---
title: "IRR by AW, both BCR by FW, and ERR"
subject: "Engineering Economics"
chapter: 3
key: "semester_6/engineering_economics:irr-aw-bcr-fw-err-250000"
source: https://www.bctengineer.com/q/semester_6/engineering_economics/irr-aw-bcr-fw-err-250000
site: BCT Engineering
---

# IRR by AW, both BCR by FW, and ERR

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

- Asked in: 2069 Bhadra, 4+4+4 marks
- Video: [Rate of return [IRR ,ERR ] | Benefit cost ratio](https://www.youtube.com/watch?v=t2MpiusaZhA) — 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-aw-bcr-fw-err-250000
- Whole chapter: https://www.bctengineer.com/md/semester_6/engineering_economics/chapter-3

## Question

Equipment costs 2,50,000 and has salvage value of 50,000 at the end of its expected life of 5 years. Annual expenses will be 40,000. It will produce a revenue of 120,000 per year. MARR = 20% = ε.
i) Evaluate IRR using AW formulation.
ii) Evaluate both types of B/C ratio with FW formulation.
iii) Find ERR.

## Short answer

IRR by AW formulation is 21.58%. Using FW at 20%, conventional B/C = 1.03, modified B/C = 1.04. ERR = 20.88%. All exceed MARR, so accept.

## Answer

Net annual inflow = 1,20,000 − 40,000 = Rs 80,000. I = 2,50,000, S = 50,000, N = 5, MARR = ε = 20%.

```text
 yr 0: -2,50,000     yr 1..5: +80,000 (net)      yr 5: +50,000
```
#### i) IRR using AW
AW(i) = −2,50,000 (A/P, i, 5) + 50,000 (A/F, i, 5) + 80,000 = 0

| i | (A/P, i, 5) | (A/F, i, 5) | AW(i) |
|---|---|---|---|
| 20% | 0.33438 | 0.13438 | +3,124 |
| 25% | 0.37185 | 0.12185 | -6,869 |
| 22% | 0.34921 | 0.12921 | -841 |
| 21% | 0.34177 | 0.13177 | +1,147 |

Interpolating between 21% and 22%: IRR = 21 + 1147/(1147 − (-841)) = **21.58%** > 20%. Accept.

#### ii) B/C ratios with FW (i = 20%)
(F/A, 20%, 5) = 7.4416; (F/P, 20%, 5) = 2.4883

- FW(B) = 1,20,000 × 7.4416 = Rs 892,992
- FW(O&M) = 40,000 × 7.4416 = Rs 297,664
- FW(I) = 2,50,000 × 2.4883 = Rs 622,080

Conventional B/C = 892,992 / (622,080 + 297,664 − 50,000) = **1.027**

Modified B/C = (892,992 − 297,664) / (622,080 − 50,000) = **1.041**

Both are greater than 1, so accept.

#### iii) ERR (ε = 20%)
- PW of outflows at ε = 2,50,000 (only at year 0)
- FW of inflows at year 5 = 80,000 (F/A, 20%, 5) + 50,000 = 80,000 × 7.4416 + 50,000 = Rs 645,328
- (1 + ERR)^5 = 645,328 / 2,50,000 = 2.5813
- ERR = (2.5813)^(1/5) − 1 = **20.88%**

ERR = 20.88% > MARR = 20%, so accept.

**Answer: IRR ≈ 21.58%; conventional B/C = 1.03, modified B/C = 1.04; ERR ≈ 20.88%. All give the same decision: accept.**

## Check yourself

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

1. Which expression is set to zero?
   A. -250,000(A/P, i, 5) + 50,000(A/F, i, 5) + 80,000
   B. -250,000(P/A) + 80,000
   C. 250,000(A/P) - 80,000
   D. 50,000(A/P) + 80,000(A/F)
   Answer: A. AW of investment, salvage and net annual inflow. (Key point: AW equation for IRR)

2. What is the IRR?
   A. 20.00%
   B. 25.00%
   C. 21.58%
   D. 22.00%
   Answer: C. Interpolating between 21% (+1,147) and 22% (-841). (Key point: IRR)

3. What are the FW B/C ratios?
   A. 0.97 and 0.96
   B. 1.03 conventional and 1.04 modified
   C. 1.50 and 1.60
   D. 1.04 and 1.03
   Answer: B. 892,992/869,744 and 595,328/572,080. (Key point: B/C ratios)

4. How is ERR found?
   A. (1+ERR)^5 = (80,000 x 7.4416 + 50,000)/250,000, ERR = 20.88%
   B. ERR = MARR = 20%
   C. (1+ERR)^5 = 250,000/645,328
   D. ERR = IRR - MARR
   Answer: A. Inflows compound to year 5 at 20%; the outflow is only at year 0. (Key point: ERR)

## Related in Engineering Economics

- [Both BCR with FW, gradient revenue](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Abcr-fw-gradient-revenue-250000) — Engineering Economics, chapter 3
- [PW, AW and FW decision for equipment](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Apw-aw-fw-equipment-500000) — Engineering Economics, chapter 3
- [Both types of BCR with FW, growing income](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Abcr-fw-growing-income) — Engineering Economics, chapter 3
- [Both types of BCR (FW) for Machine A](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Abcr-fw-machine-a-10yr) — Engineering Economics, chapter 3
- [Modified BCR with co-termination, AW](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Amodified-bcr-cotermination-aw) — Engineering Economics, chapter 4
- [Financial vs economic analysis; both BCR](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Afinancial-vs-economic-bcr-300000) — Engineering Economics, chapter 3

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