In comparing the alternatives by the annual worth method, the AW of X is determined by the following equation: (a) 200,000(0.10) 60,000 + 20,000(0.10) (b)

In comparing the alternatives by the annual worth method, the AW of X is determined 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 -200,000 -800,000 -60,000 -10,000 20,000 5 150,000