- 2068 Jestha · 10 marks
IRR to select best of A, B, C (15 years)
Short answer
Compute IRR as annual revenue divided by initial investment because salvage equals investment. Project C has highest own IRR (15%), but incremental IRR of A over C is 10.29% exceeding 10% MARR, so choose Project A.
Select the best project using IRR method. Useful life of all projects is 15 years. MARR = 10%.
| Particulars | Project A | Project B | Project C |
|---|---|---|---|
| Initial investment | 7500,000 | 5500,000 | 4000,000 |
| Annual revenue | 960,000 | 720,000 | 600,000 |
| Salvage value | 7500,000 | 5500,000 | 4000,000 |
Each project returns its full investment as salvage at year 15, so the cash flow is: -I at year 0, revenue every year, +I at year 15. For such a flow the IRR equals Revenue / I exactly (the investment is never consumed). MARR = 10%.
Step 1: IRR of each project
| Project | Investment | Revenue/yr | IRR = Rev/I | Result |
|---|---|---|---|---|
| A | 7,500,000 | 960,000 | 12.80% | accept |
| B | 5,500,000 | 720,000 | 13.09% | accept |
| C | 4,000,000 | 600,000 | 15.00% | accept |
All exceed MARR, so incremental analysis is needed. Order by investment: C < B < A.
Step 2: incremental IRR
Salvage of increments equals the extra investment, so incremental IRR = extra revenue / extra investment.
| Increment | Extra investment | Extra revenue | Incremental IRR | Decision |
|---|---|---|---|---|
| C → B | 1,500,000 | 120,000 | 8.00% | < 10%, keep C |
| C → A | 3,500,000 | 360,000 | 10.29% | > 10%, A better than C |
(B → A: extra investment 2,000,000, extra revenue 240,000, IRR 12.00% > 10%, A better than B.)
The step C → B fails, but A must still be compared with the current best (C), and A beats it. A also beats B.
Answer: Select Project A. Its extra investment over C earns 10.29%, slightly above MARR 10%. (C has the highest own IRR, but the highest IRR is not the selection rule.)
✓ Answer checked by Claude Opus 5.5 (high effort)
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