Question 1,3,9,11,14,15,16. need 7 questions typed Answer and Calculation sheet

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trovfnafuh

Economics

University of California - Davis

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Answer:

1.?

2.1,467,143

3.?

4.No

5.20%

6.Yes

7.379,548.864

8.56,932.33

9.?

10.24,263.26

11.?

12.6,554

13.5,774.3

14.?

15.?

16.?

17.Option 2

Questions:

Question 1

You are given the following financial data about an assembly machine to be implemented at a company:

- Investment cost at year 0 (n=0) is $22,000

- Investment cost at the end of the first year (n=1) is $18,500

- Useful life: 15 years

- Salvage value (at the end of 15 years): $7,000

- Annual revenues: $18,000 per year

- Annual expenses: $5,000 per year

- MARR: 10%

Assuming the first revenues and expenses will occur starting from the end of year 2, determine the conventional (without considering the time value of money or non-discounted) payback period.

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Question 2

A large food-processing company is considering using laser technology to speed up and eliminate waste in the potato-peeling process. To implement the system, the company anticipates needing $3.2 million to purchase the industrial-strength lasers. The system will save $1,500,000 per year in labor and materials. However, it will incur an additional operating and maintenance cost of $300,000 per year. Annual income taxes will be $170,000. The system is expected to have a 10-year service life and a salvage value of $200,000. If the company’s MARR is 18%, calculate the net present worth of this investment.

Question 3

You have been asked to evaluate investment of purchasing a parking lot under the following conditions:

- The proposal is for a parking lot costing $4,000,000. The deck has an expected useful life of 15 years and a net salvage value of $625,000 (after tax adjustment).

- The tenants have recently signed long-term leases, which leads you to believe that the current rental income of $250,000 per year will remain constant for the first five years, then the rental income will increase by 10% for every five-year interval over the remaining asset life.

- The estimated operating expenses, including income taxes, will be $65,000 for the first year and will increase by $6,000 each year thereafter.

Considering an annual interest rate of 15%, what is the net present worth (NPW)?

Exercise3-HW#4Cashflow_Diagram.JPG

Question 4

Would you accept the investment of the previous question? In other words, is it profitable?

Group of answer choices

Question 5

Consider the following project balances for a typical investment project with a service life of four years:

n (end of year)Cashflow AmountProject Balance
0-$1,000-$1,000
1$100-$1,100
2$520-$800
3$460-$500
4$600$0

Determine the interest rate used in computing the project balance.

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Question 6

Would the project of the previous question be acceptable at a MARR of 12%?

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Question 7

Maintenance money for a new building at a college is being solicited from potential alumni donors. You would like to make a donation to cover all future expected maintenance costs of the building. These maintenance costs are expected to be $48,000 per year for the first five years, $60,000 per year for each of years 6 through 10, and $72,000 annually after that (toward infinity under the assumption that the building has an indefinite service life). If the money is placed in an account that will pay 15% annual interest, how large should your gift be?

HW#4Cashflow_Diagram-1.JPG

Question 8

In the previous question, what is the equivalent annual maintenance cost over the infinite service life?

Question 9

Consider the following two mutually exclusive projects:

nProject AProject B
0-$15,000-$25,000
1$5,000$0
2$12,000$X
3$8,000$X
PW (15%)?$9,600

The firm’s MARR is known to be 15%.

Compute the PW (15%) for Project A.

Question 10

In question 9, compute the unknown cash flow X in years 2 and 3 for Project B.

Question 11

In question 9, which project would you select?

Group of answer choices

Question 12

Consider the following two mutually exclusive investment projects:

nProject AProject B
0-$15,000-$25,000
1$5,000$14,500
2$8,000$9,000
3$5,000$10,000
4$6,000

Using the least common multiple analysis period, determine the present worth of project A. Assume that i=12%.

Question 13

Using the least common multiple analysis period, determine the present worth of project B. Assume that i=12%.

Question 14

For questions 12 and 13, which project would you select?

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Question 15

Two options of feed-water storage installation are being considered to serve over 20 years of 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 $170,000. The annual operating and maintenance cost after tax adjustment is estimated to be $2,000. The salvage value is estimated to be negligible.

Option 2: Place a 20,000-gallon tank of equal capacity on a hill near 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. The annual operating and maintenance cost of the water tank is estimated to be $1,500 after tax adjustment. Because of the location, an additional investment of 13,000 in pumping equipment is required. The pumping equipment is expected to have a service life of 10 years with a salvage value of $2,000 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%, what is the net present worth of option 1? (Use negative sign if needed)

Question 16

In the previous question, what is the net present worth of option 2? (Use negative sign if needed)

Question 17

For questions 15 and 16, which option would you choose?

Group of answer choices

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Engineering Economics ENG106 Lecture 6:Inflation Lecture 6: Outline •Inflation •Consumer Price Index •Constant Dollars •Actual Dollars Inflation (or Deflation) increase (or decrease) in the cost of goods or services over time. Loss (or gain) in the purchasing power of money over time Year Cost of First Class stamp 1980 $0.15 1990 2000 2010 2019 $0.25 $0.33 $0.44 $0.55 What are some possible causes for inflation? •Changes in supply or demand (oil during war) •Money supply is increasing (deficit financing/print more cash) How to measure Inflation? There are several ways of measuring inflation 1. Consumer Price Index (CPI) 2. Producers’ Price Index (PPI) 1- Consumer price Index (CPI) Typical market basket of goods and services required by average consumer: 1. Food and alcoholic beverages 2. Housing 3. Apparel 4. Transportation 5. Medical care 6. Entertainment 7. Personal care 8. Other goods and services Consumer price indexes often are used to escalate or adjust payments for rents, wages, alimony, child support and other obligations that may be affected by changes in the cost of living Consumer Price Index http://www.bls.gov/cpi/data.htm • The CPI represents changes in prices of all goods and services purchased for consumption by urban households. Consumer Price Index 2007 1995 1975 BASE Period = 1982-1984 2- Producers’ price Index (PPI) https://www.bls.gov/data/ CPI is is good in measuring general price increase of consumer products but not good enough to measure industrial price increase Example: cost of industrial commodities (raw materials, lumber, steel, chemicals…), energy (electricity, gasoline…)…. Average Inflation Rate Each year (N) the inflation rate (fN)is based on the previous year’s (N-1) rate: fN = (CN-CN-1)/CN-1 How to determine an average Inflation rate? Calculate the average inflation rate of a three year period. The first year’s rate is 2.2%, the second is 7.5% and the third is 4.8% with a base price of $100. t=0  C0 = $100 t=1  C1 = $100(1+2.2%) t=2  C2 = C1 (1+7.5%) = $100(1+2.2%)(1+7.5%) t=3  C3 =C2(1+4.8%) = $100(1+2.2%)(1+7.5%)(1+4.8%) How to determine an average Inflation rate? Calculate the average inflation rate of a three year period. The first year’s rate is 2.2%, the second is 7.5% and the third is 4.8% with a base price of $100. t=3  C3 = $100(1+2.2%)(1+7.5%)(1+4.8%) 𝑓 ҧ = average inflation rate t=3  C3 = $100(1+ḟ)(1+ ḟ )(1+ ḟ ) = C0(1+ḟ)3 Inflation rate over a period of time (years) have compounding effect t=n  Cn = C0(1+ḟ)n = C0(F/P, ḟ ,N) 𝑓ҧ = 𝐶𝑁 𝐶𝑜 1 𝑁 −1 How to determine an average Inflation rate? Calculate the average inflation rate of a three year period. The first year’s rate is 2.2%, the second is 7.5% and the third is 4.8% with a base price of $100. t=3  C3 = $100(1+2.2%)(1+7.5%)(1+4.8%) = $115.14 𝑓ҧ = $115.14 $100 𝐶𝑁 𝑓ҧ = 𝐶𝑜 1 3 −1 1 𝑁 −1 𝑓 ҧ = average inflation rate = 0.04811 = 4.811% (𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑖𝑛𝑓𝑖𝑙𝑡𝑟𝑎𝑡𝑖𝑜𝑛 𝑟𝑎𝑡𝑒 𝑜𝑣𝑒𝑟 𝑡ℎ𝑒 3 𝑦𝑒𝑎𝑟𝑠) Inflation rate • For the postage stamp example: f = (0.44/0.15)1/30 - 1 = 0.0365 = 3.65% 𝑓ҧ = 𝐶𝑁 𝐶𝑜 1 𝑛 −1 Consumer Price Index CPI2010 = 218.1 𝐶𝑃𝐼2010 𝐶𝑃𝐼1980 𝑓ҧ = CPI1980 = 82.4 = Source: Bureau of Labor Statistics Data 218.1 82.4 = 3.3% 1 30 1 30 −1 −1 Inflation rate • For the postage stamp example: f = (0.44/0.15)1/30 - 1 = 0.0365 = 3.65% • The average inflation rate of the CPI (General Inflation rate) over the same period (1980-2010): f = (218.1/82.4)1/30 - 1 = 0.0330 = 3.3% The price of postage rose faster than the typical market basket of goods purchased by a typical family. Calculate the average inflation rate (2005 to 2011) P2005= 100$ Year Inflation rate (%) 2006 4.2 P2006= 100$(1+0.042) =104.2 $ 2007 9.03 P2007= 104.2$(1+0.0903) =113.6 $ 2008 2.19 P2008= 1013.6$(1+0.0219) =116.09 $ 2009 1.79 P2009= 116.09$(1+0.0179) =118.17 $ 2010 2.55 P2010= 118.17$(1+0.0255) =121.18 $ 2011 1.26 P2011= 121.18$(1+0.0126) =122.7 $ f(2005-2011) = (122.7/100)1/6 - 1 = 0.03468 = 3.468% (Average inflation rate from 2005 to 2011) Inflation • There are two basic methods of accounting for inflation in engineering economic analyses. • Constant Dollar Approach(“Real” Purchasing Power Dollars) In the Constant Dollar system all cash flows are stated in dollars that have the purchasing power at a given point in time called the Base Year. • Actual Dollar Approach (Future Dollars, or actual green pieces of paper) In the Actual Dollar system all cash flows are stated in the amount of dollars that would actually be exchanged at that point in time Inflation Example • A new UCD student has an Aunt who sends a $2,000 gift each year to buy pizza. • The cost of a pizza the student’s freshman year is $20. • The average inflation rate for pizza during the students time at UCD is expected to be 10%/year. Inflation Example • Over the next 5 years the student purchases the following: Year 0 1 Gift ($) 2000 2000 Pizza 20.00 Cost ($) 2 3 4 2000 2000 2000 22.00 24.20 26.62 29.28 No. of Pizzas 100.0 90.91 82.64 75.13 68.31 Const. $ 2000. 1818.2 1652.91502.61366.1 Fair Gift ($) 2000. 2200 2420 2662 2928 ENG 106 - D.C. Slaughter UC Davis Pizza example Actual $ constant $ $2,000 $2,000 0 1 2 $1,366 3 year 4 P = FA(1+i’)-N (1+f)-N 0 1 2 3 4 year P = F’0(1+i’)-N FA(1+i’)-N (1+f)-N = F’0(1+i’)-N Convert from constant $ to Actual $ Convert from Actual $ to constant $ FA= F’0(1+f)N F’0 = FA(1+f)-N i’: inflation free interest rate f: avearge inflation rate FA: Actual future cashflow F’0: constant future cashflow Pizza example Actual $ constant $ $2,000 $2,000 0 1 2 $1,366 3 4 year 0 1 2 3 4 year P = FA(1+i’)-N (1+f)-N P = FA(1+ i’+f+i’*f)-N if = i’+ f + i’*f We define if as the market interest rate per interest period, P = FA (1+ if)-N i’: inflation free interest rate f: avearge inflation rate FA: Actual future cashflow F’0: constant future cashflow Inflation • When using Constant Dollars you must use the inflationless interest rate, i’ P = F’0(P/F, i’,N) • When using Actual Dollars you must use the market interest rate, if P = FA(P/F, if, N) Example A father wants to save for his son’s college expenses. The son will enter the college 10 years from now. An annual amount of $40,000 in constant $ will be required for four years of college. The future general inflation rate is estimated to be 6% per year, and the market interest rate on the saving account will average 8% compounded annually. 1) What is the amount of the son’s college expense in terms of actual $? FA= F’0(1+f)N EOY Constant $ Actual $ 10 $40,000 $40,000(1+0.06)10 = $71,633 11 $40,000 $40,000(1+0.06)11 = $75,931 12 $40,000 $40,000(1+0.06)12= $80,487 13 $40,000 $40,000(1+0.06)13= $85,317 if =8% f=6% Constant $ Example A father wants to save for his son’s college expenses. The son will enter the college 10 years from now. An annual amount of $40,000 in constant $ will be required for four years of college. The future general inflation rate is estimated to be 6% per year, and the market interest rate on the saving account will average 8% compounded annually. 2) What is the equivalent single sum amount at the present time for these college expenses? Actual $ if =8% f=6% Constant $ A=$40,000 0 10 11 12 13 P P = F (1+if)n P = F (1+i’)n if =8% f=6% Actual $ P = $71,633 (P/F,8%,10) + $75,931 (P/F,8%,11) + $80,487 (P/F,8%,12) + $85,317 (P/F,8%,13) = $129,081 Constant $ i’= (if – f)/(1+f) =(0.08-0.06)/(1+0.06) = 1.887% (inflation free) P = $40,00 (P/A,1.887%,4) (P/F,1.887%,9) = $129,081 Example A father wants to save for his son’s college expenses. The son will enter the college 10 years from now. An annual amount of $40,000 in constant $ will be required for four years of college. The future general inflation rate is estimated to be 6% per year, and the market interest rate on the saving account will average 8% compounded annually. 3) What is the equal amount the father must save each year until his son goes to college? Actual $ A= $129,081 (A/P, 8%,9) = $20,663 actual $ A= $129,081 (A/P, 1.887%,9) = $15,729 constant $ $129,081 Constant $ $129,081 Engineering Economics ENG106 Lecture 7: Payback Period Lecture Outline • Inflation Recap • Typical economic assessment of an investment project • MARR • Payback period (Invalid Analysis) • Payback period (Discounted) Recap Inflation rate over a period of time (years) have compounding effect 𝑓 ҧ = average inflation rate 𝑓ҧ = 𝐶𝑁 𝐶𝑜 1 𝑁 1 𝑓ҧ = 260.474 (11𝑥12) 216.687 −1 =0.136% per month 1 𝑓ҧ = 258.811 11 218.576 −1 =1.57% per year −1 Inflation • There are two basic methods of accounting for inflation in engineering economic analyses. • Constant Dollar Approach(“Real” Purchasing Power Dollars) In the Constant Dollar system all cash flows are stated in dollars that have the purchasing power at a given point in time called the Base Year. • Actual Dollar Approach (Future Dollars, or actual green pieces of paper) In the Actual Dollar system all cash flows are stated in the amount of dollars that would actually be exchanged at that point in time Inflation Example • A new UCD student has an Aunt who sends a $2,000 gift each year to buy pizza. • The cost of a pizza the student’s freshman year is $20. • The average inflation rate for pizza during the students time at UCD is expected to be 10%/year. Inflation Example Over the next 5 years the student purchases the following: ҧ 𝑓 = 10% Year 0 1 Gift ($) 2000 2000 Pizza 20.00 Cost ($) 2 3 4 2000 2000 2000 22.00 24.20 26.62 29.28 No. of Pizzas 100.0 90.91 82.64 75.13 68.31 Const. $ 2000. 1818.2 1652.91502.61366.1 Fair Gift ($) 2000 2200 2420 2662 2928 Pizza example constant $ Actual $ $2,000 $1,818 $1,652 $1,502 $2,000 0 1 2 3 year 4 P = FA(1+i’)-N (1+f)-N 0 1 2 3 4 $1,366 year P = F’0(1+i’)-N FA(1+i’)-N (1+f)-N = F’0(1+i’)-N Convert from constant $ to Actual $ Convert from Actual $ to constant $ FA= F’0(1+f)N F’0 = FA(1+f)-N i’: inflation free interest rate f: avearge inflation rate FA: Actual future cashflow F’0: constant future cashflow Pizza example Actual $ Constant $ $2,000 $2,000 0 1 2 $1,366 3 4 year 0 1 P = FA(1+i’)-N (1+f)-N P = FA(1+ i’+f+i’*f)-N if = i’+ f + i’*f 2 3 4 year P = F’0(1+i’)-N P = F’0(P/F, i’,N) We define if as the market interest rate per interest period, P = FA (1+ if)-N i’: inflation free interest rate f: avearge inflation rate FA: Actual future cashflow F’0: constant future cashflow Summary • When using Constant Dollars you must use the inflationless interest rate, i’ P = F’0(P/F, i’,N) • When using Actual Dollars you must use the market interest rate, if P = FA(P/F, if, N) Typical economic assessment of an investment project 1) Identify costs and benefits • usually the most difficult portion 2) Identify interest & inflation rates • Use the Minimum Attractive Rate of Return (MARR), set by management, as the interest rate. 3) Identify project life 4) Resolve cash flows to equivalent basis NPW = PWbenefits – PWcosts NAW = AWbenefits – AWcosts NFW = FWbenefits – FWcosts 5) NPW/ NAW/FAW > 0 (Accept the investment) Minimum Attractive Rate of Return (MARR) Return: What you get back for the amount you invested Minimum rate of return on an investment that a decision maker is willing to accept given the associated risk and the opportunity cost of other forgone investments. Return is function of three components: • Risk free (real) return + • Inflation factor + • Risk premium Invalid Analysis Methods In this course, any method that results in the selection of an alternative not selected by the economic equivalence method is defined as: Non-valid Method Payback Period Method (Invalid method) The time required for accumulated benefits to offset accumulated costs 1. For the case where there is an initial investment and a uniform series of benefits the PP is: PP = | Initial Investment | Annual Benefit 2. Otherwise, the PP is determined by summing the investment and benefits to determine the time it takes to achieve a zero balance, IGNORING interest. Payback Period Example Compare the following two ME Alts. EOP Alt. A Alt. B 0 -1,000 -1,000 1 0 2 600 750 750 3 1,200 750 4 1,800 2,400 750 750 5 Payback Period Example Payback Period for Alt. A EOP Alt. A 0 -1,000 • The remaining $400 is recovered in the first 1/3 ($400/$1,200 = 1/3) of the third year. 1 0 2 600 • Thus the Payback Period for A is 2.33 years 3 1,200 4 1,800 5 2,400 • In the first 2 years, $600 of the $1,000 invested is recovered. These are ignored Payback Period Example Payback Period for Alt. B • In the first year, $750 of the $1,000 invested is recovered. • The remaining $250 is recovered in the first 1/3 ($250/$750 = 1/3) of the second year. • Thus the Payback Period for B is 1.33 years. or PP= $1000/$750 = 1.33 year EOP Alt. B 0 -1,000 1 750 2 750 3 750 4 750 5 750 (Initial invest +uniform series) These are ignored Payback Period Example PPA = 2.33 and PPB = 1.33 We would select Alt. B because our investment is paid back more quickly. • If we compare PP with NPW (10%) we find: NPWA = -$1,000 + $600(P/G, 10%, 5) = $3,117 NPWB = -$1,000 + $750(P/A, 10%, 5) = $1,843 So Alt. A is more profitable. Payback Period Method (Invalid method) There are two main disadvantages with the Payback Period (PP) method: 1. It does not account for the time value of money 2. It ignores benefits or costs after the payback period Payback Period Method • Payback period is often used to minimize risk and to get investment cash back as fast as possible • Payback period is a financial decision not a worthbased decision like NPW • Because the payback period decision does not always agree with decisions based upon NPW it is called Invalid method Payback Period (Discounted) Discounted Payback Period takes into account TVoM. Calculate the PP of Alt.A at an i =10% EOY Alt A. PW (10%) (P/F,10%,n) Project Balance 0 -$1000 -$1,000.00 -$1,000.00 1 0 $0.00 -$1,000.00 2 $600 $495.87 -$504.13 3 $1200 $901.58 $397.45 4 $1800 $1,229.42 $1,626.87 5 $2400 $1,490.21 $3,117.08 504.13 / 901.58 = 0.56 Discounted PP = 2.56 years Engineering Economics ENG106 Lecture 8: Comparison of Alternatives Lecture Outline •Comparison of Alternatives • Independent vs Mutually Exclusive • Equal Service Requirement •Analysis Methods • Invalid • Dollar-based • Dimensionless Types of alternatives • Independent • Accept or reject each project individually based on MARR (Min. attractive) evaluated using economic equivalence methods • Decision maker can choose any subset including the null set • Mutually Exclusive (ME) • The best alternative is selected and the others are automatically excluded • ME alternatives must be compared over an equal analysis period (fundamental principle) Analysis Methods •Non-valid methods • Payback Period (PP) •Dollar valued • Net Present worth (NPW) • Net Future worth (NFW) • Net Annual worth (NAW) •Dimensionless • Benefit/cost ratio (BC) • Internal Rate of Return (IRR) Present Worth Analysis • Net present worth (NPW) is defined as: NPW = PWbenefits - PWcosts • Objective is to maximize the NPW • Selection Criteria • Independent Alternatives • Select All with NPW > 0 • ME Alternatives • Select the Alternative with the greatest NPW Equal Service Requirement • A student is considering a beverage purchase. • The cost of a single serving of Pepsi is less than the cost of a 2 liter bottle of Coke. • Can the student make a fair comparison based upon their price alone? NO They do not provide equal service The 2 liter bottle will last 5 times longer than the single serving Equal Service Requirement • Equal service can be provided 2 ways •Least common multiple analysis period. • Set the analysis period equal to the least common multiple of the lives of the alternatives •Study period method • Set the analysis period equal to a known period of need or the shorter of the two useful lives and determine the salvage value of the longer lived alternative if it were sold at that time Equal Service Requirement • Using the Least common multiple analysis period • The student should compare the purchase of 5 cans of Pepsi to a 2 liter bottle of Coke • Using the Study Period Method • The student should determine the salvage value of selling the Coke remaining after one serving has been used. • Then compare the price of 1 can of Pepsi to the price of the 2 liter bottle of Coke minus the salvage value Equal Service Example • A pharmaceutical firm is examining two ME weighing systems • Alt. 1: Initial cost of $2000, a uniform annual benefit of $550 and a useful life of 5 years • Alt. 2: Initial cost of $3000, a uniform annual benefit of $800 and a useful life of 8 years •Assume constant dollars and MARR' = 9% • Do these alternatives provide equal service? No Equal Service Example • A pharmaceutical firm is examining two ME weighing systems • Alt. 1: Initial cost of $2000, a uniform annual benefit of $550 and a useful life of 5 years • Alt. 2: Initial cost of $3000, a uniform annual benefit of $800 and a useful life of 8 years •Assume constant dollars and MARR' = 9% • Use the Least Common Multiple (LCM) Analysis Period Method to determine the preferred alternative Two ME weighing systems (LCM) • Alt. 1: useful life = 5 years. • Alt. 2: useful life = 8 years. • What is the least common multiple of 5 and 8? 5 * 8 = 40 years Each Alternative must provide 40 years of service • Alt. 1: 8 cycles of 5 years each • Alt. 2: 5 cycles of 8 years each Two ME weighing systems (LCM) • Alt. 1: Initial cost of $2000, a uniform annual benefit of $550 and a useful life of 5 years (MARR’ = 9%) • Determine the NPW of Alt. 1 • To provide 40 years of service we must purchase 8 of Alt. 1. A=550 A=550 A=550 A=550 A=550 0 1 2000 5 6 10 11 2000 2000 15 31 35 36 2000 2000 40 Years Two ME weighing systems (LCM) • Alt. 1: Initial cost of $2000, a uniform annual benefit of $550 and a useful life of 5 years (MARR’ = 9%) • Determine the NPW of Alt. 1 • To provide 40 years of service we must purchase 8 of Alt. 1. • Simplifying the cash flow diagram A=550 0 5 10 15 30 35 1 2000 40 Years 2000 2000 2000 2000 Two ME weighing systems (LCM) • Alt. 1: Initial cost of $2000, a uniform annual benefit of $550 and a useful life of 5 years (MARR’ =9%) PW of Benefits = $550( P/A, 9%, 40) = $550(10.757) = $5916.35 Two ME weighing systems (LCM) • Alt. 1: Initial cost of $2000, a uniform annual benefit of $550 and a useful life of 5 years (MARR’ = 9%) PW of Costs ?? Method 1: Using effective interest rate ie = (1+0.09) 5 -1= 0.5386 = 53.86%/ 5 years PW of Costs = $2000 + $2000( P/A, 53.86%, 7) = $2000 + $2000(1.7657) = $5531.40 Two ME weighing systems (LCM) • Alt. 1: Initial cost of $2000, a uniform annual benefit of $550 and a useful life of 5 years (MARR’ = 9%) PW of Costs ?? Method 2:Transform costs to equivalent annual series A = $2000 (A/P, 9%, 5) = $2000(0.2571)=$514.20 • Repeat this process 7 times, once for each of the 8 systems purchased Two ME weighing systems (LCM) • Alt. 1: Initial cost of $2000, a uniform annual benefit of $550 and a useful life of 5 years (MARR’ = 9%) PW of Costs ?? Method 2:Transform costs to equivalent annual series Two ME weighing systems (LCM) • Alt. 1: Initial cost of $2000, a uniform annual benefit of $550 and a useful life of 5 years (MARR’ = 9%) PW of Costs ?? PW of Costs = $514.20( P/A, 9%, 40) = $514.20(10.757) = $5531.25 Two ME weighing systems (LCM) • Alt. 1: Initial cost of $2000, a uniform annual benefit of $550 and a useful life of 5 years (MARR’ = 9%) NPW of Alt1?? NPW = PW of Benefits - PW of Costs NPW = $5916.35 - $5531.25 = $385.10 • Is Alt. 1 profitable ? $385.10 > 0 = YES Two ME weighing systems (LCM) • Alt. 2: Initial cost of $3000, a uniform annual benefit of $800 and a useful life of 8 years (MARR’ = 9%) • Determine the NPW of Alt. 2 • To provide 40 years of service we must purchase 5 of Alt. 2. A=800 A=800 A=800 A=800 A=800 0 1 3000 8 9 16 17 24 25 32 33 3000 3000 3000 3000 40 Years Two ME weighing systems (LCM) • Alt. 2: Initial cost of $3000, a uniform annual benefit of $800 and a useful life of 8 years (MARR’ = 9%) • Determine the NPW of Alt. 2 • To provide 40 years of service we must purchase 5 of Alt. 2. • Simplifying the cash flow diagram 0 8 A=800 16 24 32 1 3000 40 Years 3000 3000 3000 3000 Two ME weighing systems (LCM) • Alt. 2: Initial cost of $3000, a uniform annual benefit of $800 and a useful life of 8 years (MARR’ = 9%) PW of benefits ?? PW of Benefits = $800( P/A, 9%, 40) = $800(10.757) = $8605.60 Two ME weighing systems (LCM) • Alt. 2: Initial cost of $3000, a uniform annual benefit of $800 and a useful life of 8 years (MARR’ = 9%) PW of Costs ?? Method 1: Using effective interest rate ie = (1+0.09) 8 -1= 0.99256 = 99.256%/ 8 years PW of Costs = $3000 + $3000( P/A, 99.256%, 4) = $3000 + $3000(0.94358) = $5830.74 Two ME weighing systems (LCM) • Alt. 2: Initial cost of $3000, a uniform annual benefit of $800 and a useful life of 8 years (MARR’ = 9%) PW of Costs ?? Method 2:Transform costs to equivalent annual series A = $3000 (A/P, 9%, 8) = $3000(0.1807)=$542.10 • Repeat this process 4 times, once for each of the 5 systems purchased. Two ME weighing systems (LCM) • Alt. 2: Initial cost of $3000, a uniform annual benefit of $800 and a useful life of 8 years (MARR’ = 9%) PW of Costs ?? Method 2:Transform costs to equivalent annual series Two ME weighing systems • Alt. 2: Initial cost of $3000, a uniform annual benefit of $800 and a useful life of 8 years (MARR’ = 9%) PW of Costs ?? PW of Costs = $542.10( P/A, 9%, 40) = $542.10(10.757) = $5831.37 Two ME weighing systems (LCM) • Alt. 2: Initial cost of $3000, a uniform annual benefit of $800 and a useful life of 8 years (MARR’ = 9%) NPW = PW of Benefits - PW of Costs NPW = $8605.60 - $5831.37 = $2774.23 Is Alt. 2 profitable ? $2774.23 > 0 = YES Conclusion (LCM) • Assume constant dollars and MARR' = 9%. • Alt. 1: Initial cost of $2000, a uniform annual benefit of $550 and a useful life of 5 years. • Alt. 2: Initial cost of $3000, a uniform annual benefit of $800 and a useful life of 8 years. Which alternative is preferred? NPW2 = $2774.23 > $385.10 = NPW1 Thus Alternative 2 is preferred Two ME weighing systems • Assume constant dollars and MARR' = 9%. • Alt. 1: Initial cost of $2000, a uniform annual benefit of $550 and a useful life of 5 years. • Alt. 2: Initial cost of $3000, a uniform annual benefit of $800 and a useful life of 8 years. Using the Study Period (SP) Method, determine the preferred alternative  Set the study period to 5 years and assume that weighing system 2 would have a salvage value of $700 if it were sold at the end of 5 years Two ME weighing systems (SP) • Determine the NPW of Alt. 1 • Alt. 1: Initial cost of $2000, a uniform annual benefit of $550 and a useful life of 5 years. PW of Benefits = $550( P/A, 9%, 5) = $550(3.890) = $2139.50 PW of costs = $2000 NPW = PW of Benefits - PW of Costs NPW = $2139.50 - $2000 = $139.50 Two ME weighing systems (SP) • Determine the NPW of Alt. 2 • Alt. 2: Initial cost of $3000, a uniform annual benefit of $800 and a useful life of 8 years. • For a 5 year study period we determine the salvage value of a 5 year old system: $700 PW of Benefits = $800(P/A, 9%, 5) + $700( P/F, 9%, 5) = $800(3.890) + $700(0.6499) = $3566.93 PW of costs = $3000 NPW = PW of Benefits - PW of Costs NPW = $3566.93 - $3000 = $566.93 Conclusion (SP) • Assume constant dollars and MARR' = 9%. • Alt. 1: Initial cost of $2000, a uniform annual benefit of $550 and a useful life of 5 years. • Alt. 2: Initial cost of $3000, a uniform annual benefit of $800 and a useful life of 8 years (SV5 = $700) Which alternative is preferred? NPW2 = $566.93 > $139.50 = NPW1 Thus Alternative 2 is preferred Engineering Economics ENG106 Lecture 8: Net Annual Worth & Project Balance Types of alternatives • Independent • Accept or reject each project individually based on MARR (Min. attractive) evaluated using economic equivalence methods • Decision maker can choose any subset including the null set • Mutually Exclusive (ME) • The best alternative is selected and the others are automatically excluded • ME alternatives must be compared over an equal analysis period (fundamental principle) Analysis Methods •Non-valid methods • Payback Period (PP) •Dollar valued • Net Present worth (NPW) • Net Future worth (NFW) • Net Annual worth (NAW) •Dimensionless • Benefit/cost ratio (BC) • Internal Rate of Return (IRR) Annual Worth Analysis • Net Annual Worth (NAW) is defined as: •NAW = AW of benefits - AW of costs • Objective is to maximize the NAW • Selection Criteria • Independent Alternatives • Select All with NAW > 0 • ME Alternatives • Select the Alternative with the greatest NAW Annual Worth Analysis One advantage of annual worth analysis is that it is better suited to comparing alternatives that have different useful lives if Constant Dollar Analysis is used Re-evaluate the weighing system problem using NAW Equal Service Example • A pharmaceutical firm is examining two ME weighing systems • Alt. 1: Initial cost of $2,000, a uniform annual benefit of $550 and a useful life of 5 years • Alt. 2: Initial cost of $3,000, a uniform annual benefit of $800 and a useful life of 8 years •Assume constant dollars and MARR' = 9% • Use the Least Common Multiple (LCM) Analysis Period Method to determine the preferred alternative Two ME weighing systems (LCM) • Alt. 1: useful life = 5 years. • Alt. 2: useful life = 8 years. • What is the least common multiple of 5 and 8? 5 * 8 = 40 years Each Alternative must provide 40 years of service • Alt. 1: 8 cycles of 5 years each • Alt. 2: 5 cycles of 8 years each Two ME weighing systems (LCM) • Alt. 1: Initial cost of $2000, a uniform annual benefit of $550 and a useful life of 5 years (MARR’ = 9%) • Determine the NAW of Alt. 1 • To provide 40 years of service we must purchase 8 of Alt. 1. A=550 A=550 A=550 A=550 A=550 0 1 5 6 10 11 2000 2000 2000 15 31 35 36 2000 2000 40 Years Two ME weighing systems (LCM) • Alt. 1: Initial cost of $2000, a uniform annual benefit of $550 and a useful life of 5 years (MARR’ = 9%) • Determine the NAW of Alt. 1 • To provide 40 years of service we must purchase 8 of Alt. 1. • Simplifying the cash flow diagram A=550 0 5 10 15 30 35 1 2000 40 Years 2000 2000 2000 2000 Two ME weighing systems (LCM) • Alt. 1: Initial cost of $2000, a uniform annual benefit of $550 and a useful life of 5 years (MARR’ = 9%) AW of Costs ?? Using effective interest rate ie = (1+0.09) 5 -1= 0.5386 = 53.86%/ 5 years AW of Costs =($2000 + $2000( P/A, 53.86%, 7) ) (A/P,9%,40) = ($2000 + $2000(1.7657)) (0.093) = $514.20 Two ME weighing systems • Determine the NAW for Alt. 1 AW of Benefits = $550.00 AW of Costs = $514.20 A=550 0 A=514.20 NAW = $550 - $514.20 = $35.80 40 Years Two ME weighing systems (LCM) • Alt. 1: Initial cost of $2000, a uniform annual benefit of $550 and a useful life of 5 years (MARR’ = 9%) NAW of one cycle AW of costs = $2000 (A/P, 9%, 5) = $2000(0.2571)=$514.20 NAW = $550 - $514.20 = $35.80 Two ME weighing systems The NAW for all eight cycles of an alternative And the NAW for one cycle of an alternative Are the same, NAW = $35.80 When using CONSTANT dollars Two ME weighing systems • Determine the NAW of Alt. 2 • Use the one-cycle method • Calculate the AW of both costs & benefits  AW of benefits = $800  AW of costs = $3000(A/P, 9%, 8) = $542.10 A=800 A=800 0 1 8 Years 8 1 A=542.10 Years  NAW = $800 - $542.10 = $257.90 3000 Two ME weighing systems • Using CONSTANT dollars • Comparing one cycle of each alternative is equivalent to multiple cycles of the alternatives. A=550 A=800 1 5 A=514.20 Years 8 1 A=542.10 Years • Compare the NAWs to determine which alternative is best.  NAW2 = 257.90 > 35.80 = NAW1  Select Alternative 2. Special Features of NAW • NAW has a computational advantage over NPW or NFW When comparing mutually exclusive alternatives with unequal service lives • However, it only works with CONSTANT $ • The results are IDENTICAL to NAW for the complete least common multiple lives analysis. Infinite Life Example • Determine the Annual Cost required to construct and maintain a railroad tunnel. • Initial Cost = $1,000,000 • Annual Maintenance = $10,000 • Retimbering Cost = $200,000/10 years • Assume Constant $ and MARR’ = 5% 0 1 10 20 30 40 A=10K Years $200k $1M $200k $200k $200k Infinite Life Example • The annual maintenance is already an annuity. • Need to convert the initial cost and the retimbering costs into annuities. 0 1 10 20 30 40 A=10000 $200k $1M Years $200k $200k $200k Infinite Life Example • For the initial $1,000,000 construction cost • Remember A = P(A/P, i, ) = Pi • A = $1,000,000 * 0.05 = $50,000 Infinite Life Example • Calculate an equivalent uniform annual series cost for the $200k /10 year cost. 0 1 10 10 Years Years A=? 200k A = $200k (A/F, 5%, 10) = $200k(0.0795)=$15.9k • Repeat this process every 10 years. Infinite Life Example • Determine the NAW of the tunnel AWcosts =$1M(A/P, 5%, ) + $200K(A/F, 5%, 10) + AWcosts = $50k + $15.9k + $10k = $75,900 $10k Special Features of NAW • When comparing mutually exclusive alternatives with • unequal service lives • a long-term need for this service • NAW has a computational advantage over NPW or NFW if the one cycle method is used • However, it only works with CONSTANT $ • The results are IDENTICAL to NAW for the complete least common multiple lives analysis. Dollar-based Methods • For Independent Alternatives • Choose all Alternatives with Net Present Future Annual Worth > 0 • For Mutually Exclusive Alternatives  Choose the Alternative with the Greatest Net Present Future Annual Worth Project Balance • The Project Balance provides insight into the Current level of a corporation’s Investment in each Design Alternative. Project Balance $1,268,000 Project Balance at time t: 𝑃𝐵𝑡 𝑖% = 𝑁𝐹𝑊𝑡 $908,000 i= 15% $681,000 𝐹 𝐹 $454,000 𝑃𝐵𝑡=3 15% = −$1,8𝑀(𝑃 , 15%, 3) + $454K(𝑃 , 15%, 2) 1 +$681K(𝑃 , 15%, 1)+ $908𝐾 = -$446,010 𝐹 0 2 3 4 5 6 7 FW = $4,113,910 $4,000,000 $3,000,000 $1,800,000 $2,845,911 $2,474,705 $2,000,000 Year Cashflow ($) Interest ($) Project balance ($) 0 -1,800,000 - -1,800,000 1 454,000 -270,000 -1,616,000 2 681,000 -242,400 -1,177,400 3 908,000 -176,610 -446,010 4 908,000 -66,902 395,089 5 908,000 59,263 1,362,352 6 908,000 204,353 2,474,705 7 1,268,000 371,206 4,113,910 $1,362,352 $1,000,000 $395,089 0 1 2 3 4 5 6 7 -$446,010 $-1,000,000 -$1,800,000 -$1,616,000 -$1,117,400 Payback period (discounted) $-2,000,000 Discounted Payback period (PP): PP = NFW = 0
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