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Repeatability assumption, select best of three projects
Short answer
Project A selected with lowest equivalent annual cost Rs 56,444/year at 14% MARR using repeatability assumption. AW calculated for each project's life cycle: A (3 yr), B (5 yr), C (7 yr). All have only costs, so minimum AW wins.
Use repeatability assumption to select the best project from the following three projects.
| Project | A | B | C |
|---|---|---|---|
| Initial Investment | 100000 | 200000 | 250000 |
| Annual Expenditure | 25000 | 20000 | 15000 |
| Useful Life, years | 3 | 5 | 7 |
| Salvage Value | 40000 | 50000 | 60000 |
| MARR | 14% | 14% | 14% |
Repeatability assumption: each project is repeated, with the same cash flows, as many times as needed, so all projects can be compared over a common (very long) period. Equal annual worth (AW) over one life cycle then represents every cycle, and lives need not be equal. All three projects have only costs, so the best is the one with the smallest annual cost (largest, i.e. least negative, AW).
AW = -I(A/P, i, N) - annual expenditure + S(A/F, i, N), MARR = 14%.
Project A (N = 3): (A/P,14%,3) = 0.4307, (A/F,14%,3) = 0.2907 AW = -100,000(0.4307) - 25,000 + 40,000(0.2907) = -43,073 - 25,000 + 11,629 = -56,444
Project B (N = 5): (A/P,14%,5) = 0.2913, (A/F,14%,5) = 0.1513 AW = -200,000(0.2913) - 20,000 + 50,000(0.1513) = -58,257 - 20,000 + 7,564 = -70,693
Project C (N = 7): (A/P,14%,7) = 0.2332, (A/F,14%,7) = 0.0932 AW = -250,000(0.2332) - 15,000 + 60,000(0.0932) = -58,298 - 15,000 + 5,592 = -67,707
| Project | Life | AW (Rs./year) |
|---|---|---|
| A | 3 | -56,444 |
| B | 5 | -70,693 |
| C | 7 | -67,707 |
Answer: Select Project A, the one with the lowest equivalent annual cost (Rs. 56,444 per year). B and C are costlier by about Rs. 14,249 and Rs. 11,263 per year respectively. (No revenue is given, so the projects are compared on cost alone, assuming they give equal service.)
✓ Answer checked by Claude Opus 5.5 (high effort)
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