Hw#3 EGME-401 Problem 2 (20 pts) Consider three alternatives: S50 $150 $110 Firs
ID: 2566019 • Letter: H
Question
Hw#3 EGME-401 Problem 2 (20 pts) Consider three alternatives: S50 $150 $110 First cost Uniform annual 28.8 39.6 39.6 benefit Useful life, in years26 Rate of return 10% 15% 16.4% At the end of its useful life, an identical alternative (with the same cost, benefits, and useful life) may be installed All the alternatives have no salvage value. If the MARR is 12%, which alternative should be selected? (a) Solve the problem by payback period. (5 pts) (b) Solve the problem by future worth analysis . (5 pts) (c) Solve the problem by benefit-cost ratio analysis. (5 pts) (d) If the answers in parts (a). (b), and (e) differ, explain why this is the case. (5 pts)Explanation / Answer
a. Payback period
Alternative A = First cost/ Uniform annual benefit = $50/ 28.8 = 1.74 years
Alternative B = First cost/ Uniform annual benefit = $150/ 39.6 = 3.79 years
Alternative C = First cost/ Uniform annual benefit = $110/ 39.6 = 2.78 years
Alternative A should be selected.
b.
Future worth = [First cost x (1+MARR)n] + [Uniform annual benefit x FVADMARR,n]
Alternative A:
= -[$50 x (1+0.10)2] + [$28.8 x FVAD10%,2]
= -$60.50 + ($28.8 x 2.10)
= -$0.02
Alternative B:
= -[$150 x (1+0.15)6] + [$39.6 x FVAD15%,6]
= -$346.96 + $346.65
= -$0.31
Alternative C:
= -[$110 x (1+0.164)4] + [$39.6 x FVAD16.4%,4]
= -$201.93 + $201.8
= -$0.13
Alternative A should be selected
c. Benefit-cost ratio = Present value of future benefits/ Initial cost
Alternative A = ($28.8 x PVA10%,2) / $50 = 50 / 50 = 1
Alternative B = ($39.6 x PVA15%,6) / $150 = 150 / 150 = 1
Alternative C = ($39.6 x PVA16.4%,4) / $110 = 110 / 110 = 1
Any alternative can be selected.
d. No difference between answers is parts a, b, and c.
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