---
title: "ERR with MARR 12% and reinvestment 14%"
subject: "Engineering Economics"
chapter: 3
key: "semester_6/engineering_economics:err-reinvestment-14"
source: https://www.bctengineer.com/q/semester_6/engineering_economics/err-reinvestment-14
site: BCT Engineering
---

# ERR with MARR 12% and reinvestment 14%

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

- Asked in: 2076 Shrawan, 6 marks
- Video: [External Rate of Return](https://www.youtube.com/watch?v=lGqZvO3JHJ4) — Perennial Pride
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/engineering_economics/err-reinvestment-14
- Whole chapter: https://www.bctengineer.com/md/semester_6/engineering_economics/chapter-3

## Question

Calculate the ERR of the following cash flow. MARR = 12%, reinvestment rate = 14%.
| EOY | 0 | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|---|
| Cash flow | -90,000 | 25,000 | 40,000 | -30,000 | 50,000 | 75,000 |

## Short answer

Outflows are discounted to present worth at 14% and inflows compounded to future worth at the same rate; the rate equating them is the ERR. ERR ≈ 16.19%, exceeding the MARR of 12%, so the project is acceptable.

## Answer

The ERR method moves all net outflows to time 0 and all net inflows to the end of the project at the external reinvestment rate ε. ERR is the rate that equates the two. Accept if ERR ≥ MARR. Here ε = 14% and MARR = 12% (used only for the decision).

```text
 EOY :  0        1       2       3       4       5
 CF  : -90,000 +25,000 +40,000 -30,000 +50,000 +75,000
```
**Step 1: PW of outflows at ε = 14%** (years 0 and 3)

PW = 90,000 + 30,000 (P/F, 14%, 3) = 90,000 + 30,000 × 0.67497 = Rs 110,249.15

**Step 2: FW of inflows at ε = 14% (at year 5)**

FW = 25,000 (F/P, 14%, 4) + 40,000 (F/P, 14%, 3) + 50,000 (F/P, 14%, 1) + 75,000
- 25,000 × 1.68896 = 42,224.00
- 40,000 × 1.48154 = 59,261.76
- 50,000 × 1.14 = 57,000.00
- 75,000
- Total = Rs 233,485.76

**Step 3: ERR**

(1 + ERR)^5 = 233,485.76 / 110,249.15 = 2.1178

ERR = (2.1178)^(1/5) − 1 = **16.19%**

**Answer: ERR ≈ 16.19% > MARR 12%, so the project is acceptable.** (The cash flow changes sign three times, so IRR may be unreliable; ERR avoids this problem.)

## Check yourself

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

1. How does ERR treat cash flows?
   A. All flows moved to year 3
   B. Outflows moved to time 0 and inflows to the end at the reinvestment rate
   C. Inflows at MARR, outflows at IRR
   D. Flows are left where they are
   Answer: B. ERR equates PW of outflows and FW of inflows at epsilon. (Key point: ERR principle)

2. Which rate is used to move flows and which for the decision?
   A. Epsilon = 14% for moving flows; MARR = 12% for the decision
   B. MARR for moving; epsilon for decision
   C. Both use 14%
   D. Both use 12%
   Answer: A. MARR is only compared with the final ERR. (Key point: Rates' roles)

3. Which flows count in the PW of outflows?
   A. Only -90,000
   B. All positive flows
   C. -90,000 (year 0) and -30,000 (year 3)
   D. -90,000 and 75,000
   Answer: C. PW = 90,000 + 30,000 x 0.67497 = 110,249. (Key point: Outflows)

4. Result: ERR = 16.19%. What is the decision?
   A. Reject, since ERR exceeds 14%
   B. Reject, since flows change sign
   C. Accept only if IRR agrees
   D. Accept, since ERR exceeds MARR 12%
   Answer: D. ERR >= MARR means acceptable; ERR also avoids multiple-root problems. (Key point: Result)

## 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
- [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
- [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
- [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
- [After-tax cash flow, sinking fund depreciation](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Aaftertax-cash-flow-sinking-fund-25000) — Engineering Economics, chapter 7

## Related in other subjects

- [Porosity at 50%, 75%, 90% hydration](https://www.bctengineer.com/question.md?key=semester_5%2Fconcrete_technology_and_masonry_structure%3Aporosity-at-hydration-stages) — Concrete Technology and Masonry Structure, chapter 4
- [Brayton cycle pressure ratio 12: mass flow for 90 MW](https://www.bctengineer.com/question.md?key=semester_2%2Ffundamental_of_thermodynamics_and_heat_transfer%3Abrayton-pressure-ratio-12-90mw) — Fundamental of Thermodynamics & Heat Transfer, chapter 6
- [Drawing hall 40 × 25 × 6 m, 90 lm/m² LED](https://www.bctengineer.com/question.md?key=semester_6%2Findustrial_power_distribution_and_illumination%3Adrawing-hall-40x25-90lux-led) — Industrial Power Distribution and Illumination, chapter 7, asked 2×
- [EVA; 60 days, 25% work, Rs 12,00,000](https://www.bctengineer.com/question.md?key=semester_7%2Fproject_engineering%3Aeva-numerical-60-days-25-percent) — Project Engineering, chapter 4