I just started a course on queue theory, yet equations are given for granted without any demonstrations, which is very frustrating... Thus
- Why is the mean number of people in a queue system following an $M/M/1$ system
$$E(L)=\frac{\rho}{1-\rho}$$
with $\rho=\frac{\lambda}{\mu}$ with $\lambda$ the clients arrival rate and $\mu$ the service rate.
- And the mean number of people waiting in the queue:
$$E(L^q)=\frac{\rho^2}{1-\rho}$$
For every common queue system you can follow this route:
Set up the balance equations: Inflow equal outflow in steady state.
Solve the balance equations. Always straight forward.
Calculate $p_0$ and $E[L]$
And its not hard to see that this gives $E[L]$, right?
The fact that the first sum to 1 gives $p_0$
$$\sum_ n P_n = 1$$
$$E[L] = \sum_ n nP_n$$
The range of the sum depends on which kind of queue you have. In your case its from $0$ to $\infty$ because you have unlimited places in the queueing system. If its limited to K places you sum to K.
E.g. $E[L^{q}]$ follows from:
(omitting the E[] from here on as im used to)
$L = L_q + L_s$ (average number of people in the queue plus average number of people getting service)
$L_s = W_s*\lambda$ (average waiting time getting service times the inflow rate. $\lambda =\lambda_e = \lambda*(1-p_k)$ when you have a limited number of places in the queue)
$W_s = 1 / \mu$ ($\mu$ mean service time)
And with those equations you can solve for $L_q$