Question 4 (e)
Machine A or Machine B: present value of labour savings at 12%
6 marks
- Time to write
- About 12 minutes
Question
Question 4 (e) · 6 marksMachine A costs ₹12,000 and has a useful life of 9 years. Machine B costs ₹9000 and its useful life is 6 years. Machine A generates annual labour savings of ₹2400, whereas machine B generates an annual labour savings of ₹2100. Assuming the time value of money is 12% per annum, which machine is preferable? (3×6)
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
- Assuming the time value of money…The paper fixes this: use it as given, and say so in your answer.
- which machine is preferableEnd with a one-line decision that follows from your figures.
What the paper leaves open
Part of "Solve any three"; the marker printed here, (3×6), is read as 6 marks for each of the three parts attempted.
Write it in this order
- Read the data; note exactly what is asked 1 min
- 1 Present value of savings, A and B 3 marks 5 min
- 2 Comparison allowing for different lives 2 marks 4 min
- 3 Decision 1 mark 1 min
- Check: arithmetic, units (₹, %), final answer underlined 1 min
Short on time? End with step 3's one-line decision, whatever happens: never leave the figures without it.
Try it first. About 12 minutes for 6 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.
Each machine is judged by comparing the present value of its labour savings, at the 12% time value of money, with its cost (a capital expenditure decision by present values). Savings arise at the end of each year: an ordinary annuity with factor P(n, i) = [1 − (1 + i)⁻ⁿ]/i.
P(9, 0.12) = 5.328250; P(6, 0.12) = 4.111407.
| Machine A | Machine B | |
|---|---|---|
| Annual labour saving (₹) | 2,400 | 2,100 |
| Life (years) | 9 | 6 |
| Annuity factor at 12% | 5.328250 | 4.111407 |
| Present value of savings (₹) | 12,787.80 | 8,633.96 |
| Cost (₹) | 12,000.00 | 9,000.00 |
| Net present value (₹) | 787.80 | −366.04 |
Machine A's savings are worth ₹787.80 more than its cost; Machine B's savings fall ₹366.04 short of its cost.
Allowing for the different lives: the equal annual cost of each machine's capital at 12% is 12,000/5.328250 = ₹2,252.15 for A (against savings of ₹2,400 a year) and 9,000/4.111407 = ₹2,189.03 for B (against savings of only ₹2,100). A gains ₹147.85 a year; B loses ₹89.03 a year.
Machine A is preferable (and Machine B is not worth buying at 12%).
Where the 6 marks sit
| # | Component | Marks |
|---|---|---|
| 1 | Present value of savings, A and B | 3 |
| 2 | Comparison allowing for different lives | 2 |
| 3 | Decision | 1 |
| Total | 6 |
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