queuing systems

Question

Suppose we have a $M/M/s$ queuing system :

$L$ : is the mean number of the clients on the system

$L_q$ : is the mean number of the clients on the queue

$s$ : The number of the servers

$n$ : The number of the clients at a given moment

$P_n$ : The probability that we have exactly $n$ clients on the system

how can I prove the validity of this equation :

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

Any ideas????

Answer 1

So I just figured out the answer :

We have

$ L=\sum_{n=0}^\infty n.P_n$

$ L_q= \sum_{n=s}^\infty (n-s).P_n$

so $L=\sum_{n=0}^\infty (n-s+s).P_n = \sum_{n=0}^\infty (n-s).P_n+P_n.s$

$L=\sum_{n=0}^\infty (n-s).P_n+\sum_{n=0}^\infty P_n.s =\sum_{n=0}^\infty (n-s).P_n+s.\sum_{n=0}^\infty P_n $

We have $\sum_{n=0}^\infty P_n = 1$ because $Pn$ represents a conditional probability.

$L=\sum_{n=0}^\infty (n-s).P_n+s$

$L-L_q=(\sum_{n=0}^\infty (n-s).P_n) - (\sum_{n=s}^\infty (n-s).P_n) +s$

$L-L_q =(\sum_{n=0}^{s-1} (n-s).P_n)+(\sum_{n=s}^\infty (n-s).P_n) - (\sum_{n=s}^\infty (n-s).P_n) +s$

Wich gives us the result we wanted :

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