- 2069 Bhadra · 2+10 marks
ESL and replacement strategy for printing machines
Short answer
Old machine ESL = 2 years (AEC Rs 510,000); new machine ESL = 4 years (AEC Rs 583,000). For the 8-year contract, the optimal strategy is running the old machine for 2 years, then replacing with new machines for 3+3 years, giving PW ≈ Rs 2,994,366 (AEC ≈ Rs 561,276), which is cheaper than other combinations.
A firm has a contract to provide printing service to IOE for next 8 years. It can provide the service using its old printing machine (the current defender) or the newly bought machine (the challenger). After the contract work neither the old machine nor the new machine will be retained. Considering the annual equivalent costs of the old machine and new machine as follows, what are their economic service life? And what is the best replacement strategy?
| Number of years (n) | Old machine (Rs) | New machine (Rs) |
|---|---|---|
| 1 | 515,000 | 750,000 |
| 2 | 510,000 | 615,000 |
| 3 | 550,000 | 586,000 |
| 4 | 596,000 | 583,000 |
| 5 | 644,000 | 590,000 |
MARR is not given; assume i = 10% (the ranking is the same at 8%, 12% and 15%). The AEC values are the costs per year if each machine is kept n years.
Economic service life (minimum AEC)
| n | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| Old machine | 515,000 | 510,000 | 550,000 | 596,000 | 644,000 |
| New machine | 750,000 | 615,000 | 586,000 | 583,000 | 590,000 |
- ESL of old machine = 2 years (AEC = Rs 5,10,000).
- ESL of new machine = 4 years (AEC = Rs 5,83,000).
Replacement strategy for the 8-year contract
Neither machine is kept after the contract, so the 8 years must be covered by "old machine for n years, then new machine(s) for the rest". PW of a plan = AEC_old(n)(P/A,10%,n) + Σ AEC_new(m)(P/A,10%,m)(P/F,10%,start year).
Example, old 2 years then new 3 + 3 years:
PW = 510,000(1.7355) + 586,000(2.4869)(0.8264) + 586,000(2.4869)(0.6209)
= 885,124 + 1,204,376 + 904,866 = 2,994,366
| Old machine kept | New machine(s) for | PW of cost (Rs) | Equivalent AEC over 8 yr |
|---|---|---|---|
| Old 2 yr | 3 + 3 | 2,994,366 | 561,276 |
| Old 3 yr | 5 | 3,048,132 | 571,354 |
| Old 1 yr | 4 + 3 | 3,053,076 | 572,281 |
| New only | 4 + 4 | 3,110,262 | 583,000 |
| Old 4 yr | 4 | 3,151,470 | 590,724 |
| Old 5 yr | 3 | 3,346,132 | 627,212 |
Answer: Economic lives: old machine 2 years, new machine 4 years. The cheapest plan is to run the old machine for 2 years, then replace it with a new machine, which is itself replaced after 3 years (3 + 3), giving PW ≈ Rs 2,994,366 (AEC ≈ Rs 561,276). Using the old machine for 3 years and then a new machine for 5 years is only about 2% dearer; running only new machines (4 + 4 years) costs about 4% more.
✓ Answer checked by Claude Opus 5.5 (high effort)
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