Problem 1 · Time value of money
Two mutually exclusive alternatives each have an 8-year life and no salvage value. Alternative A costs $40,000 and returns $11,000 per year; its rate of return is about 21.8%. Alternative B costs $60,000 and returns $15,500 per year; its rate of return is about 19.7%. The MARR is 12% per year. Which alternative should be selected?
- A.Alternative B, because the incremental rate of return on the extra $20,000 is about 15.3%, which exceeds the MARR
- B.Alternative A, because its rate of return is higher than B's
- C.Neither alternative, because the incremental rate of return is lower than both alternatives' own rates of return
- D.Alternative A, because the incremental rate of return of B over A is lower than A's own rate of return
Show the answer and solution
Answer: A: Alternative B, because the incremental rate of return on the extra $20,000 is about 15.3%, which exceeds the MARR
The increment B − A costs $20,000 and returns $4,500 per year, so (P/A, i, 8) = 4.444, giving i ≈ 15.3% > 12%; the extra investment is justified and B is selected. Present worth confirms it: PW_A = −40,000 + 11,000(4.9676) = $14,644 and PW_B = −60,000 + 15,500(4.9676) = $16,998 at 12%.
Time value of money · Reference Handbook: Engineering Economics
