Question on M/M/2 queue variation I have the following question:
Two workers handle three machines(i.e. we can at most repair two machines at a time). The time until the machine breaks down is exponential distributed with expectation value $1/2$ and independent of other repair and break down times. Every repair time is exponential distributed with expectation value $1/3$ and independent of other repair and break down times. Let $X_t$ be the amount of broken down machines at time t.
a) Determine the birth and death frequencies in the birth and death chain $(X_t)_{t \geq 0}$.
b) Determine $\lim_{t \to \infty}\mathbf{P}_i(X_t =0)$ for $i=0,1,2,3$.
Attempt
a) The jump times are minimum of the exponential times which we know have the parameter the sum of each of them. Thus I got the Q-matrix (I assumed that only one worker could repair a machine):
$$
Q =
 \begin{pmatrix}
 -9 & 9 & 0 & 0 \\
  2 & -8 & 6 & 0 \\
  0  & 4  & -7 & 3  \\
  0 & 0 & 4 & -4
 \end{pmatrix}$$
So the frequency of going from 0 to 1 is 9, from 1 to 2 is 6, 2 to 3 is 3. To the opposite direction I get from 1 to 0 is 2, 2 to 1 is 4 and from 3 to 2 is 4.
b) Since this is irreducible, aperiodic and finite we know that this limit is just $\pi_0$ where $\pi$ is the invariant distribution. From this I get $\pi_0=16/277$ (With my $\pi$,  $\pi Q=0$) .
Is this this right or am I making some wrong conclusions in my calculations?
Regards, Raxel. 
 A: $X_t$ is the number of broken machines. I think your $Q$ matrix should be
$$Q=\begin{pmatrix}
-6 & 6 & 0 & 0 \\
3 & -7 & 4 & 0 \\
0 & 6 & -8 & 2 \\
0 & 0 & 6 & -6
\end{pmatrix}$$
because


*

*repairs happen at rate 3

*failures happen at rate 2


and in state


*

*0: all machines are working and can independently fail at rate 2 (so total failure rate 6)

*1: two machines can fail (total failure rate 4) and a single machine can be repaired (at rate 3)

*2: the one remaining machine can fail (rate 2) or the two broken machines can be repaired. They are being repaired in parallel so repairs happen at total rate 6.

*3: All machines are broken, but only two are being repaired (there are only two workers) so the total repair rate remains at 6.


Computing the stationary distribution using $\pi Q = 0$ we get
$$ \pi = \left(\frac{9}{43},\frac{18}{43},\frac{12}{43},\frac{4}{43}\right)$$
so $\pi_0 = \frac{9}{43}$ (which you have correctly identified is the answer irrespective of starting state $i$).
