--- title: "ESL of an electric lift truck" subject: "Engineering Economics" chapter: 5 key: "semester_6/engineering_economics:esl-electric-lift-truck" source: https://www.bctengineer.com/q/semester_6/engineering_economics/esl-electric-lift-truck site: BCT Engineering --- # ESL of an electric lift truck Engineering Economics · Chapter 5: Replacement Analysis - Asked in: 2069 Bhadra, 2+10 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/esl-electric-lift-truck - Whole chapter: https://www.bctengineer.com/md/semester_6/engineering_economics/chapter-5 ## Question What is the economic service life of an asset? Find the economic service life of a new electric lift truck which costs \$20,000, has an operating cost of \$1000 in the first year and has salvage value of \$12,000 at the end of the first year. For the remaining years, operating costs increase each year by 10% over the previous year's operating costs. Similarly the salvage value declines each year by 20% from the previous year's salvage value. The lift truck has a maximum life of 7 years. An overhaul costing \$3000 and \$5000 will be required during the fifth and seventh year of service. The firm's required rate of return is 15% per year. ## Short answer Economic service life minimizes equivalent annual cost considering capital recovery, rising operating costs, and falling salvage value. For the $20,000 lift truck with $1,000 first-year operating cost increasing 10% yearly, $12,000 first-year salvage declining 20% yearly, overhauls at years 5 and 7, and 15% MARR, the ESL is 6 years with minimum AEC of $6,467. ## Answer #### Economic service life The economic service life of an asset is the life (number of years kept) that minimises its equivalent annual cost (AEC), taking into account the capital recovery cost, the rising operating cost and the falling salvage value. #### Calculation (i = 15%) Data: P = 20,000; O&M_k = 1,000(1.1)^(k−1); salvage MV_n = 12,000(0.8)^(n−1); overhauls of 3,000 (year 5) and 5,000 (year 7). **Assumption:** an overhaul is incurred if the truck is kept through that year, so the year-5 overhaul is charged for n ≥ 5 and the year-7 overhaul for n = 7. PW(n) = 20,000 + PW(O&M) + PW(overhauls) − MV_n (P/F,15%,n); AEC(n) = PW(n)(A/P,15%,n) | n | PW of O&M | PW of overhaul | MV_n | Total PW | AEC(n) | |---|---|---|---|---|---| | 1 | 870 | 0 | 12,000 | 10,435 | 12,000 | | 2 | 1,701 | 0 | 9,600 | 14,442 | 8,884 | | 3 | 2,497 | 0 | 7,680 | 17,447 | 7,641 | | 4 | 3,258 | 0 | 6,144 | 19,745 | 6,916 | | 5 | 3,986 | 1,492 | 4,915 | 23,034 | 6,871 | | 6 | 4,682 | 1,492 | 3,932 | 24,474 | 6,467 | | 7 | 5,348 | 3,371 | 3,146 | 27,537 | 6,619 | Example, n = 4: PW(O&M) = 1,000(0.8696) + 1,100(0.7561) + 1,210(0.6575) + 1,331(0.5718) = 3,258; MV₄ = 6,144; PW = 20,000 + 3,258 − 6,144(0.5718) = 19,745; AEC = 19,745(0.35027) = 6,916. The AEC falls to a minimum at n = 6 (\$6,467) and rises again at n = 7 because of the \$5,000 overhaul. Because of the year-5 overhaul, stopping at n = 5 (\$6,871) costs more per year than keeping the truck one more year, which spreads that overhaul over 6 years. **Answer: Economic service life = 6 years**, with minimum AEC ≈ \$6,467 per year. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The economic service life of an asset is the number of years of ownership that: A. Minimises the equivalent annual cost (AEC) B. Maximises the salvage value C. Equals the maximum physical life of the asset D. Makes the operating cost lowest Answer: A. ESL is the keeping period with minimum AEC, balancing capital recovery, rising operating cost and falling salvage value. (Key point: Definition of economic service life) 2. How is AEC(n) obtained for each possible life n? A. Average the operating costs over n years B. AEC(n) = PW(n) x (A/P, 15%, n), where PW includes first cost, PW of O&M and overhauls, less PW of salvage C. AEC(n) = first cost / n D. Add the first cost and the n-th year operating cost Answer: B. The total PW of costs for n years, net of discounted salvage, is converted to an equivalent annual amount with the capital recovery factor. (Key point: PW and AEC method) 3. From the calculated AEC values, what is the economic service life of the truck and its minimum AEC? A. 5 years, about 6,871 B. 7 years, about 6,619 C. 1 year, about 12,000 D. 6 years, about 6,467 Answer: D. AEC falls to a minimum of about 6,467 at n = 6, so the ESL is 6 years. (Key point: Result of the comparison) 4. Why does AEC rise again at n = 7? A. The salvage value becomes negative B. The interest rate changes after year 6 C. The 5,000 overhaul in year 7 is added to the cost D. Operating cost stops increasing Answer: C. The year-7 overhaul of 5,000 raises PW of costs more than the extra year reduces the annual average, so AEC goes up from 6,467 to 6,619. 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