A job shop consists of three machines and two repairmen. The amount of time a machine works before breaking down is exponentiall

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A job shop consists of three machines and two repairmen. The amount of time a machine works before breaking down is exponentiall

y distributed with mean 10. If the amount of time it takes a single repairman to fix a machine is exponentially distributed with mean 8, then(a) what is the average number of machines not in use?(b) what proportion of time are both repairmen busy?

Mathematics

1 answer:

tester [3.9K] · Answer 1 · 2026-03-04T14:51:18+03:00

Answer:

Step-by-step explanation:

Define X(t) as the count of machines that are out of order at time t.

The specified issue is modeled with a birth-death process within the confined set

S={0, 1, 2, 3} alongside

$\lambda_0=\frac{3}{10}, \mu_1=\frac{1}{8}\\\\ \lambda_1=\frac{2}{10}, \mu_2=\frac{2}{8}\\\\ \lambda_2=\frac{1}{10}, \mu_3=\frac{2}{8}$

The balancing equations for the birth-death process are given by $\lambda_sP_i=\mu_{s+1}P_{i+1},i=0,1,2$

because the exiting rate = the entering rate

0 $\lambda_0P_0=\mu_1P_1$

1 $(\lambda_1+\mu_1)P_1= \mu_2P_2 + \lambda_0P_0$

2 $(\lambda_2+\mu_2)P_2= \mu_3P_3 + \lambda_1P_1$

$P_1=\frac{12}{5}P_0=P_0=\frac{5}{12}P_1\\\\P_2=\frac{48}{25}P_0=P_0=\frac{25}{48}P_2\\\\P_3=\frac{192}{250}P_0=P_0=\frac{250}{192}P_3$

Since $\sum\limits^3_{i=0} {P_i=1}\\\\p_0=[1+\frac{5}{12}+\frac{48}{25}+\frac{192}{250}]^{-1}=\frac{250}{1522}$

a)

The average number of machines not in operation is reflected in the mean of the stationary distribution $P_1+2P_2+3P_3=\frac{2136}{751}$

b)

The fraction of time both repair technicians are engaged $P_2+P_3=\frac{672}{1522}=\frac{336}{761}$