---
title: "ESL and replacement strategy for printing machines"
subject: "Engineering Economics"
chapter: 5
key: "semester_6/engineering_economics:printing-machine-old-new-aec-esl"
source: https://www.bctengineer.com/q/semester_6/engineering_economics/printing-machine-old-new-aec-esl
site: BCT Engineering
---

# ESL and replacement strategy for printing machines

Engineering Economics · Chapter 5: Replacement Analysis

- Asked in: 2069 Bhadra, 2+10 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/engineering_economics/printing-machine-old-new-aec-esl
- Whole chapter: https://www.bctengineer.com/md/semester_6/engineering_economics/chapter-5

## Question

A firm has a contract to provide printing service to IOE for next 8 years. It can provide the service using its old printing machine (the current defender) or the newly bought machine (the challenger). After the contract work neither the old machine nor the new machine will be retained. Considering the annual equivalent costs of the old machine and new machine as follows, what are their economic service life? And what is the best replacement strategy?
| Number of years (n) | Old machine (Rs) | New machine (Rs) |
|---|---|---|
| 1 | 515,000 | 750,000 |
| 2 | 510,000 | 615,000 |
| 3 | 550,000 | 586,000 |
| 4 | 596,000 | 583,000 |
| 5 | 644,000 | 590,000 |

## Short answer

Old machine ESL = 2 years (AEC Rs 510,000); new machine ESL = 4 years (AEC Rs 583,000). For the 8-year contract, the optimal strategy is running the old machine for 2 years, then replacing with new machines for 3+3 years, giving PW ≈ Rs 2,994,366 (AEC ≈ Rs 561,276), which is cheaper than other combinations.

## Answer

**MARR is not given; assume i = 10%** (the ranking is the same at 8%, 12% and 15%). The AEC values are the costs per year if each machine is kept n years.

#### Economic service life (minimum AEC)
| n | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| Old machine | 515,000 | **510,000** | 550,000 | 596,000 | 644,000 |
| New machine | 750,000 | 615,000 | 586,000 | **583,000** | 590,000 |

- **ESL of old machine = 2 years** (AEC = Rs 5,10,000).
- **ESL of new machine = 4 years** (AEC = Rs 5,83,000).

#### Replacement strategy for the 8-year contract
Neither machine is kept after the contract, so the 8 years must be covered by "old machine for n years, then new machine(s) for the rest". PW of a plan = AEC_old(n)(P/A,10%,n) + Σ AEC_new(m)(P/A,10%,m)(P/F,10%,start year).

Example, old 2 years then new 3 + 3 years:
```text
PW = 510,000(1.7355) + 586,000(2.4869)(0.8264) + 586,000(2.4869)(0.6209)
   = 885,124 + 1,204,376 + 904,866 = 2,994,366
```
| Old machine kept | New machine(s) for | PW of cost (Rs) | Equivalent AEC over 8 yr |
|---|---|---|---|
| Old 2 yr | 3 + 3 | 2,994,366 | 561,276 |
| Old 3 yr | 5 | 3,048,132 | 571,354 |
| Old 1 yr | 4 + 3 | 3,053,076 | 572,281 |
| New only | 4 + 4 | 3,110,262 | 583,000 |
| Old 4 yr | 4 | 3,151,470 | 590,724 |
| Old 5 yr | 3 | 3,346,132 | 627,212 |

**Answer:** Economic lives: old machine 2 years, new machine 4 years. The cheapest plan is to **run the old machine for 2 years, then replace it with a new machine, which is itself replaced after 3 years (3 + 3)**, giving PW ≈ Rs 2,994,366 (AEC ≈ Rs 561,276). Using the old machine for 3 years and then a new machine for 5 years is only about 2% dearer; running only new machines (4 + 4 years) costs about 4% more.

## Check yourself

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

1. Using the AEC table, what are the economic service lives of the old and new machines?
   A. Old 1 year, new 1 year
   B. Old 5 years, new 5 years
   C. Old 2 years, new 4 years
   D. Old 3 years, new 3 years
   Answer: C. ESL is where AEC is minimum: old Rs 510,000 at 2 years; new Rs 583,000 at 4 years. (Key point: ESL from AEC table)

2. Why is the economic life alone not enough to decide the best strategy for the 8-year contract?
   A. The contract needs a fixed 8-year coverage, so the old machine must be run for some years and the new machine(s) cover the rest, and the plans are compared by PW
   B. Because the old machine cannot be used at all
   C. Because AEC values are always wrong
   D. Because the new machine must be kept for exactly 8 years
   Answer: A. Neither machine is kept after the contract, so different combinations covering 8 years must be compared by total PW. (Key point: Why a replacement plan is needed)

3. How is the PW of a plan 'old n years, then new machine(s)' computed?
   A. Add the AECs of all machines without discounting
   B. Take the AEC of the new machine times 8
   C. AEC_old(n)(P/A,i,n) plus each new-machine AEC x (P/A) x (P/F) for its start year
   D. Use only the first-year costs
   Answer: C. Each stage's AEC is converted to PW over its duration and discounted back to time zero. (Key point: PW of a plan)

4. Which plan has the lowest PW of cost?
   A. New machine only, 4 + 4 years
   B. Old for 2 years, then new for 3 + 3 years
   C. Old for 5 years, then new for 3 years
   D. Old for 4 years, then new for 4 years
   Answer: B. Old 2 yr then new 3 + 3 has the lowest PW of about Rs 2,994,366. (Key point: Best strategy)

## Related in Engineering Economics

- [Defender, challenger, ESL of router](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Adefender-challenger-esl-router) — Engineering Economics, chapter 5
- [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×
- [Replacement analysis, defender vs challenger by AEC](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Areplacement-analysis-defender-challenger-aec) — Engineering Economics, chapter 5
- [ESL of new machine; when to replace old](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Aeconomic-service-life-new-machine-old) — Engineering Economics, chapter 5
- [Reasons for replacement; strategy from AEC table](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Areasons-replacement-aec-table-12) — Engineering Economics, chapter 5, asked 2×
- [Replace existing machine now; AEC and ESL](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Areplace-existing-machine-cash-flow-aec) — Engineering Economics, chapter 5

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