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5. Model-3 example ( Enter your problem )
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1. Example-1

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

This material is intended as a summary. Use your textbook for detail explanation.
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