they cannot be changed no
matter what actions are taken.
Marginal costs: Added costs that result from increasing rates of outputs, usually by
single units.
Marginal analysis: In economic analysis, we need to answer the apparently trivial
question, “Is it worthwhile?”—whether the action in question will add sufficiently to the
benefits enjoyed by the decision maker to make performing the action worth the cost.
This is the heart of marginal-decision making—the statement that an action merits performance if, and only if, as a result, we can expect to be better off than we were before.
At the level of plant operations, engineers must make decisions involving materials,
production processes, and the in-house capabilities of company personnel. Most of
these operating decisions do not require any investments in physical assets; therefore,
they depend solely on the cost and volume of business activity, without any consideration of the time value of money.
Engineers are often asked to prepare the production budgets related to their operating
division. Doing this requires a knowledge of budgeting scarce resources, such as labor
and materials, and an understanding of the overhead cost. The same budgeting practice
will be needed in preparing the estimates of costs and revenues associated with undertaking a new project.
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Problems 421
PROBLEMS
Classifying Costs
8.1 Identify which of the following transactions and events are product costs and
which are period costs:
• Storage and material handling costs for raw materials.
• Gains or losses on the disposal of factory equipment.
• Lubricants for machinery and equipment used in production.
• Depreciation of a factory building.
• Depreciation of manufacturing equipment.
• Depreciation of the company president’s automobile.
• Leasehold costs for land on which factory buildings stand.
• Inspection costs of finished goods.
• Direct labor cost.
• Raw-materials cost.
• Advertising expenses.
Cost Behavior
8.2 Identify which of the following costs are fixed and which are variable:
• Wages paid to temporary workers.
• Property taxes on a factory building.
• Property taxes on an administrative building.
• Sales commission.
• Electricity for machinery and equipment in the plant.
• Heating and air-conditioning for the plant.
• Salaries paid to design engineers.
• Regular maintenance on machinery and equipment.
• Basic raw materials used in production.
• Factory fire insurance.
8.3 The accompanying figures are a number of cost behavior patterns that might be
found in a company’s cost structure. The vertical axis on each graph represents
total cost, and the horizontal axis on each graph represents level of activity (volume). For each of the situations that follow, identify the graph that illustrates the
cost pattern involved. Any graph may be used more than once5.
(a) Electricity bill—a flat-rate fixed charge, plus a variable cost after a certain
number of kilowatt-hours are used.
(b) City water bill, which is computed as follows:
First 1,000,000 gallons $1,000 flat or less rate
Next 10,000 gallons $0.003 per gallon used
Next 10,000 gallons $0.006 per gallon used
Next 10,000 gallons $0.009 per gallon used
Etc. etc.
5
Adapted originally from a CPA exam, and the same materials are also found in R. H. Garrison and E. W.
Noreen, Managerial Accounting, 8th ed. Irwin, 1997, copyright © Richard D. Irwin, p. 271.
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422 CHAPTER 8 Cost Concepts Relevant to Decision Making
(c) Depreciation of equipment, where the amount is computed by the straight-line
method. When the depreciation rate was established, it was anticipated that the
obsolescence factor would be greater than the wear-and-tear factor.
(d) Rent on a factory building donated by the city, where the agreement calls for a
fixed fee payment unless 200,000 labor-hours or more are worked, in which
case no rent need be paid.
(e) Cost of raw materials, where the cost decreases by 5 cents per unit for each of
the first 100 units purchased, after which it remains constant at $2.50 per unit.
(f) Salaries of maintenance workers, where one maintenance worker is needed
for every 1,000 machine-hours or less (that is, 0 to 1,000 hours requires one
maintenance worker, 1,001 to 2,000 hours requires two maintenance workers,
etc.).
(g) Cost of raw materials used.
(h) Rent on a factory building donated by the county, where the agreement calls for
rent of $100,000, less $1 for each direct labor-hour worked in excess of 200,000
hours, but a minimum rental payment of $20,000 must be paid.
(i) Use of a machine under a lease, where a minimum charge of $1,000 must be
paid for up to 400 hours of machine time. After 400 hours of machine time, an
additional charge of $2 per hour is paid, up to a maximum charge of $2,000
per period.
1
2
3
4
5
6
7
8
9
10
11
12
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Problems 423
8.4 Harris Company manufactures a single product. Costs for the year 2001 for output
levels of 1,000 and 2,000 units are as follows:
Units produced
1,000
2,000
$30,000
$30,000
20,000
40,000
Variable portion
12,000
24,000
Fixed portion
36,000
36,000
5,000
10,000
22,000
22,000
Direct labor
Direct materials
Overhead:
Selling and administrative costs:
Variable portion
Fixed portion
At each level of output, compute the following:
(a) Total manufacturing costs.
(b) Manufacturing costs per unit.
(c) Total variable costs.
(d) Total variable costs per unit.
(e) Total costs that have to be recovered if the firm is to make a profit.
Cost–Volume–Profit Relationships
8.5 Bragg & Stratton Company manufactures a specialized motor for chain saws. The
company expects to manufacture and sell 30,000 motors in year 2001. It can manufacture an additional 10,000 motors without adding new machinery and equipment. Bragg & Stratton’s projected total costs for the 30,000 units are as follows:
Direct Materials
$150,000
Direct labor
300,000
Manufacturing overhead:
Variable portion
Fixed portion
100,000
80,000
Selling and administrative costs:
Variable portion
Fixed portion
180,000
70,000
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424 CHAPTER 8 Cost Concepts Relevant to Decision Making
The selling price for the motor is $80.
(a) What is the total manufacturing cost per unit if 30,000 motors are produced?
(b) What is the total manufacturing cost per unit if 40,000 motors are produced?
(c) What is the break-even price on the motors?
8.6 The accompanying chart shows the expected monthly profit or loss of Cypress
Manufacturing Company within the range of its monthly practical operating
capacity. Using the information provided in the chart, answer the following
questions:
(a) What is the company’s break-even sales volume?
(b) What is the company’s marginal contribution rate?
(c) What effect would a 5% decrease in selling price have on the break-even point
in (a)?
(d) What effect would a 10% increase in fixed costs have on the marginal contribution rate in (b)?
(e) What effect would a 6% increase in variable costs have on the break-even point
in (a)?
(f) If the chart also reflects $20,000 monthly depreciation expenses, compute the
sale at the break-even point for cash costs.
$60,000
40,000
Profit/Loss
20,000
0
20,000
40,000
60,000
$100,000
$200,000
$300,000
$400,000
$500,000
Sales (monthly)
8.7 The accompanying graph is a cost–volume–profit graph. In the graph, identify the
following line segments or points:
(a) Line EF represents ______________.
(b) The horizontal axis AB represents ______________, and the vertical axis AD
represents ______________.
(c) Point V represents ______________.
(d) The distance CB divided by the distance AB is ______________.
(e) The point Vb is a break-even ______________.
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Problems 425
C
D
V
E
A
F
B
Vb
Volume
8.8 A cost–volume–profit (CVP) graph is a useful technique for showing relationships between costs, volume, and profits in an organization.
Relevant
range
8
10
1
6
5
3
9
4
7
2
(a) Identify the numbered components in the accompanying CVP graph.
No.
Description
No.
1
6
2
7
3
8
4
9
5
10
Description
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426 CHAPTER 8 Cost Concepts Relevant to Decision Making
(b) Using the typical CVP relationship shown, fill in the missing amounts in each
of the following four situations (each case is independent of the others):
Case Units Sold
A
9,000
B
Sales
Contribution
Variable Margin per
Fixed Net Income
Expenses
Unit
Expenses
(Loss)
$270,000 $162,000
$350,000
C
20,000
D
5,000
$ 90,000
$15
$280,000
$170,000
$ 6
$100,000
$ 40,000
$ 35,000
$ 82,000
($12,000)
Cost Concepts Relevant to Decision Making
8.9 An executive from a large merchandising firm has called your vice-president for
production to get a price quote for an additional 100 units of a given product. The
vice-president has asked you to prepare a cost estimate. The number of hours required to produce a unit is 5. The average labor rate is $12 per hour. The materials
cost $14 per unit. Overhead for an additional 100 units is estimated at 50% of the
direct labor cost. If the company wants to have a 30% profit margin, what should
be the unit price to quote?
8.10 The Morton Company produces and sells two products: A and B. Financial data
related to producing these two products are summarized as follows:
Product A
Product B
Selling price
$ 10.00
$12.00
Variable costs
$ 5.00
$10.00
Fixed costs
$ 2,000
$ 600
(a) If these products are sold in the ratio of 4A for 3B, what is the break-even point?
(b) If the product mix has changed to 5A for 5B, what would happen to the breakeven point?
(c) In order to maximize the profit, which product mix should be pushed?
(d) If both products must go through the same manufacturing machine and there
are only 30,000 machine hours available per period, which product should
be pushed? Assume that product A requires 0.5 hour per unit and B requires
0.25 hour per unit.
8.11 Pearson Company manufactures a variety of electronic printed circuit boards
(PCBs) that go into cellular phones. The company has just received an offer from
an outside supplier to provide the electrical soldering for Pearson’s Motorola product line (Z-7 PCB, slimline). The quoted price is $4.80 per unit. Pearson is interested in this offer, since its own soldering operation of the PCB is at its peak capacity.
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Short Case Study
• Outsourcing option. The company estimates that if the supplier’s offer were
accepted, the direct labor and variable overhead costs of the Z-7 slimline would
be reduced by 15% and the direct material cost would be reduced by 20%.
• In-house production option. Under the present operations, Pearson manufactures all of its own PCBs from start to finish. The Z-7 slimlines are sold through
Motorola at $20 per unit. Fixed overhead charges to the Z-7 slimline total $20,000
each year. The further breakdown of producing one unit is as follows:
Direct materials
$ 7.50
Direct labor
5.00
Manufacturing overhead
4.00
Total cost
$16.00
The manufacturing overhead of $4.00 per unit includes both variable and fixed
manufacturing overhead, based on a production of 100,000 units each year.
(a) Should Pearson Company accept the outside supplier’s offer?
(b) What is the maximum unit price that Pearson Company should be willing to
pay the outside supplier?
Short Case Study
ST8.1 The Hamilton Flour Company is currently operating its mill six days a week, 24 hours
a day, on three shifts. At current prices, the company could easily obtain a sufficient volume of sales to take the entire output of a seventh day of operation each week. The
mill’s practical capacity is 6,000 hundredweight of flour per day. Note that
• Flour sells for $12.40 a hundredweight (cwt.) and the price of wheat is $4.34 a
bushel. About 2.35 bushels of wheat are required per cwt. of flour. Fixed costs
now average $4,200 a day, or $0.70 per cwt. The average variable cost of mill
operation, almost entirely wages, is $0.34 per cwt.
• With Sunday operation, wages would be doubled for Sunday work, which would
bring the variable cost of Sunday operation to $0.66 per cwt. Total fixed costs per
week would increase by $420 (or $29,820) if the mill were to operate on Sunday.
(a) Using the information provided, compute the break-even volumes for sixday and seven-day operation.
(b) What are the marginal contribution rates for six-day and seven-day operation?
(c) Compute the average total cost per cwt. for six-day operation and the net
profit margin per cwt. before taxes.
(d) Would it be economical for the mill to operate on Sundays? (Justify your answer numerically.)
427
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NINE
CHAPTER
Depreciation and
Corporate Taxes
Know What It Costs to Own a Piece of Equipment:
A Hospital Pharmacy Gets a Robotic Helper1 When most
patients at Kirkland’s Evergreen Hospital Medical Center are asleep,
the robot comes to life. Its machinery thumps like a heart beating
as it moves around the hospital pharmacy, preparing prescriptions.
In a little more than an hour, he’ll ready 1,500 doses of medication.
Ernie, or Evergreen Robot Noticeably Improving Efficiency, is a
new $3 million addition to [the] pharmacy staff. The robot uses bar
codes to match each drug dispensed with an electronic patient profile,
helping prevent errors, said Bob Blanchard, pharmacy director.
1
“Know What It Costs to Own a Piece of Equipment: A Hospital Pharmacy Gets a Robotic Helper,”
Katherine Sather, Seattle Times Eastside bureau, Copyright © 2004 The Seattle Times Company.
428
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“It’s the future,” he said. “Safety is the main benefit.” Efficiency is
another plus.The robot can prepare a 24-hour medication supply for
in-house patients, about 1,500 doses, in a little more than an hour. The
same task used to take three people about three hours to complete,
Blanchard said.
Now, staff [members] only label the medications with bar codes. Ernie
does the rest.
The machine looks like a mini space ship. A door opens into an octagonshaped room, 12 feet long diagonally, stocked with more than 400 racks
of medicine. Each dose is labeled with a bar code that tells Ernie what it
is and where it should be stored. Affixed to the center of the room is a
mechanical arm that scans the bar-coded medicine and places it on its
designated rack.
When staff [members] give Ernie a computerized order, his arm buzzes
to the correct row of medication, grabs it with suction cups and drops it
into an envelope that is bar coded with the patient’s profile.
“Research shows using it decreases certain predictable errors,”
Blanchard said.“We’re very excited about it—we’ve really led the drive to
move to automation.”
At Evergreen, [the] pharmacy staff hope[s] for Ernie to eventually
dispense 93 percent of the medication that is distributed to patients in the
244-bed hospital. Medication that needs care such as refrigeration is
prepared by staff.
“The technology is such that it’s been tested and it’s reliable. Given
the volume of patients we see, it makes sense,” said Amy Gepner, a
spokesperson for the hospital.“It’s a safety initiative.” Ernie is being
purchased with a seven-year lease along with 23 automated medical
cabinets placed throughout the hospital. But [the] staff [is] fonder of the
robot.
Now ask yourself, How does the cost of this robot ($3 million) affect
the financial position of the hospital? In the long run, the system promises
to create greater cost savings for the hospital by enhancing productivity,
429
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430 CHAPTER 9 Depreciation and Corporate Taxes
improving safety, and cutting down lead time in filling orders. In the short run, however,
the high initial cost of the robot will adversely affect the organization’s “bottom line,”
because that cost is only gradually rewarded by the benefits the robot offers.
Another consideration should come to mind as well: This state-of-the-art robot
must inevitably wear out over time, and even if its productive service extends over
many years, the cost of maintaining its high level of functioning will increase as the individual pieces of hardware wear out and need to be replaced. Of even greater concern
is the question of how long this robot will be the state of the art. When will the competitive advantage the hospital has just acquired become a competitive disadvantage
through obsolescence?
One of the facts of life that organizations must deal with and account for is that fixed
assets lose their value—even as they continue to function and contribute to the engineering
projects that use them. This loss of value, called depreciation, can involve deterioration
and obsolescence.
The main function of depreciation accounting is to account for the cost of fixed assets in a pattern that matches their decline in value over time. The cost of the robot we
have just described, for example, will be allocated over several years in the hospital’s financial statements, so that the pattern of the robot’s costs roughly matches its pattern of
service. In this way, as we shall see, depreciation accounting enables the firm to stabilize
the statements about its financial position that it distributes to stockholders and the outside world.
On a project level, engineers must be able to assess how the practice of depreciating
fixed assets influences the investment value of a given project. To do this, the engineers
need to estimate the allocation of capital costs over the life of the project, which requires
an understanding of the conventions and techniques that accountants use to depreciate assets. In this chapter, we will review the conventions and techniques of asset depreciation
and income taxes.
We begin by discussing the nature and significance of depreciation, distinguishing its
general economic definition from the related, but different, accounting view of depreciation. We then focus our attention almost exclusively on the rules and laws that govern
asset depreciation and the methods that accountants use to allocate depreciation expenses.
Knowledge of these rules will prepare you to apply them in assessing the depreciation of
assets acquired in engineering projects. Then we turn our attention to the subject of depletion, which utilizes similar ideas, but specialized techniques, to allocate the cost of the
depletion of natural-resource assets.
Once we understand the effect of depreciation at the project level, we need to address the effect of corporate taxes on project cash flows. There are many forms of government taxation, including sales taxes, property taxes, user taxes, and state and federal
income taxes. In this chapter, we will focus on federal income taxes. When you are operating a business, any profits or losses you incur are subject to income tax consequences.
Therefore, we cannot ignore the impact of income taxes in project evaluation. The chapter
will give you a good idea of how the U.S. tax system operates and of how federal income
taxes affect economic analysis. Although tax law is subject to frequent changes, the analytical procedures presented here provide a basis for tax analysis that can be adapted
to reflect future changes in tax law. Thus, while we present many examples based on
current tax rates, in a larger context we present a general approach to the analysis of
any tax law.
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Section 9.1 Asset Depreciation 431
CHAPTER LEARNING OBJECTIVES
After completing this chapter, you should understand the following concepts:
How to account for the loss of value of an asset in business.
The meaning and types of depreciation.
The difference between book depreciation and tax depreciation.
The effects of depreciation on net income calculation.
The general scheme of U.S. corporate taxes.
How to determine ordinary gains and capital gains.
How to determine the appropriate tax rate to use in project analysis.
The relationship between net income and net cash flow.
9.1 Asset Depreciation
ixed assets, such as equipment and real estate, are economic resources that are
acquired to provide future cash flows. Generally, depreciation can be defined as
a gradual decrease in the utility of fixed assets with use and time. While this
general definition does not capture the subtleties inherent in a more specific definition of depreciation, it does provide us with a starting point for examining the variety of underlying ideas and practices that are discussed in this chapter. Figure 9.1 will
serve as a road map for understanding the different types of depreciation that we will
explore here.
F
Physical
depreciation
Economic depreciation
The gradual decrease in
utility in an asset with
use and time
Functional
depreciation
Depreciation
Accounting depreciation
The systematic allocation
of an asset’s value in
portions over its
depreciable life — often
used in engineering
economic analysis
Figure 9.1
Classification of types of depreciation.
Book
depreciation
Tax
depreciation
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432 CHAPTER 9 Depreciation and Corporate Taxes
We can classify depreciation into the categories of physical or functional depreciation. Physical depreciation can be defined as a reduction in an asset’s capacity to perform its intended service due to physical impairment. Physical depreciation can occur in
any fixed asset in the form of (1) deterioration from interaction with the environment, including such agents as corrosion, rotting, and other chemical changes, and (2) wear and
tear from use. Physical depreciation leads to a decline in performance and high maintenance costs.
Functional depreciation occurs as a result of changes in the organization or in
technology that decrease or eliminate the need for an asset. Examples of functional
depreciation include obsolescence attributable to advances in technology, a declining
need for the services performed by an asset, and the inability to meet increased quantity
or quality demands.
9.1.1 Economic Depreciation
This chapter is concerned primarily with accounting depreciation, which is the form of
depreciation that provides an organization with the information it uses to assess its financial position. It would also be useful, however, to discuss briefly the economic ideas upon
which accounting depreciation is based. In the course of the discussion, we will develop
a precise definition of economic depreciation that will help us distinguish between various conceptions of depreciation.
If you have ever owned a car, you are probably familiar with the term depreciation
as it is used to describe the decreasing value of your vehicle. Because a car’s reliability
and appearance usually decline with age, the vehicle is worth less with each passing
year. You can calculate the economic depreciation accumulated for your car by subtracting the current market value, or “blue book” value, of the car from the price you originally paid for it. We can define economic depreciation as follows:
Economic depreciation = Purchase price - market value.
Physical and functional depreciation are categories of economic depreciation. The measurement of economic depreciation does not require that an asset be sold: The market
value of the asset can be closely estimated without actually testing it in the marketplace.
The need to have a precise scheme for recording the ongoing decline in the value of an
asset as a part of the accounting process leads us to an exploration of how organizations
account for depreciation.
9.1.2 Accounting Depreciation
The acquisition of fixed assets is an important activity for a business organization,
whether the organization is starting up or acquiring new assets to remain competitive.
Like other disbursements, the cost of these fixed assets must be recorded as expenses on
a firm’s balance sheet and income statement. However, unlike costs such as maintenance,
material, and labor costs, the costs of fixed assets are not treated simply as expenses to be
accounted for in the year that they are acquired. Rather, these assets are capitalized; that
is, their costs are distributed by subtracting them as expenses from gross income, one part
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Section 9.2 Factors Inherent in Asset Depreciation 433
at a time over a number of periods. The systematic allocation of the initial cost of an asset
in parts over a time, known as the asset’s depreciable life, is what we mean by accounting
depreciation. Because accounting depreciation is the standard of the business world, we
sometimes refer to it more generally as asset depreciation.
Accounting depreciation is based on the matching concept: A fraction of the cost of
the asset is chargeable as an expense in each of the accounting periods in which the asset
provides service to the firm, and each charge is meant to be a percentage of the whole
cost that “matches” the percentage of the value utilized in the given period. The matching
concept suggests that the accounting depreciation allowance generally reflects, at least to
some extent, the actual economic depreciation of the asset. In engineering economic
analysis, we use the concept of accounting depreciation exclusively. This is because accounting depreciation provides a basis for determining the income taxes associated with
any project undertaken.
9.2 Factors Inherent in Asset Depreciation
The process of depreciating an asset requires that we make several preliminary determinations: (1) What is the cost of the asset? (2) What is the asset’s value at the end of
its useful life? (3) What is the depreciable life of the asset? and, finally, (4) What
method of depreciation do we choose? In this section, we will examine each of these
questions.
9.2.1 Depreciable Property
As a starting point, it is important to recognize what constitutes a depreciable
asset—that is, a property for which a firm may take depreciation deductions against
income. For the purposes of U.S. tax law, any depreciable property has the following
characteristics:
1. It must be used in business or must be held for the production of income.
2. It must have a definite service life, and that life must be longer than 1 year.
3. It must be something that wears out, decays, gets used up, becomes obsolete, or loses
value from natural causes.
Depreciable property includes buildings, machinery, equipment, and vehicles. Inventories are not depreciable property, because they are held primarily for sale to customers
in the ordinary course of business. If an asset has no definite service life, the asset cannot
be depreciated. For example, you can never depreciate land.2
As a side note, we mention the fact that, while we have been focusing on depreciation within firms, individuals may also depreciate assets, as long as they meet the conditions we have listed. For example, an individual may depreciate an automobile if the
vehicle is used exclusively for business purposes.
2
This also means that you cannot depreciate the cost of clearing, grading, planting, and landscaping. All four
expenses are considered part of the cost of the land.
Accounting
depreciation:
Amount
allocated during
the period to
amortize the
cost of acquiring
long term assets
over the useful
life of the assets.
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9.2.2 Cost Basis
The cost basis of an asset represents the total cost that is claimed as an expense over the
asset’s life (i.e., the sum of the annual depreciation expenses). The cost basis generally includes the actual cost of the asset and all other incidental expenses, such as freight, site
preparation, and installation. This total cost, rather than the cost of the asset only, must be
the depreciation basis charged as an expense over the asset’s life.
Besides being used in figuring depreciation deductions, an asset’s cost basis is used
in calculating the gain or loss to the firm if the asset is ever sold or salvaged. (We will discuss
these topics in Section 9.8.)
Cost basis: The
cost of an
asset used to
determine the
depreciation
base.
EXAMPLE 9.1 Cost Basis
Lanier Corporation purchased an automatic hole-punching machine priced at $62,500.
The vendor’s invoice included a sales tax of $3,263. Lanier also paid the inbound
transportation charges of $725 on the new machine, as well as the labor cost of $2,150
to install the machine in the factory. In addition, Lanier had to prepare the site at a cost
of $3,500 before installation. Determine the cost basis for the new machine for depreciation purposes.
SOLUTION
Book value: The
value of an asset
as it appears
on a balance
sheet, equal
to cost minus
accumulated
depreciation.
Given: Invoice price = $62,500, freight = $725, installation cost = $2,150, and
site preparation = $3,500.
Find: The cost basis.
The cost of machine that is applicable for depreciation is computed as follows:
Cost of new hole-punching machine
Freight
$62,500
725
Installation labor
2,150
Site preparation
3,500
Cost of machine (cost basis)
$68,875
COMMENTS: Why do we include all the incidental charges relating to the acquisition
of a machine in its cost? Why not treat these incidental charges as expenses incurred
during the period in which the machine is acquired? The matching of costs and revenue is the basic accounting principle. Consequently, the total costs of the machine
should be viewed as an asset and allocated against the future revenue that the machine
will generate. All costs incurred in acquiring the machine are costs of the services to
be received from using the machine.
If the asset is purchased by trading in a similar asset, the difference between the
book value (the cost basis minus the total accumulated depreciation) and the trade-in allowance must be considered in determining the cost basis for the new asset. If the trade-in
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Section 9.2 Factors Inherent in Asset Depreciation 435
allowance exceeds the book value, the difference (known as unrecognized gain)
needs to be subtracted from the cost basis of the new asset. If the opposite is true
(unrecognized loss), the difference should be added to the cost basis for the new
asset.
EXAMPLE 9.2 Cost Basis with Trade-In Allowance
In Example 9.1, suppose Lanier purchased the hole-punching press by trading in a
similar machine and paying cash for the remainder. The trade-in allowance is $5,000,
and the book value of the hole-punching machine that was traded in is $4,000. Determine the cost basis for this hole-punching press.
SOLUTION
Given: Accounting data from Example 9.1; trade allowance = $5,000.
Find: The revised cost basis.
Old hole-punching machine (book value)
Less: Trade-in allowance
$ 4,000
5,000
Unrecognized gains
$ 1,000
Cost of new hole-punching machine
$62,500
Less: Unrecognized gains
Freight
Installation labor
Site preparation
Cost of machine (cost basis)
(1,000)
725
2,150
3,500
$67,875
9.2.3 Useful Life and Salvage Value
Over how many periods will an asset be useful to a company? What do published statutes
allow you to choose as the life of an asset? These are the central questions to be answered
in determining an asset’s depreciable life (i.e., the number of years over which the asset is
to be depreciated).
Historically, depreciation accounting included choosing a depreciable life that was
based on the service life of an asset. Determining the service life of the asset, however,
was often very difficult, and the uncertainty of these estimates often led to disputes between taxpayers and the Internal Revenue Service (IRS). To alleviate the problems, the
IRS published guidelines on lives for categories of assets. The guidelines, known as asset
depreciation ranges, or ADRs, specified a range of lives for classes of assets based on
historical data, and taxpayers were free to choose a depreciable life within the specified
range for a given asset. An example of ADRs for some assets is given in Table 9.1.
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TABLE 9.1
Asset Depreciation: Some Selected Asset Guideline Classes
Asset Depreciation Range (Years)
Lower
Limit
Midpoint
Life
Upper
Limit
Office furniture, fixtures,
and equipment
8
10
12
Information systems (computers)
5
6
7
Assets Used
Airplanes
5
6
7
Automobiles, taxis
2.5
3
3.5
Buses
7
9
11
Light trucks
3
4
5
Heavy trucks (concrete ready-mixer)
5
6
7
12
15
18
5
6
7
Vessels, barges, tugs, and water
transportation systems
14.5
18
21.5
Industrial steam and electrical
generation and/or distribution systems
17.5
22
26.5
Manufacturer of electrical
and nonelectrical machinery
8
10
12
Manufacturer of electronic
components, products, and systems
5
6
7
Manufacturer of motor vehicles
9.5
12
14.5
35
42
Railroad cars and locomotives
Tractor units
Telephone distribution plant
28
Source: IRS Publication 534. Depreciation. Washington, DC: U.S. Government Printing Office, 1995.
The salvage value is an asset’s estimated value at the end of its life—the amount
eventually recovered through sale, trade-in, or salvage. The eventual salvage value of an
asset must be estimated when the depreciation schedule for the asset is established. If this
estimate subsequently proves to be inaccurate, then an adjustment must be made. We will
discuss these specific issues in Section 9.6.
9.2.4 Depreciation Methods: Book and Tax Depreciation
Most firms calculate depreciation in two different ways, depending on whether the calculation is (1) intended for financial reports (the book depreciation method), such as for
the balance sheet or income statement, or (2) for the Internal Revenue Service (IRS), for
the purpose of determining taxes (the tax depreciation method). In the United States,
this distinction is totally legitimate under IRS regulations, as it is in many other countries.
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Section 9.3 Book Depreciation Methods 437
Calculating depreciation differently for financial reports and for tax purposes allows the
following benefits:
• It enables firms to report depreciation to stockholders and other significant outsiders
on the basis of the matching concept. Therefore, the actual loss in value of the assets
is generally reflected.
• It allows firms to benefit from the tax advantages of depreciating assets more quickly
than would be possible with the matching concept. In many cases, tax depreciation
allows firms to defer paying income taxes. This does not mean that they pay less tax
overall, because the total depreciation expense accounted for over time is the same
in either case. However, because tax depreciation methods usually permit a higher
depreciation in earlier years than do book depreciation methods, the tax benefit of
depreciation is enjoyed earlier, and firms generally pay lower taxes in the initial
years of an investment project. Typically, this leads to a better cash position in early
years, and the added cash leads to greater future wealth because of the time value of
the funds.
As we proceed through the chapter, we will make increasing use of the distinction
between depreciation accounting for financial reporting and depreciation accounting
for income tax calculation. Now that we have established the context for our interest in
both tax and book depreciation, we can survey the different methods with an accurate
perspective.
9.3 Book Depreciation Methods
Three different methods can be used to calculate the periodic depreciation allowances:
(1) the straight-line method, (2) accelerated methods, and (3) the unit-of-production
method. In engineering economic analysis, we are interested primarily in depreciation in
the context of income tax computation. Nonetheless, a number of reasons make the study
of book depreciation methods useful. First, tax depreciation methods are based largely on
the same principles that are used in book depreciation methods. Second, firms continue to
use book depreciation methods for financial reporting to stockholders and outside parties.
Third, book depreciation methods are still used for state income tax purposes in many
states and even for federal income tax purposes for assets that were put into service before 1981. Finally, our discussion of depletion in Section 9.5 is based largely on one of
these three book depreciation methods.
9.3.1 Straight-Line Method
The straight-line (SL) method of depreciation interprets a fixed asset as an asset that offers
its services in a uniform fashion. The asset provides an equal amount of service in each
year of its useful life.
The straight-line method charges, as an expense, an equal fraction of the net cost of
the asset each year, as expressed by the relation
Dn =
1I - S2
,
N
(9.1)
Straight-line
depreciation:
An equal dollar
amount of
depreciation in
each accounting
period.
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where Dn = Depreciation charge during year n,
I = Cost of the asset, including installation expenses,
S = Salvage value at the end of the asset’s useful life,
N = Useful life.
The book value of the asset at the end of n years is then defined as
Book value in a given year = Cost basis - total depreciation charges made to date
or
Bn = I - 1D1 + D2 + D3 + Á + Dn2.
(9.2)
EXAMPLE 9.3 Straight-Line Depreciation
Consider the following data on an automobile:
Cost basis of the asset, I = $10,000,
Useful life, N = 5 years,
Estimated salvage value, S = $2,000.
Use the straight-line depreciation method to compute the annual depreciation allowances
and the resulting book values.
SOLUTION
Given: I = $10,000, S = $2,000, and N = 5 years.
Find: Dn and Bn for n = 1 to 5.
The straight-line depreciation rate is 15, or 20%. Therefore, the annual depreciation
charge is
Dn = 10.2021$10,000 - $2,0002 = $1,600.
The asset would then have the following book values during its useful life:
n
Bn 1
Dn
Bn
1
$10,000
$1,600
$8,400
2
8,400
1,600
6,800
3
6,800
1,600
5,200
4
5,200
1,600
3,600
5
3,600
1,600
2,000
Here, Bn - 1 represents the book value before the depreciation charge for year n. This
situation is illustrated in Figure 9.2.
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Section 9.3 Book Depreciation Methods 439
$10,000
Book value
$8,000
D2
B1
$6,000
B2
$4,000
D3
B3
D4
B4
$2,000
D5
Total depreciation at end of life
Annual depreciation
D1
B5
0
0
Figure 9.2
1
2
3
Year
4
5
Straight-line depreciation methods (Example 9.3).
9.3.2 Accelerated Methods
The second concept of depreciation recognizes that the stream of services provided by
a fixed asset may decrease over time; in other words, the stream may be greatest in the
first year of an asset’s service life and least in its last year. This pattern may occur because the mechanical efficiency of an asset tends to decline with age, because maintenance costs tend to increase with age, or because of the increasing likelihood that better
equipment will become available and make the original asset obsolete. This kind of
reasoning leads to a method that charges a larger fraction of the cost as an expense of
the early years than of the later years. Any such method is called an accelerated
method. The most widely used accelerated method is the double-declining-balance
method.
Declining-Balance Method
The declining-balance (DB) method of calculating depreciation allocates a fixed fraction of the beginning book balance each year. The fraction, a, is obtained as follows:
a = a
1
b 1multiplier2.
N
(9.3)
The most commonly used multipliers in the United States are 1.5 (called 150% DB) and
2.0 (called 200%, or double-declining balance, DDB). As N increases, a decreases, resulting in a situation in which depreciation is highest in the first year and then decreases
over the asset’s depreciable life.
Accelerated
nethod: Any
depreciation
method that
produces larger
deductions for
depreciation in
the early years
of a project’s life.
Doubledeclining
balance
method: A
depreciation
method, in which
double the
straight-line
depreciation
amount is taken
the first year, and
then that same
percentage is
applied to the
undepreciated
amount in
subsequent
years.
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The fractional factor can be utilized to determine depreciation charges for a given year,
Dn , as follows:
D1 = aI,
D2 = a1I - D12 = aI11 - a2,
D3 = a1I - D1 - D22 = aI11 - a22,
Thus, for any year n, we have a depreciation charge
Dn = aI11 - a2n - 1.
(9.4)
We can also compute the total DB depreciation (TDB) at the end of n years as
follows:
TDB = D1 + D2 + Á + Dn
= aI + aI11 - a2 + aI11 - a22 + Á + aI11 - a2n - 1
= aI[1 + 11 - a2 + 11 - a22 + Á + 11 - a2n - 1]
= I[1 - 11 - a2n].
(9.5)
The book value Bn at the end of n years will be the cost I of the asset minus the total depreciation at the end of n years:
Bn = I - TDB
= I - I[1 - 11 - a2n]
Bn = I11 - a2n.
(9.6)
EXAMPLE 9.4 Declining-Balance Depreciation
Consider the following accounting information for a computer system:
Cost basis of the asset, I = $10,000,
Useful life, N = 5 years,
Estimated salvage value, S = $778.
Use the double-declining-depreciation method to compute the annual depreciation
allowances and the resulting book values (Figure 9.3).
SOLUTION
Given: I = $10,000, S = $778, N = 5 years
Find: Dn and Bn for n = 1 to 5
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Section 9.3 Book Depreciation Methods 441
$10,000
Annual depreciation
Total depreciation at end of life
Book value
D1
$8,000
$6,000
D2
$4,000
B1
D3
$2,000
B2
D4
B3
D5
B4
$778
B5
0
0
Figure 9.3
1
2
3
Year
4
5
Double-declining-balance method (Example 9.4).
The book value at the beginning of the first year is $10,000, and the declining-balance
rate 1a2 is A 15 B 122 = 40%. Then the depreciation deduction for the first year will be
$4,000 140% * $10,000 = $4,0002. To figure the depreciation deduction in the
second year, we must first adjust the book value for the amount of depreciation we deducted in the first year. The first year’s depreciation from the beginning book value is
subtracted 1$10,000 - $4,000 = $6,0002, and the resulting amount is multiplied by
the rate of depreciation 1$6,000 * 40% = $2,4002. By continuing the process, we
obtain the following table:
n
Bn
Dn
Bn
1
10,000
4,000
6,000
2
6,000
2,400
3,600
3
3,600
1,440
2,160
4
2,160
864
1,296
5
1,296
518
778
The declining balance is illustrated in terms of the book value of time in Figure 9.3.
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The salvage value (S) of the asset must be estimated at the outset of depreciation analysis. In Example 9.4, the final book value 1BN2 conveniently equals the estimated salvage
value of $778, a coincidence that is rather unusual in the real world. When BN Z S, we
would want to make adjustments in our depreciation methods.
• Case 1: BN>S
When BN 7 S, we are faced with a situation in which we have not depreciated the
entire cost of the asset and thus have not taken full advantage of depreciation’s taxdeferring benefits. If you would prefer to reduce the book value of an asset to its
salvage value as quickly as possible, it can be done by switching from DB to SL
whenever SL depreciation results in larger depreciation charges and therefore a more
rapid reduction in the book value of the asset. The switch from DB to SL depreciation
can take place in any of the n years, the objective being to identify the optimal year to
switch. The switching rule is as follows: If depreciation by DB in any year is less than
(or equal to) what it would be by SL, we should switch to and remain with the SL
method for the duration of the project’s depreciable life. The straight-line depreciation
in any year n is calculated by
Dn =
Book value at beginning of year n - salvage value
Remaining useful life at beginning of year n
(9.7)
EXAMPLE 9.5 Declining Balance with Conversion to
Straight-Line Depreciation (BN>S )
Suppose the asset given in Example 9.4 has a zero salvage value instead of $778; that is,
Cost basis of the asset, I
Useful life, N
Salvage value, S
a
=
=
=
=
$10,000
5 years,
$0,
11>52122 = 40%.
Determine the optimal time to switch from DB to SL depreciation and the resulting
depreciation schedule.
SOLUTION
Given: I = $10,000, S = 0, N = 5 years, and a = 40%.
Find: Optimal conversion time, Dn and Bn for n = 1 to 5.
We will first proceed by computing the DDB depreciation for each year, as before:
Year
Dn
Bn
1
$4,000
$6,000
2
2,400
3,600
3
1,440
2,160
4
864
1,296
5
518
778
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Section 9.3 Book Depreciation Methods 443
Then, using Eq. (9.7), we compute the SL depreciation for each year. We compare SL
with DDB depreciation for each year and use the aforementioned decision rule for
when to change:
If Switch to SL at
Beginning of Year
SL
Depreciation
DDB
Depreciation
Switching
Decision
2
1$6,000 - 02/4 = $1,500
6 $2,400
Do not switch
1,200
6 1,440
1,080
7
13,600 - 02/3 =
3
4
12,160 - 02/2 =
I
Do not switch
864
Switch to SL
BN > S
BN
S
S
0
n'
N
(a)
I
BN < S
S
S
BN
0
n"
N
(b)
Figure 9.4 Adjustments to the declining-balance method:
(a) Switch from declining balance to straight line after n¿;
(b) no further depreciation allowances are available after n–
(Examples 9.5 and 9.6).
The optimal time (year 4) in this situation corresponds to n¿ in Figure 9.4(a). The
resulting depreciation schedule is
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DDB with
End-of-Year
Year
Switch to SL
Book Value
1
$ 4,000
$6,000
2
2,400
3,600
3
1,440
2,160
4
1,080
1,080
5
1,080
0
$10,000
• Case 2: BN~~1.5 MBF
= $80,000>MBF
Depletion allowance for the year = 0.5 MBF * $80,000>MBF
= $40,000.
9.5.2 Percentage Depletion
Percentage depletion is an alternative method of calculating the depletion allowance for
certain mineral properties. For a given mineral property, the depletion allowance calculation is based on a prescribed percentage of the gross income from the property during the
tax year. Notice the distinction between depreciation and depletion: Depreciation is the
allocation of cost over a useful life, whereas percentage depletion is an annual allowance
of a percentage of the gross income from the property.
Since percentage depletion is computed on the basis of the income from, rather than
the cost of, the property, the total depletion on a property may exceed the cost of the property. To prevent this from happening, the annual allowance under the percentage method
cannot be more than 50% of the taxable income from the property (figured without the
deduction for depletion). Table 9.5 shows the allowed percentages for selected mining
properties.
TABLE 9.5
Percentage Depletion Allowances for Mineral Properties
Deposits
Percentage Allowed
Oil and gas wells
(only for certain domestic and gas production)
15
Sulfur and uranium and, if from deposits in the United States,
asbestos, lead, zinc, nickel, mica, and certain other
ores and minerals
22
Gold, silver, copper, iron ore, and oil shale, if from deposits
in the United States
15
Coal, lignite, and sodium chloride
10
Clay and shale to be used in making sewer pipe or bricks
7.5
Clay (used for roofing tile), gravel, sand, and stone
5
Most other minerals; includes carbon dioxide produced
from a well and metallic ores
14
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Section 9.5 Depletion 455
EXAMPLE 9.11 Percentage Depletion versus Cost Depletion
A gold mine with an estimated deposit of 300,000 ounces of gold has a basis of $30
million (cost minus land value). The mine has a gross income of $16,425,000 for the
year from selling 45,000 ounces of gold (at a unit price of $365 per ounce). Mining
expenses before depletion equal $12,250,000. Compute the percentage depletion
allowance. Would it be advantageous to apply cost depletion rather than percentage
depletion?
SOLUTION
Given: Basis = $30 million, total recoverable volume = 300,000 ounces of gold,
amount sold this year = 45,000 ounces, gross income = $16,425,000, and this
year’s expenses before depletion = $12,250,000.
Find: Maximum depletion allowance (cost or percentage).
Percentage depletion: Table 9.5 indicates that gold has a 15% depletion allowance.
The percentage depletion allowance is computed from the gross income:
Gross income from sale of 45,000 ounces
$16,425,000
* 15%
Depletion percentage
Computed percentage depletion
$ 2,463,750
Next, we need to compute the taxable income. The percentage depletion allowance
is limited to the computed percentage depletion or 50% of the taxable income,
whichever is smaller:
Gross income from sale of 45,000 ounces
Less mining expenses
$16,425,000
12,250,000
Taxable income from mine
4,175,000
* 50%
Deduction limitation
Maximum depletion deduction
$ 2,088,000
Since the maximum depletion deduction ($2,088,000) is less than the computed percentage depletion ($2,463,750), the allowable percentage deduction is $2,088,000.
Cost depletion: It is worth computing the depletion allowance with the cost depletion method:
Cost depletion = a
$30,000,000
b 145,0002
300,000
= $4,500,000.
Note that percentage depletion is less than the cost depletion. Since the law allows
the taxpayer to take whichever deduction is larger in any one year, in this situation it
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456 CHAPTER 9 Depreciation and Corporate Taxes
A
50% of net
income
A
B
Statutory
percentage
of gross
income
If
B
True
Allowed
percentage
depletion
A
Cost
depletion
Figure 9.6
purposes.
False
Allowed
percentage
depletion
B
Greater is
allowable
depletion
Calculating the allowable depletion deduction for federal tax
would be advantageous to apply the cost depletion method. Figure 9.6 illustrates the
steps to be taken to apply that depletion method.
9.6 Repairs or Improvements Made
to Depreciable Assets
If any major repairs (e.g., an engine overhaul) or improvements (say, an addition) are
made during the life of the asset, we need to determine whether these actions will extend
the life of the asset or will increase the originally estimated salvage value. When either of
these situations arises, a revised estimate of the useful life of the asset should be made,
and the periodic depreciation expense should be updated accordingly. We will examine
how repairs or improvements affect both book and tax depreciations.
9.6.1 Revision of Book Depreciation
Recall that book depreciation rates are based on estimates of the useful lives of assets.
Such estimates are seldom precise. Therefore, after a few years of use, you may find that
the asset could last for a considerably longer or shorter period than was originally estimated. If this happens, the annual depreciation expense, based on the estimated useful
life, may be either excessive or inadequate. (If repairs or improvements do not extend the
life or increase the salvage value of the asset, these costs may be treated as maintenance
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Section 9.6 Repairs or Improvements Made to Depreciable Assets 457
expenses during the year in which they were incurred.) The procedure for correcting the
book depreciation schedule is to revise the current book value and to allocate this cost
over the remaining years of the asset’s useful life.
9.6.2 Revision of Tax Depreciation
For tax depreciation purposes, repairs or improvements made to any property are treated
as separate property items. The recovery period for a repair or improvement to the initial
property normally begins on the date the repaired or improved property is placed in service. The recovery class of the repair or improvement is the recovery class that would
apply to the property if it were placed in service at the same time as the repair or improvement. Example 9.12 illustrates the procedure for correcting the depreciation schedule for an asset that had repairs or improvements made to it during its depreciable life.
EXAMPLE 9.12 Depreciation Adjustment for an Overhauled
Asset
In January 2001, Kendall Manufacturing Company purchased a new numerical
control machine at a cost of $60,000. The machine had an expected life of 10 years
at the time of purchase and a zero expected salvage value at the end of the 10 years.
• For book depreciation purposes, no major overhauls had been planned over the
10-year period, and the machine was being depreciated toward a zero salvage
value, or $6,000 per year, with the straight-line method.
• For tax purposes, the machine was classified as a 7-year MACRS property.
In December 2003, however, the machine was thoroughly overhauled and rebuilt at a
cost of $15,000. It was estimated that the overhaul would extend the machine’s useful life by 5 years.
(a) Calculate the book depreciation for the year 2006 on a straight-line basis.
(b) Calculate the tax depreciation for the year 2006 for this machine.
SOLUTION
Given: I = $60,000, S = $0, N = 10 years, machine overhaul = $15,000, and extended life = 15 years from the original purchase.
Find: D6 for book depreciation and D6 for tax depreciation.
(a) Since an improvement was made at the end of the year 2003, the book value of
the asset at that time consisted of the original book value plus the cost added to
the asset. First, the original book value at the end of 2003 is calculated:
B3 1before improvement2 = $60,000 - 31$6,0002 = $42,000.
After the improvement cost of $15,000 is added, the revised book value is
B3 1after improvement2 = $42,000 + $15,000 = $57,000.
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To calculate the book depreciation in the year 2006, which is 3 years after the improvement, we need to calculate the annual straight-line depreciation amount
with the extended useful life. The remaining useful life before the improvement
was made was 7 years. Therefore, the revised remaining useful life should be 12
years. The revised annual depreciation is then $57,000/12 = $4,750. Using the
straight-line depreciation method, we compute the depreciation amount for 2006
as follows:
D6 = $4,750.
(b) For tax depreciation purposes, the improvement made is viewed as a separate
property with the same recovery period as that of the initial asset. Thus, we
need to calculate both the tax depreciation under the original asset and that of
the new asset. For the 7-year MACRS property, the 6th-year depreciation allowance is 8.92% of $60,000, or $5,352. The 3rd-year depreciation for the
improved asset is 17.49% of $15,000, or $2,623. Therefore, the total tax depreciation in 2006 is
D6 = $5,352 + $2,623 = $7,975.
Figure 9.7 illustrates how additions or improvements are treated in revising depreciation amounts for book and tax purposes.
Additions or improvements made
Book depreciation (original asset)
0
1
2
3
4
3
4
5
6
7
8
9
10
11
12
13
New cost basis = (Undepreciated cost + improved cost)
5
6
7
8
9
10
Extended useful life
14
15
MACRS (original asset)
0
1
2
3
3
4
5
6
7
8
4
5
6
7
8
9
10
11
MACRS (improved portion)
Figure 9.7 Revision of depreciation amount as additions or improvements are made as
described in Example 9.12.
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9.7 Corporate Taxes
Now that we have learned what elements constitute taxable income, we turn our attention to the process of computing income taxes. The corporate tax rate, is applied to the
taxable income of a corporation. As we briefly discussed in Section 8.5, the allowable
deductions include the cost of goods sold, salaries and wages, rent, interest, advertising,
depreciation, amortization,3 depletion, and various tax payments other than federal income tax. The following table is illustrative:
Item
Gross income
Expenses:
Cost of goods sold
Depreciation
Operating expenses
Taxable operating income
Income taxes
Net income
9.7.1 Income Taxes on Operating Income
The corporate tax rate structure for 2006 is relatively simple. As shown in Table 9.6, there
are four basic rate brackets (15%, 25%, 34%, and 35%), plus two surtax rates (5% and
3%), based on taxable incomes. U.S. tax rates are progressive; that is, businesses with
lower taxable incomes are taxed at lower rates than those with higher taxable incomes.
TABLE 9.6
3
Corporate Tax Schedule for 2006
Taxable Income (X)
Tax Rate
$0–$50,000
15%
50,001–75,000
25%
75,001–100,000
34%
Tax Computation Formula
$0 + 0.15X
7,500 + 0.251X - $50,0002
13,750 + 0.341X - 75,0002
100,001–335,000
34% + 5%
335,001–10,000,000
34%
113,900 + 0.341X - 335,0002
22,250 + 0.391X - 100,0002
10,000,001–15,000,000
35%
3,400,000 + 0.351X - 10,000,0002
15,000,001–18,333,333
35% + 3%
5,150,000 + 0.381X - 15,000,0002
18,333,334 and up
35%
6,416,666 + 0.351X - 18,333,3332
The amortization expense is a special form of depreciation for an intangible asset, such as patents, goodwill, and franchises. More precisely, the amortization expense is the systematic write-off to expenses of the
cost of an intangible asset over the periods of its economic usefulness. Normally a straight-line method is used
to calculate the amortization expense.
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Marginal tax
rate: The
amount of tax
paid on an
additional dollar
of income.
Effective
(average) tax
rate: The rate a
taxpayer would
be taxed at if
taxing was done
at a constant
rate, instead of
progressively.
Marginal Tax Rate
The marginal tax rate is defined as the rate applied to the last dollar of income earned.
Income of up to $50,000 is taxed at a 15% rate (meaning that if your taxable income is
less than $50,000, your marginal tax rate is 15%); income between $50,000 and $75,000
is taxed at 25%; and income over $75,000 is taxed at a 34% rate.
An additional 5% surtax (resulting in 39%) is imposed on a corporation’s taxable
income in excess of $100,000, with the maximum additional tax limited to $11,750
1235,000 * 0.052. This surtax provision phases out the benefit of graduated rates for
corporations with taxable incomes between $100,000 and $335,000. Another 3% surtax is imposed on corporate taxable income in the range from $15,000,001 to
$18,333,333.
Corporations with incomes in excess of $18,333,333 in effect pay a flat tax of 35%.
As shown in Table 9.6, the corporate tax is progressive up to $18,333,333 in taxable income, but essentially is constant thereafter.
Effective (Average) Tax Rate
Effective tax rates can be calculated from the data in Table 9.6. For example, if your corporation had a taxable income of $16,000,000 in 2006, then the income tax owed by the
corporation would be as follows:
Taxable Income
Tax Rate
Taxes
Cumulative Taxes
First $50,000
15%
$7,500
$7,500
Next $25,000
25%
6,250
13,750
Next $25,000
34%
8,500
22,250
Next $235,000
39%
91,650
113,900
Next $9,665,000
34%
3,286,100
3,400,000
Next $5,000,000
35%
1,750,000
5,150,000
Remaining $1,000,000
38%
380,000
$5,530,000
Alternatively, using the tax formulas in Table 9.6, we obtain
$5,150,000 + 0.381$16,000,000 - $15,000,0002 = $5,530,000.
The effective (average) tax rate would then be
$5,530,000
= 0.3456, or 34.56,
$16,000,000
as opposed to the marginal rate of 38%. In other words, on the average, the company paid
34.56 cents for each taxable dollar it generated during the accounting period.
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EXAMPLE 9.13 Corporate Taxes
A mail-order computer company sells personal computers and peripherals. The company leased showroom space and a warehouse for $20,000 a year and installed
$290,000 worth of inventory-checking and packaging equipment. The allowed depreciation expense for this capital expenditure ($290,000) amounted to $58,000
using the category of 5-year MACRS. The store was completed and operations began
on January 1. The company had a gross income of $1,250,000 for the calendar year.
Supplies and all operating expenses, other than the lease expense, were itemized as
follows:
Merchandise sold in the year
$600,000
Employee salaries and benefits
150,000
Other supplies and expenses
90,000
$840,000
Compute the taxable income for this company. How much will the company pay in
federal income taxes for the year?
SOLUTION
Given: Income, preceding cost information and depreciation.
Find: Taxable income and federal income taxes.
First we compute the taxable income as follows:
Gross revenues
Expenses
$1,250,000
–840,000
Lease expense
–20,000
Depreciation
–58,000
Taxable income
$332,000
Note that capital expenditures are not deductible expenses. Since the company is in
the 39% marginal tax bracket, its income tax can be calculated by using the formula
given in Table 9.6, namely, 22,250 + 0.39 1X - 100,0002:
Income tax = $22,250 + 0.391$332,000 - $100,0002
= $112,730.
The firm’s current marginal tax rate is 39%, but its average corporate tax rate is
$112,730
= 33.95%.
$332,000
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COMMENTS: Instead of using the tax formula in Table 9.6, we can compute the federal
income tax in the following fashion:
First $50,000 at 15%
$
7,500
Next $25,000 at 25%
6,250
Next $25,000 at 34%
8,500
Next $232,000 at 39%
Income tax
90,480
$112,730
9.8 Tax Treatment of Gains or Losses
on Depreciable Assets
As in the disposal of capital assets, gains or losses generally are associated with the sale
(or exchange) of depreciable assets. To calculate a gain or loss, we first need to determine
the book value of the depreciable asset at the time of its disposal.
9.8.1 Disposal of a MACRS Property
For a MACRS property, one important consideration at the time of disposal is whether
the property is disposed of during or before its specified recovery period. Moreover, with
the half-year convention, which is now mandated by all MACRS depreciation methods,
the year of disposal is charged one-half of that year’s annual depreciation amount, if it
should occur during the recovery period.
EXAMPLE 9.14 Book Value in the Year of Disposal
Consider a five-year MACRS asset purchased for $10,000. Note that property belonging
to the five-year MACRS class is depreciated over six years due to the half-year convention. The applicable depreciation percentages, shown in Table 9.3, are 20%, 32%,
19.20%, 11.52%, 11.52%, and 5.76%. Compute the allowed depreciation amounts and
the book value when the asset is disposed of (a) in year 3, (b) in year 5, and (c) in year 6.
SOLUTION
Given: Five-year MACRS asset, cost basis = $10,000, and MACRS depreciation
percentages as shown in Figure 9.8.
Find: Total depreciation and book value at disposal if the asset is sold in year 3, 5, or 6.
(a) If the asset is disposed of in year 3 (or at the end of year 3), the total accumulated
depreciation amount and the book value would be
Total depreciation = $10,00010.20 + 0.32 + 0.192/22
= $6,160,
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Recovery period = 5 years (for 5-year MACRS asset)
20%
32%
1
19.2%
2
11.52%
3
11.52%
4
5.76%
5
6
Asset disposal before recovery period
Asset disposal
after recovery
period
19.2%/2
(a)
20%
32%
1
2
3
11.52%/2
(b)
20%
32%
1
(c)
20%
19.2%
2
32%
1
11.52%
3
19.2%
2
4
11.52%
3
5
5.76%
11.52%
4
5
6
Year of disposal (half-year convention)
Figure 9.8 Disposal of a MACRS property and its effect on depreciation allowances
(Example 9.14).
Book value = $10,000 - $6,160
= $3,840.
(b) If the asset is disposed of during4, year 5, the depreciation and book values will
be
Total depreciation = $10,00010.20 + 0.32 + 0.192 + 0.1152 + 0.1152/22
= $8,848,
Book value = $10,000 - $8,848
= $1,152.
(c) If the asset is disposed of after the recovery period, there will be no penalty due
to early disposal. Since the asset is depreciated fully, we have
Total depreciation = $10,000,
Book value = $0.
4
Note that even though you dispose of the asset at the end of the recovery period, the half-year convention still
applies.
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9.8.2 Calculations of Gains and Losses on MACRS Property
When a depreciable asset used in business is sold for an amount that differs from its book
value, the gain or loss has an important effect on income taxes. The gain or loss is found
as follows:
• Case 1: Salvage valueCost basis
In the unlikely event that an asset is sold for an amount greater than its cost basis, the
gains 1salvage value - book value2 are divided into two parts for tax purposes:
Gains = Salvage value - book value
= (''''')'''''*
1Salvage value - cost basis2
Capital gains
+ (''''')'''''*
1Cost basis - book value2 .
(9.10)
Ordinary gains
Recall from Section 9.2.2 that the cost basis is the cost of purchasing an asset, plus
any incidental costs, such as freight and installation costs. As illustrated in Figure 9.9,
Capital gain:
An increase in
the value of a
capital asset that
gives it a higher
worth than the
purchase price,
when the asset
is sold.
Capital gains = Salvage value - cost basis,
Ordinary gains = Cost basis - book value.
This distinction is necessary only when capital gains are taxed at the capital gains tax rate
and ordinary gains (or depreciation recapture) at the ordinary income tax rate. Current tax
Capital gains
Total gains
Ordinary gains
or
depreciation recapture
Cost basis
Book value
Salvage value
Figure 9.9 Capital gains and ordinary gains (or depreciation recapture)
when the salvage value exceeds the cost basis.
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law does not provide a special low rate of taxation for capital gains. Instead, capital gains
are treated as ordinary income, but the maximum tax rate is set at the U.S. statutory rate
of 35%. Nevertheless, the statutory structure for capital gains has been retained in the tax
code. This provision could allow Congress to restore capital gains’ preferential treatment
at some future time.
EXAMPLE 9.15 Gains and Losses on Depreciable Assets
A company purchased a drill press costing $230,000 in year 0. The drill press, classified as seven-year recovery property, has been depreciated by the MACRS method. If it
is sold at the end of three years, compute the gains (losses) for the following four salvage values: (a) $150,000, (b) $120,693, (c) $100,000, and (d) $250,000. Assume that
both capital gains and ordinary income are taxed at 34%.
SOLUTION
Given: Seven-year MACRS asset, cost basis = $230,000, and the asset is sold three
years after its purchase.
Find: Gains or losses, tax effects, and net proceeds from the sale if the asset is sold
for $150,000, $120,693, $100,000, or $250,000.
In this example, we first compute the current book value of the machine. From the
MACRS depreciation schedule in Table 9.3, the allowed annual depreciation percentages for the first three years of a seven-year MACRS property are 14.29%,
24.49%, and 17.49%. Since the asset is disposed of before the end of its recovery period, the depreciation amount in year 3 will be reduced by half. The total depreciation and the final book value will be
Total allowed depreciation = $230,000 a0.1429 + 0.2449 +
0.1749
b
2
= $109,308,
Book value = $230,000 - $109,308
= $120,693.
(a) Case 1: Book value 6 Salvage value 6 Cost basis
In this case, there are no capital gains to consider. All gains are ordinary gains. Thus,
we have
Ordinary gains = Salvage value - book value
= $150,000 - $120,693 = $29,308,
Gains tax 134%2 = 0.341$29,3082 = $9,965,
Net proceeds from sale = Salvage value - gains tax
= $150,000 - $9,965 = $140,035.
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Cost basis = $230,000
Book value = $120,693
Ordinary
gain
Case 1
Salvage value = $150,000
Case 2
Salvage value = $120,693
Case 3
Salvage value = $100,000
Loss
Ordinary gains
Case 4
Capital
gains
Salvage value = $250,000
$0
Figure 9.10
$50,000
$100,000
$150,000
$200,000
$250,000
Calculations of gains or losses on MACRS property (Example 9.15).
This situation (in which the asset’s salvage value exceeds its book value) is denoted as Case 1 in Figure 9.10.
(b) Case 2: Salvage value = Book value
In Case 2, the book value is again $120,693. Thus, if the drill press’s salvage value
equals $120,693—its book value—no taxes are levied on that salvage value.
Therefore, the net proceeds equal the salvage value.
(c) Case 3: Salvage value 6 Book value
Case 3 illustrates a loss, when the salvage value (say, $100,000) is less than the
book value. We compute the net salvage value after tax as follows:
Gain 1loss2 = Salvage value - book value
= $100,000 - $120,693
= 1$20,6932,
Tax savings = 0.341$20,6932
= $7,036,
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Net proceeds from sale = $100,000 + $7,036
= $107,036.
(d) Case 4: Salvage value 7 Cost basis
This situation is not likely for most depreciable assets (except real property). But
it is the only situation in which both capital gains and ordinary gains can be observed. Nevertheless, the tax treatment on this gain is as follows:
Capital gains = Salvage value - cost basis
= $250,000 - $230,000
= $20,000,
Capital gains tax = $20,00010.342
= $6,800,
Ordinary gains = $230,000 - $120,693
= $109,307,
Gains tax = $109,30710.342
= $37,164,
Net proceeds from sale = $250,000 - 1$6,800 + $37,1642
= $206,036.
COMMENTS: Note that in (c) the reduction in tax, due to the loss, actually increases the
net proceeds. This is realistic when the incremental tax rate (34% in this case) is
positive, indicating the corporation is still paying tax, but less than if the asset had
not been sold at a loss. The incremental tax rate will be discussed in Section 9.9.
9.9 Income Tax Rate to Be Used in Economic Analysis
As we have seen in earlier sections, average income tax rates for corporations vary with
the level of taxable income from 0 to 35%. Suppose that a company now paying a tax rate
of 25% on its current operating income is considering a profitable investment. What tax
rate should be used in calculating the taxes on the investment’s projected income?
9.9.1 Incremental Income Tax Rate
The choice of a corporation’s rate depends on the incremental effect of an investment on
the company’s taxable income. In other words, the tax rate to use is the rate that applies
to the additional taxable income projected in the economic analysis.
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To illustrate, consider ABC Corporation, whose taxable income from operations is
expected to be $70,000 for the current tax year. ABC management wishes to evaluate the
incremental tax impact of undertaking a project during the same tax year. The revenues,
expenses, and taxable incomes before and after the project are estimated as follows:
Before
After
Incremental
$200,000
$240,000
$40,000
Salaries
100,000
110,000
10,000
Wages
30,000
40,000
10,000
$ 70,000
$ 90,000
$20,000
Gross revenue
Taxable income
Because the income tax rate is progressive, the tax effect of the project cannot be isolated from the company’s overall tax obligations. The base operations of ABC without
the project are expected to yield a taxable income of $70,000. With the new project, the
taxable income increases to $90,000. From the tax computation formula in Table 9.6, the
corporate income taxes with and without the project are as follows:
Income tax without the project = $7,500 + 0.251$70,000 - $50,0002
= $12,500,
Income tax with the project = $13,750 + 0.341$90,000 - $75,0002
= $18,850.
The additional income tax is then $18,850 - $12,500 = $6,350. This amount, on
the additional $20,000 of taxable income, is based on a rate of 31.75%, which is an incremental rate. This is the rate we should use in evaluating the project in isolation from
the rest of ABC’s operations. As shown in Figure 9.11, the 31.75% is not an arbitrary figure, but a weighted average of two distinct marginal rates. Because the new project pushes
ABC into a higher tax bracket, the first $5,000 it generates is taxed at 25%; the remaining
$15,000 it generates is taxed in the higher bracket, at 34%. Thus, we could have calculated the incremental tax rate with the formula
0.25 a
$15,000
$5,000
b + 0.34 a
b = 31.75%.
$20,000
$20,000
$20,000 incremental
taxable income due to
undertaking project
Regular income from operation
Marginal tax rate
15%
$0
Figure 9.11
$20,000
$40,000
$5,000
at 25%
$15,000
at 34%
25%
34%
$60,000
Illustration of incremental tax rate.
$80,000
$100,000
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Section 9.9 Income Tax Rate to Be Used in Economic Analysis 469
The average tax rates before and after the new project being considered is undertaken
are as follows:
Taxable income
Income taxes
Average tax rate
Incremental tax rate
Before
After
Incremental
$70,000
$90,000
$20,000
12,500
18,850
6,350
17.86%
20.94%
31.75%
Note that neither 17.86% nor 20.94% is a correct tax rate to use in evaluating the new
project.
A corporation with continuing base operations that place it consistently in the highest tax bracket will have both marginal and average federal tax rates of 35%. For such
firms, the tax rate on an additional investment project is, naturally, 35%. But for corporations in lower tax brackets, and for those which fluctuate between losses and profits,
the marginal and average tax rates are likely to vary. For such corporations, estimating a
prospective incremental tax rate for a new investment project may be difficult. The only
solution may be to perform scenario analysis, in which we examine how much the income tax fluctuates due to undertaking the project. (In other words, we calculate the
total taxes and the incremental taxes for each scenario.) A typical scenario is presented
in Example 9.16.
EXAMPLE 9.16 Scenario Analysis for a Small Company
EverGreen Grass Company expects to have an annual taxable income of $320,000
from its regular grass-sodding business over the next two years. EverGreen has just
won a contract to sod grasses for a new golf complex for just those years. This twoyear project requires a purchase of new equipment costing $50,000. The equipment
falls into the MACRS five-year class, with depreciation allowances of 20%, 32%,
19.2%, 11.52%, 11.52%, and 5.76% in each of the six years, respectively, during
which the equipment will be depreciated. After the contract is terminated, the equipment will be retained for future use (instead of being sold), indicating no salvage
cash flow, gain, or loss on this asset. The project will bring in an additional annual
revenue of $150,000, but it is expected to incur additional annual operating costs of
$90,000. Compute the incremental (marginal) tax rates applicable to the project’s operating profits for years 1 and 2.
SOLUTION
Given: Base taxable income = $320,000 per year, incremental income, expenses,
and depreciation amounts as stated.
Find: Incremental tax rate for this new project in years 1 and 2.
First, we compute the additional taxable income from the golf course project over the
next two years:
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Year
1
2
Gross Revenue
$150,000
$150,000
Expenses
90,000
90,000
Depreciation
10,000
16,000
$ 50,000
$ 44,000
Taxable income
Next, we compute the income taxes. To do this, we need to determine the applicable
marginal tax rate, but because of the progressive income tax rate on corporations, the
project cannot be isolated from the other operations of EverGreen.
We can solve this problem much more efficiently by using the incremental tax
rate concept discussed at the beginning of this section. Because the golf course project pushes EverGreen into the 34% tax bracket from the 39% bracket, we want to
know what proportion of the incremental taxable income of $50,000 in year 1 is
taxed at 39% and what proportion is taxable at 34%. Without the project, the firm’s
taxable income is $320,000 and its marginal tax rate is 39%. With the additional taxable income of $50,000, EverGreen’s tax bracket reverts to 34%, as its combined taxable income changes from $320,000 to $370,000. Since the rate changes at
$335,000, the first $15,000 of the $50,000 taxable income will still be in the 39%
bracket, and the remaining $35,000 will be in the 34% bracket. In year 2, we can divide the additional taxable income of $44,000 in a similar fashion. Then we can calculate the incremental tax rates for the first two years as follows:
0.39 a
$15,000
$35,000
b + 0.34 a
b = 0.3550,
$50,000
$50,000
0.39 a
$29,000
$15,000
b + 0.34 a
b = 0.3570.
$44,000
$44,000
Note that these incremental tax rates vary slightly from year to year. Much larger
changes could occur if a company’s taxable income fluctuates drastically from its continuing base operation.
9.9.2 Consideration of State Income Taxes
For large corporations, the top federal marginal tax rate is 35%. In addition to federal
income taxes, state income taxes are levied on corporations in most states. State income taxes are an allowable deduction in computing federal taxable income, and two
ways are available to consider explicitly the effects of state income taxes in an economic
analysis.
The first approach is to estimate explicitly the amount of state income taxes before
calculating the federal taxable income. We then reduce the federal taxable income by the
amount of the state taxes and apply the marginal tax rate to the resulting federal taxes.
The total taxes would be the sum of the state taxes and the federal taxes.
The second approach is to calculate a single tax rate that reflects both state and federal income taxes. This single rate is then applied to the federal taxable income, without
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subtracting state income taxes. Taxes computed in this fashion represent total taxes. If
state income taxes are considered, the combined state and federal marginal tax rate may
be higher than 35%. Since state income taxes are deductible as expenses in determining
federal taxes, the marginal rate for combined federal and state taxes can be calculated
with the expression
tm = tf + ts - 1tf21ts2,
(9.11)
where
tm = combined marginal tax rate,
tf = federal marginal tax rate.
ts = state marginal tax rate.
This second approach provides a more convenient and efficient way to handle taxes
in an economic analysis in which the incremental tax rates are known. Therefore, incremental tax rates will be stated as combined marginal tax rates, unless indicated otherwise.
(For large corporations, these would be about 40%, but they vary from state to state.)
EXAMPLE 9.17 Combined State and Federal Income Taxes
Consider a corporation whose revenues and expenses before income taxes are as follows:
Gross revenue
All expenses
$1,000,000
400,000
If the marginal federal tax rate is 35% and the marginal state rate is 7%, compute the
combined state and federal taxes, using the two methods just described.
SOLUTION
Given: Gross income = $1,000,000, deductible expenses = $400,000, tf = 35%,
and ts = 7%.
Find: Combined income taxes tm .
(a) Explicit calculation of state income taxes:
Let’s define FT as federal taxes and ST as state taxes. Then
State taxable income = $1,000,000 - $400,000
and
ST = 10.0721$600,0002
= $42,000.
Also,
Federal taxable income = $1,000,000 - $400,000 - ST
= 1$558,0002,
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472 CHAPTER 9 Depreciation and Corporate Taxes
so that
FT = 10.3521$558,0002
= $195,300.
Thus,
Combined taxes = FT + ST
= $237,300.
(b) Tax calculation based on the combined tax rate:
Compute the combined tax rate directly from the formula:
Combined tax rate 1tm2 = 0.35 + 0.07 - 10.35210.072
= 39.55%.
Hence,
Combined taxes = $600,00010.39552
= $237,300.
As expected, these two methods always produce exactly the same results.
9.10 The Need for Cash Flow in Engineering
Economic Analysis
Traditional accounting stresses net income as a means of measuring a firm’s profitability,
but it is desirable to discuss why cash flows are relevant in project evaluation. As seen in
Section 8.2, net income is an accounting measure based, in part, on the matching concept. Costs become expenses as they are matched against revenue. The actual timing of
cash inflows and outflows is ignored.
9.10.1 Net Income versus Net Cash Flow
Over the life of a firm, net incomes and net cash inflows will usually be the same. However, the timing of incomes and cash inflows can differ substantially. Given the time value
of money, it is better to receive cash now rather than later, because cash can be invested to
earn more cash. (You cannot invest net income.) For example, consider two firms and
their income and cash flow schedules over two years:
Year 1
Year 2
Company A
Company B
Net income
$1,000,000
$1,000,000
Cash flow
1,000,000
0
Net income
1,000,000
1,000,000
Cash flow
1,000,000
2,000,000
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Section 9.10 The Need for Cash Flow in Engineering Economic Analysis 473
Both companies have the same amount of net income and cash over two years, but
Company A returns $1 million cash yearly, while Company B returns $2 million at the end
of the second year. If you received $1 million at the end of the first year from Company A,
you could, for example, invest it at 10%. While you would receive only $2 million in total
from Company B at the end of the second year, you would receive $2.1 million in total
from Company A.
9.10.2 Treatment of Noncash Expenses
Apart from the concept of the time value of money, certain expenses do not even require
a cash outflow. Depreciation and amortization are the best examples of this type of expense. Even though depreciation (or amortization expense) is deducted from revenue on
a daily basis, no cash is paid to anyone.
In Example 9.13, we learned that the annual depreciation allowance has an important
impact on both taxable and net income. However, although depreciation has a direct impact on net income, it is not a cash outlay; hence, it is important to distinguish between
annual income in the presence of depreciation and annual cash flow.
The situation described in Example 9.13 serves as a good vehicle to demonstrate the
difference between depreciation costs as expenses and the cash flow generated by the
purchase of a fixed asset. In that example, cash in the amount of $290,000 was expended
in year 0, but the $58,000 depreciation charged against the income in year 1 was not a
cash outlay. Figure 9.12 summarizes the difference.
Net income (accounting profit) is important for accounting purposes, but cash flows
are more important for project evaluation purposes. However, as we will now demonstrate,
net income can provide us with a starting point to estimate the cash flow of a project.
0
1
2
3
4
5
6
3
4
5
6
$33,408
$33,408
Capital expenditure
(actual cash flow)
$290,000
0
1
2
$16,704
$58,000
$55,680
$92,800
Allowed depreciation expenses
(not cash flow)
Figure 9.12 Cash flows versus depreciation expenses for
an asset with a cost basis of $290,000, which was placed in
service in year 0.
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474 CHAPTER 9 Depreciation and Corporate Taxes
The procedure for calculating net cash flow (after tax) is identical to that used to obtain net income from operations, with the exception of depreciation, which is excluded
from the net cash flow computation. (It is needed only for computing income taxes.) Assuming that revenues are received and expenses are paid in cash, we can obtain the net
cash flow by adding the noncash expense (depreciation) to net income, which cancels
the operation of subtracting it from revenues:
Item
Gross income
Expenses:
Cost of goods sold
Depreciation
Operating expenses
Taxable operating income
Income taxes
Net income
Net cash flow Net income Depreciation
Example 9.18 illustrates this relationship.
EXAMPLE 9.18 Cash Flow versus Net Income
A company buys a numerically controlled (NC) machine for $28,000 (year 0) and
uses it for five years, after which it is scrapped. The allowed depreciation deduction
during the first year is $4,000, as the equipment falls into the category of seven-year
MACRS property. (The first-year depreciation rate is 14.29%.) The cost of the goods
produced by this NC machine should include a charge for the depreciation of the machine. Suppose the company estimates the following revenues and expenses, including depreciation, for the first operating year:
Gross income
Cost of goods sold
Depreciation on NC machine
Operating expenses
=
=
=
=
$50,000,
$20,000,
$4,000,
$6,000.
(a) If the company pays taxes at the rate of 40% on its taxable income, what is its
net income from the project during the first year?
(b) ~~