AW1 = -65,000(A/P,10%,1) – 50,000 + 30,000(A/F,10%,1) = $-91,500 — construction company bought 180 metric ton earth sifter cost The expects keep

A construction company bought a 180,000 metric ton earth sifter at a cost of $65,000. The company expects to keep the equipment a maximum of 7 years. The operating cost is expected to follow the series described by 40,000 + 10,000 k, where k is the number of years since it was purchased ( k = 1, 2, . . . , 7). The salvage value is estimated to be $30,000 for years 1 and 2 and $20,000 for years 3 through 7. At an interest rate of 10% per year, determine the economic service life and the associated equivalent annual cost of the sifter.

SOLUTION

AW1 = -65,000(A/P,10%,1) – 50,000 + 30,000(A/F,10%,1) = $-91,500

AW2 = -65,000(A/P,10%,2) – [50,000 + 10,000(A/G,10%,2)] + 30,000(A/F,10%,2) = $-77,929

AW3 = -65,000(A/P,10%,3) – [50,000 + 10,000(A/G,10%,3)] + 20,000(A/F,10%,3) = $-79,461