A firm is having a large piece of equipment overhauled. It expects that the mach
ID: 1245370 • Letter: A
Question
A firm is having a large piece of equipment overhauled. It expects that the machine will be needed for the next 12 years. The firm has an 8% MARR. The contractor has suggested three alternatives:
A. A complete overhaul for $6,000 that should permit 12 years of operation
B. A major overhaul for $4,500 that can be expected to provide 8 years of service. At the end of 8 years, a minor overhaul would be needed.
C. A minor overhaul now. At the end of 4 and 8 years, additional minor overhauls would be needed.
If minor overhauls cost $2,500, which alternative should the firm select? If minor overhauls, which now cost $2,500, increase in cost at 5% per year, but other costs remain unchanged, which alternative should the firm select?
Explanation / Answer
I'm inclined to disagree with this analysis, and I will provide what I believe to be the correct solution.
(Pre-5% inflation)
Cost for A: $6,000 present cost
Cost for B:$4,500 + $2,500(P/F,8%,8) = $5,850.67 present cost
Cost for C:$2,500+$2,500(P/F,8%,4) + $2,500(P/F,8%,8) = $5,688.24 present cost
Alternative C should be chosen without the 5% increase in cost. The book even says Alt C should be chosen in this instance so I'm not sure why people are giving the OP an average of 4.2 stars for an answer that is wrong.
(With a 5% yearly cost increase)
In this instance you need to take the present worth of the future overhauls that you found in the previous part of the problem, and add in the 5% yearly increase in the form of the compound interest equation.
Cost for A: $6,000 present cost
Cost for B:$4,500+($2,500(P/F,8%,8))(1+0.05)^5 = $6,223.835 present cost
Cost for C:$2,500+($2,500(P/F,8%,4))(1+.05)^5+($2,500(P/F,8%,8))(1+.05)^5 = $6,569.092 present cost
This is what I believe to be the correct solution. Please comment if you disagree, I would like to make sure I'm doing this right.
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