How to find the service rate $\rho$? 
The engineer of a small atelier observes. $6$ workers employed in this workshop are versatile, so that any order can be done by any of them. Nevertheless, the engineer is stressed because he noticed that workers are always busy and that its booking record has, as a mean, 20 ongoing orders (booked but not satisfied).
In order to have a better understanding of the situation, the engineer would like to estimate

*

*the mean time passed by each workers with each command. (I think it's $E(W)-E(W^q)$)

*when booking, the delivering delay to be waited by the client.


I said that

*

*$\lambda = 5$

*$E(L)=20$
By Little's law
$E(L)=\lambda E(W)$
thus $E(W)=\frac{20}{5}=4$
Then we can calculate $E(W^q)$, still by Little's law:
$E(W^q)=\frac{E(L^q)}{\lambda}$
yet I'm stuck here, because, I don't know what $\rho$ is...
Indeed $E(L^q)=\pi_0*\frac{\rho^s}{s!}*\frac{\rho'}{(1-\rho')^2}$... which I cannot calculate.
 A: I will write $\mu$ for the service rate and $\rho=\frac\lambda{6\mu}$ for the server utilization. 
This is a $M/M/6$ queue and so assuming $\rho<1$, the average number of customers in the system is given by $$\mathbb E[L]=\frac\rho{1-\rho}C\left(6,\frac\lambda\mu\right)+6\rho, $$ where $C(\cdot,\cdot)$ denotes the equilibrium probability that an arriving customer must wait in the queue (known as Erlang's $C$ formula) and is given by
$$C\left(6,\frac\lambda\mu\right) = \frac1{1+(1-\rho)\left(\frac{6!}{(6\rho)^6}\right)\sum_{k=0}^{6-1}\frac{(6\rho)^k}{k!}}. $$ Solving for $\rho$ analytically is not trivial, so let us use the approximation $\rho\approx 0.94402$. The mean service time $\mathbb E[S]$ is the reciprocal of the service rate, so $$\mathbb E[S] = \frac1\mu =\frac{6\rho}{\lambda}\approx\frac{5.66412}\lambda. $$
If we assume $\lambda=5$ (which does not follow from the problem statement!), then we have $$\mathbb E[S]\approx\frac{5.66412}{5}\approx 1.13282.$$
The expected number in queue can be computed by subtracting the long-run fraction of time $\pi_0$ that the queue is empty from the expected number in system. We have $$\pi_0 = \left[\sum_{k=0}^{6-1}\frac{(6\rho)^k}{k!} +\frac{(6\rho)^6}{6!}\left(\frac1{1-\rho}\right)\right]^{-1}\approx 0.00103763,$$ and so $$\mathbb E[L_q]\approx 20-0.00103763\approx19.999.$$
We may then use Little's Law to determine the expected waiting time in queue:
$$\mathbb E[W_q] = \frac{\mathbb E[L_q]}\lambda \approx\frac{19.999}{5}\approx 3.99979.$$
