What we deal with when talking about performance measures of this systems is mostly average values. But how can I get the distributions of this values, i.e. the distribution of the number of customers in the waiting queue in an M/M/1/S system? Here is how I tried to answer:
The distribution of number of clients in the entire system is:
$P[N = k] = \rho^k P_0$
then, normalizing this equation and sums to S:
$P_0 \sum_{k=0}^S\rho^k = 1 => P_0 = \frac{1 - \rho}{1-\rho^{S+1}}$
So, the distribution of the number of clients in the entire system is:
$P[N = k] = \rho^k \frac{1 - \rho}{1-\rho^{S+1}}$, with $0 \leq k \leq S$.
Considering that we have clients in the waiting queue only if we have 1 client in service, the probability that we have $k$ clients in queue is equal to the probability that we have $k + 1$ clients in the entire system, so:
$P[N_q = k] = P[N = k + 1] = \rho^{k+1} \frac{1 - \rho}{1-\rho^{S+1}}$, with $1 \leq k \leq S - 1$.
For $k=0$ we have: $P[N_q = 0] = P[N = 0] + P[N = 1] = \frac{1 - \rho}{1-\rho^{S+1}} + \rho\frac{1 - \rho}{1-\rho^{S+1}} = \frac{1 - \rho}{1-\rho^{S+1}}(1 + \rho)$.
What do you think? Thank you all