an engineer uses an economic analysis to determine which of two different machin
ID: 1101435 • Letter: A
Question
an engineer uses an economic analysis to determine which of two different machines to purchase. All the machines that are being considered are capable of performing the same task. Assume that the minimum attractive rate of return is 12% compounded monthly. what is the annual worth (EAW)? which is the best economic investment?
machine x; initial cost=8000, estimated life=3years, salvage value=1000, monthly maintanace cost=1500, monthly income=3550
machine y; initial cost=11000, estimated life=3years, salvage value=1500, monthly maintanace cost=1475, monthly income=3475
a) EAWx=669.51, EAWy=807.53; Y is the best investment
b) EAWx=807.53., EAWy=669.51; X is the best investment
c) EAWx=1784.76, EAWy=1646.83; X is the best investment
c) EAWx=1784.76, EAWy=1646.83; Y is the best investment
Explanation / Answer
effective annual rate = (1+12/1200)^12-1 = 12.6825%
monthly interest rate = 12/12 = 1%
for machine X,
to calculate EAW, first we have to calculate the NPV
annual net monthly cashflow = 3550-1500 = $2050
NPV = -8000+2050*PVIFA(1%,36)+(1000/1.01^36)
NPV = -8000+2050*30.1075+(1000/1.01^36) = $54419.30
EAWx = NPV/PVIFA(1,36) = 54419.30/30.1075 = $1784.76
monthly cash flow for Project Y = 3475-1475 = 2000
NPVy = -11000+2000*PVIFA(1%,36)+(1500/1.01^36)
NPVy = -11000+2000*30.1075+(1500/1.01^36) = $50263.387
EAWy = 50263.387/30.1075
EAWy = $1646.83
c) EAWx=1784.76, EAWy=1646.83; X is the best investment
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