Telecommunication TrafficII

Telecommunication Traffic

Trunk: In telecommunication engineering the term

trunk is used to describe any entity that will carry one call.

Erlang: Is the unit of traffic intensity, which is

defined as the average calls in progress.

Busy Hour

The average weekday reading over one or two weeks in the known busy season.

The average of the BH traffic on 30 busiest days of the year.

The average of the BH traffic on the 10 busiest days of the year

The average BH traffic on the 5 busiest days of the year.

Holding time: Is the duration of a call, the call holds the

trunk for that time.CCS: In North America , traffic is sometime

expressed in hundreds of call seconds per hour

(CCS). 1 Erlang =36 CCS

Traffic carried by a group of trunks is A=Ch/TC: average no. of call arrivals during time T.h: average call holding time. Since a single trunk can not carry more

than one call, A<=1 The traffic is a fraction of an erlang equal to

the average proportion of time for which the trunk is busy.

Example

On average, during the busy hour, a company makes 120 outgoing calls of average duration 2 minutes. It receives 200 incoming calls of average duration 3 minutes. Find

(1)The outgoing traffic. (2)The incoming traffic. (3)The total traffic

Solution

The outgoing traffic is 120x2/60=4E.The incoming traffic is 200x3/60=10EThe total traffic is 4+10=14E.

Example

During the busy hour, on average, a customer with a single telephone line makes 3 calls and receive 3 calls. The average call duration is 2 minutes. What is the probability that a caller will find the line engaged.

Solution

Occupancy of line =(3+3)x2/60=0.1E =probability of finding the line engaged

Congestion

*The situation that all trunks in a group of trunks are busy is known as congestion.

*In circuit switched systems, all attempts to make calls over a congested group of trunks are unsuccessful.

*Such systems are called lost-call systems. *In a lost-call system, the result of congestion

is that the traffic actually carried is less than the traffic offered to the system

* Traffic carried=traffic offered-traffic lost

Grade of Service

The proportion of calls that is lost or delayed due to congestion is a measure of the service provided. It is called the grade of service.

For a lost-call system, the grade of service, B, may be defined as

Number of calls lostB =Number of calls offered

Hence, also: B=Proportion of the time for which the

congestion exists.=probability of congestion.=probability that a call will be lost due to

congestionThus if traffic A erlang is offered to a group

of trunks having a grade of service B, the traffic lost is AB and the traffic carried is A(1-B)

The larger the GOS, the worse is the service given.

The GOS is normally specified for the traffic at the busy hour.

It varies from 1 in 1000 for cheap trunks inside an exchange to 1 in 100 for inter-exchange connections and 1 in 10 for expensive international routes

Dimensionality Problem

The basic problem of determining the size of a telecommunication systems, known as dimensionality problem is, : Given the offered traffic, A, and the specifies GOS, B, find the No. of trunks, N, that is required

Example During the busy hour, 1200 calls were

offered to a group of trunks and six calls were lost. The average call duration was 3 minutes. Find:

1. The traffic offered. 2. The traffic carried. 3. The traffic lost. 4. The GOS. 5. The total duration of the period of congestion.

Solution

1. A=Ch/T=1200x3/60=60 E. 2. 1194x3/60=59.7 E. 3. 6x3/60=0.3 E. 4. B=6/1200=0.005 5. 0.005x3600= 18 seconds.

Traffic Measurement

It is important for an operating company to know how much BH-traffic its systems are handling.

In particular, it needs to know when a system is becoming overloaded and additional equipment should be installed.

The traffic should be measured regularly and records kept. In modern SPC systems, the central processors generate records of the calls they set up.

Example

Observations were made of the No. of busy lines in a group of junctions at intervals of 5 minutes during the BH. The results obtained were:

11, 13, 10, 14, 12, 7, 9, 15, 17, 16, 12 It is therefore estimated that the traffic

carried in erlangs, was:

11+13+10+14+12+7+9+15+17+16+12=12 E 12

A mathematical Model

A simple model for the traffic offered to telecommunication systems is based on the assumptions:

. Pure-chance traffic. . Statistical equilibrium. Pure-chance traffic assumption means

random call arrivals and terminations. This leads to the following results:

1-The No of call arrivals in a given time has a Poisson distribution, i.e.:

( )!

x

P x ex

Where x is the No of call arrivals in time T and μ is the mean No of call arrivals in time T . For this reason pure chance traffic is called Poissonian traffic

2-The intervals T, between call arrivals have negative exponential distributions

( )t

TP T t e

( )t

hP T t e

where T is the mean interval between call arrivals.

3- call durations T are also have a negative exponential distribution, i.e.:

where h is the mean call duration (holding time)

Example

On average, one call arrives every 2 seconds. During a period of 10 seconds, what is the probability that:

1-No call arrivals? 2-One call arrives? 3-Two calls arrive? 4-More than two calls arrive?

Solution

0

2 2

2

22

( ) , where 2!

21. (0) 0.135

0!2

2. (1) 0.2701!

23. (2) 0.270

2!4. ( 2) 1 (0) (1) (2)

1 0.135 0.270 0.270 0.325

x

P x ex

P e e

P e

P e

P P P P

Example

In a telephone system the average call duration is 2 minutes. A call has already lasted 4 minutes. What is the probability that:

The call will last at least another 4 minutes?The call will end within the next 4 minutes?

Solution

2( ) 0.135

( ) 1 ( ) 1 0.135 0.865

t

hP T t e e

P T t P T t

Simple Markov Chain

For a group of N trunks, the No of calls in progress varies randomly in a process similar to a birth death process or a simple Markov Chain.

P(0) P(1) P(j) P(k) P(N)

State diagram of N trunks

P01 PN-1,N

Nj k

The No of calls in progress is alwaysbetween 0 and N.It thus has N+1 states.P(j) is the prob. of state j;P(k) is the prob.of the next higher state k;

Pj,k is the prob.of state increase to k,given that the present state is j.P(0), P(1), …P(N) are state probabilities.Pj,k; Pk,j are transitional probabilities.

For a very small interval of time δt, the prob.of something happen is small.The prob. of two or more events during δt is

negligible.The events that can happen during δt are 1.One call arriving with prob. P(a). 2.One call ending with prob. P(e). 3.No change, with prob. 1-P(a)-P(e).

The mean No of arriving during δt is A(δt)/h<<1

, ( )j kP P a A t h

If the mean holding time is h and the No of calls in progress is k, one expects an average of k calls to end during a period h.

The average No of calls ending during δt is

k(δt)/h<<1

, ( )k jP P e k t h

If the prob. of j calls in progress at time t is P(j), then the prob. of transition form j to k busy trunks during δt is

( ) ( ) ( ) ( )P j k P j P a P j A t h

If the prob. of k calls at time t is P(k), then the prob. of transition form k to j busy trunks during δt is

( ) ( ) ( ) ( )P k j P k P e P k k t h

Statistical equilibrium requires that

2

3

( ) ( )

( ) ( )

( ) ( )

Hence, (1) (0)1

(2) (1) (0)2 2 1

(3) (2) (0)3 3 2 1

ingenral, ( ) (0)!

x

P j k P k j

kP k t h AP j t h

AP k P j

kA

P P

A AP P P

A AP P P

AP x P

x

Pure chance traffic implies very large No of sources, x can have any value between 0 and infinity and

0 0

1 ( ) (0) (0)!

(0)

and ( )!

xA

x x

A

xA

AP x P e P

x

P e

AP x e

x

Thus if call arrivals have a Poisson distribution, so does the no of calls in progress.

This requires an infinite no of trunks to carry the calls.

If the no of trunks available is finite, some calls can be lost or delayed and the distribution is no longer Poissonian.

Lost-call systems

Erlang determined the GOS of a lost-call system having N trunks, when offered traffic A, as shown in Fig.

His solution depends on the following assumption:

Pure chance traffic. Statistical equilibrium. Full availability. Calls which encounter congestion are lost.

N Outgoing trunks

Traffic Offered A erlangs

Lost-call system

If there are x calls in progress, then

( ) (0)!

xAP x P

x

However, there cannot be a negative No of calls and there cannot be more than N.

Thus, we know with certainty that 0 ≤ x ≤ N

0 0

0

0

( ) 1 (0)!

Hence

(0) 1!

by substitution

!( )

!

xN N

x x

xN

x

x

Nk

x

AP x P

x

AP

x

A xP x

A k

This is the first Erlang distribution.P(N) is the probability of congestion, i.e:The probability of a lost call, which is theGOS, B.For a full availability group of N trunksoffered A erlangs GOS is denoted as

1,

0

!( )

!

N

N Nk

k

A NB E A

A k

It may be computed directly or iteratively1 1

1, 10

11

0 1, 1

1, 11,

1, 1

1,0 1,

( 1)! !

( 1)

! ( ) !

by substitution;

( )( )

( )

Since 1, this iterative formula enables ( ) to be computed

for all values of

N kN

Nk

k N NN

k N

NN

N

N

A AE

N k

A A N A

k E A N

AE AE A

N AE A

E E A

N

Example

A group of five trunks is offered 2 E oftraffic. Find: 1.The GOS 2.The prob. that only one trunk is busy 3.The prob. that only one trunk is free 4.The prob. that at least one trunk is free

Solution

1,

32 1201. ( )

2 4 8 16 321

1 2 6 24 1200.2667

0.0377.2667

2. (1) 2 7.2667 0.275

16 243. (4) 0.917

7.26674. ( 5) 1 (5) 1 1 .037 .963

NB E A

P

P

P P B

Example

A group of 20 trunks provides a GOS of0.01 when offered 12 E of traffic. 1.How much is the GOS improved if one

extra trunk is added to the group? 2.How much does the GOS deteriorate if

one trunk out of service

Solution

1,201,21

1,20

1,191,20

1,19

1,19 1,19

1,19

12 (12) 12 0.011. (12) 0.0057

21 12 (12) 21 12 0.01

12 (12)2. (12) 0.01

20 12 (12)

0.2 0.12 (12) 12 (12);

(12) 0.017

EE

E

EE

E

E E

E