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Solution
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Solution provided by AtoZmath.com
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Replacement Model-1.3 calculator
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1. Machine A costs Rs 45,000 and its operating costs are estimated to be Rs 1,000 for the first year
increasing by Rs 10,000 per year in the second and subsequent years.
Machine B costs Rs 50,000 and operating costs are Rs 2,000 for the first year, increasing by Rs 4,000 in the second and subsequent years.
If at present we have a machine of type A, should we replace it with B? if so when?
Assume that both machines have no resale value and their future costs are not discounted.
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Example1. Machine A costs Rs 45,000 and its operating costs are estimated to be Rs 1,000 for the first year
increasing by Rs 10,000 per year in the second and subsequent years.
Machine B costs Rs 50,000 and operating costs are Rs 2,000 for the first year, increasing by Rs 4,000 in the second and subsequent years.
If at present we have a machine of type A, should we replace it with B? if so when?
Assume that both machines have no resale value and their future costs are not discounted.
Solution:In order to determine the optimal time n when the machine A should be replaced, we first calculate the average cost per year during the life of the machine A, 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 | 1,000 `1000 = 1000`
RunCost = OperatingCost | 1,000 `1000 = 0 + 1000`
(3) = Previous(3) + (2) | 45,000 | 46,000 `46000 = 1000 + 45000`
`(5)=(3)+(4)` | 46,000 `46000 = 46000 / 1`
(6)=(5)/(1) | | 2 | 11,000 `11000 = 1000 + 10000`
RunCost = Previous RunCost + OperatingCost | 12,000 `12000 = 1000 + 11000`
(3) = Previous(3) + (2) | 45,000 | 57,000 `57000 = 12000 + 45000`
`(5)=(3)+(4)` | 28,500 `28500 = 57000 / 2`
(6)=(5)/(1) | | 3 | 21,000 `21000 = 11000 + 10000`
RunCost = Previous RunCost + OperatingCost | 33,000 `33000 = 12000 + 21000`
(3) = Previous(3) + (2) | 45,000 | 78,000 `78000 = 33000 + 45000`
`(5)=(3)+(4)` | 26,000 `26000 = 78000 / 3`
(6)=(5)/(1) | | 4 | 31,000 `31000 = 21000 + 10000`
RunCost = Previous RunCost + OperatingCost | 64,000 `64000 = 33000 + 31000`
(3) = Previous(3) + (2) | 45,000 | 109,000 `109000 = 64000 + 45000`
`(5)=(3)+(4)` | 27,250 `27250 = 109000 / 4`
(6)=(5)/(1) | | 5 | 41,000 `41000 = 31000 + 10000`
RunCost = Previous RunCost + OperatingCost | 105,000 `105000 = 64000 + 41000`
(3) = Previous(3) + (2) | 45,000 | 150,000 `150000 = 105000 + 45000`
`(5)=(3)+(4)` | 30,000 `30000 = 150000 / 5`
(6)=(5)/(1) | | 6 | 51,000 `51000 = 41000 + 10000`
RunCost = Previous RunCost + OperatingCost | 156,000 `156000 = 105000 + 51000`
(3) = Previous(3) + (2) | 45,000 | 201,000 `201000 = 156000 + 45000`
`(5)=(3)+(4)` | 33,500 `33500 = 201000 / 6`
(6)=(5)/(1) |
The calculations in table show that the average cost of machine A is lowest during the `3^(rd)` year (Rs 26,000). Hence, the machine A should be replaced after every `3^(rd)` years. In order to determine the optimal time n when the machine B should be replaced, we first calculate the average cost per year during the life of the machine B, 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 | 2,000 `2000 = 2000`
RunCost = OperatingCost | 2,000 `2000 = 0 + 2000`
(3) = Previous(3) + (2) | 50,000 | 52,000 `52000 = 2000 + 50000`
`(5)=(3)+(4)` | 52,000 `52000 = 52000 / 1`
(6)=(5)/(1) | | 2 | 6,000 `6000 = 2000 + 4000`
RunCost = Previous RunCost + OperatingCost | 8,000 `8000 = 2000 + 6000`
(3) = Previous(3) + (2) | 50,000 | 58,000 `58000 = 8000 + 50000`
`(5)=(3)+(4)` | 29,000 `29000 = 58000 / 2`
(6)=(5)/(1) | | 3 | 10,000 `10000 = 6000 + 4000`
RunCost = Previous RunCost + OperatingCost | 18,000 `18000 = 8000 + 10000`
(3) = Previous(3) + (2) | 50,000 | 68,000 `68000 = 18000 + 50000`
`(5)=(3)+(4)` | 22,666.67 `22666.67 = 68000 / 3`
(6)=(5)/(1) | | 4 | 14,000 `14000 = 10000 + 4000`
RunCost = Previous RunCost + OperatingCost | 32,000 `32000 = 18000 + 14000`
(3) = Previous(3) + (2) | 50,000 | 82,000 `82000 = 32000 + 50000`
`(5)=(3)+(4)` | 20,500 `20500 = 82000 / 4`
(6)=(5)/(1) | | 5 | 18,000 `18000 = 14000 + 4000`
RunCost = Previous RunCost + OperatingCost | 50,000 `50000 = 32000 + 18000`
(3) = Previous(3) + (2) | 50,000 | 100,000 `100000 = 50000 + 50000`
`(5)=(3)+(4)` | 20,000 `20000 = 100000 / 5`
(6)=(5)/(1) | | 6 | 22,000 `22000 = 18000 + 4000`
RunCost = Previous RunCost + OperatingCost | 72,000 `72000 = 50000 + 22000`
(3) = Previous(3) + (2) | 50,000 | 122,000 `122000 = 72000 + 50000`
`(5)=(3)+(4)` | 20,333.33 `20333.33 = 122000 / 6`
(6)=(5)/(1) |
The calculations in table show that the average cost of machine B is lowest during the `5^(th)` year (Rs 20,000). Hence, the machine B should be replaced after every `5^(th)` years. The lowest average running cost of (Rs 20,000) per year for machine B is less than the lowest average running cost of (Rs 26,000) per year for machine A. Hence machine A should be replaced by machine B.
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