---
title: "Replace existing machine now; AEC and ESL"
subject: "Engineering Economics"
chapter: 5
key: "semester_6/engineering_economics:replace-existing-machine-cash-flow-aec"
source: https://www.bctengineer.com/q/semester_6/engineering_economics/replace-existing-machine-cash-flow-aec
site: BCT Engineering
---

# Replace existing machine now; AEC and ESL

Engineering Economics · Chapter 5: Replacement Analysis

- Asked in: 2076 Shrawan, 4+8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/engineering_economics/replace-existing-machine-cash-flow-aec
- Whole chapter: https://www.bctengineer.com/md/semester_6/engineering_economics/chapter-5

## Question

An existing machine has market value of Rs 10000 and can be used for five more years. Its value will decline linearly and will be Rs 2000 at the end. The operation and maintenance cost will be Rs 4000 at the end of this year and will increase by Rs 1000 per year for remaining years. A new machine is available in market at Rs 25000 now and can be used for five years with annual operating cost of Rs 2000 per year. Its market value will decline by Rs 1500 each year. Use MARR = [value not printed] %.
a) Is it worthwhile to replace the existing machine now? Use cash flow approach.
b) Find AEC and ESL of both defender and challenger.

## Short answer

Assume MARR = 10%. Incremental cash flow PW is Rs 9,068 positive, so replace now. Defender ESL is 1 year with AEC Rs 6,600; challenger ESL is 5 years with AEC Rs 5,728.

## Answer

**MARR is not printed in the question; assume MARR = 10%.** Study period = 5 years (the remaining life of the defender and the life of the challenger).

Defender (existing): market value now 10,000, falling by 1,600 per year to 2,000 at year 5 (market value = 10,000 − 1,600n). O&M = 4,000, 5,000, 6,000, 7,000, 8,000.
Challenger (new): cost 25,000; O&M 2,000 each year; market value = 25,000 − 1,500n (17,500 at year 5).

#### a) Cash-flow approach (incremental: challenger − defender)
Replacing now means selling the old machine for 10,000 (a receipt of 10,000 against the 25,000 purchase), saving O&M each year, and giving up the old machine's salvage at year 5 for the new machine's.
| Year | Defender O&M | Defender MV | Challenger O&M | Challenger MV | Incremental CF (replace) |
|---|---|---|---|---|---|
| 1 | 4000 | 8,400 | 2000 | 23,500 | 2,000 |
| 2 | 5000 | 6,800 | 2000 | 22,000 | 3,000 |
| 3 | 6000 | 5,200 | 2000 | 20,500 | 4,000 |
| 4 | 7000 | 3,600 | 2000 | 19,000 | 5,000 |
| 5 | 8000 | 2,000 | 2000 | 17,500 | 21,500 |

Year 0: −25,000 + 10,000 = −15,000. Years 1–5: savings in O&M = defender O&M − 2,000 = 2,000, 3,000, 4,000, 5,000, 6,000. Year 5: also + (17,500 − 2,000) = +15,500.
```text
Cash flow: 0: −15,000 | 1: +2,000 | 2: +3,000 | 3: +4,000 | 4: +5,000 | 5: +6,000 + 15,500 = +21,500
PW = −15,000 + 2,000(P/F,10%,1) + 3,000(P/F,10%,2) + 4,000(P/F,10%,3) + 5,000(P/F,10%,4) + 21,500(P/F,10%,5)
   = 9,068
```
PW of replacing = Rs 9,068 > 0. **Yes, it is worthwhile to replace the existing machine now.**

#### b) AEC and economic service life (ESL)
For a service life of n years: AEC(n) = [P + Σ O&M_k (P/F,10%,k) − MV_n (P/F,10%,n)] (A/P,10%,n), where P is the investment (for the defender, its present market value 10,000).

**Defender**
| n | PW of cost | AEC(n) |
|---|---|---|
| 1 | 6,000 | 6,600 |
| 2 | 12,149 | 7,000 |
| 3 | 18,370 | 7,387 |
| 4 | 24,599 | 7,760 |
| 5 | 30,783 | 8,121 |

**Challenger**
| n | PW of cost | AEC(n) |
|---|---|---|
| 1 | 5,455 | 6,000 |
| 2 | 10,289 | 5,929 |
| 3 | 14,572 | 5,860 |
| 4 | 18,362 | 5,793 |
| 5 | 21,715 | 5,728 |

Example (defender, n = 1): (10,000 + 4,000/1.1 − 8,400/1.1) × 1.1 = 6,600.

- Defender: AEC is lowest at n = 1, so **ESL(defender) = 1 year**, AEC = Rs 6,600.
- Challenger: AEC keeps falling to n = 5, so **ESL(challenger) = 5 years**, AEC = Rs 5,728.

The challenger's minimum AEC (5,728) is below the defender's (6,600), and at the full 5 years AEC(defender) = 8,121 against 5,728, which agrees with the cash-flow result.

**Answer:** Replace now. AEC and ESL: defender Rs 6,600 at 1 year; challenger Rs 5,728 at 5 years.

## Check yourself

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

1. In the incremental cash-flow approach, how is the sale of the old machine treated at year 0?
   A. Ignored
   B. Added to the cost of the new machine
   C. Deducted as 10,000 from the 25,000 purchase, giving -15,000
   D. Treated as a year-5 receipt
   Answer: C. Replacing now yields the old machine's market value of 10,000, so net year 0 flow = -25,000 + 10,000 = -15,000. (Key point: Defender's investment base)

2. What are the annual incremental inflows in years 1-4?
   A. Challenger O&M minus defender O&M
   B. Defender O&M minus challenger O&M (savings of 2,000 to 5,000)
   C. Challenger market value
   D. Defender market value
   Answer: B. Replacement saves the difference between the defender's and challenger's O&M each year. (Key point: Annual incremental flows)

3. Why is the year 5 incremental flow 21,500?
   A. Savings of 6,000 plus the extra salvage (17,500 - 2,000)
   B. Savings of 6,000 plus 17,500
   C. Savings of 8,000 plus 2,000
   D. Only the new machine's salvage
   Answer: A. 6,000 O&M saving plus 15,500 extra market value gives 21,500. (Key point: Year 5 flow)

4. How is the ESL of an asset found from AEC(n)?
   A. The n with the highest AEC
   B. The last year of the asset's life
   C. The n with the lowest AEC (defender 1 yr, challenger 5 yr)
   D. The n where O&M equals salvage
   Answer: C. ESL is the service life giving minimum AEC; defender's is lowest at n = 1 (6,600), the challenger's at n = 5 (5,728). (Key point: Economic service life)

## Related in Engineering Economics

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- [Sunk cost, ESL, replacement strategy from AEC table](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Asunk-cost-esl-replacement-aec-table) — Engineering Economics, chapter 5, asked 2×
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