Proposed capital projects can be evaluated in several ways.

2. The objective of Chapter 5 is to discuss and critique contemporary methods for determining project profitability.

3. Proposed capital projects can be evaluated in several ways. • Present worth (PW) • Future worth (FW) • Annual worth (AW) • Internal rate of return (IRR) • External rate of return (ERR) • Payback period (generally not appropriate as a primary decision rule)

4. To be attractive, a capital project must provide a return that exceeds a minimum level established by the organization. This minimum level is reflected in a firm’s Minimum Attractive Rate of Return (MARR).

5. Many elements contribute to determining the MARR. • Amount, source, and cost of money available • Number and purpose of good projects available • Perceived risk of investment opportunities • Type of organization

6. The most-used method is the present worth method. The present worth (PW) is found by discounting all cash inflows and outflows to the present time at an interest rate that is generally the MARR. A positive PW for an investment project means that the project is acceptable (it satisfies the MARR).

7. Formulating Alternatives Two types of economic proposals • Mutually Exclusive (ME) Alternatives: Onlyone can be selected; • Compete against each other • Independent Projects: More than one can be selected; • Compete only against DN Do Nothing (DN) – An ME alternative or independent project to maintain the current approach; no new costs, revenues or savings

8. Formulating Alternatives Two types of cash flow estimates Revenue: Alternatives include estimates of costs (cash outflows) and revenues (cash inflows) Cost: Alternatives include only costs; revenues and savings assumed equal for all alternatives; also called service alternatives

9. PW Analysis of Alternatives Convert all cash flows to PW using MARR Precede costs by minus sign; receipts by plussign EVALUATION For one project, if PW > 0, it is justified For mutually exclusive alternatives, select one with numericallylargest PW For independent projects, select all with PW > 0

10. For the alternatives shown below, which should be selected selected selected if they are (a) mutually exclusive; (b) independent? Project ID Present Worth A $30,000 B $12,500 C $-4,000 D $ 2,000 Selection of Alternatives by PW (a) Select numerically largest PW; alternative A Solution: (b) Select all with PW > 0; projects A, B & D

11. Example: PW Evaluation of Equal-Life ME Alts. Alternative X has a first cost of $20,000, an operating cost of $9,000 per year, and a $5,000 salvage value after 5 years. Alternative Y will cost $35,000 with an operating cost of $4,000 per year and a salvage value of $7,000 after 5 years. At an MARR of 12% per year, which should be selected? Solution: Find PW at MARR and select numerically larger PW value PWX = -20,000 - 9000(P/A,12%,5) + 5000(P/F,12%,5) = -$49,606 PWY = -35,000 - 4000(P/A,12%,5) + 7000(P/F,12%,5) = -$45,447 Select alternative Y

12. Present Worth Example Consider a project that has an initial investment of $50,000 and that returns $18,000 per year for the next four years. If the MARR is 12%, is this a good investment? PW = -50,000 + 18,000 (P/A, 12%, 4) PW = -50,000 + 18,000 (3.0373) PW = $4,671.40  This is a good investment!

13. PW of Different-Life Alternatives Mustcompare alternatives for equal service (i.e., alternatives must end at the same time) Two ways to compare equal service: Least common multiple (LCM) of lives Specified study period (The LCM procedure is used unless otherwise specified)

14. Assumptions of LCM approach • Service provided is needed over the LCM or more years • Selected alternative can be repeated over each life cycle of LCM in exactly the same manner • Cash flow estimates are the same for each life cycle (i.e., change in exact accord with the inflation or deflation rate)

15. Example: Different-Life Alternatives Compare the machines below using present worth analysis at i = 10% per year Machine B Machine A First cost, $ 30,000 20,000 Annual cost, $/year 9000 7000 Salvage value, $ 4000 6000 Life, years 3 6 Solution: LCM = 6 years; repurchase A after 3 years PWA = -20,000 – 9000(P/A,10%,6) – 16,000(P/F,10%,3) + 4000(P/F,10%,6) = $-68,961 20,000 – 4,000 in year 3 PWB = -30,000 – 7000(P/A,10%,6) + 6000(P/F,10%,6) = $-57,100 Select alternative B

16. PW Evaluation Using a Study Period • Once a study period is specified, all cash flows after this time are ignored • Salvage value is the estimated market value at the end of study period Short study periods are often defined by management when business goals are short-term Study periods are commonly used in equipment replacement analysis

17. Example: Study Period PW Evaluation Compare the alternatives below using present worth analysis at i = 10% per year and a 3-year study period Machine B Machine A First cost, $ -30,000 -20,000 Annual cost, $/year -9,000 -7,000 Salvage/market value, $ 4,000 6,000 (after 6 years) 10,000 (after 3 years) Life, years 3 6 Study period = 3 years; disregard all estimates after 3 years Solution: PWA = -20,000 – 9000(P/A,10%,3) + 4000(P/F,10%,3) = $-39,376 PWB = -30,000 – 7000(P/A,10%,3) + 10,000(P/F,10%,3) = $-39,895 Marginally, select A; different selection than for LCM = 6 years

18. Capitalized Cost (CC) Analysis CC refers to the present worth of a project with a very long life, that is, PW as n becomes infinite Basic equation is: CC = P = A/i “A” essentially represents the interest on a perpetual investment For example, in order to be able to withdraw $50,000 per year forever at i = 10% per year, the amount of capital required is 50,000/0.10 = $500,000 For finite life alternatives, convert all cash flows into an A value over one life cycle and then divide by i

19. Example: Capitalized Cost Compare the machines shown below on the basis of their capitalized cost. Use i = 10% per year Machine 1 Machine 2 First cost,$ -20,000 -100,000 Annual cost,$/year -9000 -7000 Salvage value, $ ----- 4000 Life, years ∞ 3 Solution: Convert machine 1 cash flows into A and then divide by i A1 = -20,000(A/P,10%,3) – 9000 + 4000(A/F,10%,3) = $-15,834 CC1 = -15,834 / 0.10 = $-158,340 CC2 = -100,000 – 7000/ 0.10 = $-170,000 Select machine 1

20. Capitalized Cost (CC) • Capitalized Cost- the present worth of a project that lasts forever. • Examples are (Government Projects that have perpetual life) : • Roads, • Dams, • Bridges, • Charitable organizations and endowments • Infinite analysis period

21. Capitalized Cost (CC) Ex : Assume you are asked to maintain a cemetery site forever; if the interest rate = 4% and $50/year is required to maintain the site? P0 = $50[1/0.04] P0 = $50[25] = $1250.00

22. Capitalized Cost (CC): Endowments • Assume a wealthy donor wants to endow a chair in an engineering department. • The fund should supply the department with $200,000 per year for a deserving faculty member. • How much will the donor have to come up with to fund this chair if the interest rate = 8% / yr. • P = $200,000/0.08 = $2,500,000 • If $2,500,000 is invested at 8% then the interest per year = $200,000

23. Procedure for calculating the CC for an infinite sequence of cash flows 1. Draw a cash flow diagram showing all nonrecurring (one-time) cash flows and at least two cycles of all recurring (periodic) cash flows. Drawing the cash flow diagram (step 1) is more important in CC calculations than elsewhere, because it helps separate nonrecurring and recurring amounts. 2. Find the present worth of all nonrecurring amounts. This is the CC value. 3. Find the equivalent uniform annual worth (A value) through one life cycle of all recurring amounts. Add this to all other uniform amounts occurring in years 1 through infinity and the result is the total equivalent uniform annual worth (AW). 4. Divide the AW obtained in step 3 by the interest rate i to obtain a CC value. This is an application of Equation [5.2]. 5. Add the CC values obtained in steps 2 and 4.

24. Examples of capitalized cost analyses • Example -- Finds the CC and Annual Worth values of a county government software system expected to be used for the indefinite future • Example -- Compares the CC values for 2 indefinite-life bridge designs – suspension and truss

25. The property appraisal district for Marin County has just installed new software to track residential market values for property tax computations. The manager wants to know the total equivalent cost of all future costs incurred when the three county judges agreed to purchase the software system. If the new system will be used for the indefinite future, find the equivalent value (a) now and (b) for each year hereafter. The system has an installed cost of $150,000 and an additional cost of $50,000 after 10 years. The annual software maintenance contract cost is $5000 for the first 4 years and $8000 thereafter. In addition, there is expected to be a recurring major upgrade cost of $15,000 every 13 years. Assume that i = 5% per year for county funds. • Find the PW at t=0.

26. Using the 5-step procedure: • Draw a cash flow diagram for two cycles. 2. Find the present worth of the nonrecurring costs of $150,000 now and $50,000 in year 10 at i = 5%. Label this CC1.

27. Convert the recurring cost of $15,000 every 13 years into an annual worth A1 for the first 13 years. A1 = -15,000 (A/F, 5%, 13) = -$847 The same value, A1 = -$847, applies to the other 13 year periods as well. • The capitalized cost for the two annual maintenance cost series may be determined in either of two ways: • (1) consider a series of $5000 from now to infinity and find the present worth of • - $8000 - (-$5000) = - $3000 from year 5 on; or • (2) find the CC of $5000 for 4 years and the present worth of $8000 from year 5 to infinity. • Using the first method, the annual cost (A2) is $5000 forever. The capitalized cost CC2 of -$3000 from year 5 to infinity is found using Equation times the P/F factor. The two annual cost series are converted into a capitalized cost CC3.

28. 5. The total capitalized cost CCT is obtained by adding the three CC values. (b) Equation determines the A value forever. Correctly interpreted, this means Marin County officials have committed the equivalent of $17,350 forever to operate and maintain the property appraisal software. The CC2 value is calculated using n = 4 in the P/F factor because the present worth of the annual $3000 cost is located in year 4, since P is always one period ahead of the first A.

29. Summary of Important Points PW method converts all cash flows to present value at MARR Alternatives can be mutuallyexclusive or independent Cash flow estimates can be for revenue or cost alternatives PW comparison must always be made for equal service Equal service is achieved by using LCM or study period Capitalized cost is PW of project with infinite life; CC = P = A/i

30. Capitalized worth is a special variation of present worth. • Capitalized worth is the present worth of all revenues or expenses over an infinite length of time. • If only expenses are considered this is sometimes referred to as capitalized cost. • The capitalized worth method is especially useful in problems involving endowments and public projects with indefinite lives.

31. The application of CW concepts. The CW of a series of end-of-period uniform payments A, with interest at i% per period, is A(P/A, i%, N). As N becomes very large (if the A are perpetual payments), the (P/A) term approaches 1/i. So, CW = A(1/i).

32. Engineering Economy Chapter 5: Evaluating a Single Project

33. Future Worth (FW) method is an alternative to the PW method. • Looking at FW is appropriate since the primary objective is to maximize the future wealth of owners of the firm. • FW is based on the equivalent worth of all cash inflows and outflows at the end of the study period at an interest rate that is generally the MARR. • Decisions made using FW and PW will be the same.

34. Future Worth Analysis FW exactly like PW analysis, except calculate FW Must compare alternatives for equal service (i.e. alternatives must end at the same time) Two ways to compare equal service: Least common multiple (LCM) of lives Specified study period (The LCM procedure is used unless otherwise specified)

35. FW of Different-Life Alternatives Compare the machines below using future worth analysis at i = 10% per year Machine B Machine A First cost, $ -30,000 -20,000 Annual cost, $/year -9000 -7000 Salvage value, $ 4000 6000 Life, years 3 6 Solution: LCM = 6 years; repurchase A after 3 years FWA = -20,000(F/P,10%,6) – 9000(F/A,10%,6) – 16,000(F/P,10%,3) + 4000 = $-122,168 FWB = -30,000(F/P,10%.6) – 7000(F/A,10%,6) + 6000 = $-101,157 Select B (Note: PW and FW methods will always result in same selection)

36. Future worth example. A $45,000 investment in a new conveyor system is projected to improve throughput and increasing revenue by $14,000 per year for five years. The conveyor will have an estimated market value of $4,000 at the end of five years. Using FW and a MARR of 12%, is this a good investment? FW = -$45,000(F/P, 12%, 5)+$14,000(F/A, 12%, 5)+$4,000 FW = -$45,000(1.7623)+$14,000(6.3528)+$4,000 FW = $13,635.70  This is a good investment!

37. Annual Worth (AW) is another way to assess projects. • Annual worth is an equal periodic series of dollar amounts that is equivalent to the cash inflows and outflows, at an interest rate that is generally the MARR. • The AW of a project is annual equivalent revenue or savings minus annual equivalent expenses, less its annual capital recovery (CR) amount.

38. Capital recovery reflects the capital cost of the asset. • CR is the annual equivalent cost of the capital invested. • The CR covers the following items. • Loss in value of the asset. • Interest on invested capital (at the MARR). • The CR distributes the initial cost (I) and the salvage value (S) across the life of the asset.

39. A project requires an initial investment of $45,000, has a salvage value of $12,000 after six years, incurs annual expenses of $6,000, and provides an annual revenue of $18,000. Using a MARR of 10%, determine the AW of this project. Since the AW is positive, it’s a good investment.