Elmazi-Shkumbin-Week7-8A

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New Jersey Institute Of Technology *

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492

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Industrial Engineering

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Oct 30, 2023

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docx

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Shkumbin Elmazi Professor Tchouassi IE 492 10/13/23 8A Assigned Questions-Schaums Chapters FE Problems Team 4 #16 16) A company is considering two alternatives for manufacturing a certain part. Method R will have a first cost of $40,000, an annual operating cost of $25,000, and a $10,000 salvage value after its five year life. Method S will have an initial cost of $100,000, an annual operating cost of $15,000, and a $12,000 salvage value after its 10 year life. At an interest rate of 12% per year, the present worth values of the two alternatives are closest to: a. $124,446/$180,889 b. $195,057/$180,889 c. $124,446/$147,263 d. $195,057/$147,263 Answer is D. $195,057/$147,263 Method R First Cost = $40,000 Annual Operating Cost = $25,000 Salvage Value = $10,000 Life, years = 5 Interest Rate = 12% Present Worth = -First Cost – Annual Operating Cost*(P/A, i, n) – (First Cost – Salvage Value) * (P/F, i, n) + Salvage Value*(P/F, i, n) Where: P/A = Uniform series present worth P/F = Single payment present worth i = interest rate n = life, years Present Worth = -$40,000 - $25,000 (P/A, 0.12, 10) – ($40,000 - $10,000)(P/F, 0.12, 5) + $10,000(P/F, 0.12,10) Present Worth = -$40,000 - $25,000(5.6502) -$30,000(0.5674) + $10,000(0.3220) Present Worth = -$40,000 - $141,255 - $17,022 + $3,220 Present Worth= -$195,057

Method S First Cost = $100,000 Annual Operating Cost = $15,000 Salvage Value = $12,000 Life, years = 10 Interest Rate = 12% Present Worth = -First Cost – Annual Operating Cost*(P/A, i, n) + Salvage Value*(P/F, i, n) Where: P/A = Uniform series present worth P/F = Single payment present worth i = interest rate n = life, years Present Worth = -$100,000 - $15,000 (P/A, 0.12, 10) + $12,000(P/F, 0.12,10) Present Worth = -$100,000 - $15,000(5.6502) + $12,000(0.3220) Present Worth = -$100,000 - $84,753 + $3,864 Present Worth= -$180,889

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