Home > Operation Research calculators > Replacement Model-1.3 example

3. Model-1.3 example ( Enter your problem )
  1. Examples
Other related methods
  1. Model-1.1
  2. Model-1.2
  3. Model-1.3
  4. Model-2
  5. Model-3

2. Model-1.2
(Previous method)
4. Model-2
(Next method)

1. Examples



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.

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.



This material is intended as a summary. Use your textbook for detail explanation.
Any bug, improvement, feedback then Submit Here





2. Model-1.2
(Previous method)
4. Model-2
(Next method)




 
Copyright © 2020. All rights reserved. Terms, Privacy
 
This site is protected by reCAPTCHA and the Google Privacy Policy and Terms of Service apply.
 





Adblocker detected!

Dear user,

We've detected that you are using AdBlock Plus or some other adblocking software which is preventing the page from fully loading.

We don't have any banner, Flash, animation, obnoxious sound, or popup ad. We do not implement these annoying types of ads!

We need money to operate the site, and almost all of it comes from our online advertising.

Please add atozmath.com to your ad blocking whitelist or disable your adblocking software.

After unblocking website please refresh the page and click on find button again.

Thanks for your support

×
  .