---
title: "Define ESL; find ESL of 50,000 asset"
subject: "Engineering Economics"
chapter: 5
key: "semester_6/engineering_economics:esl-define-50000-15-percent-growth"
source: https://www.bctengineer.com/q/semester_6/engineering_economics/esl-define-50000-15-percent-growth
site: BCT Engineering
---

# Define ESL; find ESL of 50,000 asset

Engineering Economics · Chapter 5: Replacement Analysis

- Asked in: 2069 Poush
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/engineering_economics/esl-define-50000-15-percent-growth
- Whole chapter: https://www.bctengineer.com/md/semester_6/engineering_economics/chapter-5

## Question

Define economic service life of an asset. From the following information find the economic service life of an asset.
| Initial investment (Rs.) | 50,000 |
|---|---|
| Annual operating cost (Rs.) | 10,000 for the first year and increasing by 15% over the previous year |
| Salvage value (Rs.) | Declining each year by 20% from the previous year's salvage value |
| Useful life (year) | 7 |
| MARR | 15% |

## Short answer

Economic service life is the ownership period minimizing equivalent annual cost (capital recovery plus equivalent annual operating cost). For the asset with Rs 50,000 investment, operating costs starting at Rs 10,000 increasing 15% yearly, salvage declining 20% yearly, and 15% MARR, the ESL is 5 years with minimum AEC of Rs 25,456.

## Answer

#### Economic service life
The economic service life (ESL) of an asset is the number of years of ownership that gives the lowest equivalent annual cost (AEC), where AEC = capital recovery cost (investment less salvage) + equivalent annual operating cost. It differs from physical life and is the right period to use when evaluating replacements.

#### Calculation
P = 50,000; O&M in year k = 10,000(1.15)^(k−1); MV_n = 50,000(0.8)^n; MARR = 15%.
AEC(n) = [50,000 + Σ O&M_k (P/F,15%,k) − MV_n (P/F,15%,n)] (A/P,15%,n)

| n | MV_n | O&M year n | PW of cost | AEC(n) |
|---|---|---|---|---|
| 1 | 40,000 | 10,000 | 23,913 | 27,500 |
| 2 | 32,000 | 11,500 | 43,195 | 26,570 |
| 3 | 25,600 | 13,225 | 59,255 | 25,952 |
| 4 | 20,480 | 15,209 | 73,073 | 25,595 |
| 5 | 16,384 | 17,490 | 85,333 | 25,456 |
| 6 | 13,107 | 20,114 | 96,507 | 25,501 |
| 7 | 10,486 | 23,131 | 106,928 | 25,701 |

Example, n = 3: PW = 50,000 + 10,000(0.8696) + 11,500(0.7561) + 13,225(0.6575) − 25,600(0.6575) = 59,255; AEC = 59,255(0.43798) = 25,952.

AEC falls until n = 5 (Rs 25,456) and then rises.

**Answer: Economic service life = 5 years**, with minimum AEC ≈ Rs 25,456 per year.

## Check yourself

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

1. What is the economic service life?
   A. The physical life of the asset
   B. The number of years of ownership that gives the lowest AEC
   C. The years until salvage value is zero
   D. The years until O&M doubles
   Answer: B. ESL minimises AEC (capital recovery plus equivalent annual operating cost) and differs from physical life. (Key point: ESL definition)

2. How is the market value in year n computed in this problem?
   A. 50,000 - 20% x n
   B. 50,000(0.8)^n, declining 20% each year
   C. 50,000(1.2)^n
   D. 10,000(0.8)^n
   Answer: B. Salvage falls 20% from the previous year: MV_n = 50,000(0.8)^n. (Key point: Salvage value model)

3. Why does AEC first fall and then rise?
   A. Capital cost per year falls with longer use, but rising O&M eventually dominates
   B. MARR changes yearly
   C. Salvage value increases
   D. Investment grows
   Answer: A. Spreading the investment over more years lowers capital cost per year, but increasing O&M eventually outweighs it. (Key point: Shape of AEC)

4. AEC values: 27,500; 26,570; 25,952; 25,595; 25,456; 25,501; 25,701. What is the ESL?
   A. 4 years
   B. 7 years
   C. 6 years
   D. 5 years
   Answer: D. Minimum AEC (Rs 25,456) occurs at n = 5. (Key point: ESL result)

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