Question

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

Answer 1

Answer:

Step-by-step explanation:

Let X(t) denote the number of machines breakdown at time t.

The givenn problem follows birth-death process with finite space

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

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

The birth-death process having balance equations $\lambda_sP_i=\mu_{s+1}P_{i+1},i=0,1,2$

since, state  rate at which leave = rate at which enter

            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)

Average number not in use equals the mean of the stationary distribution $P_1+2P_2+3P_3=\frac{2136}{751}$

b)

Proportion of time both repairmen are busy $P_2+P_3=\frac{672}{1522}=\frac{336}{761}$