IE463 Chapter 4. Objective: COMPARING INVESTMENT AND COST ALTERNATIVES

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1 IE463 Chapter 4 COMPARING INVESTMENT AND COST ALTERNATIVES Objective: To learn how to properly apply the profitability measures described in Chapter 3 to select the best alternative out of a set of mutually exclusive alternatives (MEA). CHAPTER 4 2 1

2 Mutually Exclusive The selection of one alternative excludes the consideration of any other alternative. Example: suppose you are shopping for a used automobile. You consider several cars, but will only buy one from a mutually exclusive set of choices. CHAPTER 4 3 Basic Rule: Spend the least amount of capital possible unless the extra capital can be justified by the extra savings or benefits. In other words, any increment of capital spent (above the minimum) must be able to pay its own way. CHAPTER 4 4 2

3 Investment Alternatives Each alternative has an initial investment producing positive cash flows resulting from increased revenues, reduced costs, or both "Do nothing" (DN) is usually an implicit investment alternative If positive cash flows) > negative cash flows) then IRR 0 If EW(MARR) 0, investment is profitable if EW(MARR) 0, do nothing (DN) is better CHAPTER 4 5 Cost Alternatives Each alternative has all negative cash flows except for the salvage value (if applicable) These alternatives represent must do situations, and DN is not an option IRR not defined for cost alternatives. Can you explain why? IRR can be used for comparing cost alternatives. Can you explain why? CHAPTER 4 6 3

4 The study period must be appropriate for the decision being made Study Period: The time interval over which service is needed to fulfill a specified function. Useful Life: The period over time during which an asset is kept in productive operation. Case 1: Study period Case 2: Study period Useful life Useful life Fundamental Principle: Compare MEAs over the same period of time. CHAPTER 4 7 Comparing MEAs using EW Methods-Example A Four mutually exclusive alternatives are being evaluated, and their costs and revenues are given in the following table. If the MARR is 12% per year, and the analysis period is 10 years, use the PW method to determine which alternatives are economically acceptable and which one should be selected. CHAPTER 4 8 4

5 Costs and revenues Investment cost (I) Net annual receipts (A) Salvage value (S) MEAs I II III IV $100,000 $152,000 $184,000 $220,000 15,200 31,200 35,900 41,500 10, ,000 20,000 Useful life CHAPTER 4 9 Comparing MEAs using PW They are investment alternatives Study Period = Useful Life PW(12%) = I + A(P/A, 12%, 10) + S(P/F, 12%, 10) PW DN (12%) = 0 CHAPTER

6 Comparing MEAs using PW Costs & Revenues Investment cost (I) Net annual receipts (A) Salvage value (S) MEA I $100,000 15,200 10,000 PW I (12%) = 100, ,200(P/A, 12%, 10) + 10,000(P/F, 12%, 10) PW I (12%) = 10,897 Useful life 10 CHAPTER 4 11 Comparing MEAs using PW Costs & Revenues Investment cost (I) Net annual receipts (A) Salvage value (S) MEA II $152,000 31,200 0 PW II (12%) = 152, ,200(P/A, 12%, 10) PW II (12%) = 28,241 Useful life 10 CHAPTER

7 Comparing MEAs using PW Costs & Revenues Investment cost (I) Net annual receipts (A) Salvage value (S) MEA III $184,000 35,900 15,000 PW III (12%) = 184, ,900(P/A, 12%, 10) + 15,000(P/F,12%,10) PW III (12%) = 23,672 Useful life 10 CHAPTER 4 13 Comparing MEAs using PW Costs & Revenues Investment cost (I) Net annual receipts (A) Salvage value (S) MEA IV $220,000 41,500 20,000 PW IV (12%) = 220, ,500(P/A, 12%, 10) + 20,000(P/F,12%,10) PW IV (12%) = 20,923 Useful life 10 CHAPTER

8 Summary of comparison PW DN (12%) = 0 PW I (12%) = 10,897 PW II (12%) = +28,241 PW III (12%) = +23,672 PW IV (12%) = +20,923 Select Alternative II to maximize PW. CHAPTER 4 15 Comparing MEAs using EW Methods-Example B The net cash flows are shown in the following table for three preliminary design alternatives. The MARR = 12% per year, and the study period is seven years. Which preliminary design is economically preferred based on the AW method. CHAPTER

9 Costs of MEAs A B C Initial cost (I) $85,600 $63,200 $71,800 Annual expenses, years 1 7 (E) 7,400 12,100 10,050 CHAPTER 4 17 Comparing MEAs using AW They are cost alternatives Study Period = Useful Life = 7 years AW = I(A/P, 12%, 7) + E DN is not an option CHAPTER

10 Comparing MEAs using AW Costs Initial cost (I) $85,600 Annual expenses, years 1 7 (E) 7,400 AW A = $85,600(A/P, 12%, 7) $7,400 = 26,155 A CHAPTER 4 19 Comparing MEAs using AW Costs Initial cost (I) $63,200 Annual expenses, years 1 7 (E) 12,100 AW B = $63,200(A/P, 12%, 7) $12,100 = 25,947 B CHAPTER

11 Comparing MEAs using AW Costs Initial cost (I) $71,800 Annual expenses, years 1 7 (E) 10,050 AW C = $71,800(A/P, 12%, 7) $10,050 = 25,781 C CHAPTER 4 21 Summary of comparison AW A = 26,155 AW B = 25,947 AW C = 25,781 Select alternative C to minimize AW of costs. CHAPTER

12 Comparing MEAs using IRR Method-Example A Consider 2 alternatives: Investment $100 $10,000 Lump-Sum Receipt Next Year $1,000 $15,000 IRR 900% 50% If comparable risk is involved and MARR = 20%, would you rather have A or B? A B CHAPTER 4 23 Comparing MEAs using IRR Method-Example A Investment $100 $10,000 Lump-Sum Receipt Next Year $1,000 $15,000 IRR 900% 50% PW A = (P/F,20%,1) = $733 PW B = 10, ,000(P/F,20%,1) = $2,500 A B CHAPTER

13 Comparing MEAs using IRR Method-Example A Investment $100 $10,000 Lump-Sum Receipt Next Year $1,000 $15,000 IRR 900% 50% Never simply maximize the IRR. Never compare the IRR to anything except the MARR. A B CHAPTER 4 25 Comparing MEAs using IRR Method-Example A Investment $100 $10,000 Lump-Sum Receipt Next Year A B $1,000 $15,000 IRR 900% 50% B A ( ) $9,900 $14, % IRR A B PW A B =0= 9, ,000(P/F, i'%, 1) 9,900 / 14,000 = (P/F, i'%, 1) i' = 41.4% (MARR = 20%) CHAPTER

14 Comparing MEAs using IRR Method-Example B A small investment project involves two alternatives, A and B. The cash flow for each alternative are stated here. Alternative Capital investment - $60,000 - $73,000 Annual revenue less expenses 22,000 26,225 A B The useful life of each alternative and the study period is four years. Also, assume that the MARR = 10% per year. Use the IRR method and decide on which one will be selected. CHAPTER 4 27 Ranking Inconsistency with the IRR Method IRR A : 0 = -$60,000 + $22,000(P/A, i A '%, 4) i A ' = 17.3% IRR B : 0 = -$73,000 + $26,225(P/A, i B '%, 4) i B ' = 16.3% NEVER simply select the MEA that MAXIMIZES the IRR. CHAPTER

15 We don't maximize rate of return. Look at the increment Alternative Difference A B (B A) Capital investment - $60,000 - $73,000 - $13,000 Annual revenue less expenses 22,000 26,225 4,225 Useful life 4 years 4 years 4 years MARR 10% 10% 10% IRR A B : 0 = -$13,000 + $4,225(P/A, i A B ', 4) i A B ' = 11.4% The rate of return on the increment, 11.4%, >MARR. It is worth the additional investment to select Alternative B CHAPTER 4 29 Never compare IRR to anything except MARR so as to be consistent with EW methods CHAPTER

16 Comparing MEAs using IRR Method-Example C In the design of a new facility, the mutually exclusive alternatives in the following table are under consideration. Assume that the MARR is 15% per year and the analysis period is 10 years. Use the incremental IRR procedure to choose the best alternative. CHAPTER 4 31 Costs and revenues Capital investment $28,000 $16,000 $23,500 Net cash flow /year 5,500 3,300 4,800 Salvage value 1, Useful life 10 yrs 10 yrs 10 yrs CHAPTER

17 The incremental IRR procedure to choose the best alternative: Step1: Rank order alternatives from low capital investment to high capital investment Step2: Compare MEAs in order Step3: Determine the winner from comparison Step4: Continue steps 2 and 3 until all alternatives are considered Step5: The last winner is selected from among all CHAPTER 4 33 Step 1: Rank order alternatives from low capital investment to high capital investment Alternative Capital Investment Rank DN , , ,500 3 DN CHAPTER

18 Compare DN 2: (2 - DN) cash flows Investment 16,000 0 = $16,000 Annual Receipts 3,300 0 = 3,300 Salvage Value 0 0 = 0 Compute IRR DN 2 (i.e., IRR 2 ) PW( i') = 0 = -$16,000 + $3,300(P/A, i'%, 10) i' DN % Since i' > MARR, keep alt. 2 as current best alternative. Drop DN from further consideration CHAPTER 4 35 Compare 2 3: (3-2) cash flows Investment -23,500 (-16,000) = $7,500 Annual Receipts 4,800 3,300 = 1,500 Salvage Value = 500 Compute IRR 2 3 PW( i') = 0 = -$7,500 + $1,500(P/A, i'%, 10) + $500(P/F, i'%, 10) i' % Since i' > MARR, keep alt. 3 as current best alternative. Drop 2 from further consideration CHAPTER

19 Compare 3 1: (1-3) cash flows Investment -28,000 (-23,500) = $4,500 Annual Receipts 5,500 4,800 = 700 Salvage Value 1, = 1,000 Compute IRR 3 1 PW( i') = 0 = -$4,500 + $700(P/A, i'%, 10) + $1,000(P/F, i'%, 10) i' % Since i' < MARR, keep alt. 3 as the best of all CHAPTER 4 37 Comparing Cost Alternatives with IRR ( ) method Study Period =Useful Life =7 years MARR =12% A B C Investment cost $85,600 $63,200 $71,800 Annual costs 7,400 12,100 10,050 CHAPTER

20 Compare B C: B C (C B) Investment cost $63,200 $71,800 8,600 Annual costs 12,100 10,050 2,050 AW( i') = 0 = -$8,600(A/P, i'%, 7) + $2,050 i' B C 14.7% > MARR =12% Keep C, Reject B Next comparison: C A CHAPTER 4 39 Compare C A: C A (A C) Investment cost $71,800 $85,600 13,800 Annual costs 10,050 7,400 2,650 AW( i') = 0 = -13,800(A/P, i'%, 7) + 2,650 i' C A 8% < MARR =12% Keep C, Reject A C is the least cost of all CHAPTER

21 Study Period Useful Life Up until now, study periods and useful lives have been the same length. The study period is frequently taken to be a common multiple of the alternatives lives when study period useful life. CHAPTER 4 41 Repeatability Assumption It assumes that study period is either indefinitely long or equal to a common multiple of the lives of the alternative. The cash flows associated with an alternative's initial life span are representative of what will happen in succeeding life spans. CHAPTER

22 Cotermination Assumption It is preferred if the study period is not a common multiple of the alternatives' lives or repeatability is not applicable Cost alternatives: Without repeatability, we must purchase/lease the service/asset for the remaining years Investment alternatives: Assume all cash flows will be reinvested at the MARR to the end of the study period CHAPTER 4 43 Comparing procedure based on Repeatability Assumption - Example A piece of production equipment is to be replaced immediately because it no longer meets quality requirements for the end product. The two best alternatives are a used piece of equipment (E1) and a new automated model (E2). The economic estimates for each are shown in the accompanying table. CHAPTER

23 Cash flows of alternatives Alternative E1 E2 Capital investment $14,000 $65,000 Annual expenses 14,000 9,000 Useful life (years) 5 20 Market value (at end of useful life) 8,000 13,000 CHAPTER 4 45 Common Multiple of Useful Lives Useful life of E1 = 5 Useful life of E2 = 20 Study period = 20 years E1 0 5 E1 E1 E E CHAPTER

24 Cash flow diagram of four times repeated E1 CHAPTER 4 47 PW of alternatives PW E1 = -14,000-6,000(P/F, 15%, 5) - 6,000(P/F, 15%, 10) - 6,000(P/F, 15%, 15) + 8,000(P/F, 15%, 20) - 14,000(P/A, 15%, 20) = -$106,345 PW E2 = -65,000-9,000(P/A, 15%, 20) + 13,000(P/F, 15%, 20) = -$120,539 Select E1 to minimize costs CHAPTER

25 AW of alternatives AW E1 = -$106,345(A/P, 15%, 20) = -$16,990 AW E2 = -$120,539(A/P, 15%, 20) = -$19,262 CHAPTER 4 49 Consider the AW over the useful life of Alternative E1 AW E1 = -14,000(A/P, 15, 5) - 14, ,000(A/F, 15, 5) = -16,990 CHAPTER

26 Comparing procedure based on Cotermination Assumption When Study Period > Useful Life: Calculate FW at end of useful life and move this to the end of the study period using the MARR When Study Period < Useful Life: Truncate the alternative at the end of the study period using an estimated or imputed Market Value CHAPTER 4 51 Study Period > Useful Life R A E A MV A R B E B MV B FW B (F/P,i,6) CHAPTER

27 Study Period < Useful Life MV A R A E A R B E B MV B CHAPTER 4 53 Study Period < Useful Life R A E A MV ı A MV A R B E B MV B CHAPTER

28 Study Period < Useful Life R A E A MV ı A = imputed or estimated MV A EW A is computed with MV ı A R B E B MV B CHAPTER 4 55 Example of Cotermination Investment cost $50,000 $120,000 Useful life 20 yrs. 40 yrs. Salvage value 10,000 20,000 Annual cost 9,000 6,000 a) Study period = 40 yrs. b) Study period = 20 yrs. Assume MV EOY 20 = $50,000 Which alternative would you recommend? The MARR is 10% per year A B CHAPTER

29 Solution to (a): Project A 10, ,000 Project B 9,000 20, , ,000 CHAPTER 4 57 Solution to (a): Project A 10,000 FW A FW A ,000 9,000 FW = 50,000(F/P,10%,20) 9,000(F/A,10%,20) + 10,000 = 841,850 Same service will be leased or purchased for 20 years: FW = 841,850(F/P, 10%, 20) = 5,663,546 CHAPTER

30 Solution to (a): Project B 20, ,000 6,000 FW = 120,000(F/P,10%,40) 6,000(F/A,10%,40) + 20,000 = 8,066,666 Select A to minimize costs (Cost of B = 8,066,666 Cost of A = 5,663,546) CHAPTER 4 59 Solution to (b): Project A 10, ,000 9,000 Project B ,000 20, ,000 6,000 CHAPTER

31 Solution to (b): Project A 10,000 Project B 50, ,000 9, ,000 6,000 AW A (10%) = [50,000(A/P,10%,20) -10,000(A/F,10%,20)] = -14,700 AW B (10%) = [120,000(A/P,10%,20) -50,000(A/F,10%,20)] = -19,225 Still select A to minimize costs CHAPTER 4 61 Sensitivity question? Project A 10,000 Project B X? ,000 9, ,000 6,000 What would the market value of EOY 20 have to be in order to select B instead of A? CHAPTER

32 Solve for X? Project A 10,000 Project B X? ,000 9, ,000 6,000 Set AW A = AW B, let X be the unknown market value -14,700 = -6,000 - {120,000(A/P, 10%, 20) - X(A/F,10%, 20)} -8,700 = -{14, X} 5,400 = X X = $308,571; therefore, MV B > $308,571 to favor B Such a value is very unlikely because X is more than the initial cost of B CHAPTER 4 63 Imputed Market Value R-E MV T R-E S T-1 T T+1 T+2... N-1 N MV T = CR (i %) (P/A, i%, N T) + S (P/F, i%, N T) Imputed (implied) market value is often used to truncate an alternative at end of time T. CHAPTER

33 Imputed Market Value - Example Alternative E1 E2 Capital investment $14,000 $65,000 Annual expenses 14,000 9,000 Useful life (years) 5 20 Market value 8,000 13,000 If the study period = 10 years which alternative is preferred? Repeatability is valid. MARR = 15% CHAPTER 4 65 Repeatibility for E1 study period = 10 years E1: 8,000 8, ,000 14,000 14,000 14,000 AW E1 = -14,000(A/P, 15, 5) - 14, ,000(A/F, 15, 5) = -16,990 CHAPTER

34 Imputed market value for E2 study period = 10 years E2: 54,705 13, ,000 9,000 MV 10 = 13,000(P/F, 15%, 10) + [65,000(A/P, 15%, 20) 13,000(A/F, 15%, 20)] (P/A, 15%, 10) = 54,705 CHAPTER 4 67 Imputed market value for E2 study period = 10 years E2: 54, ,000 9,000 AW E2 = -65,000(A/P,15%,10) - 9, ,705(A/F,15%,10) = -$19,258 Alternative E1 is preferred since cost of E1 < cost of E2 CHAPTER

35 COMBINATIONS OF PROJECTS Mutually Exclusive: At most one project out of the group can be chosen. Independent: The choice of a project is independent of the choice of any other project in the group, so that all or none of the projects may be selected or some number in between. Contingent: The choice of a project is conditional on the choice of one or more other projects. CHAPTER 4 69 Mutually exclusive alternatives Mutually Exclusive Project Combinations A B C Explanation Accept none Accept A Accept B Accept C CHAPTER

36 Independent alternatives Mutually Exclusive Project Combinations A B C Explanation Accept none Accept A Accept B Accept C Accept A and B Accept A and C Accept B and C Accept A, B and C CHAPTER 4 71 A is contingent on the acceptance of B and C. C is contingent on the acceptance of B Mutually Exclusive Project Combinations A B C Explanation Accept none Accept B Accept B and C Accept A, B and C CHAPTER

37 Combination of projects Example A Project Investment Annual Revenue Useful Life A, B, and C are independent projects MARR = 10% per year No budget limitation Market Value A $10,000 $2,300 5 years $10,000 B 12,000 2, C 15,000 4, CHAPTER 4 73 Calculation CR(10%) of projects Project Investment Annual Revenue Useful Life Market Value A $10,000 $2,300 5 years $10,000 B 12,000 2, C 15,000 4, CR(10%) A = 10,000(A/P,10%,5) 10,000(A/F,10%,5) = $1,000 CR(10%) B = 12,000(A/P,10%,5) = $3,166 CR(10%) C = 15,000(A/P,10%,5) = $3,957 CHAPTER

38 Calculation AW(10%) of projects Project R CR (10%) AW = R CR(10%) A $2,300 $1,000 1,300 B 2,800 3, C 4,067 3, Reject project B. Accept A and C. Because no budget limitation; Recommend A + C for implementation to maximize AW CHAPTER 4 75 Combination of projects Example B There are five projects (B1, B2, C1, C2, and D) to be considered B1 and B2 are mutually exclusive alternatives C1 and C2 are mutually exclusive alternatives and contingent on the acceptance of B2 D is an individual alternative and contingent on the acceptance of C1 MARR = 10% per year CHAPTER

39 PW(10%) of projects Project Investment Useful life PW B1 $50,000 4 years $13,400 B2 30, ,000 C1 14, ,300 C2 15, D 10, ,000 What is the best portfolio if; a) budget is unlimited b) budget is limited to $48,000 CHAPTER 4 77 Solution to (a) Mutually Exclusive Projects Invested Combinations B1 B2 C1 C2 D Capital PW $0 $ ,000 13, ,000 8, ,000 6, ,000 8, ,000 15,700 Recommend portfolio 6 (B2 + C1 + D) with PW of $15,700 CHAPTER

40 Solution to (b) Mutually Exclusive Projects Invested Combinations B1 B2 C1 C2 D Capital PW $0 $ ,000 13, ,000 8, ,000 6, ,000 8, ,000 15,700 Budget = $48,000 Portfolio 5 is the best one CHAPTER 4 79 Combination of projects Example C Project Investment, I Annual Revenue, A Useful Life, N PW(10%) = - I + A(P/A, 10%, N) X $93,000,000 $13,000, years $5,879,034 Y 55,000,000 9,500, years 3,373,388 Z 71,000,000 10,400, years 27,039,910 X, Y, and Z are three independent; nonrepeating projects MARR = 10% per year Investment budget is $200 million Study period = 30 years (the longest-lived project) CHAPTER

41 Under cotermination assumption PW of alternatives for their initial life span FW X = $5,879,034 (F/P, 10%, 30) FW Y = $3,373,388 (F/P, 10%, 30) FW Z = $27,039,910(F/P, 10%, 30) This will result in the same selection as that of PW based on alternative s useful life CHAPTER 4 81 MEC 6 (X+Z) is selected Mutually Exclusive Project ($million) Invested Combinations X Y Z Capital PW (10%) $0 $ ,879, ,373, ,039, ,252, ,918, ,413, > ,292,332 CHAPTER