---
title: "ERR with MARR = ε = 10%"
subject: "Engineering Economics"
chapter: 3
key: "semester_6/engineering_economics:err-six-year-cash-flow"
source: https://www.bctengineer.com/q/semester_6/engineering_economics/err-six-year-cash-flow
site: BCT Engineering
---

# ERR with MARR = ε = 10%

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

- Asked in: 2067 Mangsir, 6 marks
- Video: [External Rate of Return - Engineering Economics Lightboard](https://www.youtube.com/watch?v=jV6LVt9D1fY) — 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/err-six-year-cash-flow
- Whole chapter: https://www.bctengineer.com/md/semester_6/engineering_economics/chapter-3

## Question

Use ERR method to evaluate the project with following cash flow. MARR = ε = 10%.
| Year | 1st | 2nd | 3rd | 4th | 5th | 6th |
|---|---|---|---|---|---|---|
| Cash flow | -8,00,000 | 2,00,000 | 2,00,000 | -50,000 | 4,00,000 | 4,00,000 |

## Short answer

ERR = 10.81% > MARR 10%, so project is acceptable. PW of outflows Rs 837,566, FW of inflows Rs 1,399,020 over 5 periods.

## Answer

The periods are labelled 1st to 6th year, so the first cash flow is taken as the start (time 0) and the last is 5 intervals later (N = 5). ERR is not affected by shifting every flow by the same period count. ε = MARR = 10%.

```text
 period: 1st      2nd     3rd     4th     5th     6th
 CF    : -8,00,000 +2,00,000 +2,00,000 -50,000 +4,00,000 +4,00,000
 time  :  0         1        2        3        4        5
```
#### Step 1: PW of outflows at ε = 10% (to time 0)
PW = 8,00,000 + 50,000 (P/F, 10%, 3) = 8,00,000 + 50,000 × 0.7513 = Rs 837,566

#### Step 2: FW of inflows at ε = 10% (to time 5)
- 2,00,000 (F/P, 10%, 4) = 2,00,000 × 1.4641 = 292,820
- 2,00,000 (F/P, 10%, 3) = 2,00,000 × 1.3310 = 266,200
- 4,00,000 (F/P, 10%, 1) = 4,00,000 × 1.1 = 440,000
- 4,00,000 = 4,00,000
- Total FW = Rs 1,399,020

#### Step 3: ERR
(1 + ERR)^5 = 1,399,020 / 837,566 = 1.6703

ERR = (1.6703)^(1/5) − 1 = **10.81%**

**Answer: ERR ≈ 10.81% > MARR 10%, so the project is acceptable (only just).**

## Check yourself

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

1. How are the 1st to 6th year flows indexed?
   A. Time 0 to 5, so N = 5
   B. Time 1 to 6, N = 6
   C. All at time 0
   D. N = 3
   Answer: A. The first flow is the start; ERR is unaffected by a uniform shift. (Key point: Time indexing)

2. Which flows are in the PW of outflows?
   A. Only -8,00,000
   B. 2,00,000 flows
   C. -8,00,000 and -50,000 (discounted from time 3)
   D. -50,000 only
   Answer: C. PW = 800,000 + 50,000 x 0.7513 = 837,566. (Key point: Outflows)

3. To where are inflows compounded?
   A. Time 0
   B. Time 5 at 10%
   C. Time 3
   D. Time 6
   Answer: B. Total FW = 1,399,020. (Key point: Inflows)

4. ERR = 10.81% and MARR = 10%. Decision?
   A. Rejected
   B. Cannot decide
   C. Rejected because epsilon equals MARR
   D. Acceptable, just
   Answer: D. ERR > MARR. (Key point: Result)

## Related in Engineering Economics

- [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
- [Define IRR; find IRR and ERR of project](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Airr-err-define-project-15) — Engineering Economics, chapter 3, asked 2×
- [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
- [ERR from separate inflow and outflow rows](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Aerr-cash-inflow-outflow-480000) — 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
- [After-tax cash flow, DDB with switchover](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Aafter-tax-cash-flow-ddb-switchover) — Engineering Economics, chapter 7

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