Question 1 OR (b)
Replacement decision with a maintenance rise: NPV under two salvage values
11 marks
- Time to write
- About 22 minutes
From an older CBCS paper
Question
Question 1 OR (b) · 11 marksNoble Electronics is considering a proposal to replace a machine bought three years ago for ₹ 12,00,000. Its remaining life is 5 years, after which its salvage value will be nil. If sold now, however, it will realize ₹ 7,20,000. From the sixth year of its installation, its maintenance cost is expected to rise by ₹ 60,000 p.a. A new, more efficient machine is available for ₹ 18,00,000, and can be sold for ₹ 7,20,000 when its life of 5 years is over. The new machine will raise productivity, increasing the revenue by ₹ 1,80,000 p.a., and will cut operating expenses by ₹ 1,20,000 p.a. The company's tax rate is 50%, and its cost of capital is 10%. It follows the straight line method of depreciation. Advise the company on the replacement.
If the new machine's expected salvage value is ₹ 2,40,000, what will your decision be ? 11
Reworded from the university paper: names and figures differ, the task is the same. The original paper belongs to the University of Delhi. See DU's question bank.
Before you write
Read off the question and its marks
Read from the question
- replaceCash flows year by year in one table, then discount, then a one-line decision.
What the paper leaves open
The old machine's maintenance rise from the sixth year of its installation falls in years 3 to 5 of the comparison (it was bought three years ago). No tax is charged on the sale of either machine unless stated (C7).
Write it in this order
- Read the data; note exactly what is asked 1 min
- 1 Initial outlay; old machine's depreciation and maintenance rise 2 marks 4 min
- 2 Case 1: incremental cash flows 3 marks 6 min
- 3 Case 1: NPV 2 marks 4 min
- 4 Case 2: cash flows and NPV 3 marks 5 min
- 5 Advice 1 mark 1 min
- Check: arithmetic, units (₹, %), final answer underlined 1 min
Short on time? End with step 5's one-line decision, whatever happens: never leave the figures without it.
Try it first. About 22 minutes for 11 marks. Write it out, then open the solution.
Solution
Hover or tap a marked figure, like 72,000, to see how it is made and where else it appears.
Initial outlay. New machine ₹18,00,000 − sale of the old machine ₹7,20,000 = ₹10,80,000 (no tax on the sale, paper note). For a replacement, the outflow is the new cost less the net proceeds of the old asset.
Old machine. Bought three years ago for ₹12,00,000 with 5 years left, an 8-year life, nil salvage: depreciation ₹1,50,000 a year (book value now ₹7,50,000). Its maintenance would rise by ₹60,000 a year from the sixth year of installation, that is in years 3 to 5 here (paper note); replacing it saves that rise.
Incremental operating gain: revenue +1,80,000 and expenses −1,20,000 = ₹3,00,000 a year, plus ₹60,000 in years 3–5. Only the incremental depreciation affects tax; its tax saving is a cash inflow.
Case 1: new machine's salvage ₹7,20,000. New depreciation = (18,00,000 − 7,20,000) ÷ 5 = 2,16,000; incremental = 2,16,000 − 1,50,000 = ₹66,000.
| Year | Incremental gain | Incremental depreciation | EBT | Tax @ 50% | EAT | Add: depreciation | Cash inflow after tax |
|---|---|---|---|---|---|---|---|
| 1 | 3,00,000 | 66,000 | 2,34,000 | 1,17,000 | 1,17,000 | 66,000 | 1,83,000 |
| 2 | 3,00,000 | 66,000 | 2,34,000 | 1,17,000 | 1,17,000 | 66,000 | 1,83,000 |
| 3 | 3,60,000 | 66,000 | 2,94,000 | 1,47,000 | 1,47,000 | 66,000 | 2,13,000 |
| 4 | 3,60,000 | 66,000 | 2,94,000 | 1,47,000 | 1,47,000 | 66,000 | 2,13,000 |
| 5 | 3,60,000 | 66,000 | 2,94,000 | 1,47,000 | 1,47,000 | 66,000 | 2,13,000 |
| Year | Cash inflow | PV factor @ 10% | Present value |
|---|---|---|---|
| 1 | 1,83,000 | 0.909 | 1,66,347 |
| 2 | 1,83,000 | 0.826 | 1,51,158 |
| 3 | 2,13,000 | 0.751 | 1,59,963 |
| 4 | 2,13,000 | 0.683 | 1,45,479 |
| 5 (2,13,000 + salvage 7,20,000) | 9,33,000 | 0.621 | 5,79,393 |
| Total present value | 12,02,340 | ||
| Less: initial outlay | 10,80,000 | ||
| NPV | 1,22,340 |
Case 2: new machine's salvage ₹2,40,000. New depreciation = (18,00,000 − 2,40,000) ÷ 5 = 3,12,000; incremental = ₹1,62,000.
| Year | Incremental gain | Incremental depreciation | EBT | Tax @ 50% | EAT | Add: depreciation | Cash inflow after tax |
|---|---|---|---|---|---|---|---|
| 1 | 3,00,000 | 1,62,000 | 1,38,000 | 69,000 | 69,000 | 1,62,000 | 2,31,000 |
| 2 | 3,00,000 | 1,62,000 | 1,38,000 | 69,000 | 69,000 | 1,62,000 | 2,31,000 |
| 3 | 3,60,000 | 1,62,000 | 1,98,000 | 99,000 | 99,000 | 1,62,000 | 2,61,000 |
| 4 | 3,60,000 | 1,62,000 | 1,98,000 | 99,000 | 99,000 | 1,62,000 | 2,61,000 |
| 5 | 3,60,000 | 1,62,000 | 1,98,000 | 99,000 | 99,000 | 1,62,000 | 2,61,000 |
| Year | Cash inflow | PV factor @ 10% | Present value |
|---|---|---|---|
| 1 | 2,31,000 | 0.909 | 2,09,979 |
| 2 | 2,31,000 | 0.826 | 1,90,806 |
| 3 | 2,61,000 | 0.751 | 1,96,011 |
| 4 | 2,61,000 | 0.683 | 1,78,263 |
| 5 (2,61,000 + salvage 2,40,000) | 5,01,000 | 0.621 | 3,11,121 |
| Total present value | 10,86,180 | ||
| Less: initial outlay | 10,80,000 | ||
| NPV | 6,180 |
Advice. A project is accepted when its NPV is positive. Replace the machine: the NPV is ₹1,22,340 with a salvage of ₹7,20,000. With a salvage of only ₹2,40,000 the NPV falls to ₹6,180, still positive, so the decision stands, though the margin is now small.
Where the 11 marks sit
| # | Component | Marks |
|---|---|---|
| 1 | Initial outlay; old machine's depreciation and maintenance rise | 2 |
| 2 | Case 1: incremental cash flows | 3 |
| 3 | Case 1: NPV | 2 |
| 4 | Case 2: cash flows and NPV | 3 |
| 5 | Advice | 1 |
| Total | 11 |
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