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 6% per year, compounded annually.
Alternative A Alternative B
Initial Cost 350 985
Annual Benefit 80 226
Salvage Value 160 186
Useful Life (yrs) 2 3
A. Alternative B, because it only incurs the initial cost once every three years instead of every two years
B. Alternative B, because it costs $250.00 more than Alternative A, in terms of present worth
C. Alternative A, because its present worth is positive
D. Alternative A, because it costs $250.00 less than Alternative B, in terms of present worth

Answer 1

Answer:

D. Alternative A, because it costs $250.00 less than Alternative B, in terms of present worth.

Explanation:

Net Present Worth of Alternative A:

-350 + 80 * (P/A, 6%, 6) - (350 - 160) * (P/F, 6%, 2) - (350 - 160) * (P/F, 6% , 4) + 160 * (P/F, 6% , 6)

= -350 + 80 * 5.41791 - (340 - 160) * 0.942596 - (350 - 160) * 0.888487 + 160 * 0.837484

NPW = $ -429.39

Net Present Worth of Alternative B:

-985 + 226 * (P/A, 6%, 6) - (985 - 226) * (P/F, 6%, 3) - (985 - 186) * (P/F, 6% , 4) + 186 * (P/F, 6% , 6)

= -985 + 226 * 5.41791 - (985 - 186) * 0.942596 - (985 - 186) * 0.888487 + 186 * 0.837484

NPW = $ -657.24