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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 )
( Enter your problem )
  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)
3. Replacement of items that deteriorates with time (Model-1.3)
(Previous method)		2. Example-2
(Next example)

1. Example-1


1. An engineering company is offered a material handling equipment A. It is priced at Rs 60,000 includeing cost of installation. The costs for operation and maintenance are estimated to be Rs 10,000 for each of the first five years, increasing every year by Rs 3,000 in the sixth and subsequent years. The company expects a return of 10 percent on all its investment. What is the optimal replacement period?
Year12345678910
Running Cost10,00010,00010,00010,00010,00013,00016,00019,00022,00025,000
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 60,000

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)=60,000+(5)		Summation of Discount
`Sigma d^(i-1)`
(7)=`Sigma`(3)		Weighted Average Cost
`W(n)`
(8)=(6)/(7)
110,000 1 `1 = 0.9091^(1-1)` 10,000 `10000 = 10000 xx 1`
`(4)=(2)xx(3)`		 10,000 `10000 = 0 + 10000`
(5) = Previous(5) + (4)		 70,000 `70000 = 60000 + 10000`
(6) = 60000 + (5)		 1 `1 = 0 + 1`
(7) = Previous(7) + (3)		 70,000 `70000 = 70000 / 1`
(8)=(6)/(7)
210,000 0.9091 `0.9091 = 0.9091^(2-1)` 9,090.91 `9090.91 = 10000 xx 0.9091`
`(4)=(2)xx(3)`		 19,090.91 `19090.91 = 10000 + 9090.91`
(5) = Previous(5) + (4)		 79,090.91 `79090.91 = 60000 + 19090.91`
(6) = 60000 + (5)		 1.9091 `1.9091 = 1 + 0.9091`
(7) = Previous(7) + (3)		 41,428.57 `41428.57 = 79090.91 / 1.9091`
(8)=(6)/(7)
310,000 0.8264 `0.8264 = 0.9091^(3-1)` 8,264.46 `8264.46 = 10000 xx 0.8264`
`(4)=(2)xx(3)`		 27,355.37 `27355.37 = 19090.91 + 8264.46`
(5) = Previous(5) + (4)		 87,355.37 `87355.37 = 60000 + 27355.37`
(6) = 60000 + (5)		 2.7355 `2.7355 = 1.9091 + 0.8264`
(7) = Previous(7) + (3)		 31,933.53 `31933.53 = 87355.37 / 2.7355`
(8)=(6)/(7)
410,000 0.7513 `0.7513 = 0.9091^(4-1)` 7,513.15 `7513.15 = 10000 xx 0.7513`
`(4)=(2)xx(3)`		 34,868.52 `34868.52 = 27355.37 + 7513.15`
(5) = Previous(5) + (4)		 94,868.52 `94868.52 = 60000 + 34868.52`
(6) = 60000 + (5)		 3.4869 `3.4869 = 2.7355 + 0.7513`
(7) = Previous(7) + (3)		 27,207.5 `27207.5 = 94868.52 / 3.4869`
(8)=(6)/(7)
510,000 0.683 `0.683 = 0.9091^(5-1)` 6,830.13 `6830.13 = 10000 xx 0.683`
`(4)=(2)xx(3)`		 41,698.65 `41698.65 = 34868.52 + 6830.13`
(5) = Previous(5) + (4)		 101,698.65 `101698.65 = 60000 + 41698.65`
(6) = 60000 + (5)		 4.1699 `4.1699 = 3.4869 + 0.683`
(7) = Previous(7) + (3)		 24,388.95 `24388.95 = 101698.65 / 4.1699`
(8)=(6)/(7)
613,000 0.6209 `0.6209 = 0.9091^(6-1)` 8,071.98 `8071.98 = 13000 xx 0.6209`
`(4)=(2)xx(3)`		 49,770.63 `49770.63 = 41698.65 + 8071.98`
(5) = Previous(5) + (4)		 109,770.63 `109770.63 = 60000 + 49770.63`
(6) = 60000 + (5)		 4.7908 `4.7908 = 4.1699 + 0.6209`
(7) = Previous(7) + (3)		 22,912.86 `22912.86 = 109770.63 / 4.7908`
(8)=(6)/(7)
716,000 0.5645 `0.5645 = 0.9091^(7-1)` 9,031.58 `9031.58 = 16000 xx 0.5645`
`(4)=(2)xx(3)`		 58,802.21 `58802.21 = 49770.63 + 9031.58`
(5) = Previous(5) + (4)		 118,802.21 `118802.21 = 60000 + 58802.21`
(6) = 60000 + (5)		 5.3553 `5.3553 = 4.7908 + 0.5645`
(7) = Previous(7) + (3)		 22,184.21 `22184.21 = 118802.21 / 5.3553`
(8)=(6)/(7)
819,000 0.5132 `0.5132 = 0.9091^(8-1)` 9,750 `9750 = 19000 xx 0.5132`
`(4)=(2)xx(3)`		 68,552.22 `68552.22 = 58802.21 + 9750`
(5) = Previous(5) + (4)		 128,552.22 `128552.22 = 60000 + 68552.22`
(6) = 60000 + (5)		 5.8684 `5.8684 = 5.3553 + 0.5132`
(7) = Previous(7) + (3)		 21,905.77 `21905.77 = 128552.22 / 5.8684`
(8)=(6)/(7)
922,000 0.4665 `0.4665 = 0.9091^(9-1)` 10,263.16 `10263.16 = 22000 xx 0.4665`
`(4)=(2)xx(3)`		 78,815.38 `78815.38 = 68552.22 + 10263.16`
(5) = Previous(5) + (4)		 138,815.38 `138815.38 = 60000 + 78815.38`
(6) = 60000 + (5)		 6.3349 `6.3349 = 5.8684 + 0.4665`
(7) = Previous(7) + (3)		 21,912.71 `21912.71 = 138815.38 / 6.3349`
(8)=(6)/(7)
1025,000 0.4241 `0.4241 = 0.9091^(10-1)` 10,602.44 `10602.44 = 25000 xx 0.4241`
`(4)=(2)xx(3)`		 89,417.82 `89417.82 = 78815.38 + 10602.44`
(5) = Previous(5) + (4)		 149,417.82 `149417.82 = 60000 + 89417.82`
(6) = 60000 + (5)		 6.759 `6.759 = 6.3349 + 0.4241`
(7) = Previous(7) + (3)		 22,106.42 `22106.42 = 149417.82 / 6.759`
(8)=(6)/(7)


The calculations in table show that the average cost is lowest during the `8^(th)` year (Rs 21,905.77).

Hence, the machine should be replaced after every `8^(th)` years, otherwise the average cost per year for running the machine would start increasing.




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3. Replacement of items that deteriorates with time (Model-1.3)
(Previous method)		2. Example-2
(Next example)


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