Switching, Part I Switching, Part I Circuit Circuit - - Switching Switching Chapter 5: Chapter 5: Space Space - - Division Division Switching Switching David Larrabeiti David Larrabeiti Jos Jos é é F F é é lix Kukielka lix Kukielka Piotr Pacyna Piotr Pacyna M M ó ó nica Cort nica Cort é é s s
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Switching, Part ISwitching, Part I
CircuitCircuit--SwitchingSwitching
Chapter 5:Chapter 5:
SpaceSpace--Division Division SwitchingSwitching
David LarrabeitiDavid Larrabeiti
JosJoséé FFéélix Kukielkalix Kukielka
Piotr PacynaPiotr Pacyna
MMóónica Cortnica Cortééss
Switching, Chapter 5 2
Space-Division Switches
Basic Scheme: Rectangular N x M matrix of crosspoints
Nx: Number of crosspoints
Special case: M = N: Square matrix
Configuration (control)
N in
pu
ts
M outputs (N)
Crosspoint:set (1,..3) of electronic contacts which are activated / deactivated simultaneously
2matrix) (square NNNMN
X
X
=
⋅=
Switching, Chapter 5 3
Basic Structure for Two-/Four-Wire Switches I
Using a square matrix for four-wire transmission:
Connection between i and j requires:
Activation of two crosspoints: (i, j) and (j, i)
123
i
j
4
N-1
N
1 2 3 4 i j N-1 N
(i,j)
(j,i)
.... .... ....
........
....
Example: Square matrix for four-wire transmission
Switching, Chapter 5 4
Basic Structure for Two-/Four-Wire Switches II
For two-wire transmissionOnly a single crosspoint necessaryCost of implementation for a square matrix:
Roughly proportional to the number of crosspoints Nx = N2
Highly inefficient scheme
E.g. if N = 1.000, Nx = 1.000.000
At maximum: only 1.000 points in use simultaneously
More efficient schemes necessarySolution: Reduce the number of crosspoints
Tradeoff: Increased control complexityImportant for electromechanical switches
Tradeoff: Scalability more complex to achieve
Switching, Chapter 5 5
Example: Two-Wire Switch
Square Matrix for two-wire switches
(SIMPLE CONTROL)
Requires activating only one crosspoint (by definition,
e.g., the one of the source row)
1 2 3 4 i j N-1N
123
i
j
4
N-1N
Switching, Chapter 5 6
Improving Utilization: Folding
Remove crosspoints for (i, i) connectionsRemove crosspoints for (j, i) connections
If j > i
Need more control logicTest whether i < j
123
i
j
4
N-1N
1 2 3 4 i j N-1 N
No matter if the call is originated by i or
j, the same crosspoint is
activated
Triangular or folded matrix for two-wire switching
( )22
22)1( NONONNNX <
=
−=
External cabling
Switching, Chapter 5 7
Extension of Basic Switches I
Easy extension of square matricesHigh scalability
123
i
j
4
N-1
N
1 2 3 4 i j N-1 N
1
2
3
i
j
4
N-1
N
1 2 3 4 i j N-1 N
1
2
3
i
j
4
N-1
N
1 2 3 4 i j N-1 N
1
2
3
i
j
4
N-1
N
1 2 3 4 i j N-1 N
1
2
3
i
j
4
N-1
N
1 2 3 4 i j N-1 N
Extending NxNswitch to a
(2N)x(2N) switch
Switching, Chapter 5 8
Extension of Basic Switches II
Difficult extension of triangular matrices
Used in switching centrals which are assumed not
to be extended (e.g., PBX)
123
i
j
4
N-1
N
1 2 3 4 i j N-1 N
1
2
3
i
j
4
N-1
N
1 2 3 4 i j N-1 N1
2
3
i
j
4
N-1
N
1 2 3 4 i j N-1 N
1
2
3
i
j
4
N-1
N
1 2 3 4 i j N-1 N
1
2
3
i
j
4
N-1
N
1 2 3 4 i j N-1 N
Extending a NxN switch to a (2N)x(2N) switch?
??
Must re-cable all connections and build
a new triangular (2N)x(2N) matrix
Switching, Chapter 5 9
Reduction of Crosspoints in Transit Switches
Assumption: In transit switches, not necessary that each incoming link can be connected to each outgoing link
1. Grading Construction of a switch with limited access to save crosspoints
Sharing of links within the matrix
2. Division and tradeoff crosspoints vs. linkIncrease the number of matrices at the expense of requiring more links
With / without sharing links between matrices, which makes control and extension of switches more complex
Switching, Chapter 5 10
Example: Grading
Here: 50% reduction of number of crosspoints
A well-chosen selection of the paths can minimize access restrictions (= internal blocking)
1 2 3 4
1
2
3
4
5
6
7
8
Input
Output
Switching, Chapter 5 11
Example: Multiple Matrices I
200 subscribers, A0 = 10 erlangTarget GoS = 0.2%How many crosspoints in the local central?Simple solution: rectangular matrix
B = 0.002 = E (m, 10) m = 20Nx = 200 * 20 = 4000
Advanced solution: Two matrices with 100 subscribers eachB = 0.002 = E(m, 5) m = 13 (26 in total)Nx = 2 * (100 * 13) = 2600
More advanced solution: Four matrices with 50 subscribers each
B = 0.002 = E(m, 2,5) m = 8 (32 in total)Nx = 4 * (50 * 8) = 1800
Tradeoff: Number of crosspoints vs. number of links
In single-stage switches:Only one specific crosspoint for a certain input / output pair
Inefficiency
Low robustness: if the crosspoint fails, no alternative
connection
Multiple-stage switches:Sharing of crosspoints
Several internal paths through the matrices using
different crosspoints
Design principle: Use of small matrices
Arranged in two or more stages
Interconnected by a mesh structure between the stages
Switching, Chapter 5 15
Discussion: Multiple-Stage Switching
Advantage:Reduction of crosspoints
Try to make usage of crosspoints “more efficient”
Use one crosspoint for several internal paths
Robustness: Have several path for one input / output pair
Use of regular and uniform structures to alleviate the design and possible extensions
But: Potentially limited accessibility (i.e. from all input links to all output links)
Might be restricted, even though there is a free output link (but no free internal path)
INTERNAL BLOCKING
Switching, Chapter 5 16
Two-Stage Switches: Example I
Continuation of previous example:200 subscribers, A0 = 10 erlang
Nx = 50x(4x4) + 4x(50x6) = 2.000 crosspoints
Problem: Internal blocking possible
4 x 4(1)
4 x 4(50)
200 input links
50 x 6(1)
50 x 6(2)
50 x 6(3)
50 x 6(4)
24 output links
...1st stage, 50 matrices 2nd stage, 4 matrices
Mes
hed
M
esh
ed
inte
rco
nn
ecti
on
inte
rco
nn
ecti
on
Switching, Chapter 5 17
Two-Stage Switches: Internal Blocking
Internal Blocking:Only a single path available in the meshed interconnection for one input and a given output:
Example: Connect input 2 and output 5Input 1 and 3 can not be connected to output 4 or 6
3x3
123
456
789
123
4
6
789
5
Switching, Chapter 5 18
Two-Stage Switches: Example II
Specify internal blocking in previous example:One connection from 1st-stage switch 1 to 2nd-stage switch 2
No other connection possible
Bint = a = 0,05 for a concrete output link
Probability that the necessary path is busy
Btot < [ a + E (6, 0,05 * 50) ]4 << 0.0001
Btot < 0,002 = GoS
Btot: congestion and internal blocking to any output
Only a rough calculation, not easy to separate Bint
and congestion blocking
Switching, Chapter 5 19
Gain of Crosspoint Reduction in Two-Stage Switches
Reduction of crosspoints:E.g. for an NxN switch of n matrices nxn in each stage:
If N = 1000, Nx = 2.000.000In contrast: Nx = 100.000.000 for a single square
matrix
At minimum three stages necessary to avoid internal blocking
23
2)(2 NnnnN
NnnnN
X =⋅=
=⇒⋅=
Switching, Chapter 5 20
Multiple Stage Switches Without Internal Blocking
To avoid internal blocking: Have more than one path for each pair input - outputBased on the number of possible paths, the blocking probability can be reduced or even eliminated
Example: Three stages with blocking123
456
789
123
456
789
When adding the third stage, connect input 2 with exit 5 and then connection
1—6 and 3—4 are possible
1
2
3
4
5
6
7
8
9
1
2
3
4
5
6
7
8
9
Having the connections 1—1, 2—2, and 5—8, connection 3—7 is no
longer possible.
Switching, Chapter 5 21
Generic Structure of a Switch with Three Stages I
NxM switch
b = N/nmatrices
k matricesN
ninputs
mM
nN×
kn ×
mM
nN×
kn ×
...
...
...m
outputs
......
...... m
outputsn
inputs
p = M/mmatrices
mk ×
mk ×
M
...
k internal paths
possible
1st stage 3rd stage
2nd stage
Switching, Chapter 5 22
Generic Structure of a Switch with Three Stages II
Abbreviated notation:
Number of crosspoints
Special case: Symmetric NxN switch:
Number of crosspoints
N n
b=N/n p=M/m
M
k
M/mN/n kk m
( ) MNnmNMkkMNpkm
mM
nNkbnkN X *<++=++=
N n
b=N/n b=N/n
N
k
N/nN/n kk n
222
2 NnNkNkbkn
nNkbnkN X <
+=+
+=
Switching, Chapter 5 23
Theorem of Closfor Three-Stage Switches
Basic question: Possible to construct a three-stage switch without internal blocking?I.e. is there a minimum value for k which always allows to find an internal path for a given input / output pair?
k: number of matrices in the 2nd stage
Theorem of Clos (1953), looking at the worst case:
Given an input and an output, find a 2nd-stage matrix which has
a free connection to the 1st-stage matrix and the 3rd-stage matrix
In the 1st-stage matrices of the input and the 3rd-stage matrices
of the output, in summary n-1 and m-1 connections are
occupiedAssumption: The given input link and the given output link are idle
Need at least one more 2nd-stage matrix
Clos condition:11)1()1(min −+=+−+−= nmmnk
Switching, Chapter 5 24
Theorem of Clos for Three-Stage Switches: The Worst Case
N
......
M
BusyFreeAvailable
n-1 busy paths
n-1 free paths
m-1 free paths
m-1 busy paths
Available path Available
path
mM
nN×
kn×
kn×
kn× mk ×
mk ×
mk ×
mM
nN×
mM
nN×
mM
nN×
mM
nN×
11)1()1(min −+=+−+−= nmmnk
Clos condition
Switching, Chapter 5 25
Theorem of Clos for Symmetric Three-Stage SwitchesM=N:
Clos condition:
Crosspoints:
Optimization to minimize the number of crosspoints:
12min −= nk
( ) ( )
+−=
+=−
22
3 2122,nNNn
nNkNknNN ClosX
( )
( )
=−=
≈⇒=
∂∂
−
−
)(124
20,
233
3
NONNN
Nn
nnNN
mínimalClosX
optimalClosX
n>>1N>>1
Switching, Chapter 5 26
Comparison: Three-Stage Clos Switch vs. Single-Stage Square Matrix
N n Crosspoints ClosCrosspoints
square matrix16 2 288 25627 3 675 729
128 8 7.680 16.3842048 32 516.096 4.194.304
131072 256 267.911.168 17.179.869.184
Switching, Chapter 5 27
Multistage Switches based on Clos
Given a three-stage switch complying with the Closcondition (Clos switches): Reduce the number of crosspoints even more by substituting the 2nd-stage switches by three-stage Clos switches
Example: Symmetric Closswitch with 5 stages:
n
n
n
n
...
...
...
k=2n-1
nN
nN×)12( −× nn
)12( −× nn nN
nN×
nn ×− )12(
nn ×− )12(
N/n N/n
N N
m
m
m
m
......
...
k’ =2m-1
mnN
mnN
⋅×
⋅)12( −× mm mm ×− )12(
N/(nm) N/(nm)
N/n N/n
mnN
mnN
⋅×
⋅)12( −× mm mm ×− )12(
Switching, Chapter 5 28
Multistage Switches based on Clos:Abbreviated Notation
N n
N/n N/n
N
2n-1
N/nN/n 2n-12n-1 n
N/n m
N/(nm) N/(nm)
N/n
2m-1
N/(nm)2m-1 N/(nm) m2m-1
( )2
3 )12(122,
−+−=
− mnNmm
nNm
nNN ClosX
Switching, Chapter 5 29
Multistage Switches based on Clos
Crosspoints for a Clos switch with 5 stages:
Optimization of the number of crosspoints:
( ) ( )
⋅+−+−=
=
−+−= −−
2
35
2)12(2)12(
,12)12(2,,
mnN
nNmNn
mnNNnnn
nNmnNN ClosXClosX
( )
( )( )
≈
≈
⇒
=∂
∂
=∂
∂
−
−
31
31
5
5
2
40,,
0,,
Nn
Nm
mmnNN
nmnNN
opt
opt
ClosX
ClosX m>>1n>>1N>>1
Switching, Chapter 5 30
Multistage Switches based on Clos
Generalization:With the same procedure, can create switches of 7, 9, … stages.
Finally, Cantor demonstrated in 1972:
Upper limit for reducing the number of crosspoints at the expense of increasing the interconnection effort in the meshed interconnection network
( )NLog
NstagesClos eNON 22⋅→
∞→∞→
Switching, Chapter 5 31
Exercise: Clos Switches
Design an optimal 3-stage switch without blocking for 1200 lines
Solution:
1200 24
50 50
N
47
5050 4747 24
Switching, Chapter 5 32
Switches with Internal Blocking
Strictly non-blocking switches: Still too large number of crosspoints
Reduce further, tolerate a certain (small) blocking probability
Example: Typical residence telephoneBusy: 5-10% of the time in the busy hourBlocking: Appr. 1% in the busy hour
Peer might be busy in anyway...
Can reduce the number of crosspoints significantly
Example: Local centers / PBX (less loaded)
Two methods for evaluating blocking probabilities: Lee: very easy, not that accurate, very useful to compare different structures in generalJacobaeus: more accurate
Switching, Chapter 5 33
Method of Lee: Lee Graphs
To calculate the blocking probability in a multi-stage switch:
Draw a graph with all possible paths
Denote the link utilization for each link
Called “loading” p = probability that link is busy
Also: utilization of the link or % of time it is busy
q = 1 – p : probability the link is idle
Example (three stages)...
1
2
k
pp’
p’p’ p’
p’
p’p
Switching, Chapter 5 34
Blocking Probability Calculation (Lee)
n parallel links, each with a utilization of p
Assuming: Independence between links
Blocking probability: B = pn
n links in series, each with a utilization of pBlocking probability: B = 1 – qn
1
2
n
p
pp
1 2 n
Switching, Chapter 5 35
Lee: Three-Stage Example
Here: B = (1 – (q’)2 )k
q’: probability for an interstage link to be idle
k parallel links
2 links in series
Problem: What is q’ ( = 1 – p’)?Look at stage 1 (n input links, k output links)
p’ = (p * n ) / k = p / β,
β = k / n: expansion factor
• > 1: More 2nd stage matrices than input links• < 1: Concentration (for local centrals only)
q’ = 1- p / β
Then: B = (1 – (1 – (p / β))2)k
...1
2
k
pp’
p’p’ p’
p’
p’p
Switching, Chapter 5 36
Example Design for Three-Stage Switches (Inlet utilization: 0.1)
For local centrals / PBXConcentrators (n > k, β < 1)
4. Based on problem 1, compare the necessary number of crosspoints for
1. Square matrices2. Matrices based on Clos3. For problem 2 and B < 10-2
4. For problem 3 and B < 10-2
Solution:1. Square Matrix (N = 512): Nx = 262.1442. Clos: 63.488 (cf. problem 1)3. Problem 2 (A = 0.1 erlang): Nx = 40960
According to Jacobaeus (Lee overestimates B)
4. Problem 3 (A = 0.7 erlang): Nx = 12.288
According to Lee (Jacobaeus overestimates B)
Switching, Chapter 5 47
Blocking Probabilities in Transit Switches
Up to now two assumptions:Blocking probabilities between two specific input/output
Service requests independent from each other
Ok for line-to-line switching (local centrals)Well, some approximations necessary...
Not ok for transit switchesWhere any output from a trunk group is needed, not a specific one
Lower blocking probability than for line-to-line switching
If independent blocking probabilities: Just the product of the individual blocking probabilities of each output
But: Not independent because of common links
• E.g.: Between 1st stage and 2nd stage
Switching, Chapter 5 48
Assigning Output Lines in a Transit Switch
Bad: Assign all links of a trunk to one output matrix (e.g. 7, 8, 9):
Same path to all links of the trunk
Three paths in total
Better: Distribute links of a trunk over the output matrices (e.g., 1, 6, 9)
Decreases the probability of internal blocking
Three paths for each output link
1
2
3
4
5
6
7
8
9
1
2
3
4
5
6
7
8
9
1
2
3
4
5
6
7
8
9
1
2
3
4
5
6
7
8
9
Switching, Chapter 5 49
Further Issues on Transit Switches
Blocking probability p of single lines not independent
If some busy, p for the remaining links increases
Average traffic intensity varies:Business subscriber loops have higher activity than private subscriber loops
Try to distribute highly active subscribers uniformly over the 1st stages of a switch
Decreases the internal blocking probability
So-called “Line Management”
Switching, Chapter 5 50
Folded Four-Wire Switches
Remember for four-wire connections: 4 wires for digital transmissionsNeed two paths through the switch
Useful approach: Find one path as for two-wire connectionUse the mirror image of the path for second path
Folded operation
3Input 6of Matrix 3
4
(11,7)
(15,4)
15Input 11of Matrix 15
7
(6,4)
(3,7)......
...3
Output 6of Matrix 3
4
(11,7)
(15,4)
15Output 11of Matrix 15
7
(6,4)
(3,7)
......
...
Switching, Chapter 5 51
Discussion: Folded Four-Wire Switches
Advantages:Only one pathfinding operation
Reverse path automatically available (pairs of
crosspoints)
Half the information for control status needed
Simpler control
Half the blocking probability compared to finding two path independently
For switches with an even number of stagesFor odd numbers: center stage must have an even number of arrays so that it can be folded in the middle
Switching, Chapter 5 52
Pathfinding in Multiple-Stage Switches
In multiple-stage switches: More than one path from a certain input to a certain output
Need control logic to find a path
Two important components:Store state information about established paths
To be updated on each request / release of a pathAmount depends on the number of established paths
Need some time to search the state information to find a free path (pathfinding process)
Necessary time depends on the amount of state information!Determine the time necessary to process a request!Can take long in large switches!
Switching, Chapter 5 53
Pathfinding Time
Assumption: All paths through a switch busy independently with probability p (free with prob. q = 1 – p)
Define:pi: probability that exactly i paths are tested before an idle one is found
Is equal to: probability that first i-1 paths are busy and
path i is free: pi = p(i-1)q
Np: Expected number of path tested before an idle path is found
Sums up to:
Last term: probability that all paths k are busy
Closed form:ppNk
p −−
=1
1
kkp pkqpkqppqqN )()(...)3()2()1( 12 +++++= −
Switching, Chapter 5 54
Example: Pathfinding Time
Expected number of paths to be tested to find an idle path in the following switch:
3-stage, 8192 lines, n = 64, k = 15, A=0,1 erlang
Target GoS: 0,002
Solution: 3 out of 15 potential paths to be tested on average
Switching, Chapter 5 55
Example Exam: June 1999 I
The university has in one of its campuses a small switching central based on a square-matrix architecture. It gives service to 100 subscriber lines and has 12 outgoing lines. As a new building has been built, this central is too small now and must be replaced by a larger one, which shall be a Clos-based switch of three stages supporting 300 subscriber lines.For dimensioning the new switch, a traffic measurement has led to the following results:
Each subscriber line is busy on average 40 minutes per day.The hour with the most traffic is between 11:00 and 12:00 in the morning, where 20% of the total traffic occurs.40% of the calls are internal (within the University) and the remaining 60% are external (go to outside).
Switching, Chapter 5 56
Example Exam: June 1999 II
Suppose that the traffic from the 300 lines has the same characteristic as measured for the 100 subscriber lines:
1. Calculate the number of output lines necessary for the new central to provide the same GoS.
Solution: 30 output lines
Switching, Chapter 5 57
Example Exam: June 1999 III
2. Show the internal switching architecture of the new central and compare the necessary number of crosspoints and the internal blocking probability with the old central.
Solution: Square (new): Nx = 108.900, Clos: Nx = 32.760, Square (old) Nx= 12.544Blocking: Zero for the old and the new central
Switching, Chapter 5 58
Example Exam: September 2000 I
A company has a Clos switch of 3 stages with 10 output lines, giving service to 500 subscribers.The measured traffic is:
In the busy hour, the output lines are not available for 6 minutes on average.80% of the calls are internal.The service time of internal calls is twice as much as for external calls.
Calculate for the busy hour:1. The traffic intensity of the outgoing traffic.
Solution: Aext = 7,51 erlangs2. The total traffic intensity
Solution: Atotal = 67,59 erlangs
Switching, Chapter 5 59
Example Exam: September 2000 II
3. The minutes each subscriber line is busy on average.
Solution: 8.11 minutes
4. The probability that the communicating partner is busy in case of an internal call.
Solution: 13,5 %
5. The optimal design of the internal switch architecture.
Solution: N = 510, n = 15, k = 29, N/n = 34, Nx = 63.104
Switching, Chapter 5 60
Bibliography
J. Bellamy, “Digital Telephony”, 3rd edition, John Wiley & Sons, ISBN 0-471-34571-7.