# Problem with two server Queue

I am practicing for an exam on queuing theory and I found this question somewhat confusing. Appreciate if somebody can shed some light on it.

There are 2 facilities, A and B, which provide the same type of service. Each facility contains a single server with service times distributed as negative exponential random variables. The mean service times are $1/\mu_1=1$ minute and $1/\mu_2 = 4$ minutes at facilities A and B, respectively.

A combined total of 60 customers per hour wish to avail themselves of the service provided by facilities A and B. Arrivals of customers at the critical point are Poisson. At that point each customer makes a choice, independently of all others, on which facility he is going to use. This choice is made without any knowledge of the status of the queue in front of either facility. Let $p$ denote the probability that a random customer chooses facility A (and $1 - p$ that he or she chooses B).

Each user is only concerned with minimizing his expected time spent in a service facility, (= average waiting time + average service time). It can be reasonably expected that in the long run (system in steady state and in equilibrium-as far as the distribution of customers between the two facilities is concerned) the customers will "distribute" themselves among the two facilities, in the sense that p will stabilize around a specific value. What is that value of p?

Let $\lambda=60$, then the arrival rate to station $A$ is $60p$ and the arrival rate to station $B$ is $60(1-p)$, so the utilization at station $A$ is $\rho_A=p$ and the utilization at station $B$ is $\rho_B=4(1-p)$ (so we must have $p>\frac34$ for the system to be stable). The expected size in system is \begin{align} \mathbb E[L_A] &= \sum_{n=1}^\infty n(1-\rho_A)\rho_A^n= \frac{\rho_A}{1-\rho_A} = \frac p{1-p}\\ \mathbb E[L_B] &= \sum_{n=1}^\infty n(1-\rho_B)\rho_B^n= \frac{\rho_B}{1-\rho_B} = \frac{4(1-p)}{4p-3}. \end{align} By Little's law, the mean sojourn times are \begin{align} \mathbb E[W_A] &= \frac{\mathbb E[L_A]}{\lambda_A} = \left(\frac p{1-p}\right)\frac1p = \frac1{1-p}\\ \mathbb E[W_B] &= \frac{\mathbb E[L_B]}{\lambda_B} = \left(\frac{4(1-p)}{4p-3}\right)\frac1{4(1-p)} = \frac1{4p-3}. \end{align} Over time we can reasonably expect the distribution of customers to tend so that the expected waiting time is the same at each station. Therefore $$\mathbb E[W_A]=\mathbb E[W_B]\implies \frac1{1-p}=\frac1{4p-3}\implies p=\frac45.$$