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Home > Operation Research calculators > Replacement Model-2 example
4. Replacement of items that fail completely (Model-2) example ( Enter your problem )
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  1. Example-1
  2. Example-2
 		Other related methods
  1. Replacement of items that deteriorates with time (Model-1.1)
  2. Replacement of items that deteriorates with time (Model-1.2)
  3. Replacement of items that deteriorates with time (Model-1.3)
  4. Replacement of items that fail completely (Model-2)
  5. Group replacement policy (Model-3)
1. Example-1
(Previous example)		5. Group replacement policy (Model-3)
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2. Example-2


2. A company is buying mini computers. It costs Rs 5 lakh, and its running and maintenance costs are Rs 60,000 for each of the first five years, increasing by Rs 20,000 per year in the sixth and subsequent years. If the money is worth 10 percent per year, What is the optimal replacement period?
Year12345678910
Running Cost60,00060,00060,00060,00060,00080,0001,00,0001,20,0001,40,0001,60,000
Solution:
Since money is worth 10 per cent per year, the discounted factor over a period of one year is given by:
`d = 1/(1 + 10 / 100) = 0.9091`

It is also given that `C` = Rs 500,000

The optimum replacement age must satisfy the condition `R_n < W(n) < R_(n+1)`

Year of Service
`n`
(1)		Running Cost
`R_n`
(2)		Discounted factor
`d^(n-1)`
`(3) = 0.9091^(n-1)`		Discounted Cost
`R_n * d^(n-1)`
`(4)=(2)xx(3)`		Summation of Discounted Cost
`Sigma R_i d^(i-1)`
(5)=`Sigma`(4)
`C+Sigma R_i d^(i-1)`
(6)=500,000+(5)		Summation of Discount
`Sigma d^(i-1)`
(7)=`Sigma`(3)		Weighted Average Cost
`W(n)`
(8)=(6)/(7)
160,000 1 `1 = 0.9091^(1-1)` 60,000 `60000 = 60000 xx 1`
`(4)=(2)xx(3)`		 60,000 `60000 = 0 + 60000`
(5) = Previous(5) + (4)		 560,000 `560000 = 500000 + 60000`
(6) = 500000 + (5)		 1 `1 = 0 + 1`
(7) = Previous(7) + (3)		 560,000 `560000 = 560000 / 1`
(8)=(6)/(7)
260,000 0.9091 `0.9091 = 0.9091^(2-1)` 54,545.45 `54545.45 = 60000 xx 0.9091`
`(4)=(2)xx(3)`		 114,545.45 `114545.45 = 60000 + 54545.45`
(5) = Previous(5) + (4)		 614,545.45 `614545.45 = 500000 + 114545.45`
(6) = 500000 + (5)		 1.9091 `1.9091 = 1 + 0.9091`
(7) = Previous(7) + (3)		 321,904.76 `321904.76 = 614545.45 / 1.9091`
(8)=(6)/(7)
360,000 0.8264 `0.8264 = 0.9091^(3-1)` 49,586.78 `49586.78 = 60000 xx 0.8264`
`(4)=(2)xx(3)`		 164,132.23 `164132.23 = 114545.45 + 49586.78`
(5) = Previous(5) + (4)		 664,132.23 `664132.23 = 500000 + 164132.23`
(6) = 500000 + (5)		 2.7355 `2.7355 = 1.9091 + 0.8264`
(7) = Previous(7) + (3)		 242,779.46 `242779.46 = 664132.23 / 2.7355`
(8)=(6)/(7)
460,000 0.7513 `0.7513 = 0.9091^(4-1)` 45,078.89 `45078.89 = 60000 xx 0.7513`
`(4)=(2)xx(3)`		 209,211.12 `209211.12 = 164132.23 + 45078.89`
(5) = Previous(5) + (4)		 709,211.12 `709211.12 = 500000 + 209211.12`
(6) = 500000 + (5)		 3.4869 `3.4869 = 2.7355 + 0.7513`
(7) = Previous(7) + (3)		 203,395.82 `203395.82 = 709211.12 / 3.4869`
(8)=(6)/(7)
560,000 0.683 `0.683 = 0.9091^(5-1)` 40,980.81 `40980.81 = 60000 xx 0.683`
`(4)=(2)xx(3)`		 250,191.93 `250191.93 = 209211.12 + 40980.81`
(5) = Previous(5) + (4)		 750,191.93 `750191.93 = 500000 + 250191.93`
(6) = 500000 + (5)		 4.1699 `4.1699 = 3.4869 + 0.683`
(7) = Previous(7) + (3)		 179,907.95 `179907.95 = 750191.93 / 4.1699`
(8)=(6)/(7)
680,000 0.6209 `0.6209 = 0.9091^(6-1)` 49,673.71 `49673.71 = 80000 xx 0.6209`
`(4)=(2)xx(3)`		 299,865.63 `299865.63 = 250191.93 + 49673.71`
(5) = Previous(5) + (4)		 799,865.63 `799865.63 = 500000 + 299865.63`
(6) = 500000 + (5)		 4.7908 `4.7908 = 4.1699 + 0.6209`
(7) = Previous(7) + (3)		 166,959.14 `166959.14 = 799865.63 / 4.7908`
(8)=(6)/(7)
7100,000 0.5645 `0.5645 = 0.9091^(7-1)` 56,447.39 `56447.39 = 100000 xx 0.5645`
`(4)=(2)xx(3)`		 356,313.03 `356313.03 = 299865.63 + 56447.39`
(5) = Previous(5) + (4)		 856,313.03 `856313.03 = 500000 + 356313.03`
(6) = 500000 + (5)		 5.3553 `5.3553 = 4.7908 + 0.5645`
(7) = Previous(7) + (3)		 159,901.28 `159901.28 = 856313.03 / 5.3553`
(8)=(6)/(7)
8120,000 0.5132 `0.5132 = 0.9091^(8-1)` 61,578.97 `61578.97 = 120000 xx 0.5132`
`(4)=(2)xx(3)`		 417,892 `417892 = 356313.03 + 61578.97`
(5) = Previous(5) + (4)		 917,892 `917892 = 500000 + 417892`
(6) = 500000 + (5)		 5.8684 `5.8684 = 5.3553 + 0.5132`
(7) = Previous(7) + (3)		 156,412.15 `156412.15 = 917892 / 5.8684`
(8)=(6)/(7)
9140,000 0.4665 `0.4665 = 0.9091^(9-1)` 65,311.03 `65311.03 = 140000 xx 0.4665`
`(4)=(2)xx(3)`		 483,203.03 `483203.03 = 417892 + 65311.03`
(5) = Previous(5) + (4)		 983,203.03 `983203.03 = 500000 + 483203.03`
(6) = 500000 + (5)		 6.3349 `6.3349 = 5.8684 + 0.4665`
(7) = Previous(7) + (3)		 155,203.55 `155203.55 = 983203.03 / 6.3349`
(8)=(6)/(7)
10160,000 0.4241 `0.4241 = 0.9091^(10-1)` 67,855.62 `67855.62 = 160000 xx 0.4241`
`(4)=(2)xx(3)`		 551,058.65 `551058.65 = 483203.03 + 67855.62`
(5) = Previous(5) + (4)		 1,051,058.65 `1051058.65 = 500000 + 551058.65`
(6) = 500000 + (5)		 6.759 `6.759 = 6.3349 + 0.4241`
(7) = Previous(7) + (3)		 155,504.5 `155504.5 = 1051058.65 / 6.759`
(8)=(6)/(7)


The calculations in table show that the average cost is lowest during the `9^(th)` year (Rs 155,203.55).

Hence, the machine should be replaced after every `9^(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.
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