Assume that the number of arrivals of cells from slot to slot is an independent and identically distributed (i.i.d) random variable with an arbitrary distribution.
Let
Ai
be the number of cells arriving during the time slot
i.
Let
ni
be the number of the cells in the system (both in the buffer and the server).
Let us define the variable si as:
Then, we have the formula (1):
Since {Ai } has an independent identical distribution, ni is a discrete time homogeneous Markov Chain. A simple way of viewing a Markov Chain is that everything one can tell about the future from complete knowledge of the entire past, one can also tell just based on the most recent past.
The formal definition of Markov Chain is:
Now, let we use the state status to analyze the system. Let: aj = P{ Ai = j } be the probability of j cells arriving during slot i ;
= E{Ai } .
If
0 < < 1
(i.e. the arrival rate is less than the service rate),
we can show that a stationary (steady state) distribution
{ pj }
exist. Here:
In particular, we want to find:
Let time go to infinity,
The average number of services per slot is equal to the average number of arrivals per slot. This makes intuitive sense!
But, we still did not find the wanted value of E{ n } .
Let's try squaring the formula (1) before taking the expectation:
} =
E{ (ni -
si +
Ai )
}
} +
E{ si
} +
E{ Ai
}
-2E{ ni
si }
-2E{ si
Ai }
+2E{ ni
Ai }
We know that
si
=
si
since si is a binary random variable.
Thus,
E{ si
} =
E{ si }
.
Also,
Since Ai is independent of ni and Ai is independent of si , we have
}
=
E{ n
} +
+
-2E{ n}
-2()
+2E{ n
}
Thus, we have the Pollaczek-Khinchine formula for discrete time:
Let Tk be the average time spent in system up to time k, then
Let 
, we know that
Finally, we have Little's Formula
E{ T }