Question 7.27: Consider a machine that can be in one of three states: good ......
Consider a machine that can be in one of three states: good condition, fair condition, or broken down. Suppose that a machine in good condition will remain this way for a mean time μ_{1} and then will go to either the fair condition or the broken condition with respective probabilities \frac{3}{4} and \frac{1}{4}. A machine in fair condition will remain that way for a mean time μ_{2} and then will break down. A broken machine will be repaired, which takes a mean time μ_{3}, and when repaired will be in good condition with probability \frac{2}{3} and fair condition with probability \frac{1}{3} What proportion of time is the machine in each state?
Letting the states be 1, 2, 3, we have by Equation (7.23)
\pi_i=\sum\limits_{j=1}^N \pi_j P_{j i}, \quad i=1,2, \ldots, N (7.23)
that the π_{i} satisfy
\begin{aligned}\pi_1+\pi_2+\pi_3 & =1, \\\pi_1 & =\frac{2}{3} \pi_3, \\\pi_2 & =\frac{3}{4} \pi_1+\frac{1}{3} \pi_3, \\\pi_3 & =\frac{1}{4} \pi_1+\pi_2\end{aligned}
The solution is
\pi_1=\frac{4}{15}, \quad \pi_2=\frac{1}{3}, \quad \pi_3=\frac{2}{5}
Hence, from Equation (7.24)
P_i=\frac{\pi_i \mu_i}{\sum_{j=1}^N \pi_j \mu_j}, \quad i=1,2, \ldots, N (7.24)
we obtain that P_i, the proportion of time the machine is in state i, is given by
\begin{aligned}& P_1=\frac{4 \mu_1}{4 \mu_1+5 \mu_2+6 \mu_3}, \\& P_2=\frac{5 \mu_2}{4 \mu_1+5 \mu_2+6 \mu_3}, \\& P_3=\frac{6 \mu_3}{4 \mu_1+5 \mu_2+6 \mu_3}\end{aligned}
For instance, if \mu_1=5, \mu_2=2, \mu_3=1, then the machine will be in good condition\frac{5}{9} of the time, in fair condition \frac{5}{18} of the time, in broken condition \frac{1}{6} of the time.