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- 1 -

Case Study 1 : Route Optimisation

For given values of

A = traffic offered to the direct route

CD = cost per circuit on the direct route

CT = cost per erlang on the overflow path

B = required grade of service for the traffic case

find the optimal number of circuits, N, to be installed on the direct route, as well as the resulting

BR = congestion on the direct route

m = overflow traffic mean

v = overflow traffic variance.

Note : Finding the optimal number of circuits, N, on the direct route corresponds to find the N-value for whichthe cost function,

C(N) = N * CD + m * CT

has a minimum.

The value of m for given A and N can be found from an Erlang-table, or from the attached diagram.

A CD CT B Ð N BR m v Cost

A CD CT B N BR m v Cost

20. 1. 2. 0.10 Ð

20. 1. 2. 0.05 Ð

20. 1. 2. 0.01 Ð

20. 1. 1.2 0.01 Ð

- 2 -

Case Study 2 : Optimisation and Dimensioning of Junction Network

Consider a network with 6 exchanges. The tables below give the traffic interest between all exchanges, and thecost per circuit for each pair of exchanges. Find the number of circuits between all pairs of exchanges, calculate the totalnetwork cost for the 3 cases described below, and decide on the most economic way of realizing the network.

Case 1 : all traffic between any exchange pair (i,j) is carried on a direct, low-loss route.

Case 2 : all traffic between any exchange pair (i,j) is routed through a transit exchange, T.

Case 3 : optimize the number of circuits on each route i to j. The optimal number of circuits, N(i,j), depends on

• the cost of a circuit on the direct route; • the traffic offered: • the cost per erlang on the overflow routes, which in turn depends on the cost per circuit and the efficiency

of the route; as this efficiency, F, can be approximately expressed as the ratio between circuits and offeredtraffic, it is in turn dependent on the traffic offered, which is not known before the lower-level routes havebeen optimised; therefore an iterative procedure has to be established:

Step 1 : use Rapp’s approximation, and the attached diagram, to find the optimal number of circuits on all high-usage routes;

Step 2 : find the corresponding parameters of the overflow traffic, m(i,j) and v(i,j);

Step 3 : find the traffics offered to/from the tandem exchange by summing up the relevant individual overflowtraffics;

Step 4 : find the required circuits on the tandem routes, and calculate the total network cost.

Note : For cases 2 and 3 the following should be observed :

• you will have to decide which of the exchanges should be the tandem exchange, T;

• the grade of service used for routes to/from T should be adjusted to give the same overall grade of serviceas Case 1;

• calculate overflow traffics, and routes to/from T, using the attached diagrams, and Wilkinson’s method.

Traffic interest matrix ( in Erlang) : Cost-per-circuit matrix

1 2 3 4 5 6 1 2 3 4 5 61 - 10 15 5 2 20 1 - 100 150 120 130 2002 9 - 25 6 8 10 2 110 - 100 100 110 1503 20 23 - 18 20 30 3 130 110 - 120 110 1304 6 6 20 - 10 12 4 100 90 110 - 100 1305 3 7 22 10 - 11 5 120 100 110 120 - 1506 15 12 40 10 13 - 6 180 150 130 130 150 -

Required grade of service : 0.01

You should use the following page to tabulate your results.

- 3 -

Case 1 : All traffic on low-loss routes

Circuit matrix :

1 2 3 4 5 6

1

23 Total cost =4

56

Case 2 : All traffic through a transit exchange, T

Selected tandem :

Traffics and circuits :

To Tandem T From Tandem T

Traffic Circuits Traffic Circuits

1 1

2 2

3 3

4 4

5 5

6 6

Total cost =

- 4 -

Case 3 : Optimization of Alternative Routing Network

Selected tandem :

Calculate cost relations : Calculate improvement factors :

ε = & &' 7

/ ) $1

( ) ( . . )= × + ×ε ε0 7 0 3 2

1 2 3 4 5 6 1 2 3 4 5 6

1 1

2 2

3 3

4 4

5 5

6 6

Optimize circuits on High-usage routes :

1 2 3 4 5 6

1

2

3

4

5

6

Overflow Traffic Mean : Overflow Traffic Variance :

1 2 3 4 5 6 1 2 3 4 5 6

1 1

2 2

3 3

4 4

5 5

6 6

Use attached diagrams to determinecircuits and overflow traffics!

- 5 -

Calculate traffics ( mean and variance ) offered to tandem routes, and find required number of circuits :

Traffics and circuits :

To Tandem From Tandem

Mean Variance V/M Circuits Mean Variance V/M Circuits

1 1

2 2

3 3

4 4

5 5

6 6

Total cost =

- 6 -

1 � �1 X P E H U � �R I � O LQ H V � LQ �G LU H F W �U R X W H

$ � �7 U D I I LF �R I I H U H G �W R �G LU H F W �U R X W H

) �1 �$ � � � � � � ,P S U R Y H P H Q W �I D F W R U � �& ' �& 7

5 1 0 1 5 2 0 e r l A

0 .1

0 .2

0 .3

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0 .5

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1 .0

F ( N ,A )

N = 0 1 2 3 4 5 6 7 8 9 1 0 1 1 1 2 1 3 1 4 1 5 1 6 1 7 1 8 1 9 2 0 2 1

2 2

2 3

2 4

2 5

2 6

2 7

2 8

- 7 -

1 � �1 X P E H U � �R I � O LQ H V � L Q �G LU H F W �U R X W H

$ � �7 U D I I L F �R I I H U H G � W R �G L U H F W � U R X W H

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F ( N ,A )

1 2

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1 6

1 8

2 0

2 2 2 4 2 6 2 8 3 0 3 2 3 4 3 6 3 8 4 0 4 2 4 4 4 6 4 8 5 0 5 2 5 4 5 6 5 8 6 0 6 26 4

6 6

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7 0

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V/M = 1 V/M = 1.2

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V/M = 3

V/M = 4

0

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N

V/M = 1 V/M = 1.2 V/M = 1.5

V/M = 2

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V/M = 4

V/M = 5

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