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 Solution provided by AtoZmath.com
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        | Replacement Model-1.2 calculator |  
        |  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.
 
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 Example1. 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)
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 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.
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