Question on M/M/2 queue variation

Question

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.

Answer 1

$X_t$ 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 $i$).