# 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.

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 $$\frac{53}{60}$$.