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
[tex] \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}[/tex]
The birth-death process having balance equations [tex]\lambda_sP_i=\mu_{s+1}P_{i+1},i=0,1,2[/tex]
since, state rate at which leave = rate at which enter
0 [tex]\lambda_0P_0=\mu_1P_1[/tex]
1 [tex](\lambda_1+\mu_1)P_1= \mu_2P_2 + \lambda_0P_0[/tex]
2 [tex](\lambda_2+\mu_2)P_2= \mu_3P_3 + \lambda_1P_1[/tex]
[tex]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[/tex]
Since [tex]\sum\limits^3_{i=0} {P_i=1}\\\\p_0=[1+\frac{5}{12}+\frac{48}{25}+\frac{192}{250}]^{-1}=\frac{250}{1522}[/tex]
a)
Average number not in use equals the mean of the stationary distribution [tex]P_1+2P_2+3P_3=\frac{2136}{751}[/tex]
b)
Proportion of time both repairmen are busy [tex]P_2+P_3=\frac{672}{1522}=\frac{336}{761}[/tex]
