M/M/S Queue, probability interpretation

Question

In an M/M/s queue, what does this expression mean? :

$\sum_{n=0}^{s-1}{(s-n)P_n}$

Furthermore, is it possible that the following equation holds? :

$\sum_{n=0}^{s-1}{(s-n)P_n} = (1-\rho)s$

If so, how could I demonstrate it?

EDIT: An M/M/S queue is a queue with Markovian (Poisson law) process for arrivals, Markovian (Exponential law) process for departures and $s$ servers.

Thanks,
Louis

Answer 1

If $p_n$ is the stationary distribution then the first expression is the expected number of free servers.

The equation is indeed correct and this can be argued using Little’s law: the arrival rate to each server is $\frac{\lambda}{s}$ and the expected time in service is $\frac{1}{\mu}$, so the probability each server is free is $1-\frac{\lambda}{s\mu}$. The servers are identical so the the sum is just $s$ times the probability of a free server.