---
title: "IRR by trial and error; investment decision"
subject: "Engineering Economics"
chapter: 3
key: "semester_6/engineering_economics:irr-trial-error-25000"
source: https://www.bctengineer.com/q/semester_6/engineering_economics/irr-trial-error-25000
site: BCT Engineering
---

# IRR by trial and error; investment decision

Engineering Economics · Chapter 3: Basic Methodologies of Engineering Economic Analysis

- Asked in: 2070 Magh, 4 marks
- Video: [How to Calculate IRR (using trial and error)](https://www.youtube.com/watch?v=noWg6tQnHZ0) — Edspira
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/engineering_economics/irr-trial-error-25000
- Whole chapter: https://www.bctengineer.com/md/semester_6/engineering_economics/chapter-3

## Question

Compute IRR by using trial and error process of the following project. Determine also investment decision.
Initial investment = 25,000
Annual revenue = 8,000
Salvage value = 5,000
Useful life = 5 years
MARR = 20%

## Short answer

PW equation: -25,000 + 8,000(P/A,i,5) + 5,000(P/F,i,5)=0. Trial and error yields IRR=21.6%, which exceeds MARR of 20%, so accept the project.

## Answer

```text
 yr 0: -25,000      yr 1..5: +8,000 each       yr 5: +5,000 (salvage)
```
IRR is the rate at which PW = 0:

PW(i) = −25,000 + 8,000 (P/A, i, 5) + 5,000 (P/F, i, 5)

#### Trial and error
| Trial i | (P/A, i, 5) | (P/F, i, 5) | PW(i) |
|---|---|---|---|
| 20% | 2.9906 | 0.4019 | +934 |
| 25% | 2.6893 | 0.3277 | -1,847 |
| 22% | 2.8636 | 0.3700 | -241 |
| 21% | 2.9260 | 0.3855 | +336 |

PW is positive at 21% (+336) and negative at 22% (-241). Interpolating:

IRR = 21 + 1 × 336 / (336 − (-241)) = 21 + 0.58 = **21.58%**

#### Decision
IRR = 21.58% > MARR = 20%, so the investment is acceptable.

**Answer: IRR ≈ 21.6%; accept the project.**

## Check yourself

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

1. Which equation is solved for IRR?
   A. -25,000 + 8,000(P/A, i, 5) + 5,000(P/F, i, 5) = 0
   B. -25,000 + 8,000(P/A, i, 5) = 5,000
   C. -25,000 + 13,000 x 5 = 0
   D. 8,000(P/F, i, 5) = 25,000
   Answer: A. Salvage is a year-5 inflow in addition to the annuity. (Key point: PW equation)

2. Between which trial rates does PW change sign?
   A. 20% and 21%
   B. 21% (+336) and 22% (-241)
   C. 22% and 25%
   D. 10% and 20%
   Answer: B. A sign change brackets the IRR for interpolation. (Key point: Bracketing)

3. What is the interpolated IRR?
   A. 21.00%
   B. 22.58%
   C. 21.58%
   D. 20.58%
   Answer: C. 21 + 336/(336+241) = 21.58%. (Key point: IRR)

4. What is the decision?
   A. Reject since IRR is above 20%
   B. Accept only if IRR > 25%
   C. Reject since PW at 20% is positive
   D. Accept since IRR exceeds MARR 20%
   Answer: D. IRR >= MARR means acceptable. (Key point: Decision)

## Related in Engineering Economics

- [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
- [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
- [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
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- [Modified B/C ratio for 100,000 project](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Amodified-bcr-100000-25000) — Engineering Economics, chapter 3

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