Replacement Model-3 calculator

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Solution

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t	0	1	2	3	4	5	6
P(t)	0	0.03	0.07	0.20	0.40	0.15	0.15

What is the optimal replacement plan?

2. The following mortality rates have been observed for a certain type of fuse:

t	0	1	2	3	4	5
P(t)	0	0.05	0.10	0.20	0.40	0.25

There are 1,000 fuses in use and it costs Rs 5 to replace an individual fuse. If all fuses were replaced simultaneously it would cost Rs 1.25 per fuse. It is proposed to replace all fuses at fixed intervals of time, whether or not they have burnt out, and to contiune replacing burnt out fuses as they fail. At what time intervals should the group replacement be made? Also prove that this optimal policy is superior to the straight forward policy of replacing each fuse only when it fails.

Example

1. A computer contains 10,000 resistors. When any resistor fails, it is replaced. The cost of replacing a resistor individually is Rs 1 only. If all the resistors are replaced at the same time, the cost per resistor would be reduced to 35 paise. The percentage of surviving resistors say S(t) at the end of month t and the probability of failure P(t) during the month t are as follows:

t 0 1 2 3 4 5 6
P(t) 0 0.03 0.07 0.20 0.40 0.15 0.15

What is the optimal replacement plan?

Solution:
The percentage of survivers resistors at the end of month t and the probability of failure are as follows,

t	0	1	2	3	4	5	6
P(t)	0	0.03	0.07	0.2	0.4	0.15	0.15

Let `N_i` be that number of resistors replaced at the end of the `i^(th)` month.

The different value of `N_i` can then be calculated as follows

`N_0` = number of resistors in the beginning = 10000

`N_1` = number of resistors being replaced by the end of `1^(st)` month

= `N_(0) P_(1)`

= `10000 * 0.03`

= `300`

`N_2` = number of resistors being replaced by the end of `2^(nd)` month

= `N_(0) P_(2) + N_(1) P_(1)`

= `10000 * 0.07 + 300 * 0.03`

= `709`

`N_3` = number of resistors being replaced by the end of `3^(rd)` month

= `N_(0) P_(3) + N_(1) P_(2) + N_(2) P_(1)`

= `10000 * 0.2 + 300 * 0.07 + 709 * 0.03`

= `2042`

`N_4` = number of resistors being replaced by the end of `4^(th)` month

= `N_(0) P_(4) + N_(1) P_(3) + N_(2) P_(2) + N_(3) P_(1)`

= `10000 * 0.4 + 300 * 0.2 + 709 * 0.07 + 2042 * 0.03`

= `4171`

`N_5` = number of resistors being replaced by the end of `5^(th)` month

= `N_(0) P_(5) + N_(1) P_(4) + N_(2) P_(3) + N_(3) P_(2) + N_(4) P_(1)`

= `10000 * 0.15 + 300 * 0.4 + 709 * 0.2 + 2042 * 0.07 + 4171 * 0.03`

= `2030`

`N_6` = number of resistors being replaced by the end of `6^(th)` month

= `N_(0) P_(6) + N_(1) P_(5) + N_(2) P_(4) + N_(3) P_(3) + N_(4) P_(2) + N_(5) P_(1)`

= `10000 * 0.15 + 300 * 0.15 + 709 * 0.4 + 2042 * 0.2 + 4171 * 0.07 + 2030 * 0.03`

= `2590`

The expected life of each resistor is given by
`= sum_(i=1)^6 x_i p(x_i)`

= `1 * 0.03 + 2 * 0.07 + 3 * 0.2 + 4 * 0.4 + 5 * 0.15 + 6 * 0.15`

= `4.02` months.

Average number of failures per month is given by
`N_0 / "(Mean age)" = 10000/4.02 = 2487.56`

` = 2488` resistors (approx.)

Hence, the total cost of individual replacement at the cost of Rs `1` per resistor will be

Rs `(2488 * 1)` = Rs `2488`

The cost replacemnet of all the resistors at the same time can be calculated as follows:

End of month	Total Cost of Group Replacement (Rs)	Average Cost per Month (Rs)
1	`(300) * 1 + 10000 * 0.35=``3800`	3800 `3800 = 3800 / 1`
2	`(300 + 709) * 1 + 10000 * 0.35=``4509` `3800 = 3800 / 1`	2254.5 `2254.5 = 4509 / 2`
3	`(300 + 709 + 2042) * 1 + 10000 * 0.35=``6551` `2254.5 = 4509 / 2`	2183.67 `2183.67 = 6551 / 3`
4	`(300 + 709 + 2042 + 4171) * 1 + 10000 * 0.35=``10722` `2183.67 = 6551 / 3`	2680.5 `2680.5 = 10722 / 4`
5	`(300 + 709 + 2042 + 4171 + 2030) * 1 + 10000 * 0.35=``12752` `2680.5 = 10722 / 4`	2550.4 `2550.4 = 12752 / 5`
6	`(300 + 709 + 2042 + 4171 + 2030 + 2590) * 1 + 10000 * 0.35=``15342` `2550.4 = 12752 / 5`	2557 `2557 = 15342 / 6`

Since the average cost per month of Rs 2,183.67 is obtained in the `3^(rd)` month,

it is optimal to have a group replacement after every `3^(rd)` month