---
title: "Defender, challenger, ESL of router"
subject: "Engineering Economics"
chapter: 5
key: "semester_6/engineering_economics:defender-challenger-esl-router"
source: https://www.bctengineer.com/q/semester_6/engineering_economics/defender-challenger-esl-router
site: BCT Engineering
---

# Defender, challenger, ESL of router

Engineering Economics · Chapter 5: Replacement Analysis

- Asked in: 2076 Shrawan, 2+2+8 marks
- Video: [Engineering Economics - Economic Service Life](https://www.youtube.com/watch?v=HbIIoQ4_qiE) — Sofia Explains It All
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/engineering_economics/defender-challenger-esl-router
- Whole chapter: https://www.bctengineer.com/md/semester_6/engineering_economics/chapter-5

## Question

Define defender and challenger and explain economic service life. Nepal Telecom is going to purchase a router having initial cost \$18,000 with salvage value of \$12,000 at the end of first year and decreasing by 10% each year thereafter for the remaining useful life. Annual operation and maintenance cost is \$2000 in the first year and increases by \$1500 each year. Its useful life is 8 years. Calculate economic service life of the router.

## Short answer

Defender is the existing asset; challenger is the replacement. Economic service life minimizes equivalent annual cost. For the router at 10% MARR, ESL is 3 years with minimum AEC of $7,706.

## Answer

#### Defender and challenger
- **Defender:** the asset currently in use, which is being considered for replacement. Its cost basis is its present market (opportunity) value, not its original cost.
- **Challenger:** the best available alternative asset that may replace the defender.

#### Economic service life (ESL)
ESL is the number of years of service that gives the **minimum equivalent annual cost (AEC)** of owning and operating an asset. As the asset ages, the capital cost per year falls but operating and maintenance costs rise, so total AEC first falls and then rises; the ESL is where it is lowest.

#### ESL of the router
**MARR is not given; assume i = 10%.** Data: P = \$18,000; salvage value MV₁ = 12,000, then falling 10% a year (MV_n = 12,000(0.9)^(n−1)); O&M = 2,000 + 1,500(k − 1) in year k; maximum life 8 years.

AEC(n) = [18,000 + Σ O&M_k (P/F,10%,k) − MV_n (P/F,10%,n)] (A/P,10%,n)

| n | MV_n | PW of total cost | AEC(n) |
|---|---|---|---|
| 1 | 12,000 | 8,909 | 9,800 |
| 2 | 10,800 | 13,785 | 7,943 |
| 3 | 9,720 | 19,165 | 7,706 |
| 4 | 8,748 | 24,932 | 7,865 |
| 5 | 7,873 | 30,986 | 8,174 |
| 6 | 7,086 | 37,237 | 8,550 |
| 7 | 6,377 | 43,609 | 8,958 |
| 8 | 5,740 | 50,035 | 9,379 |

Example for n = 3: PW of O&M = 2,000(0.9091) + 3,500(0.8264) + 5,000(0.7513) = 8,467; PW = 18,000 + 8,467 − 9,720(0.7513) = 19,165; AEC = 19,165(0.40211) = 7,706.

AEC is lowest at n = 3 (\$7,706) and rises afterwards.

**Answer: The economic service life of the router is 3 years** (minimum AEC ≈ \$7,706 per year at i = 10%).

## Check yourself

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

1. What is a defender and what is its cost basis?
   A. The new asset; its purchase price
   B. The current asset; its original cost
   C. The cheapest asset; its salvage value
   D. The current asset; its present market value
   Answer: D. The defender is the asset in use under replacement consideration and is valued at present market value, not original cost. (Key point: Defender)

2. What is economic service life?
   A. The number of years giving minimum equivalent annual cost
   B. The physical life of the asset
   C. The year with maximum salvage value
   D. The year when O&M is lowest
   Answer: A. ESL minimises AEC; capital cost per year falls and O&M rises with age, so total AEC has a minimum. (Key point: ESL definition)

3. Which formula gives AEC(n) for the router?
   A. [18,000 + sum O&M - MV_n](A/F,10%,n)
   B. [18,000 + sum PW of O&M(P/F) - MV_n(P/F,10%,n)](A/P,10%,n)
   C. [18,000 - sum O&M + MV_n](A/P,10%,n)
   D. 18,000(A/P) + MV_n
   Answer: B. PW of investment plus PW of O&M minus PW of salvage, converted with (A/P,i,n). (Key point: Router AEC formula)

4. AEC values are 9,800; 7,943; 7,706; 7,865; ... What is the ESL?
   A. 1 year
   B. 2 years
   C. 4 years
   D. 3 years
   Answer: D. The minimum AEC (7,706) occurs at n = 3. (Key point: Router ESL)

## Related in Engineering Economics

- [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
- [Define ESL; find ESL of 50,000 asset](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Aesl-define-50000-15-percent-growth) — 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×
- [Sensitivity on initial cost and life; interpret](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Asensitivity-30-initial-cost-life-25) — Engineering Economics, chapter 6
- [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
- [Best project for 4-year and 8-year service periods](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Aselect-best-project-fw-4yr-irr) — Engineering Economics, chapter 4

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