3. Example-2
Solve the following problem, using Replacement Model
| Year | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | | Running Cost | 100 | 250 | 400 | 600 | 900 | 1200 | 1600 | 2000 |
Cost Price = 6000 and Scrape Value = 0
Solution:
We are given the running cost, `R(n)`, the scrap value `S` = Rs 0 and the cost of machine, `C` = Rs 6,000
Depreciation Cost = cost price - scrap value = 6,000 - 0
The given running cost, `R(n)`
| Year | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
|---|
| Running Cost | 100 | 250 | 400 | 600 | 900 | 1,200 | 1,600 | 2,000 |
|---|
Depreciation Cost = 6,000
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 | 100 | 100 `100 = 0 + 100`
(3) = Previous(3) + (2) | 6,000 | 6,100 `6100 = 100 + 6000`
`(5)=(3)+(4)` | 6,100 `6100 = 6100 / 1`
(6)=(5)/(1) | | 2 | 250 | 350 `350 = 100 + 250`
(3) = Previous(3) + (2) | 6,000 | 6,350 `6350 = 350 + 6000`
`(5)=(3)+(4)` | 3,175 `3175 = 6350 / 2`
(6)=(5)/(1) | | 3 | 400 | 750 `750 = 350 + 400`
(3) = Previous(3) + (2) | 6,000 | 6,750 `6750 = 750 + 6000`
`(5)=(3)+(4)` | 2,250 `2250 = 6750 / 3`
(6)=(5)/(1) | | 4 | 600 | 1,350 `1350 = 750 + 600`
(3) = Previous(3) + (2) | 6,000 | 7,350 `7350 = 1350 + 6000`
`(5)=(3)+(4)` | 1,837.5 `1837.5 = 7350 / 4`
(6)=(5)/(1) | | 5 | 900 | 2,250 `2250 = 1350 + 900`
(3) = Previous(3) + (2) | 6,000 | 8,250 `8250 = 2250 + 6000`
`(5)=(3)+(4)` | 1,650 `1650 = 8250 / 5`
(6)=(5)/(1) | | 6 | 1,200 | 3,450 `3450 = 2250 + 1200`
(3) = Previous(3) + (2) | 6,000 | 9,450 `9450 = 3450 + 6000`
`(5)=(3)+(4)` | 1,575 `1575 = 9450 / 6`
(6)=(5)/(1) | | 7 | 1,600 | 5,050 `5050 = 3450 + 1600`
(3) = Previous(3) + (2) | 6,000 | 11,050 `11050 = 5050 + 6000`
`(5)=(3)+(4)` | 1,578.57 `1578.57 = 11050 / 7`
(6)=(5)/(1) | | 8 | 2,000 | 7,050 `7050 = 5050 + 2000`
(3) = Previous(3) + (2) | 6,000 | 13,050 `13050 = 7050 + 6000`
`(5)=(3)+(4)` | 1,631.25 `1631.25 = 13050 / 8`
(6)=(5)/(1) |
The calculations in table show that the average cost is lowest during the `6^(th)` year (Rs 1,575).
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
|
