Question

Use Present Worth Analysis to determine whether Alternative A or B should be chosen. Items are identically replaced at the end of their useful lives. Assume an interest rate of 3% per year, compounded annually. Alternative A 340 60 Alternative B 870 182 Initial Cost Annual Benefit Salvage Value Useful Life (yrs) 78 106 Alternative A, because it costs $65.43 less than Alternative B, in terms of present worth Alternative B, because it costs $65.43 more than Alternative A, in terms of present worth Alternative A, because its present worth is positive Alternative B, because it only incurs the initial cost once every three years instead of every two years

Answer 1

NPW of Alternative A = -340 + 60 * (P/A,3%,6) - (340 - 78) *(P/F,3%,2) - (340 - 78) *(P/F,3%,4) + 78 * (P/F,3%,6)

= -340 + 60 *5.417191 - (340 - 78) *0.942596 - (340 - 78) *0.888487 + 78 *0.837484

= -429.39

NPW of Alternative B = -870 + 182 * (P/A,3%,6) - (870 - 106) *(P/F,3%,3) + 106 * (P/F,3%,6)

= -870 + 182 *5.417191 - (870 - 106) *0.915142 + 106 *0.837484

= -494.47

Saving in selecting option A = NPW of A - NPW of B = -429.39 - (-494.47) = 65.08 ~ 65.43

First option is correct answer