Question 1 (b)
Depreciation schedules of two machines in matrix notation
6 marks
- Time to write
- About 12 minutes
Question
Question 1 (b) · 6 marksA firm buys two machines, at a cost of ₹41,000 and ₹45,000 respectively. Each machine has a useful life of 4 years and zero salvage values at the end of 4 years. Using matrix notation, find the depreciation of each machine for each year if:
Both machines are depreciated using the straight-line method.
First machine is depreciated by the straight-line method and the second by the sum of years digit method.
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
- findFormula first, then the working, then the answer with its unit (₹, %, units).
Write it in this order
- Read the data; note exactly what is asked 1 min
- 1 Matrix set-up cost and rate matrices 2 marks 3 min
- 2 Straight-line schedule 2 marks 3 min
- 3 Sum-of-years'-digits schedule 2 marks 3 min
- Check: arithmetic, units (₹, %), final answer underlined 2 min
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.
Write the costs as a diagonal matrix C and the depreciation rates (fraction of cost written off in each of the 4 years) as a rate matrix R; the depreciation matrix is the product D = C × R, row by column. Salvage value is zero.
C = [[41,000, 0], [0, 45,000]].
(i) Both machines by the straight-line method (¼ of cost each year):
R = [[1/4, 1/4, 1/4, 1/4], [1/4, 1/4, 1/4, 1/4]]
D = C × R = [[10,250, 10,250, 10,250, 10,250], [11,250, 11,250, 11,250, 11,250]].
| Machine | Year 1 | Year 2 | Year 3 | Year 4 | Total |
|---|---|---|---|---|---|
| First (₹41,000) | 10,250 | 10,250 | 10,250 | 10,250 | 41,000 |
| Second (₹45,000) | 11,250 | 11,250 | 11,250 | 11,250 | 45,000 |
(ii) First machine straight-line, second by the sum-of-years'-digits method
For a 4-year life the digits sum to 1 + 2 + 3 + 4 = 10, so the second machine is written off at 4/10, 3/10, 2/10 and 1/10 of cost in years 1 to 4:
R = [[1/4, 1/4, 1/4, 1/4], [4/10, 3/10, 2/10, 1/10]]
D = C × R = [[10,250, 10,250, 10,250, 10,250], [18,000, 13,500, 9,000, 4,500]].
| Machine | Year 1 | Year 2 | Year 3 | Year 4 | Total |
|---|---|---|---|---|---|
| First (straight line) | 10,250 | 10,250 | 10,250 | 10,250 | 41,000 |
| Second (sum of years' digits) | 18,000 | 13,500 | 9,000 | 4,500 | 45,000 |
Under either method each machine is written down to its zero salvage value by the end of year 4 (row totals equal cost).
Where the 6 marks sit
| # | Component | Marks |
|---|---|---|
| 1 | Matrix set-up (cost and rate matrices) | 2 |
| 2 | Straight-line schedule | 2 |
| 3 | Sum-of-years'-digits schedule | 2 |
| Total | 6 |
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