---
title: "Define IRR; find IRR and ERR of project"
subject: "Engineering Economics"
chapter: 3
key: "semester_6/engineering_economics:irr-err-define-project-15"
source: https://www.bctengineer.com/q/semester_6/engineering_economics/irr-err-define-project-15
site: BCT Engineering
---

# Define IRR; find IRR and ERR of project

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

- Asked in: 2076 Shrawan, 2+6 marks; 2071 Bhadra, 2+6 marks (asked 2 times)
- Video: [Internal Rate of Return IRR and Linear Interpolation - Engineering Economics Lightboard](https://www.youtube.com/watch?v=IP2XNmbKaFk) — Engineering Economics Guy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/engineering_economics/irr-err-define-project-15
- Whole chapter: https://www.bctengineer.com/md/semester_6/engineering_economics/chapter-3

## Question

Define IRR. Find IRR and ERR of the following project. MARR = ε = 15%.
| Year | 0 | 1 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| Cash Flow | -50 | -10 | 30 | 40 | 50 |

## Short answer

IRR is the discount rate that makes the project’s net present worth zero. For the given cash flow, IRR ≈ 19.2%, ERR ≈ 18.2%; both exceed MARR of 15%, so the project is acceptable.

## Answer

#### IRR
The internal rate of return (IRR) is the interest rate i\* at which the PW (or FW, AW) of the net cash flow of a project is zero, i.e. PW of inflows = PW of outflows. A project is accepted if IRR ≥ MARR.

#### Numerical
Assumption: the table shows no cash flow at year 2, so it is taken as zero. Amounts are in the same unit as printed (say Rs thousand).

```text
 Year :  0    1    2    3    4    5
 CF   : -50  -10   0   +30  +40  +50
```
**IRR (trial and error)**: PW(i) = −50 − 10(P/F, i, 1) + 30(P/F, i, 3) + 40(P/F, i, 4) + 50(P/F, i, 5)

| i | PW(i) |
|---|---|
| 15% | 8.759 |
| 18% | 2.271 |
| 20% | -1.588 |
| 19.16% | 0.000 |

Interpolating between 18% and 20%: IRR = 18 + 2 × 2.271/(2.271 − (-1.588)) ≈ 19.2%.

**IRR ≈ 19.16%** > MARR 15%.

**ERR (external rate of return)** with ε = 15%, N = 5:

- PW of outflows at ε: 50 + 10 (P/F, 15%, 1) = 50 + 10 × 0.8696 = 58.696
- FW of inflows at ε (year 5): 30 (F/P, 15%, 2) + 40 (F/P, 15%, 1) + 50 = 30 × 1.3225 + 40 × 1.15 + 50 = 135.675
- (1 + ERR)^5 = 135.675 / 58.696 = 2.3115
- ERR = (2.3115)^(1/5) − 1 = 18.24%

**Answer: IRR ≈ 19.2%, ERR ≈ 18.2%. Both exceed MARR = 15%, so the project is acceptable.** ERR is lower than IRR because the inflows are reinvested at 15%, not at the IRR.

## Check yourself

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

1. What is the IRR?
   A. The rate equal to MARR
   B. The rate at which PW of the net cash flow is zero
   C. The rate at which payback equals life
   D. The average annual profit percentage
   Answer: B. IRR i* makes PW of inflows equal PW of outflows. (Key point: IRR definition)

2. How is the missing year-2 cash flow treated?
   A. As equal to year 1
   B. As equal to year 3
   C. As zero
   D. The question is ignored
   Answer: C. The table shows no year-2 flow, so it is taken as zero. (Key point: Year 2 flow)

3. How is ERR computed here?
   A. (1+ERR)^5 = FW of inflows at 15% / PW of outflows at 15%
   B. ERR equals the interpolated IRR
   C. ERR = average of IRR and MARR
   D. (1+ERR)^5 = PW of inflows / FW of outflows
   Answer: A. Outflows are discounted to year 0 and inflows compounded to year 5 at epsilon = 15%. (Key point: ERR method)

4. Why is ERR (18.2%) lower than IRR (19.2%)?
   A. ERR ignores inflows
   B. MARR is higher than IRR
   C. ERR uses a longer life
   D. Inflows are reinvested at 15%, not at the IRR
   Answer: D. With reinvestment at 15% below the IRR, the overall return drops. (Key point: ERR vs IRR)

## Related in Engineering Economics

- [ERR with MARR = ε = 10%](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Aerr-six-year-cash-flow) — Engineering Economics, chapter 3
- [IRR evaluation, MARR 20%](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Airr-60000-cash-flow-20) — Engineering Economics, chapter 3
- [ERR with MARR 12% and reinvestment 14%](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Aerr-reinvestment-14) — Engineering Economics, chapter 3
- [IRR by AW, both BCR by FW, and ERR](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Airr-aw-bcr-fw-err-250000) — Engineering Economics, chapter 3
- [IRR choose best of four, 10-year study](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Airr-choose-best-abcd-study-10) — Engineering Economics, chapter 4
- [AW sensitivity to investment and net annual cash flow](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Asensitivity-aw-500000-75000) — Engineering Economics, chapter 6

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