--- title: "ESL of new machine; when to replace old" subject: "Engineering Economics" chapter: 5 key: "semester_6/engineering_economics:economic-service-life-new-machine-old" source: https://www.bctengineer.com/q/semester_6/engineering_economics/economic-service-life-new-machine-old site: BCT Engineering --- # ESL of new machine; when to replace old Engineering Economics · Chapter 5: Replacement Analysis - Asked in: 2070 Magh, 8+4 marks - Video: [Replacement Analysis | Economic Service Life | Engineering Economics](https://www.youtube.com/watch?v=r1KUkAZ30zE) — Salicylaldehyde - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/engineering_economics/economic-service-life-new-machine-old - Whole chapter: https://www.bctengineer.com/md/semester_6/engineering_economics/chapter-5 ## Question The new machine costs 10,000, operating cost 2200 in first year, then increases by 20% per year. Market value is 6000 after one year and will decline by 15% each year. N = 5 years. If required, old machine can work another 3 years. Market value now is 5000 and will decline by 25% each year. Immediate overhauling to restore to operable condition costs 1200. Operating costs 2000 in the first year, increases by 1500 per year thereafter. MARR = 15%. i) Find the economic service life of this machine (new). ii) AEC of defender is as follows: | N | 1 | 2 | 3 | 4 | |---|---|---|---|---| | AEC | 5380 | 5203 [?] | 5468 | 5845 | When should the old machine be replaced with the new machine? ## Short answer The new machine's economic service life is 4 years with minimum AEC of Rs 5,625. The old machine should be kept for 2 more years (its marginal costs for years 1-2 are Rs 5,380 and Rs 5,000, both below the challenger's AEC of Rs 5,625) and replaced at the end of year 2. ## Answer MARR = 15%. #### i) Economic service life of the new machine Market value: MV₁ = 6,000, then declines 15% a year (MV_n = 6,000(0.85)^(n−1)). Operating cost = 2,200(1.2)^(k−1). AEC(n) = [10,000 + Σ O&M_k (P/F,15%,k) − MV_n (P/F,15%,n)] (A/P,15%,n) | n | MV_n | O&M in year n | PW of cost | AEC(n) | |---|---|---|---|---| | 1 | 6,000 | 2,200 | 6,696 | 7,700 | | 2 | 5,100 | 2,640 | 10,053 | 6,184 | | 3 | 4,335 | 3,168 | 13,142 | 5,756 | | 4 | 3,685 | 3,802 | 16,059 | 5,625 | | 5 | 3,132 | 4,562 | 18,877 | 5,631 | Example, n = 4: PW = 10,000 + 2,200(0.8696) + 2,640(0.7561) + 3,168(0.6575) + 3,802(0.5718) − 3,685(0.5718) = 16,059; AEC = 16,059(0.35027) = 5,625. AEC is minimum at n = 4, so **ESL of the new machine = 4 years, AEC = Rs 5,625**. #### ii) When to replace the old machine The defender's investment = market value 5,000 + overhaul 1,200 = 6,200; MV_n = 5,000(0.75)^n; operating cost 2,000, 3,500, 5,000. | n | MV_n | PW of cost | AEC(n) | |---|---|---|---| | 1 | 3,750 | 4,678 | 5,380 | | 2 | 2,812 | 8,459 | 5,203 | | 3 | 2,109 | 12,486 | 5,469 | These agree with the given table (5,380, 5,203 and 5,468; the difference of 1 at n = 3 is rounding), so the uncertain second value, 5,203, is confirmed. Marginal cost of keeping the old machine one more year = MV_{k−1}(1.15) − MV_k + O&M_k, with MV₀ = 6,200: | Year of keeping | Marginal cost | |---|---| | 1 | 5,380 | | 2 | 5,000 | | 3 | 6,125 | The challenger's best AEC = 5,625. Keep the old machine as long as its cost for the next year is below this figure. Year 1 (5,380) and year 2 (5,000) are below 5,625; year 3 (6,125) is above. The defender's own lowest AEC is also at n = 2 (5,203). **Answer:** The new machine's ESL is 4 years (AEC ≈ Rs 5,625). Keep the old machine for **2 more years** and replace it with the new machine at the end of year 2. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. AEC of the new machine is 7,700; 6,184; 5,756; 5,625; 5,631. What is its ESL? A. 3 years B. 4 years C. 5 years D. 1 year Answer: B. The minimum AEC (5,625) is at n = 4. (Key point: ESL of new machine) 2. What is the defender's investment base here? A. Market value 5,000 only B. Market value plus overhaul: 6,200 C. Original cost of the old machine D. Overhaul cost only: 1,200 Answer: B. To keep the old machine you forgo its 5,000 market value and spend 1,200 on overhaul. (Key point: Defender investment) 3. How is the marginal cost of keeping the old machine one more year computed? A. O&M only B. Previous MV x (1 + i) - current MV + O&M C. Current MV x (1 + i) - previous MV D. AEC(n) minus AEC(n-1) Answer: B. It includes interest on the market value tied up, the loss in market value, and that year's O&M. (Key point: Marginal cost rule) 4. Challenger's best AEC is 5,625. Marginal costs for the old machine are 5,380; 5,000; 6,125. When should it be replaced? A. Now B. After 1 year C. After 3 years D. After 2 years, before the cost exceeds 5,625 Answer: D. Keep while the next-year cost is below the challenger's AEC; year 3 (6,125) exceeds it, so replace at the end of year 2. 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