chapter research

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Economics

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I am assigned to write a research paper on the following topic and the following are instructions on how to do the paper I also attached a powerpoint slides that have chapter one in it and also attached a sample of how the paper should look like

In a single Word doc (or similar editable application), respond to the next four items:

1. Restate one point/idea (from your assigned chapter) in your own words. It could be the same as you posted on the Discussions (above). In any case, the idea must have two or more possible viewpoints.

You must do this to get any credit for parts 2, 3, and 4.

1. Research to find one writer (public, that is, someone who has been published in print, or other medium) who agrees with the point. State that writer’s view in ONE paragraph; give the reference location for this writer. 10 points

2. Research to find one writer (as above) who disagrees with the point. State that writer’s view in ONE paragraph; give the reference location for this writer. 10 points

3. State your views on this topic—again in one paragraph. 16 points

Post your document to this Assignment (in Canvas) regardless of which chapter you have to cover. That is, everyone posts their work to the same Canvas Assignment. You are to do this research ONLY for your assigned chapter.

Everyone does this for their specific chapter.

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Chapter 7 Rate of Return Analysis 1 Chapter Contents  Internal Rate of Return  Rate of Return Calculations  Plot of NPW versus interest rate i  Fees or Discounts  Examples  Incremental Analysis  Using Spreadsheet Engineering Economics 2 Rate of Return Analysis  Rate of return analysis is the most frequently used exact analysis technique in industry.  Major advantages • Rate of return is a single figure of merit that is readily understood. • Calculation of rate of return is independent from the minimum attractive rate of return (MARR). Engineering Economics 3 Internal Rate of Return What is the internal rate of return (IRR)? IRR is the interest rate at which present worth or equivalent uniform annual worth is equal to 0. In other words, the internal rate of return is the interest rate at which the benefits are equivalent to the costs. Engineering Economics 4 Internal Rate of Return  Internal rate of return is commonly used to evaluate the desirability of investments or projects.  IRR can be used to rank multiple prospective projects.  Because the internal rate of return is a rate quantity, it is an indicator of the efficiency, quality, or yield of an investment.  To decide how to proceed, IRR will be compared to preselected minimum attractive rate of return (Chapter 8) Engineering Economics 5 Internal Rate of Return (IRR) Given a cash flow stream, IRR is the interest rate i which yields a zero NPW (i.e., the benefits are equivalent to the costs), or a zero worth at any point in time. This can be expressed in 5 different ways as follows.      NPW = 0 PW of benefits – PW of costs = 0 PW of benefits = PW of costs PW of benefits/PW of costs = 1 EUAB – EUAC = 0 Engineering Economics 6 Example A person invests $1000 at the end of each year. If the person would like to have $80,000 in savings at EOY 26 what interest rate should he select? Net Present Wo rth = 0 = Net Future Worth − $1,000 (F/A, i%,26) + $80,000 = 0 (F/A, i%,26) = $80,000/$1 ,000 = 80 Checking the Tables 26 yrs @ 6%, F/A = 59.156 26 yrs @ 10%, F/A = 109.182 26 yrs @ 8%, F/A = 79.954 Engineering Economics 7 Example A person invests $1000 at the end of each year. If the person would like to have $80,000 in savings at EOY 26 what interest rate should he select? Net Present Wo rth = 0 = Net Future Worth − $1,000 (F/A, i%,26) + $80,000 = 0 (F/A, i%,26) = $80,000/$1 ,000 = 80 When the compound interest tables are visited the value of i where (F/A, i%, 26)=80 is found as 8%, so i=8% Engineering Economics 8 Checking the Tables 26 yrs @ 6%, F/A = 59.156 26 yrs @ 10%, F/A = 109.182 26 yrs @ 8%, F/A = 79.954 Example A person invests $1000 at the end of each year. If the person would like to have $80,000 in savings at EOY 26 what interest rate should he select? Net Present Wo rth = 0 = Net Future Worth − $1,000 (F/A, i%,26) + $80,000 = 0 (F/A, i%,26) = $80,000/$1 ,000 = 80 When the compound interest tables are visited the value of i where (F/A, i%, 26)=80 is found as 8%, so i=8% Engineering Economics 9 Example – EXCEL solution RATE(n, A, P, F, type, guess) rate(26, 1000, 0, -80000) = 8% rate (26, -1000, 0, 80000) = 8% A, P, F must have different signs (+ or –)! IRR(value range, guess) value range = the cash flow stream Engineering Economics 10 Example Cash flows for an investment are shown in the following figure. What is the IRR to obtain these cash flows? YEAR CASH FLOW 0 ($500) 1 $100 2 $150 3 $200 4 $250 Engineering Economics 11 YEAR CASH FLOW 0 ($500) 1 $100 2 $150 3 $200 4 $250 EXAMPLE CONTINUES EUAW = EUAB − EUAC = 0 100 + 50( A / G, i %, 4) − 500( A / P, i %, 4) = 0 Try i = 5% 100 + 50( A / G,5%, 4) − 500( A / P,5%,4) 100 + 50(1.439) − 500(0.2820) = 30.95 Try i = 15% 100 + 50( A / G,15%, 4) − 500( A / P,15%, 4) 100 + 50(1.326) − 500(0.3503) = -8.85 −10.20 Nov. 2, 2011 Engineering Economics 12 QUESTION CONTINUES EUAW = EUAB − EUAC = 0 100 + 50( A / G, i %, 4) − 500( A / P, i %, 4) = 0 Try i = 5% 100 + 50( A / G,5%, 4) − 500( A / P,5%,4) 100 + 50(1.439) − 500(0.2820) = 30.95 Try i = 15% 100 + 50( A / G,15%, 4) − 500( A / P,15%, 4) 100 + 50(1.326) − 500(0.3503) = -8.85 −10.20 Nov. 2, 2011 Engineering Economics 13 INTERPOLATION 5% 30.95 30.95 5-X 10 X% 0 15% -8.85 39.80 5− X 30.95 − 0 30.95 = = 5 − 15 30.95 − (−8.85) 39.80 X = 12.78  13% Engineering Economics 14 INTERPOLATION 5% 30.95 30.95 5-X -10 X% 0 15% -8.85 39.80 5− X 30.95 − 0 30.95 = = 5 − 15 30.95 − (−8.85) 39.80 X = 12.78  13% Engineering Economics 15 INTERPOLATION 12% 3.350 3.350 12-X -3 X% 0 15% -8.850 12.200 12 − X 3.350 − 0 3.350 = = 12 − 15 3.350 − (−8.850) 12.200 X = 12.82  13% Engineering Economics 16 INTERPOLATION 12% 3.350 3.350 12-X -3 X% 0 15% -8.850 12.200 12 − X 3.350 − 0 3.350 = = 12 − 15 3.350 − (−8.850) 12.200 X = 12.82  13% Engineering Economics 17 EXCEL solution IRR(C1:C5) = 12.83% C1 ~ C5 stores the stream of the 5 cash flows: -500, 100, 150, 200, 250 Engineering Economics 18 Example A student, who will graduate after 4 years, borrows $10,000 per year at 5% interest rate at the beginning of each year. No interest is charged till graduation. If the student makes five equal annual payments after the graduation (end-of-period payments). a) What is each payment after the graduation? b) Calculate IRR of loan? (hint: use cash flow from when the student started borrowing the money to when it is all paid back) c) Is the loan attractive to the student? Engineering Economics 19 EXAMPLE CONTINUES Year Cash Flow 0 10,000 1 10,000 2 10,000 3 10,000 4 0 5 (9240) Net Present Worth = 0 6 (9240) 10,000 + 10,000(P/A, i%, 3) 7 (9240) 8 (9240) 9 (9240) a) Loan Payment = $40,000(A/P, 5%, 5) = $40,000(0.2310) b) To find IRR% − [$9,240(P/A, i%, 5)](P/F, i%,4) = 0 Engineering Economics 20 10,000 + 10,000(P/A, i%, 3) − [$9,240(P/A, i%, 5)](P/F,i%,4) = 0 Try i = 3% 10,000 + 10,000(2.829) − [$9,240(4.580)](0.8885)  690 Try i = 2% 10,000 + 10,000(2.884) − [$9,240(4.713)](0.9238)  −1,389 INTERPOLATION: 3 − IRR 690 − 0 690 = = 3− 2 690 + 1389 2079 IRR = %2.67 c) Since the rate is low, the loan looks like a good choice. Engineering Economics 21 10,000 + 10,000(P/A, i%, 3) − [$9,240(P/A, i%, 5)](P/F,i%,4) = 0 Try i = 3% 10,000 + 10,000(2.829) − [$9,240(4.580)](0.8885)  690 Try i = 2% 10,000 + 10,000(2.884) − [$9,240(4.713)](0.9238)  −1,389 INTERPOLATION: 3 − IRR 690 − 0 690 = = 3− 2 690 + 1389 2079 IRR = %2.67 c) Since the rate is low, the loan looks like a good choice Engineering Economics 22 EXCEL Solution a) pmt(5%, 5, -40000) = $9,238.99 per month. b) irr(g1:g10) = 2.66%. c) Since the rate is low, the loan looks like a good choice. Engineering Economics 23 Plot of NPW versus Interest Rate Borrowing Cases NPW Year Cash Flow $50.00 0 1 2 3 4 5 200 -50 -50 -50 -50 -50 ($25.00) : : : : ($50.00) $25.00 $0.00 0% 5% 10% 15% 20% p. 218 Engineering Economics 24 Plot of NPW versus Interest Rate Investment Cases NPW Year Cash Flow 0 1 2 3 4 5 : : -200 50 50 50 50 50 : : $50.00 $25.00 $0.00 0% 5% 10% 15% 20% ($25.00) ($50.00) Engineering Economics 25 Fees or Discounts  Question: Option 1: If a property is financed through a loan provided by a seller, its price is $200,000 with 10% down payment and five annual payments at 10%. Option 2: If a property is financed through the same seller in cash, the seller will accept 10% less. However, the buyer does not have $180,000 in cash. What is the IRR for the loan offered by seller? Engineering Economics 26 QUESTION CONTINUES Year 0 Pay Cash ($180,000) Borrow from Seller Down Payment ($20,000) = 20,000 1 ($47,484) 2 ($47,484) 3 ($47,484) 4 ($47,484) 5 ($47,484) Engineering Economics = 200,000 10% Annual Payments = (200, 000 − 20, 000)(A/P,10%,5) = 180,000(0.2638) = 47,484 27 QUESTION CONTINUES To find IRR%, setting cash flows equal in PW terms − 180,000 = −20,000 − 47,484(P/A , i%, 5) (P/A, i%, 5) = (180,000 − 20,000)/47 ,484 = 3.37 Looking in the tables for the above value, IRR% should be between 12%(3.605) and 15%(3.352) . INTERPOLATION: 12 − IRR 3.605 − 3.37 0.235 = = = 0.928 12 − 15 3.605 − 3.352 0.253 IRR = 14.79% Engineering Economics 28 QUESTION CONTINUES To find IRR%, setting cash flows equal in PW terms − 180,000 = −20,000 − 47,484(P/A , i%, 5) (P/A, i%, 5) = (180,000 − 20,000)/47 ,484 = 3.37 Looking in the tables for the above value, IRR% should be between 12%(3.605) and 15%(3.352) . INTERPOLATION: 12 − IRR 3.605 − 3.37 0.235 = = = 0.928 12 − 15 3.605 − 3.352 0.253 IRR = 14.79% This is a relatively high rate of interest, so that borrowing from a bank and paying cash to the property owner is better. 29 Engineering Economics EXCEL Solution Combined cash flows (difference between options 1 & 2): At time 0: EOY 1-5: -$160,000 $47,484 IRR = rate(5, 47484, -160000) = 14.78% per year IRR = irr(a1:a6) = 14.78% per year Engineering Economics 30 Loan and Investments are Everywhere  Question: A student will decide whether to buy weekly parking permit or summer semester parking permit from USF. The former costs $16 weekly; the latter costs $100 due May 17th 2010; in both cases the duration is 12 weeks. Assuming that the student pay the weekly fee on every Monday: a) What is the rate of return for buying the weekly permit? b) Is weekly parking attractive to student? *Total 12 weeks Engineering Economics 31 Effective annual interest = (1+.15)^52 – 1 = 1432% ! QUESTION CONTINUES Week Weekly Semester May 17 0 ($16) ($100) May 24 1 ($16) May 31 2 ($16) June 7 3 ($16) June 14 4 ($16) June 21 5 ($16) June 28 6 ($16) July 5 7 ($16) July 12 8 ($16) July 19 9 ($16) July 26 10 ($16) August 2 11 ($16) Engineering Economics a) To find IRR%, set cash flows equal in PW terms – 100 = – 16 – 16 (P/A,i%,11) (P/A,i%,11) = (100 - 16) / 16 (P/A,i%,11) = 5.25 Looking in the table for the above value: IRR = 15% b) Nominal interest rate for 52 weeks IRR ≈ 15%/week or 15*52 = 780%/yr Since the rate is high, paying the semester fee looks like a good choice. 32 EXCEL Solution Combined cash flows (difference between the 2 options): At time 0: -$84 EOM 1~11: $16 IRR = rate(11, 16, -84) = 14.92% per month IRR = irr(C1: C12) = 14.92% per month Engineering Economics 33 Rate Of Return Calculations Question: There are two options for an equipment: Buy or Lease for 24 months. The equipment might be either leased for $2000 per month or bought for $30,000. If the plan is to buy the equipment, the salvage value of the equipment at EOM 24 is $3,000. What is the IRR or cost of the lease? Engineering Economics 34 QUESTION CONTINUES Month Buy Option Lease Option 0 ($30,000) ($2,000) 1-23 24 ($2,000) $3,000 0 To find IRR%, set cash flows of Buy and Lease options equal in PW terms −30,000 + 3,000(P/F,i%,24) = −2,000 − 2,000(P/A, i%, 23) 0 = 28, 000 − 3,000(P/F,i%,24) − 2,000(P/A, i%, 23)` Engineering Economics 35 0 = 28, 000 − 3,000(P/F,i%,24) − 2,000(P/A, i%, 23)` Try i = 6% 28, 000 − 3,000(0.2470) − 2000(12.303) = 2653 Try i = 5% 28000 – 3000(0.3101) – 20000(13.489) = 91.7 Try i = 4% 28000 – 3000(0.3477) – 2000(14.148) = -1339.1 i ≈ 5% per month Engineering Economics 36 EXCEL Solution Combined cash flows (difference of buy & lease): At time 0: -$28000 EOM 1~23: $2000 EOM 24: $3000 IRR = irr(e1:e25) = 4.97% per month Engineering Economics 37 Incremental Analysis When there are two alternatives, rate of return analysis is often performed by computing the incremental rate of return, ΔIRR, on the difference between the two alternatives. Engineering Economics 38 Incremental Analysis  The cash flow for the difference between alternatives is calculated by taking the higher initial-cost alternative minus the lower initial-cost alternative.  The following decision path is made for incremental rate of return (ΔIRR) on difference between alternatives: Two -Alternative Situations Decision ΔIRR≥MARR Choose the higher-cost alternative ΔIRR
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Running Head: CHAPTER RESEARCH

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Chapter Research
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CHAPTER RESEARCH

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The internal rate of return is used to evaluate whether a project is desirable and can be used to
rank multiple projects using their profitability levels.
McAllister agrees with the statement that internal rate is used to evaluate whether a
project is desirable. The writer indicates that the internal rate of return measures the profitability
of an investment. The wri...


Anonymous
Excellent resource! Really helped me get the gist of things.

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