- 2069 Bhadra · 2+10 marks
ESL of an electric lift truck
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.
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.
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.
✓ Answer checked by Claude Opus 5.5 (high effort)
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