Comparing Mutually Exclusive Alternatives

Transcription

1 Comparing Mutually Exclusive Alternatives Comparing Mutually Exclusive Projects Mutually Exclusive Projects Alternative vs. Project Do-Nothing Alternative 2 Some Definitions Revenue Projects Projects whose revenues depend on the choice of alternatives Service Projects Projects whose revenues do not depend on the choice of alternative Analysis Period The time span over which the economic effects of an investment will be evaluated (study period or planning horizon). 3 Presented by: Dr. Magdy Akladios 1

2 Two situations: 1. Useful lives are the same for all alternatives and equal to the study period 2. Useful lives are different among the alternatives and at least one does not match the study period 4 Comparing Mutually Exclusive Projects Principle: Projects must be compared over an equal time span. Rule of Thumb: If the required service period is given, the analysis period should be the same as the required service period. 5 First, comparing alternatives w/useful lives = study period Chapter 5 - Part II Presented by: Dr. Magdy Akladios 2

3 Comparing MEAs Beginning Planning Horizon End Project A Project B 7 Case 1: Analysis Period Equals Project Lives Compute the PW for each project over its life 0 $450 $ $500 $1,400 0 $2,075 $2, $1,000 A PW (10%) = $283 A PW (10%) = $579 B $4,000 Select B B 8 Example Alternatives A and B are 2 mutually exclusive projects as shown below: A B Capital investment: $60,000 $73,000 Annual Revenues: $22,000 $26,225 Project life = 4 years MARR = 10% Which project should you invest in? 11 Presented by: Dr. Magdy Akladios 3

4 Solution PW(10%) A = -$60,000 + $22,000(P/A, 10%, N=4) = $9,738 PW(10%) B = -$73,000 + $26,225(P/A, 10%, N=4) = $10,131 Select alternative B since PW B > PW A Can also solve for the difference between the 2 alternatives, PW B-A (Alt B Alt A) = $ Example of a Cost Alternative Alternatives C and D are 2 mutually exclusive cost alternatives as shown below: EOY C D 0 -$380,000 -$415, $38,100 -$27, $39,100 -$27, $40,100 -$27,400 Salvage value $0 $26,000 MARR = 10% Which cost alternative should you adopt? 13 Solution: PW(10%) C = -$380,000 - $38,100(P/F, 10%, N=1) - $39,100(P/F, 10%, N=2) - $40,100(P/F, 10%, N=3) = -$477,077 PW(10%) D = -$415,000 - $27,400(P/A, 10%, N=3) + $26,000(P/F, 10%, N=3) = -$463,607 Select alternative D since PW D > PW C Can also solve for the difference. 14 Presented by: Dr. Magdy Akladios 4

5 Example Alternatives A, B, and C are 3 mutually exclusive projects as shown below: A B C Capital investment: $390,000 $920,000 $660,000 Annual Revenues: $69,000 $167,000 $133,500 Project life = 10 years MARR = 10% Which project should you invest in? 15 Solution (PW method) PW(10%) A = -$390,000 + $69,000(P/A, 10%, N=10) = $33,977 PW(10%) B = -$920,000 + $167,000(P/A, 10%, N=10) = $106,148 PW(10%) C = -$660,000 + $133,500(P/A, 10%, N=10) = $160,304 Therefore, we select alternative C w/the highest PW. 16 Solution (FW method) FW(10%) A = -$390,000(F/P, 10%, N=10) + $69,000(F/A, 10%, N=10) = $88,138 FW(10%) B = -$920,000(F/P, 10%, N=10) + $167,000(F/A, 10%, N=10) = $275,342 FW(10%) C = -$660,000(F/P, 10%, N=10) + $133,500(F/A, 10%, N=10) = $415,801 Therefore, we select alternative C w/the highest FW. 17 Presented by: Dr. Magdy Akladios 5

6 Can also solve using the Incremental Method Sort the projects in ascending order of initial investment A C B Capital investment: $390,000 $660,000 $920,000 Annual Revenues: $69,000 $133,500 $167, Start your Iterations A C (C-A) Capital investment: $390,000 $660,000 $270,000 Annual Revenues: $69,000 $133,500 $64,500 PW(10%) (C-A) = -$270,000 + $64,500(P/A, 10%, N=10) PW(10%) (C-A) = a positive amount, therefore, C is higher than A, then you should drop A, keep C, etc. 19 Second, useful lives are different among the alternatives Presented by: Dr. Magdy Akladios 6

7 Comparing MEAs Beginning End Planning Horizon Project A Project B 21 Case 2: Analysis Period Shorter than Project Lives Estimate the Salvage value at the end of the required service period. Compute the PW for each project over the required service period 22 Useful Life(s) > Study Period Truncate the alternative at the end of the study period using an estimated market value. This method assumes disposable assets will be sold at the end of the study period at that value 24 Presented by: Dr. Magdy Akladios 7

8 IMPUTED MARKET VALUE TECHNIQUE When current marketplace data is unavailable for an asset, it is sometimes necessary to estimate the market value of an asset Referred to as an imputed or implied market value Estimating is based on logical assumptions about the remaining life for the asset: MV T = [PW at the end of year T of remaining capital recovery amounts] + [PW at EOY T of original market value (salvage) at the end of useful life] T < useful life PW is Present worth at i = MARR 25 Example Your firm is purchasing a pump ($47,600). The pump has a useful life of 9 years. Salvage value = $5,000. Use the imputed market value technique to estimate the market value of this pump at the end of year 5. MARR = 20% 26 MV 5 = PW at EOY 5 of remaining CR amounts + PW at EOY 5 of salvage value at EOY 9 MV 5 = $? F 9 = $5, P 0 = $47, Presented by: Dr. Magdy Akladios 8

9 Solution (Part I) PW at the end of year 5 of remaining CR amounts: PW(20%) CR = [$47,600 (A/P, 20%, 9) - $5,000 (A/F, 20%, 9)] X (P/A, 20%, 4) PW(20%) CR = $29,949 MV 5 = $? F 9 = $5, P 0 = $47, Solution (Part II) PW at the end of year 5 of original market value at the end of useful life: PW(20%) Salvage = $5,000 (P/F, 20%, 4) PW(20%) Salvage = $2,412 MV 5 = $? F 9 = $5, P 0 = $47, Final Solution MV 5 = PW CR + PW Salvage MV 5 = $29,949 + $2,412 MV 5 = $32, Presented by: Dr. Magdy Akladios 9

10 Example: Investment: $10 K $15 K Annuity: $3 K $5 K Salvage $4 K $7 K Project life 7 yrs 9 yrs Study period = 5 years, MARR = 10% Which project should you invest in? A B 31 Solution: CR A = I (A/P, i, 7) - S (A/F, i, 7) CR A = 10k ( ) - 4k (0.1054) = $1, CR B = I (A/P, i, 9) - S (A/F, i, 9) CR B = 15K ( ) - 7k (0.0736) = $2, MV A = CR (P/A, i, 2) + S (P/F, i, 2) MV A = 1, (1.7355) + 4K (0.8264) = $6, MV B = CR (P/A, i, 4) + S (P/F, i, 4) MV B = 2, (3.1699) + 7K ( = $11, NPW A = -10K + 3K (P/A, i, 5) + 6, (P/F, i, 5) = $5, NPW B = -15K + 5K (P/A, i, 5) + 11,402.29(P/F,i, 5) = $11, Therefore, select B 32 Analysis Period Longer than Project Lives Presented by: Dr. Magdy Akladios 10

11 Case 3: Analysis Period Longer than Project Lives Come up with replacement projects that match or exceed the required service period. Compute the PW for each project over the required service period. 34 Useful Life(s) < Study Period a. Cost alternatives -- each cost alternative must provide same level of service as study period: 1) contract for service or lease equipment for remaining time; 2) repeat part of useful life of original alternative until study period ends b. investment alternatives -- assume all cash flows reinvested in other opportunities at MARR to end of study period 35 Example Use the co-terminated assumption to select one of the following MEA s: A B I -$3,500 -$5,000 Annual Rev. 1,255 1,480 Useful Lives 4 yrs 6 yrs Salvage values = $0 MARR = 10%/year Planning Horizon = 6 years 36 Presented by: Dr. Magdy Akladios 11

12 Solution: FW A for the useful life of the project (Yr 4): FW A:Useful = -$3,500 (F/P, 10%, 4) + $1,255 (F/A, 10%, 4) FW A:Useful = -$3,500 (1.4641) + $1,255 (4.641) FW A:Useful = $700 Then, reinvest this cash flow at the MARR for the remainder of the study period (2 yrs till Yr 6): FW A = FW A:Useful (F/P, 10%, 2) FW A = $700 (1.21) FW A = $847 FW A:Useful FW A Solution: FW B for the useful life of the project (Yr 6): FW B = -$5,000 (F/P, 10%, 6) + $1,480 (F/A, 10%, 6) FW B = -$5,000 (1.7716)+ $1,480 (7.7156) FW B1 = $2,561 FW B Comparison for Service Projects When the Required Service Period is Longer than the Individual Project Life PW(15%) $34,359 Model A PW(15%) $31,031 Model B. 39 Presented by: Dr. Magdy Akladios 12

13 Example: Analysis Period Coincides with Longest Project Life Revenue Projects PW(15%) $2,208,470 Drill PW(15%) $2,180,210 Lease 40 Case 4: Analysis Period Is Not Specified Come up with replacement projects that serve out the least common multiple period. Compute the PW for each project over the required service period. 41 REPEATABILITY ASSUMPTION The study period: Is either indefinitely long, or, Equal to a common multiple of the lives of the alternatives. The economic consequences that are estimated to happen in an alternative s initial useful life span will be repeated in all succeeding life spans 42 Presented by: Dr. Magdy Akladios 13

14 Example: Analysis Period is Not Specified: Lowest Common Multiple Method PW(15%) Model A $53,657 PW(15%) Model B $48, Example Use the repeatability assumption to select one of the following MEA s: A B I -$3,500 -$5,000 Annual Rev. 1,255 1,480 Useful Lives 4 yrs 6 yrs Salvage values = $0 MARR = 10%/year 44 Solution The least common multiple of the useful lives = 12 yrs Therefore, repeat A three times, repeat B twice. To solve this problem, find the PW of each alternative, then repeat it as many times as necessary. 45 Presented by: Dr. Magdy Akladios 14

15 Solution: PW A1 for the first term: PW A1 = -$3,500 + $1,255 (P/A, 10%, 4) PW A1 = -$3,500 + $1,255 (3.1699) PW A1 = $ Then, repeat PW A 3 times: PW A = PW A1 + PW A2 (P/F, 10%, 4) + PW A3 (P/F, 10%, 8) PW A = $ $ (P/F, 10%, 4) + $ (P/F, 10%, 8) PW A = $ $ (0.6833) + $ (0.4665) = $1,028 PW A1 PW A2 PW A Solution: PW B1 for the first term: PW B1 = -$5,000 + $1,480 (P/A, 10%, 6) PW B1 = -$5,000 + $1,480 (4.3553) PW B1 = $1, Then, repeat PW B 2 times: PW B = PW B1 + PW B2 (P/F, 10%, 6) PW B = $1, $1, (P/F, 10%, 6) PW B = $1, $1, (0.5645) = $2,262 PW B1 PW B An alternative Solution: Use the AW method to solve the previous problem AW A = PW A (A/P, 10%, 12) AW A = $1,028 (0.1468) = $ Or, by calculation: AW A1 for the first term: AW A1 = -$3,500 (A/P, 10%, 4) + $1,255 AW A1 = -$3,500 (0.3155) + $1,255 AW A1 = $ Then, repeat AW A 3 times: AW A =AW A1 = AW A2 = AW A3 = $ Presented by: Dr. Magdy Akladios 15

16 An alternative Solution: Use the AW method to solve the previous problem AW B = PW B (A/P, 10%, 12) AW B = $2,262 (0.1468) = $ Or, by calculation: AW B1 for the first term: AW B1 = -$5,000 (A/P, 10%, 6) + $1,480 AW B1 = -$5,000 (0.2296) + $1,480 AW A1 = $332 Then, repeat AW B 3 times: AW B =AW B1 = AW B2 = $ Summary Present worth is an equivalence method of analysis in which a project s cash flows are discounted to a lump sum amount at present time. MARR is the interest rate at which a firm can always earn or borrow money. MARR is generally dictated by management and is the rate at which NPW analysis should be conducted. Two measures of investment, the Net Future Worth and the capitalized Equivalent Worth, are variations to the NPW criterion. The term mutually exclusive means that, when one of several alternatives that meet the same need is selected, the others will be rejected. 50 Continue Summary Revenue projects are those for which the income generated depends on the choice of project. Service projects are those for which income remains the same, regardless of which project is selected. The analysis period (study period) is the time span over which the economic effects of an investment will be evaluated. The required service period is the time span over which the service of an equipment (or investment) will be needed. The analysis period should be chosen to cover the required service period. 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. 51 Presented by: Dr. Magdy Akladios 16