---
title: "ERR method to select best of three projects"
subject: "Engineering Economics"
chapter: 4
key: "semester_6/engineering_economics:err-select-best-abc"
source: https://www.bctengineer.com/q/semester_6/engineering_economics/err-select-best-abc
site: BCT Engineering
---

# ERR method to select best of three projects

Engineering Economics · Chapter 4: Comparative Analysis of Alternatives

- Asked in: 2072 Asoj, 6 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/err-select-best-abc
- Whole chapter: https://www.bctengineer.com/md/semester_6/engineering_economics/chapter-4

## Question

Select the best project by ERR method. Take MARR = 10% and ε = 20%.
| BOY→ | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|---|
| PROJECT A | -64,000 | 26,200 | 29,000 | 30,200 | 31,000 | 31,000 | 26,000 |
| PROJECT B | -68,000 | -4,000 | 39,200 | 38,000 | 38,000 | 38,000 | 38,000 |
| PROJECT C | -75,500 | 20,500 | 40,600 | 40,000 | 39,000 | 39,000 | 32,400 |

## Short answer

ERR at ε=20%: A=28.29%, B=25.99%, C=28.49%. All exceed 10% MARR. Incremental ERR: A→B=14.32%>10%, B→C=39.79%>10%. Project C is best.

## Answer

ERR (external rate of return) assumes net receipts are reinvested outside the project at the external rate ε. Receipts are compounded to the end of the life at ε, expenditures (negative flows) are discounted to year 0 at ε, and ERR i' solves

 Σ E_t(P/F, ε, t)(F/P, i', N) = Σ R_t(F/P, ε, N - t)

Here ε = 20%, MARR = 10%, N = 6 (cash flows in Rs.).

#### Step 1: ERR of each project
**Project A**: 
- Receipts compounded to year 6 at ε: FW(R) = 26,200(2.4883) + 29,000(2.0736) + 30,200(1.7280) + 31,000(1.4400) + 31,000(1.2000) + 26,000(1.0000) = 285,354
- Expenditures discounted to year 0 at ε: PW(E) = 64,000(1.0000) = 64,000
- ERR: 64,000(1 + i')^6 = 285,354 → (1 + i') = (4.4587)^(1/6) → **ERR = 28.29%**

**Project B**: 
- Receipts compounded to year 6 at ε: FW(R) = 39,200(2.0736) + 38,000(1.7280) + 38,000(1.4400) + 38,000(1.2000) + 38,000(1.0000) = 285,269
- Expenditures discounted to year 0 at ε: PW(E) = 68,000(1.0000) + 4,000(0.8333) = 71,333
- ERR: 71,333(1 + i')^6 = 285,269 → (1 + i') = (3.9991)^(1/6) → **ERR = 25.99%**

**Project C**: 
- Receipts compounded to year 6 at ε: FW(R) = 20,500(2.4883) + 40,600(2.0736) + 40,000(1.7280) + 39,000(1.4400) + 39,000(1.2000) + 32,400(1.0000) = 339,679
- Expenditures discounted to year 0 at ε: PW(E) = 75,500(1.0000) = 75,500
- ERR: 75,500(1 + i')^6 = 339,679 → (1 + i') = (4.4991)^(1/6) → **ERR = 28.49%**

| Project | ERR | vs MARR 10% |
|---|---|---|
| A | 28.29% | acceptable |
| B | 25.99% | acceptable |
| C | 28.49% | acceptable |

All three are acceptable, so an incremental comparison is needed (order of investment A 64,000 < B 68,000 < C 75,500). Incremental cash flows are found and ERR is applied on them in the same way.

#### Step 2: incremental ERR
Increment A → B: [-4000, -30200, 10200, 7800, 7000, 7000, 12000]
ERR(B - A) = 14.32% > 10%, so B is better than A.

Increment B → C: [-7500, 24500, 1400, 2000, 1000, 1000, -5600]
ERR(C - B) = 39.79% > 10%, so C is better than B.

(Check: A → C gives 27.10% > 10%.)

**Answer: Select Project C.** C has the highest ERR of its own (28.49%), and each extra investment, from A to B and then from B to C, earns more than the MARR of 10%.

## Check yourself

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

1. How does ERR treat receipts and expenditures?
   A. Receipts compounded to end of life at ε, expenditures discounted to year 0 at ε
   B. All discounted at MARR
   C. All compounded at IRR
   D. Ignored
   Answer: A. ERR assumes reinvestment at external rate ε; the result i' solves the equation. (Key point: ERR idea)

2. What are the individual ERRs?
   A. A 28.29%, B 25.99%, C 28.49%
   B. A 14%, B 10%, C 39%
   C. A 20%, B 20%, C 20%
   D. All 10%
   Answer: A. All exceed MARR 10%, so all are acceptable. (Key point: Own ERRs)

3. Why is incremental ERR needed?
   A. All are acceptable, so choose among them by the return on extra investment
   B. Only A is acceptable
   C. To find MARR
   D. ERR cannot be calculated
   Answer: A. With several acceptable mutually exclusive projects, extra investment must be tested against MARR. (Key point: Why incremental)

4. Which is selected?
   A. None
   B. C
   C. A
   D. B
   Answer: B. B->C increment ERR is 39.79% > 10%, so C is better. (Key point: Final choice)

## Related in Engineering Economics

- [ERR to select between projects A and B](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Aerr-project-a-b-18-12) — 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×
- [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
- [ERR to select best proposal A or B](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Aerr-proposals-a-b-25) — Engineering Economics, chapter 4
- [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
- [Best combination with independent and contingent projects](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Acombination-a-independent-b-contingent-c) — Engineering Economics, chapter 4

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