Markov process - Queuing system - What is the average number of long-term service repairers? 
The operating time of a unit has an exponential law of expectancy $2$
  while the repair time follows an exponential law of expectancy $1/2$ .
  There are $4$ equipments and $2$ repairers who can work on each device
  at a time. It is assumed that all uptime and repair are independents.
  What is the average number of long-term service repairers?

Let $F :$ the operating time of an appliance with $F \sim Exp(\frac{1}{2})$ and $R :$ the repair time  with $R \sim Exp(2)$
Honnestly, I just don't know how to proceed with this question. Is anyone could help me at this point?
 A: Let $X(t)$ be the number of broken machines at time $t$. Then $\{X(t):t\geqslant0\}$ is a continuous-time Markov chain on $\{0,1,2,3,4\}$ with generator matrix
$$Q=\begin{bmatrix}
-2 & 2 & 0 & 0 & 0\\
2 & -\frac72& \frac32 & 0 & 0\\ 
0 & 4 & -5 & 1 & 0\\
0 & 0 & 4 & -\frac92 & \frac12 \\
0 & 0 & 0 & 4 & -4 \\
\end{bmatrix}.$$
Since the process is irreducible and has a finite state space, there is a unique stationary distribution $\pi$. That is, $\pi$ satisfies $\pi P(t) = \pi$ for all $t>0$ where $P_{ij}(t) = \mathbb P(X(t)=j\mid X(0)=i)$. From the Kolmogorov backward equations $P'(t) = QP(t)$ it is straightforward to verify that $\pi$ is a stationary distribution if and only if $\pi Q=0$. So from $\pi Q=0$ and $\sum_i \pi(i)=1$ we obtain the system of equations
\begin{align}
-2\pi(0) + 2\pi(1) &= 0\\
2\pi(0) -\frac72\pi(1) + 4\pi(2) &= 0\\
\frac32\pi(1) -5\pi(2) + 4\pi(3) &= 0\\
\pi(2) - \frac92\pi(3)+ 4\pi(4) &= 0\\
\pi(0) + \pi(1) + \pi(2) + \pi(3) + \pi(4) &= 1,
\end{align}
which yields
$$
\pi = \begin{bmatrix}\dfrac{256}{635}& \dfrac{256}{635} & \dfrac{96}{635} & \dfrac{24}{635} & \dfrac{3}{635} \end{bmatrix}.
$$
