The annual worth of perpetual service for alternative X is represented by the following equation: (a) 200,000(0.10)

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The annual worth of perpetual service for alternative X is represented by the following equation:

(a) ˆ’200,000(0.10) ˆ’ 60,000 + 20,000(0.10)

(b) ˆ’200,000(Aˆ•P,10%,5) ˆ’ 60,000 + 20,000(Aˆ•F,10%,5)

(c) ˆ’200,000(Aˆ•P,10%,5) ˆ’ 60,000 ˆ’ 20,000(Aˆ•F,10%,5)

(d) ˆ’200,000(0.10) ˆ’ 60,000 + 20,000(Aˆ•F,10%,5)


Problem is based on the following cash flows and an interest rate of 10% per year.

Alternative First cost, $ Annual cost, $/year Salvage value, $ |Life, years х -200,000 -800,000 -60,000 -10,000 20,000

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Engineering Economy

ISBN: 978-0073523439

8th edition

Authors: Leland T. Blank, Anthony Tarquin

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