Expected Waiting Time in a Queuing System $(M | M | 2 | 5)$ 
In the queuing system $(M | M | 2 | 5)$, the input flow rate is $240$ requests per hour, the average service time for one request is $30$ seconds. Find the average waiting time for an application in the queue.

How do I find that?
 A: Let $\pi_k$ denote the steady-state probability that there are $k$ requests.  We have $\lambda=4\,\texttt{min}^{-1}$ and $\mu=2\,\texttt{min}^{-1}$.  Write $\rho:=\dfrac{\lambda}{\mu}=2$.  We have the balance equations
$$\lambda\,\pi_0=\mu\,\pi_1\,,$$
and
$$\lambda\,\pi_k=2\mu\,\pi_{k+1}$$
for $k=1,2,3,4$.  That is,
$$\pi_1=\rho\,\pi_0\,\text{ and }\,\pi_{k+1}=\frac{\rho}{2}\,\pi_k$$
for $k=1,2,3,4$.  If $p:=\pi_0$, then
$$\pi_1=\rho\,p\,\text{ and }\,\pi_k=\frac{\rho^k}{2^{k-1}}\,p$$
for $k=2,3,4,5$.  Since $\sum\limits_{k=0}^5\,\pi_k=1$, we obtain
$$p=\frac{1}{1+\rho+\frac{\rho^2}{2}+\frac{\rho^3}{4}+\frac{\rho^4}{8}+\frac{\rho^5}{16}}\,.$$
The expected queue size is
$$L_q=0\,\pi_0+0\,\pi_1+0\,\pi_2+1\,\pi_3+2\,\pi_4+3\,\pi_5=\frac{\frac{\rho^3}{4}+\frac{2\rho^4}{8}+\frac{3\rho^5}{16}}{1+\rho+\frac{\rho^2}{2}+\frac{\rho^3}{4}+\frac{\rho^4}{8}+\frac{\rho^5}{16}}\,.$$
From Little's Law, the expected wait time $W_q$ is
$$W_q=\frac{L_q}{\lambda}=\frac{1}{\lambda}\,\left(\frac{\frac{\rho^3}{4}+\frac{2\rho^4}{8}+\frac{3\rho^5}{16}}{1+\rho+\frac{\rho^2}{2}+\frac{\rho^3}{4}+\frac{\rho^4}{8}+\frac{\rho^5}{16}}\right)\,.$$
For $\lambda=4\,\texttt{min}^{-1}$ and $\rho=2$, we get
$$W_q=\left(\frac{1}{4}\,\texttt{min}\right)\frac{2+4+6}{1+2+2+2+2+2}=\frac{3}{11}\,\texttt{min}\,.$$
