2. Example-1
1. A firm is considering the replacement of a machine, whose cost price is Rs 12,200 and its scrap value is Rs 200. From experience the running (maintenance and operating) costs are found to be as follows:
| Year | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
|---|
| Running Cost | 200 | 500 | 800 | 1,200 | 1,800 | 2,500 | 3,200 | 4,000 |
|---|
When should the machine be replaced?
Solution:
We are given the running cost, `R(n)`, the scrap value `S` = Rs 200 and the cost of machine, `C` = Rs 12,200
Depreciation Cost = cost price - scrap value = 12,200 - 200
The given running cost, `R(n)`
| Year | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
|---|
| Running Cost | 200 | 500 | 800 | 1,200 | 1,800 | 2,500 | 3,200 | 4,000 |
|---|
Depreciation Cost = 12,000
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) | Depritiation Cost
`C-S`
(4) | Total Cost
`TC`
(5)=(3)+(4) | Average Total Cost
`ATC_n`
(6)=(5)/(1) | | 1 | 200 | 200 `200 = 0 + 200`
(3) = Previous(3) + (2) | 12,000 | 12,200 `12200 = 200 + 12000`
`(5)=(3)+(4)` | 12,200 `12200 = 12200 / 1`
(6)=(5)/(1) | | 2 | 500 | 700 `700 = 200 + 500`
(3) = Previous(3) + (2) | 12,000 | 12,700 `12700 = 700 + 12000`
`(5)=(3)+(4)` | 6,350 `6350 = 12700 / 2`
(6)=(5)/(1) | | 3 | 800 | 1,500 `1500 = 700 + 800`
(3) = Previous(3) + (2) | 12,000 | 13,500 `13500 = 1500 + 12000`
`(5)=(3)+(4)` | 4,500 `4500 = 13500 / 3`
(6)=(5)/(1) | | 4 | 1,200 | 2,700 `2700 = 1500 + 1200`
(3) = Previous(3) + (2) | 12,000 | 14,700 `14700 = 2700 + 12000`
`(5)=(3)+(4)` | 3,675 `3675 = 14700 / 4`
(6)=(5)/(1) | | 5 | 1,800 | 4,500 `4500 = 2700 + 1800`
(3) = Previous(3) + (2) | 12,000 | 16,500 `16500 = 4500 + 12000`
`(5)=(3)+(4)` | 3,300 `3300 = 16500 / 5`
(6)=(5)/(1) | | 6 | 2,500 | 7,000 `7000 = 4500 + 2500`
(3) = Previous(3) + (2) | 12,000 | 19,000 `19000 = 7000 + 12000`
`(5)=(3)+(4)` | 3,166.67 `3166.67 = 19000 / 6`
(6)=(5)/(1) | | 7 | 3,200 | 10,200 `10200 = 7000 + 3200`
(3) = Previous(3) + (2) | 12,000 | 22,200 `22200 = 10200 + 12000`
`(5)=(3)+(4)` | 3,171.43 `3171.43 = 22200 / 7`
(6)=(5)/(1) | | 8 | 4,000 | 14,200 `14200 = 10200 + 4000`
(3) = Previous(3) + (2) | 12,000 | 26,200 `26200 = 14200 + 12000`
`(5)=(3)+(4)` | 3,275 `3275 = 26200 / 8`
(6)=(5)/(1) |
The calculations in table show that the average cost is lowest during the `6^(th)` year (Rs 3,166.67).
Hence, the machine should be replaced after every `6^(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
|
