1. Examples
1. The data collected in running a machine, the cost of which is Rs 60,000 are given below:
| Year | 1 | 2 | 3 | 4 | 5 |
|---|
| Resale Value | 42,000 | 30,000 | 20,400 | 14,400 | 9,650 |
|---|
| Cost of spares | 4,000 | 4,270 | 4,880 | 5,700 | 6,800 |
|---|
| Cost of labour | 14,000 | 16,000 | 18,000 | 21,000 | 25,000 |
|---|
Determine the optimum period for replacement of the machine.
Solution:The data collected in running a machine, | Year | 1 | 2 | 3 | 4 | 5 |
|---|
| Resale Value | 42,000 | 30,000 | 20,400 | 14,400 | 9,650 |
|---|
| Cost of spares | 4,000 | 4,270 | 4,880 | 5,700 | 6,800 |
|---|
| Cost of labour | 14,000 | 16,000 | 18,000 | 21,000 | 25,000 |
|---|
The costs of spares and labour, together, determine the running cost The running costs and resale price of the machine in successive years | Year | 1 | 2 | 3 | 4 | 5 |
|---|
| Resale Value | 42,000 | 30,000 | 20,400 | 14,400 | 9,650 |
|---|
| Running Cost | 18,000 | 20,270 | 22,880 | 26,700 | 31,800 |
|---|
In order to determine the optimal time n when the machine should be replaced, we first calculate the average cost per year during the life of the machine, Year
`n`
(1) | Running Cost
`R(n)`
(2) | Cummulative Running Cost
`Sigma R(n)`
(3) | Resale Value
`S`
(4) | Depritiation Cost
`C-S`
(5)=60,000-(4) | Total Cost
`TC`
(6)=(3)+(5) | Average Total Cost
`ATC_n`
(7)=(5)/(1) | | 1 | 18,000 | 18,000 `18000 = 0 + 18000`
(3) = Previous(3) + (2) | 42,000 | 18,000 `18000 = 60000 - 42000`
`(5)=60000-(4)` | 36,000 `36000 = 18000 + 18000`
`(6)=(3)+(5)` | 36,000 `36000 = 36000 / 1`
(7)=(5)/(1) | | 2 | 20,270 | 38,270 `38270 = 18000 + 20270`
(3) = Previous(3) + (2) | 30,000 | 30,000 `30000 = 60000 - 30000`
`(5)=60000-(4)` | 68,270 `68270 = 38270 + 30000`
`(6)=(3)+(5)` | 34,135 `34135 = 68270 / 2`
(7)=(5)/(1) | | 3 | 22,880 | 61,150 `61150 = 38270 + 22880`
(3) = Previous(3) + (2) | 20,400 | 39,600 `39600 = 60000 - 20400`
`(5)=60000-(4)` | 100,750 `100750 = 61150 + 39600`
`(6)=(3)+(5)` | 33,583.33 `33583.33 = 100750 / 3`
(7)=(5)/(1) | | 4 | 26,700 | 87,850 `87850 = 61150 + 26700`
(3) = Previous(3) + (2) | 14,400 | 45,600 `45600 = 60000 - 14400`
`(5)=60000-(4)` | 133,450 `133450 = 87850 + 45600`
`(6)=(3)+(5)` | 33,362.5 `33362.5 = 133450 / 4`
(7)=(5)/(1) | | 5 | 31,800 | 119,650 `119650 = 87850 + 31800`
(3) = Previous(3) + (2) | 9,650 | 50,350 `50350 = 60000 - 9650`
`(5)=60000-(4)` | 170,000 `170000 = 119650 + 50350`
`(6)=(3)+(5)` | 34,000 `34000 = 170000 / 5`
(7)=(5)/(1) |
The calculations in table show that the average cost is lowest during the `4^(th)` year (Rs 33,362.5). Hence, the machine should be replaced after every `4^(th)` years, otherwise the average cost per year for running the machine would start increasing.
This material is intended as a summary. Use your textbook for detail explanation.Any bug, improvement, feedback then Submit Here
|
