- 2070 Magh · 8+4 marks
ESL of new machine; when to replace old
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.
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? |
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.
✓ Answer checked by Claude Opus 5.5 (high effort)
Related questions
- Replace existing machine now; AEC and ESL · Engineering Economics, chapter 5
- Replacement strategy for four-year need · Engineering Economics, chapter 5
- Define ESL; find ESL of 50,000 asset · Engineering Economics, chapter 5
- ESL and replacement strategy for printing machines · Engineering Economics, chapter 5
- ESL of an electric lift truck · Engineering Economics, chapter 5
- Value of a 20-year-old building · Construction Management, chapter 13 · asked 2×
- Value of a 10-year-old building · Construction Management, chapter 13 · asked 2×
- Fair market value of Tata Sipradi showroom · Construction Management, chapter 13
- Capitalised value from Rs 20,000 monthly rent · Construction Management, chapter 13