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Incremental IRR to select best of A to D

Short answer

All own IRRs exceed 10%, so use incremental IRR starting with lowest-investment B. Increment B to C yields IRR 10.38% above MARR, while other increments fail. Select Project C.

Select the best alternative using incremental IRR method. Useful life is 10 years and salvage value is 25% of initial investment. MARR = 10%.

ProjectABCD
Initial Investment600500800700
Annual Revenues150125175160
Annual Cost40253035

Net annual revenue = revenue - cost; salvage = 25% of investment; life 10 years; MARR = 10%.

Project (by investment)InvestmentNet revenue/yrSalvageOwn IRR
B50010012516.58%
A60011015014.56%
D70012517513.97%
C80014520014.30%

(Own IRR from -I + a(P/A, i, 10) + S(P/F, i, 10) = 0; e.g. for B at 16%: -500 + 100(4.8332) + 125(0.2267) = 11.7.)

All four exceed MARR = 10%, so B (lowest investment) is the starting defender. Incremental cash flows (year 0, years 1-9, year 10):

StepDefender → ChallengerIncremental flowsIncremental IRRDecision
1B → A-100, 10, 353.65%< 10%, keep B
2B → D-200, 25, 757.10%< 10%, keep B
3B → C-300, 45, 12010.38%> 10%, C replaces B

Check of C against the others: A → C = 13.5%, D → C = 16.6%, both above 10%.

For step 3: -300 + 45(P/A, i, 10) + 75(P/F, i, 10) = 0. At 10%: -300 + 45(6.1446) + 75(0.3855) = 5.4 > 0, and the root is 10.38%.

Answer: Select Project C. Its extra investment of 300 over B earns 10.38%, just above MARR = 10%.

✓ Answer checked by Claude Opus 5.5 (high effort)

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