---
title: "Both types of payback period for Machine A"
subject: "Engineering Economics"
chapter: 3
key: "semester_6/engineering_economics:payback-both-machine-a-5yr"
source: https://www.bctengineer.com/q/semester_6/engineering_economics/payback-both-machine-a-5yr
site: BCT Engineering
---

# Both types of payback period for Machine A

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

- Asked in: 2071 Bhadra, 4 marks
- Video: [How to Calculate Payback Period and Discounted Payback Period](https://www.youtube.com/watch?v=JQgQetvbFwQ) — Ryan O'Connell, CFA, FRM
- 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-machine-a-5yr
- Whole chapter: https://www.bctengineer.com/md/semester_6/engineering_economics/chapter-3

## Question

Machine A: initial investment Rs. 6000, annual benefits Rs. 3000, O & M cost Rs. 1000, salvage value Rs. 1500, MARR 10%. Evaluate both types of payback period. Useful life = 5 years. (Take standard payback period = 3 years.)

## Short answer

Net annual inflow = Rs 2,000. Simple payback = 6,000/2,000 = 3.00 years (marginal accept). Discounted payback using 10% = 3.75 years, exceeding the 3-year standard, so project is rejected.

## Answer

Net annual cash flow = benefit − O&M = 3,000 − 1,000 = Rs 2,000. Salvage Rs 1,500 comes only at year 5 and is not needed for payback. Standard payback = 3 years, MARR = 10%.

```text
 yr 0: -6,000    yr 1..5: +2,000 each    (yr 5 also +1,500 salvage)
```
| EOY | Net CF | Cumulative CF | PW at 10% | Cumulative PW |
|---|---|---|---|---|
| 1 | 2,000 | -4,000 | 1,818.18 | -4,181.82 |
| 2 | 2,000 | -2,000 | 1,652.89 | -2,528.93 |
| 3 | 2,000 | 0 | 1,502.63 | -1,026.30 |
| 4 | 2,000 | 2,000 | 1,366.03 | 339.73 |
| 5 | 2,000 | 4,000 | 1,241.84 | 1,581.57 |

#### Simple payback
θ = Investment / annual net inflow = 6,000 / 2,000 = **3.00 years**. This equals the standard period of 3 years, so it is acceptable (just at the limit).

#### Discounted payback
PW of 3 years of inflows = 2,000 × (P/A, 10%, 3) = 2,000 × 2.4869 = 4,973.70, which is less than 6,000. The balance of 1,026.30 is recovered in year 4, where the PW of the inflow is 2,000 × 0.6830 = 1,366.03.

θ_d = 3 + 1,026.30 / 1,366.03 = **3.75 years**

#### Decision
| Method | Payback | Standard | Result |
|---|---|---|---|
| Simple | 3.00 yr | 3 yr | Accept (marginal) |
| Discounted | 3.75 yr | 3 yr | Reject |

**Answer: simple payback = 3 years (accept, at the limit); discounted payback = 3.75 years (more than 3 years, reject). Since the discounted payback accounts for the time value of money, the machine does not meet the 3-year requirement.**

## Check yourself

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

1. What annual flow is used for payback?
   A. Benefit only, 3,000
   B. O&M only, 1,000
   C. Benefit minus O&M = 2,000
   D. 3,000 + 1,500
   Answer: C. Net inflow is 3,000 - 1,000 = 2,000. (Key point: Net annual flow)

2. What is simple payback?
   A. 3.00 years
   B. 2.00 years
   C. 3.75 years
   D. 5.00 years
   Answer: A. 6,000 / 2,000 = 3 years. (Key point: Simple payback)

3. What is discounted payback?
   A. 3.00 years
   B. 3.75 years
   C. 4.50 years
   D. 2.40 years
   Answer: B. After 3 years PW recovered is 4,973.70; 1,026.30 remains, year-4 PW is 1,366.03, so 3 + 0.75. (Key point: Discounted payback)

4. Standard period 3 years. What is the conclusion?
   A. Both accept
   B. Both reject
   C. Discounted accepts, simple rejects
   D. Simple just accepts; discounted rejects, so the machine fails
   Answer: D. Discounted payback accounts for the time value of money and exceeds 3 years. (Key point: Decision)

## Close variants set in other exams

- [Both types of BCR (FW) for Machine A](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Abcr-fw-machine-a-10yr) — Engineering Economics, chapter 3

## Related in Engineering Economics

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