---
title: "IRR to select best of A, B, C (15 years)"
subject: "Engineering Economics"
chapter: 4
key: "semester_6/engineering_economics:irr-select-abc-15yr"
source: https://www.bctengineer.com/q/semester_6/engineering_economics/irr-select-abc-15yr
site: BCT Engineering
---

# IRR to select best of A, B, C (15 years)

Engineering Economics · Chapter 4: Comparative Analysis of Alternatives

- Asked in: 2068 Jestha, 10 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/irr-select-abc-15yr
- Whole chapter: https://www.bctengineer.com/md/semester_6/engineering_economics/chapter-4

## Question

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 |

## 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.

## Answer

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.)

## Check yourself

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

1. When salvage equals the investment, IRR equals?
   A. Salvage / life
   B. Revenue / Investment
   C. Revenue x life
   D. Investment / Revenue
   Answer: B. The investment is never consumed, so IRR = Revenue / I. (Key point: IRR shortcut)

2. Which has the highest own IRR?
   A. A (12.80%)
   B. C (15.00%)
   C. All equal
   D. B (13.09%)
   Answer: B. But highest IRR is not the selection rule. (Key point: Own IRRs)

3. What happens at C -> B?
   A. 8% < 10%, keep C
   B. Above 10%, choose B
   C. Reject all
   D. Choose A
   Answer: A. Extra Rs 120,000 / Rs 1,500,000 = 8%. (Key point: Incremental C->B)

4. Which is selected?
   A. B
   B. A, since C -> A earns 10.29% > 10%
   C. None
   D. C
   Answer: B. A beats both C and B. (Key point: Final)

## Related in Engineering Economics

- [Incremental IRR to select best of A to D](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Aincremental-irr-abcd-salvage-25) — Engineering Economics, chapter 4
- [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×
- [Sensitivity on investment, revenue and life; interpret](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Asensitivity-30-initial-100000-useful-10) — Engineering Economics, chapter 6
- [Sensitivity to investment, revenue and life](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Asensitivity-30-investment-revenue-life-500000) — Engineering Economics, chapter 6
- [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
- [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

## Related in other subjects

- [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
- [Rank projects by payback, PW, IRR, B/C](https://www.bctengineer.com/question.md?key=semester_7%2Fproject_engineering%3Arank-projects-payback-pw-irr-bc) — Project Engineering, chapter 6
- [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