---
title: "Incremental IRR to select best of A to D"
subject: "Engineering Economics"
chapter: 4
key: "semester_6/engineering_economics:incremental-irr-abcd-salvage-25"
source: https://www.bctengineer.com/q/semester_6/engineering_economics/incremental-irr-abcd-salvage-25
site: BCT Engineering
---

# Incremental IRR to select best of A to D

Engineering Economics · Chapter 4: Comparative Analysis of Alternatives

- Asked in: 2067 Mangsir, 12 marks
- Video: [IRR method and Incremental Analysis | Engineering Economics |Board Exam Solution in NEPALI](https://www.youtube.com/watch?v=mibE6pNV_GA) — OPengineer
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/engineering_economics/incremental-irr-abcd-salvage-25
- Whole chapter: https://www.bctengineer.com/md/semester_6/engineering_economics/chapter-4

## Question

Select the best alternative using incremental IRR method. Useful life is 10 years and salvage value is 25% of initial investment. MARR = 10%.
| Project | A | B | C | D |
|---|---|---|---|---|
| Initial Investment | 600 | 500 | 800 | 700 |
| Annual Revenues | 150 | 125 | 175 | 160 |
| Annual Cost | 40 | 25 | 30 | 35 |

## 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.

## Answer

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

| Project (by investment) | Investment | Net revenue/yr | Salvage | Own IRR |
|---|---|---|---|---|
| B | 500 | 100 | 125 | 16.58% |
| A | 600 | 110 | 150 | 14.56% |
| D | 700 | 125 | 175 | 13.97% |
| C | 800 | 145 | 200 | 14.30% |

(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):

| Step | Defender → Challenger | Incremental flows | Incremental IRR | Decision |
|---|---|---|---|---|
| 1 | B → A | -100, 10, 35 | 3.65% | < 10%, keep B |
| 2 | B → D | -200, 25, 75 | 7.10% | < 10%, keep B |
| 3 | B → C | -300, 45, 120 | 10.38% | > 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%.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is net annual revenue found?
   A. Revenue only
   B. Revenue + cost
   C. Cost only
   D. Revenue - cost
   Answer: D. Net revenue = revenue - annual cost. (Key point: Net revenue)

2. Which project is the starting defender and why?
   A. A
   B. B, lowest investment with IRR above MARR
   C. D
   D. C
   Answer: B. Lowest-investment acceptable project is first defender. (Key point: Starting defender)

3. What happens to A and D against B?
   A. Replace B
   B. Incremental IRR 3.65% and 7.10% < 10%, B kept
   C. Both chosen
   D. Ignored
   Answer: B. Both increments fall below MARR. (Key point: Rejected challengers)

4. Which is chosen?
   A. C, incremental 10.38% > 10%
   B. B
   C. A
   D. D
   Answer: A. C's extra Rs 300 earns just above MARR. (Key point: Final)

## Related in Engineering Economics

- [IRR to select best of A, B, C (15 years)](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Airr-select-abc-15yr) — Engineering Economics, chapter 4
- [Best project for 4-year and 8-year service periods](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Aselect-best-project-fw-4yr-irr) — Engineering Economics, chapter 4
- [IRR by trial and error; investment decision](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Airr-trial-error-25000) — Engineering Economics, chapter 3
- [PW sensitivity to investment, income, life by ±25%](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Asensitivity-pw-400000-60000-25) — Engineering Economics, chapter 6
- [Repeatability assumption, select best of three projects](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Arepeatability-abc-14) — Engineering Economics, chapter 4, asked 2×
- [Best of three mutually exclusive projects, 10-year study](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Amutually-exclusive-abc-study-10-years) — Engineering Economics, chapter 4

## Related in other subjects

- [RoR best installed capacity (NRs, 45 years)](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Aror-best-capacity-nrs-10-percent) — Hydropower Engineering, chapter 3
- [Feasibility using two methods, MARR 15%](https://www.bctengineer.com/question.md?key=semester_7%2Fproject_engineering%3Afeasibility-two-methods-marr-15) — Project Engineering, chapter 6
- [Three-point problem: A 800 m N, C 600 m E (600, 200, 500)](https://www.bctengineer.com/question.md?key=semester_3%2Fengineering_geology_i%3Athreepoint-problem-a-800-m) — Engineering Geology I, chapter 5, asked 4×
- [Sources of finance; ARR, payback, IRR, PI decision](https://www.bctengineer.com/question.md?key=semester_7%2Fproject_engineering%3Aproject-finance-sources-investment-decision-c75) — Project Engineering, chapter 6