Heterogeneous two-server queue ($\mu ={\mu }_1+{\mu }_2$)

A queue with a Poisson process of arrivals and two independent service times with exponential distributions of different rates must distinguish the two possible single-customer states. If an arrival chooses either idle server with probability $1/2$, the continuous-time Markov chain is reversible. Write $a,b$ for the states with only server one or two busy, and $n\ge 2$ for total population. Its detailed balance equations give ${\pi }_a={\pi }_0\nu /(2{\mu }_1)$, ${\pi }_b={\pi }_0\nu /(2{\mu }_2)$ and ${\pi }_n={\pi }_0{\nu }^2(\nu /\mu {)}^{n-2}/(2{\mu }_1{\mu }_2)$. These weights normalize exactly when $\nu <\mu$. In equilibrium, reversed upward transitions have total rate $\nu$ in every state, so time reversal of a continuous-time Markov chain makes the departure stream a Poisson process of rate $\nu$. This output conclusion requires stationarity.