A three stage network is realized by using switching matrices of size p x s in stage 1, r x r matrices in stage 2 and s x p matrices in stage 3. Using the Lee's probability graph show that the Blocking Probability for the three stage network is given by PB = [1 - (a/k)2]s here, Where k = s / p and α = Inlet utilization factor.
LEE'S Graph for a Three Stage Network
For a three stage network;
One path from input to output=2 links in series=link among stage 1 and stage 2+link among stage 1 and stage 2
Assume that β = probability such that a link is busy
Therefore 1- β = probability such that a link is free
That is the probability such that a particular link (for example 1-2) by first stage to second stage is available = l- β.
Likewise the probability that a particular link (for example 2-3) from second stage to third stage is available = l-β
More probability that both of such links are available = (l- β) (1- β)= (1- β)2
Probability is that a particular path is busy = [1-(1- β)2]
There are now S parallel paths in between input and output, therefore so probability that all they are busy = [l-[1-β)2]S = PB
Hence,
PB = [l-[1-β)2]5
If α = Inlet utilization factor, that is probability that an inlet at first stage is busy
After that β = (α/s)p,
putting s/p = k
β = α/k therefore PB = [l-[1-( α/k))2]5