03-CS Space Division Switching 1pps

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

=

⋅=


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


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


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


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


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


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


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


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


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


Example: Multiple Matrices II

Number of

matrices

Inputs per matrix

Traffic per matrix (erl)

Output per matrix

E(m,A)Crosspoints per matrix

Total number of crosspoints

Total number of

output links

1 200 10 20 0,0019 4000 4000 202 100 5 13 0,0013 1300 2600 264 50 2,5 9 0,00086 450 1800 368 25 1,25 6 0,0015 150 1200 48

0

500

1.000

1.500

2.000

2.500

3.000

3.500

4.000

4.500

1 2 4 8

Number of matrices

Cro

ssp

oin

ts

0

10

20

30

40

50

60

To

tal

lin

ks

Total # crosspoints

Total number output links


Sharing Links between Matrices

If free links are used in a sequential search:

Last links have low probability of being busy

Share these last links between matrices

Problem: Higher control complexity

10010

12

1112

12

10010

12

1112

12


Multiple-Stage Switching

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


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


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


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


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


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 =⋅=

=⇒⋅=


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.


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


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 <

+=+

+=


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


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


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


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


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(


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



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



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⋅→

∞→∞→


Exercise: Clos Switches

Design an optimal 3-stage switch without blocking for 1200 lines

Solution:

1200 24

50 50

N

47

5050 4747 24


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


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


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


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


Example Design for Three-Stage Switches (Inlet utilization: 0.1)

For local centrals / PBXConcentrators (n > k, β < 1)

Can save a lot of crosspoints!

Switch Size, N

n k βNumber of Crosspoints

Number of Crosspoints (non-

blocking)128 8 5 0,625 2.560 7.680 (k = 15)512 16 7 0,438 14.336 63.488 (k = 31)

2048 32 10 0,313 81.920 516.096 (k = 63)8192 64 15 0,234 491.520 4,2 million (k = 127)32768 128 24 0,188 3,1 million 33 million (k = 255)

131072 256 41 0,160 21,5 million 268 million (k = 511)


Example Design for Three-Stage Switches (Inlet utilization: 0.7)

For transit switchesSpace expansion (n < k, β > 1)

Not so big savings (50% at maximum)

Switch Size, N

n k βNumber of Crosspoints

Number of Crosspoints (non-

blocking)128 8 14 1,75 7.168 7.680 (k = 15)512 16 22 1,38 45.056 63.488 (k = 31)2048 32 37 1,36 303.104 516.096 (k = 63)8192 64 64 1 2,1 million 4,2 million (k = 127)

32768 128 116 0,91 15,2 million 33 million (k = 255)131072 256 215 0,84 113 million 268 million (k = 511)


Lee: Multiple-Stage Example I

Three-stage example to be generalizedExample: 5 stages (build from k1 three-stage switches):

...

12

k2

p

p2

p2

p2p2

p2

p2 = p(n1 / k1) (n2 / k2)

p

...

12

k2

p1 = p(n1 / k1)p2

p2

p2p2

p2

p2

...

12

k2

p2

p2

p2p2

p2

p2

......

p1

p1 = p(n1 / k1)

p1

p1

1

k1

2


Lee: Multiple-Stage Example II

Then: B = (1 – (q1)2[1 – (1 – q22)k2])k1

With:q1 = 1 – p1 and p1 = p(n1 / k1)

q2 = 1 – p2 and p2 = p(n1 / k1) (n2 / k2)

k1: number of outgoing links from each 1st-stage switch

k2: number of outgoing links from each 2nd-stage switch


Discussion: Lee Graphs

Problem of Lee scheme:Assumption that the blocking probabilities p’on the interstage links are independent

This is not true if the expansion factor β is large (i.e. β > 1)

Example: Non-blocking three-stage switch with k = 2n-1 (i.e. k 2nd-stage matrices)

According to Lee: B ≠ 0

But: Non-blocking switch (i.e. B = 0!)


Example: Worst Case

Why is Lee inaccurate? Assumes that if (2n – 2) links busy, the remaining one busy with probability (1 – q’)2

But: cannot be busy...

I.e.: the more paths are busy, the lower the blocking probability for the remaining paths

n

......

m

BusyFreeAvailablen-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×


The Jacobaeus Method (1950)

More complex than LeeJust presenting the Formula:

Works only for β > 1 (else: Lee’s method is more accurate)

Take care to use the right method depending on the value of β !!

knk ppknk

nB −−−

= 22

)2()!2(!

)!(


Example Problems: 1

1. Design a 3-stage switch without blocking with a minimum number of crosspoints for 512 lines.

Solution: 1st stage: b = 32 matrices

2nd stage: k = 31 matrices

Number of crosspoints: Nx = 63.488


Example Problems: 2

2. Based on problem 1 (n=16), calculate B for k = {31, 28, 24, 20, 16}

An increasing degree of concentration and, thus, blocking

Use a traffic intensity of 0,7 (transit central)

For both methods (Lee and Jacobaeus)

Solution: k β B (Lee)B

(Jacobaeus)Number of Crosspoints

31 1,94 8,5*10-8 0,1*10-12 63.488

28 1,75 3,7*10-6 7,7*10-9 57.344

24 1,50 3,2*10-4 2,7*10-5 49.15220 1,25 0,014 0,007 40.96016 1,00 0,221 0,221 32.768


Example Problems: 3

3. Based on problem 1 (n=16), calculate B for k = {16, 14, 12, 10, 8, 6}

Also, an increasing degree of concentration and, thus, blockingUse a traffic intensity of 0,1 (local central)For both methods (Lee and Jacobaeus)

Solution:k β B (Lee)

B (Jacobaeus)

Number of Crosspoints

16 1,00 2,9*10-12 2,9*10-12 32.768

14 0,88 4*10-10 7,8*10-10 28.672

12 0,75 5,7*10-8 1,4*10-7 24.576

10 0,63 4,9*10-6 1,5*10-5 20.480

8 0,50 2,8*10-4 8,6*10-4 16.3846 0,38 0,0097 0,027 12.288


Example Problem 4

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)


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


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


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”


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)

......

...


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


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!


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 +++++= −


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


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).


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


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


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


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


Bibliography

J. Bellamy, “Digital Telephony”, 3rd edition, John Wiley & Sons, ISBN 0-471-34571-7.

Chapter 5

Documents

03-CS Space Division Switching 1pps