Consider $N$ busses, each which breaks down independently at a rate of $\mu$. Once a bus breaks down it is sent to a repair shop which can only repair one bus at a time and takes an exponentially distributed amount of time with parameter $\lambda$ to repair it.
I'm trying to find the equilibrium distribution of the number of busses in service.
Could some one explain to me how to set this up? Am I supposed to do a Q matrix? Is $\mu$ intended to be determinate, so that the first time around all buses break down at the same time, and then it takes an exponentially distributed amount of time with parameter $\lambda$ for the first bus to exit the repair shop? Then that would mean it would take twice the random variable for the next one to exit, and so on, but then as soon as one exists it comes back in $\mu$ time. Like I said I'm just lost as to how to set this up.
Thanks
If we write $N$ for the number of buses in operation, the system you describe is an M/M/$N$/$N$ queue. State $0$ represents the situation where all buses are broken and $N$ when all are working. The queue is finite capacity ($N$ can take values in $\{0, ..., N\}$), arrivals happen when a bus is repaired at rate $\lambda$ and when there are $c$ buses in operation the buses break down at rate $c\mu$. Note repairs can only happen when there is at least one broken bus so there is no $\lambda$ transition from state $N$.
The system has a $Q$ matrix $$ Q=\begin{pmatrix} -\lambda & \lambda \\ \mu & -(\mu+\lambda) & \lambda \\ &2\mu & -(2\mu+\lambda) & \lambda \\ &&3\mu & -(3\mu+\lambda) & \lambda \\ &&&&\ddots\\ &&&&(N-1)\mu & -((N-1)\mu+\lambda) & \lambda \\ &&&&&N\mu & -(N\mu) \\ \end{pmatrix} $$
and therefore stationary distribution can be found by solving $\pi Q = 0$ which gives
$$ \pi_k = \frac{(\lambda/\mu)^k}{k!} \left[\sum_{k=0}^N \frac{\lambda^k}{\mu^k k!} \right]^{-1}. $$