# Matrix Generator for M/M/1 Queue Waiting Time Distribution

I "believe" that generator, $\bf W$, of the waiting time distribution for the M/M/1 queue is given by the following (I'm not sure if this is even correct):

${\bf W} =\left( \begin{array}{ccccc} 0 & 0 & 0 & 0 & 0\\ \mu & -\mu & 0 & 0 & 0\\ 0 & \mu & -\mu & 0 & 0 \\ 0 & 0 & \mu & -\mu & \dots \end{array} \right)$

But the question I have is that I am unclear how to solve this Markov chain. That is, I'm looking for an analytic solution to

$\bf pW=0$

I think $\bf p$ should look something like

${\bf p} = [1−ρ,…],$

but again, I am unclear how to solve these problems.

Thanks for help in these matters.

Almost. The generator is ${\bf Q} =\left( \begin{array}{ccccc} -\lambda & \lambda & 0 & 0 & 0\\ \mu & -(\lambda+\mu) & \lambda & 0 & 0\\ 0 & \mu & -(\lambda+\mu) & \lambda & 0 \\ 0 & 0 & \mu & -(\lambda+\mu) & \dots \end{array} \right),$ where $\lambda$ is the arrival rate and $\mu$ the departure. If $\mu>\lambda$ (more people leave the queue than arrive), the chain is recurrent and the invariant distribution is an exponential distribution, i.e., $p_k=(1-\rho)\rho^k$, and $\rho=\frac{\mu}{\lambda}$. If $\rho> 1$ (resp. $=1$), the queue is transient (resp. null recurrent) and there is no stationary distribution.
• @PMF In any case, it seems to me that the solution you are looking for is $[1,0,0,0,\dots]$: from the equation corresponding to the first column of $W$ you get $p_2=0$, then from the second column $p_3=0$, and so on. $p_1$ remains indeterminate, and for the usual normalization $p\underline{\mathbf{1}}$ to hold we need $p_1=1$. May 24 '14 at 9:04