Expected time for two new customers in a $(\lambda_A=1,\lambda_S=2)-M/M/2$ in steady state. Two customers enter a $(\lambda_A=1,\lambda_S=2)-M/M/2$ in steady state.
Find the expected time that it will take for both to make through queue and have their service done.
My attempt: if this is a $M/M/1$ queue, then the time for the first person is $(\lambda_S-\lambda_A)=1$-exponential, and the other one is $\lambda_S=2$-exponential.
But I don't know how to proceed if this is $M/M/2$
Any help would be appreciated.
 A: Let $\lambda$ be the arrival rate and $\mu$ the service rate. From the detailed balance equations
\begin{align}
\lambda \pi_0 &= \mu \pi_1\\
\lambda \pi_{n} &= 2\mu \pi_{n+1}, n\geqslant1
\end{align}
we find by induction that $$\pi_{n+1}=\left(\frac\lambda{2\mu}\right)^n\frac\lambda\mu \pi_0,n\geqslant0 $$ stationary distribution for the number of customers in the system to be
\begin{align}
\pi_0 &= \frac1{1+\frac\lambda\mu\left(\frac1{1-\frac\lambda{2\mu}} \right)}\\
\pi_n &= \frac{\left(\frac\lambda{2\mu}\right)^{n-1}\frac\lambda\mu}{1+\frac\lambda\mu\left(\frac1{1-\frac\lambda{2\mu}} \right)},n\geqslant 1.
\end{align}
Given that two customers enter the system when there are $n$ customers in the system (and no customer finishes service between their arrivals), these two customers exit the system after $n+2$ services. The first $n+1$ are exponentially distributed with rate $2\mu$ and the last is exponentially distributed with rate $\mu$. The expected time is thus $\frac1{2\mu}(n+1) + \frac1\mu = \frac{n+3}{2\mu}$. Summing over $n$, we have
\begin{align}
\frac{\frac3{2\mu}}{1+\frac\lambda\mu\left(\frac1{1-\frac\lambda{2\mu}} \right)} + \sum_{n=1}^\infty \frac{\frac{n+3}{2\mu}\left(\frac\lambda{2\mu}\right)^{n-1}\frac\lambda\mu}{1+\frac\lambda\mu\left(\frac1{1-\frac\lambda{2\mu}} \right)} &= \frac{3}{2 \mu }-\frac{2 \lambda }{\lambda ^2-4 \mu ^2}.
\end{align}
Substituting $\lambda=1$ and $\mu=2$ gives $\frac{53}{60}$.
