This Presentation describes the two stage networks used in the telecommunication switching network applications.

2. OUTLINE
• Two stage representation of N X N network
• Connections
• Switching Elements
• Two stage network with multiple switching matrices
• Network Parameters
• Blocking Network
• Blocking probability
• Discussion
• M inlets vs N outlets
• Problem
• Solution
• Non-Blocking Network

3. 3
Two stage representation of N X N
network
• For any single stage network there exists an equivalent multistage
network.
• So, N X N single stage network with capacity k can be realized by a two
stage network of N X K and K X N stages.

4. Connections
• Any of the N inlets can be connected to any of
the K outputs of 1st stage.
• Similarly, Any of the K inputs can be connected
to any of the N outputs of 2nd stage.
• So, there are K alternative paths and 2NK
switching elements.
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Any of the N inlets can be connected to any of the N outlets.

5. Switching Elements
• Each stage has NK switching elements
• Assume only a fraction of the subscribers to be active
on an average
• K can be equal to N/16
• So, no. of switching elements,
S = 2NK = N2/8
Example:
N = 1024, K = 64
S = 131,027
So, for large N, the switching matrix NxK may still be
difficult to realize practically.
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6. Two stage network with multiple
switching matrices
• M inlets are
divided into r
blocks of p
inlets. M = pr
• N outlets are
divided into s
blocks of q
outlets. N = qs
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7. Network Parameters
• For full connectivity there must be at least one outlet
from each block in the 1st stage terminating as inlet
on every block of the 2nd stage.
• So, block sizes are p x s and r x q respectively
• So, S = psr + qrs
• Putting values for M, N
• S = Ms + Nr
• The number simultaneous calls in the network,
switching capacity, SC = rs
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8. Blocking Network
• For rs connections to be simultaneously active, the s
active inputs in one block of the 1st stage must be
uniformly distributed across all the s blocks in the 2nd
stage at the rate of one per block.
• Blocking may occur in two conditions:
I. Calls are uniformly distributed (there are rs calls in
progress and (rs + 1)th calls arrives)
II. Calls are not uniformly distributed, there is a call in
progress from I-th block from the first stage to the J-
th block in the 2nd stage and another call originates
in the I-th block destined to J-th block.
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9. Blocking probability
Let α be the probability the a given inlet is active.
Now, probability that an outlet at the i-th block is active is, β =
(pα)/s
The probability that another inlet becomes active and seeks
an outlet other than the one which already active is given by
(p - 1)α/(s - 1)
Now, Probability that an ready active outlet is sought
PB = ((pα)/s)[1 – (p-1)α/(s-1)]
Substituting, p = M/r, we have
PB = ((Mα)/rs)[1 – ((M/r)-1)α/(s-1)]
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10. Discussion
• If s and r decrease then S can be minimized
• But if we decrease s and r we are increasing
blocking probability!
• So, we have to choose values for s and r as
small as possible but giving sufficient links to
provide a reasonable grade of service.
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11. M inlets vs N outlets
• If N > M, network is expanding traffic
• If M > N, concentrating the traffic
• If N = M, matrix size is uniform
i.e. r=s, p=q
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12. Problem
• In the two stage network discussed so far,
there is only one link between a block in the
1st stage and a block in the 2nd stage.
• What will happen if this particular link fails?
• Risk of severe blocking in the network!!
• How can we improve this performance?
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13. Solution
• Increase number of links between the blocks of the
stages.
• Consider k links beings introduced between every 1st
and 2nd stage pair.
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14. Non-Blocking Network
• In order to make the network non-blocking, must
have K=√N and for M=N, p=q=√N and r=s=K√N
• Now, S =Ms+Nr=2Nr=2N2
• And, SC =rs=√Nx√N=N
• So, a two-stage non-blocking network requires twice
the number of switching elements as the single stage
non-blocking network.
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15. THANK YOU