---
title: "Both types of payback period, 3000 investment"
subject: "Engineering Economics"
chapter: 3
key: "semester_6/engineering_economics:payback-both-3000-cash-flow"
source: https://www.bctengineer.com/q/semester_6/engineering_economics/payback-both-3000-cash-flow
site: BCT Engineering
---

# Both types of payback period, 3000 investment

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

- Asked in: 2069 Bhadra, 4 marks
- Video: [#2 Payback Period - Investment Decision - Financial Management ~ B.COM / BBA / CMA](https://www.youtube.com/watch?v=hAYAdBIzMUI) — Saheb Academy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/engineering_economics/payback-both-3000-cash-flow
- Whole chapter: https://www.bctengineer.com/md/semester_6/engineering_economics/chapter-3

## Question

From the following cash flow, calculate both types of payback period. MARR = 10%.
| EOY | 0 | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|---|
| Cash flow | -3000 | 800 | 1000 | 1100 | 1210 | 1464 |

## Short answer

Simple payback is 3.08 years; discounted payback at 10% MARR is 3.75 years. The discounted period is longer because of the time value of money.

## Answer

```text
 EOY : 0      1    2     3     4     5
 CF  : -3000  800  1000  1100  1210  1464
```
| EOY | CF | Cumulative CF | PW at 10% | Cumulative PW |
|---|---|---|---|---|
| 0 | -3,000 | -3,000 | -3,000.00 | -3,000.00 |
| 1 | 800 | -2,200 | 727.27 | -2,272.73 |
| 2 | 1,000 | -1,200 | 826.45 | -1,446.28 |
| 3 | 1,100 | -100 | 826.45 | -619.83 |
| 4 | 1,210 | 1,110 | 826.45 | 206.61 |
| 5 | 1,464 | 2,574 | 909.03 | 1,115.64 |

#### Simple payback
Cumulative CF is −100 at year 3 and +1,110 at year 4.

θ = 3 + 100 / 1,210 = **3.08 years**

#### Discounted payback (MARR = 10%)
Cumulative PW is −619.83 at year 3 and +206.61 at year 4.

θ_d = 3 + 619.83 / 826.45 = **3.75 years**

**Answer: simple payback = 3.08 years; discounted payback = 3.75 years. The discounted period is longer because future inflows are worth less today. If the firm's cut-off is more than 3.75 years, the project is accepted by both.**

## Check yourself

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

1. What is it?
   A. 3.08 years
   B. 3.75 years
   C. 4.00 years
   D. 2.50 years
   Answer: A. 3 + 100/1,210. (Key point: Simple payback)

2. What is it?
   A. 3.08 years
   B. 4.25 years
   C. 3.75 years
   D. 3.00 years
   Answer: C. 3 + 619.83/826.45. (Key point: Discounted payback)

3. Which cumulative column gives discounted payback?
   A. Cumulative CF
   B. Cumulative PW at 10%
   C. Net CF
   D. Cumulative CF at 0%
   Answer: B. Flows must be discounted at MARR first. (Key point: Which table)

4. Why is discounted payback longer?
   A. Costs are larger
   B. It ignores year 5
   C. It uses simple interest
   D. Future inflows are worth less today
   Answer: D. Discounting shrinks later inflows. (Key point: Relation)

## Related in Engineering Economics

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- [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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