4 . The chief of an industrial company is planning to purchase a new device of e

Question

4 . The chief of an industrial company is planning to purchase a new device of either category X or category Y. For each day's operation, the number of repairs M that machine X requires is a Poisson random variable with mean 0.10z, where z denotes the time (in hours) of daily operation. The number of daily repairs N for machine Y is a Poisson random variable with mean 0.12z. The daily cost of operating X is CX (z-202+ 40M2; for Y, the cost is CY (z) 16z +40N2. Assume that the repairs take negligible time and that each night the devices are to be cleaned so that they operate like new devices at the start of each day. (a) Which device reduces the expected daily cost for 10 hours of daily operation? (b) Which device reduces the expected daily cost for 20 hours of daily operation?

Explanation / Answer

Question 4

(a) Here z= 10 hours of daily operation

for Category X , so E[M] = 0.10 z = 0.10 * 10 = 1

For machine X,

Expected cost of operating X = E[CX(z)] = E [20z + 40M2] = 20E[Z] + 40 E[M2] = 20 * 10 + 40 * 1* 1 = $ 240

for category Y, so E[N] = 0.12 z = 0.12 * 10 = 1.2

Expected cost of operatig Y = E[CY(z)] = E [16z + 40N2] = 16 * 10 + 40 * 1.2 * 1.2 = $217.6

Here device of category Y has lower expected cost for 10 hours of daily operation.

(b)

Here z= 20 hours of daily operation

for Category X , so E[M] = 0.10 *20 = 0.10 * 20 = 2

For machine X,

Expected cost of operating X = E[CX(z)] = E [20z + 40M2] = 20E[Z] + 40 E[M2] = 20 * 20 + 40 * 2 * 2 = $ 560

for category Y, so E[N] = 0.12 z = 0.12 * 20 = 2.4

Expected cost of operatig Y = E[CY(z)] = E [16z + 40N2] = 16 * 20 + 40 * 2.4 * 2.4 = $ 550.4

Here device of category Y has lower expected cost for 20 hours of daily operation.

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