--- title: "Sunk cost, ESL, replacement strategy from AEC table" subject: "Engineering Economics" chapter: 5 key: "semester_6/engineering_economics:sunk-cost-esl-replacement-aec-table" source: https://www.bctengineer.com/q/semester_6/engineering_economics/sunk-cost-esl-replacement-aec-table site: BCT Engineering --- # Sunk cost, ESL, replacement strategy from AEC table Engineering Economics · Chapter 5: Replacement Analysis - Asked in: 2071 Bhadra, 8+4 marks - Video: [Engineering Economics - AW & ESL Challenger](https://www.youtube.com/watch?v=3x1tKHDX3qg) — 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/sunk-cost-esl-replacement-aec-table - Whole chapter: https://www.bctengineer.com/md/semester_6/engineering_economics/chapter-5 ## Question Explain about the sunk cost, economic life and reasons for replacement of an asset. The annual equivalent cost of defender and challenger are given in the table below. What is the best replacement strategy? Use MARR = 10%. The planning horizon of the project is 8 years. | End of year (n) | 1 | 2 | 3 | 4 | 5 | 6 | |---|---|---|---|---|---|---| | (AEC)D | 5400 | 5200 | 5500 | 5700 | 6200 | 6600 | | (AEC)C | 7700 | 6200 | 5700 | 5600 | 5680 | 5900 | ## Short answer Sunk costs are past irrecoverable costs ignored in replacement decisions. Economic life minimizes AEC. With given AECs, defender ESL is 2 years (AEC 5,200), challenger ESL is 4 years (5,600). Best strategy: keep defender for 2 years, then replace with challengers used for 3 years each, total PW Rs 29,541 (AEC 5,537). ## Answer #### Sunk cost A sunk cost is a cost that has already been incurred and cannot be recovered whatever is decided now, for example the book value of an old machine above its present market value. It must be ignored in replacement analysis; only the present market value of the defender and future cash flows matter. Treating the sunk cost as part of the defender's investment would wrongly favour keeping the old asset. #### Economic life Economic life (economic service life) is the period of service that minimises the asset's equivalent annual cost (AEC). Before it, capital recovery cost is falling faster than operating cost rises; after it, rising operating and maintenance cost dominates. #### Reasons for replacement - **Physical impairment:** wear and tear raises maintenance cost and reduces output or quality. - **Obsolescence:** a new machine gives the same service at lower cost or better quality. - **Inadequacy:** capacity is no longer enough for the demand. - **Requirement changes:** changes in product, law or safety standards. - **Financial reasons:** lease or buy decisions, tax effects, or the risk of high repair bills. #### Strategy from the AEC table (MARR = 10%, 8-year horizon) From the table: defender minimum AEC = 5,200 at n = 2 (ESL = 2 years); challenger minimum AEC = 5,600 at n = 4 (ESL = 4 years). Because the defender's minimum is lower than the challenger's, keep it at least until year 2. Since the horizon is 8 years and tables stop at 6, each plan is the defender for n years followed by challengers covering the remaining years. PW of a plan = AEC_D(n)(P/A,10%,n) + AEC_C(m)(P/A,10%,m)(P/F,10%,n) (+ the same for a second challenger, if used). Worked example, defender 2 years then challenger 3 + 3 years: ```text PW = 5,200(P/A,10%,2) + [5,700(P/A,10%,3)](P/F,10%,2) + [5,700(P/A,10%,3)](P/F,10%,5) = 5,200(1.7355) + 5,700(2.4869)(0.8264) + 5,700(2.4869)(0.6209) = 9,025 + 11,714 + 8,802 = 29,541 ``` | Defender kept | Then challenger(s) for | PW of cost | Equivalent AEC (8 yr) | |---|---|---|---| | Defender 2 yr | 3 + 3 yr | 29,541 | 5,537 | | Defender 1 yr | 4 + 3 yr | 29,848 | 5,595 | | Defender 3 yr | 5 yr | 29,855 | 5,596 | | Challenger only | 4 + 4 yr | 29,876 | 5,600 | | Defender 4 yr | 4 yr | 30,193 | 5,659 | | Defender 5 yr | 3 yr | 32,304 | 6,055 | | Defender 6 yr | 2 yr | 34,819 | 6,527 | The lowest PW is Rs 29,541 (equivalent to Rs 5,537 per year). **Answer:** Keep the defender for **2 years** (its economic life), then replace it by the challenger and use challengers (each replaced after 3 years) for the remaining 6 years. The three plans "defender 1 yr, then challengers 4 + 3", "defender 3 yr, then challenger 5" and "challenger replaced after 4 yr each" are within about 1% of this, so the choice is not very sensitive. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How should a sunk cost be treated in replacement analysis? A. Added to the defender's investment B. Ignored; only present market value and future cash flows matter C. Subtracted from the challenger cost D. Used to set MARR Answer: B. Sunk costs are unrecoverable; including them would wrongly favour keeping the old asset. (Key point: Sunk cost) 2. From the table, what are the ESLs of defender and challenger? A. Defender 1 yr, challenger 6 yr B. Defender 3 yr, challenger 4 yr C. Defender 2 yr (5,200), challenger 4 yr (5,600) D. Defender 6 yr, challenger 2 yr Answer: C. ESL is at the minimum AEC: 5,200 at n = 2 for the defender, 5,600 at n = 4 for the challenger. (Key point: Economic life) 3. Which is a standard reason for replacing an asset? A. Sunk cost of the asset B. Higher book value C. Lower MARR D. Obsolescence, because a newer machine gives the same service at lower cost Answer: D. Other reasons: physical impairment, inadequacy, changed requirements and financial reasons. (Key point: Reason for replacement) 4. Which plan has the lowest PW for the 8-year horizon? A. Defender 2 years, then challengers 3 + 3 years B. Defender 6 years, then challenger 2 years C. Challengers only, 4 + 4 years D. Defender 4 years, then challenger 4 years Answer: A. Defender 2 yr then challengers 3+3 has PW 29,541, the minimum in the table. (Key point: Best strategy) ## Close variants set in other exams - [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× ## 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 - [ESL and replacement strategy for printing machines](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Aprinting-machine-old-new-aec-esl) — Engineering Economics, chapter 5 - [Defender, challenger, ESL of router](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Adefender-challenger-esl-router) — Engineering Economics, chapter 5 - [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 - [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 - [Replacement strategy for four-year need](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Areplace-machine-aec-four-years) — Engineering Economics, chapter 5