Traffic 2 by Mushtaq 12042010

Traffic Engineering

• Telecom system to service the voice traffic and data traffic . The traffic is defined as the occupancy of server .

• The basic purpose of the traffic engineering is to determine the condition under which adequate service is provided to the subscribers. While using economical use of resources providing the service.

• The functions performed by the telecom network depend on applications it handles. Some major functions are ,switching , routing , flow control, security, failure monitoring , traffic monitoring , internetworking and network management.

• To perform the above functions a telephone network composed of variety of communication equipment such as digit receiver, call processing , inter-stage switching links

Traffic Engineering

and inter office links etc.

• The telecom engineering provides basis for analysis and design of telecom networks . It provides means to determine the quantum of common equipment required.

• To provide particular level of service for given traffic pattern and volume. The developed model is capable of providing best accessibility and greater utilization of their lines and links.

• The traffic engineering also determine the ability of telecom network to carry given traffic at particular loss probability. Traffic theory and queuing theory is used to estimate the probability of occurrence of call blocking.

• Earlier traffic analysis based purely on analytical approach that involves advanced mathematical concepts .

Traffic Engineeringpresent day approach combine the advent of powerful and affordable software tools that aim to implement traffic engineering concepts and automatic engineering tasks.

Traffic Design Requirements• In the study of traffic engineering , to model a system

and to analyze the change in traffic after designing , the statistics of an exchange should be studied .

• The incoming traffic undergoes variation in many ways , due to peak hours , business hours ,seasons ,weekends , festival and location of exchange etc.

• The traffic is unpredictable and random in nature. The traffic characteristics of an exchange should be analyzed for system design .

• The grade of service and blocking probability are also important parameters for traffic study.

Traffic Design Requirements

1 Traffic Statistics• Following are the traffic statistics which are important for

analysis and design of network.

A Calling rate• This is average number of requests for connections that

are made per unit time . If the instant in time that a call request arises is random variable.

• The calling rate is stated that as the probability that a call request will occur in certain short interval of time . If n is the average number of calls to and from a terminal during T seconds. The calling rate is defined as

λ = n/T (1)• In telecom system voice traffic and data traffic are two

types of traffics . The calling rate (λ) is also referred as average arrival rate . The average calling rate is measure in calls per hour.


B Holding Time• The average holding time or service time h is average

duration of occupancy of traffic path by a call. For voice traffic it is average holding time per call in hours.

• The reciprocal of average holding time referred to as service rate (µ) is call per hour .

µ = 1/h (2)

• The probability of call lasting at least t sec is given by

P (t) = exp (- t/h ) (3)

• For mean holding time h = 100 sec the negative exponential distribution function is give by figure 8.1 . The fig shows that 50 % probability calls last longer than 70 sec.


70

50 %


C The distribution of Destination• Number of calls receiving at an exchange may be

destined to its own exchange or remote exchanges or foreign exchange . The distribution is defined as probability calls request for particular destination

• The hierarchical structure of telecom network includes many intermediate exchanges , the knowledge of this parameter helps in determining the number of trunks needed between individual exchange.

D User Behavior • The user may abandon a request if his 1st attempt to

make a call is failed. The user may make repeated attempts to make a call . The user may wait for some times to make next attempt to setup a call.

• The behavior varies person to person and also depend on situation.


E Average Occupancy• If the average number of calls to and from a terminal

during time t sec is n and average holding time is h . The average occupancy of terminal is given by.

A = nh / T = λh = λ/µ (4)

• Average occupancy is ratio of average arrival rate to the average service rate. It is measured in Erlangs . Average occupancy is referred as traffic flow or traffic intensity or traffic carried.

Traffic Design Requirements2 Traffic Pattern• The telephone traffic and its distribution with respect to time

(traffic load) , which is normally 24 hours . It helps in determining the amount of lines required to fulfill the subscribers needs.

• According to the need of subscribers the telephone traffic varies greatly. The variations are not uniform and varies season to season, month to month , day to day and hour to hour.

• The degree of hourly variation is greater than that of any other period. Fig 8.2 shows the variation of calls from 0800 hrs to mid night (00:00 hrs).

• If the behavior of the traffic shown in fig is symmetric for period of time or season ,good judgment about the design of switching system or lines or trunks or any shared equipment can be made.


Fig. 8.2: Variations of calls from 8.0 A.M. to mid night

Nu

mb

er o

f ou

tgoi

ng

calls

Traffic Design Requirements• The combination of historical records , experience ,

location of exchange (business area or residential area ) , vacations , holidays etc. decide the design of the of telecom network . Various parameters related to traffic pattern are below:

a. Busy Hour• The telecom facilities are designed on the intensity of

traffic during the busy hour . The busy hour vary from exchange to exchange , month to month and day to day etc.

• The busy hour is defined as the 60 minutes interval in a day in which traffic is highest.

b. Call Completion Rate • Based on the status of the subscribers or the design of the

switching system , the call attempt may be successful or not.

Traffic Design Requirements• The call completion rate is defined as the ratio of number

of successful calls to the number of call attempts. A CCR value of 75 % is considered as excellent and 70 % is usually expected.

c. Busy Hour Call Attempts (BHCA)• It is an important parameter in deciding the processing

capacity of an exchange . It is defined as number of calls attempt in busy hour.

d. Busy Hour Calling Rate (BHCR)• It is useful parameter in designing a local exchange to

handle the peak hour traffic . It is defined as the average number of calls originated by a subscriber during busy hour.


3 Unit of Telephone Traffic• The traffic is measured in Erlangs , its name after Danish

Mathematician Agner Erlangs . Erlang is international unit of traffic .

• A server is said to have traffic of 1 Erlang if it is occupied for entire period of observation. Simply 1 Erlang represents 1 circuit occupied for 1 complete hour . The maximum capacity of a single server (or channel) is 1 Erlang (server is always busy). Thus the maximum capacity in Erlangs of a group of servers is merely equal to the number of servers. The traffic intensity is ratio of period for which the server is occupied to the total period of observation is measured in Erlangs.

The Unit of Traffic• The traffic intensity is defined as the average number of

calls in progress per unit time.• The unit of traffic is Erlang (E) name after Agner Erlang

the pioneer of traffic theory.• On a group of trunks , the average number of calls in

progress depend on both the number of calls which arrive and their duration.

• The duration of call is always called holding time , because it holds a trunk for that period of time.

• The Erlang of traffic carried by the group of trunks is given by :A = n h/ T (1)

Where A is traffic in Erlangs, n is average number of calls arrival during time T and h is average call holding time.

The Unit of Traffic 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- Outgoing traffic

2- Incoming traffic

3- total traffic .

Solution :

1 Outgoing traffic = 120 * 2/60 = 4 E

2 Incoming traffic = 200 * 3/60 = 10 E

3 Total traffic = 4 E + 10 E = 14 E.

The Unit of Traffic

Example

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

Solution :

Occupancy of line = (3 + 3 ) * 2/60 = 0.1 E

0.1 E is the probability of finding the line engage.

Example :

If a group of 20 trunks carries 10 Erlangs and the average call duration is 3 minutes . Calculate (a) Average number of calls in progress (b) Total number of calls originated per hour.Solution : No. of trunks = 20 Traffic intensity (A) = 10 EHolding time (h) = 03 minutes Observation time (T) = 60 minutes(a) Traffic intensity per trunk = 10 E/20 = 0.5 E/TrunkAverage No. of calls per trunk for 1 E for 60 minutes = 20For 0.5 E average number of calls = 20 x 0.5 = 10 (b) Traffic intensity :A = n h/T =10 ETotal number of calls originated per hour (n = A x T/h) n = 10 x 60/3 = 200 calls

Grade of Service (GOS)• For non blocking service of an exchange , it is necessary to

provide as many lines as there are subscribers. But it is not economical . So , some calls have to be rejected and have to be retried when lines have been used by other subscribers.

• The grade of service refer to proportion of unsuccessful calls . GOS is defined as ratio of lost traffic to offered traffic.

GOS = Blocked busy hour call / Offered busy hour calls

GOS = A-A0/A

A0 = Carried traffic

A = Offered traffic

A-A0 = Lost traffic

Grade of Service (GOS)• The smaller the value of GOS ,the better is the service.

• Recommended GOS is 0.002 i.e. 2 calls per 1000 offered may lost . In system with equal number of servers and subscribers , GOS is equal to zero.

Example:

During a busy hour ,1400 calls were offered to a group of trunks and 14 calls were lost . The average call duration is 3 minutes . Find (a) Traffic offered (b) Traffic carried ( c ) GOS (d) Total duration of congestion period .

Solution :

n = 1400 calls

h = 3 minutes

T = 60 minutes

( a) Traffic offered = A = n h/T = 1400 x 3 /60 = 70 E

Grade of Service (GOS)

(b) Traffic carried A0 = 1386 x 3/60 = 69.3 E

(c ) GOS = A – A0/ A

A – A0 = 70 – 69.3 = 0.7 E (lost traffic )

GOS = 0.7 /70 = 0.01

(d ) Total duration of congestion = 0.01 x 3600 = 36 second.

Congestion • It is uneconomical to provide sufficient equipment to carry

all traffic that could possibly be offered to telecom system.

• In a telephone exchange it is theoretically possible for every subscriber to make a call simultaneously . The cost of meeting this demand is not possible ,but probability of it happening is negligible .

• The situation arises that all the trunks in a group are busy, so it can accept no further calls . This is known as congestion.

• In message switching system , calls that arrives during congestion wait in a queue until an outgoing trunks becomes free. They are delayed but not lost. Such systems are called queuing system or delay system.

Congestion

• In circuit switching system such as telephone exchange , all attempts to make calls over a congested trunks group 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 system.

Traffic carried = Traffic offered – traffic lost.

• For lost call system the grade of service B is defined as :

B = No. of calls lost ÷ No. of calls offered

= Portion of time for which congestion exist

= Probability of congestion

= Probability that a call will lost due to congestion.

Congestion

• If traffic Erlangs is offered to a group of trunks having grade of service is B the traffic lost is AB and traffic carried is A (1-B ) Erlangs .

• Larger is the grade of service , the worse is the service given , the worse service is normally specified for traffic at busy hour, for other time it is much better.

• If B is too large , user make many unsuccessful calls . If it is too small unnecessary expenditure is incurred on equipment which is rarely used.

CongestionExample

During the busy hour , 1200 calls were offered to a group of trunks and 6 calls were lost . The average call duration is 3 minutes. Find :

1- The traffic offered , 2- The traffic carried 3- The traffic lost , 4- The grade of service 5- The total duration of period of congestion.

Solution

1- A = n * h/ T = 1200 * 3/60 = 60 E

2- 1194 * 3/60 = 59.7 E

3- 6 * 3/60 = 0.3 E

4- B = 6/1200 = 0.005

5- 0.005 * 3600 = 18 Seconds

Lost Call System• Erlang determine the grade of service ( i.e. loss

probability) of a lost call system having N trunks when offered traffic A are shown in fig below. Solution depend on following assumptions.

Traffic offered A Erlang

N Outgoing trunks

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

Fig 4.5 lost call system

Lost Call System• The assumption of pure chance traffic implies that call

arrival and termination are independent random event. If call arrivals are independent random events, their occurrence is not affected by previous calls. This traffic is therefore sometimes called memory less traffic .

• Statistical equilibrium implies that the probabilities do not change.

• Full availability means that every call that arrives can be connected to any outgoing trunk which is free.

• Each switch have sufficient outlets to provide access to every outgoing trunk.

• The lost call assumption implies that any attempted call which encounter congestion is immediately cleared from system. When this happen user is likely to make another attempt shortly afterwards.

• The traffic offered during busy hours is slightly greater than that if it would have been there were no congestion.

Lost Call System• If there are x calls in progress , then

P(x) = A x x! P (0) (1)

There will not be a negative number of calls and there can not be more than N calls so 0 ≤ x ≤ N so

∑N

X = 0

P (x) = 1 =

∑N

X = 0

x!Ax

Hence P (0) =

1

∑N

X = 0

x! P (0)Ax

(2)

Lost Call System• The probability of lost call which is grade of service B .

This is given the symbol E 1.N (A) which denote the loss probability for full availability group of N trunks offered A erlangs.

B = AN/N!

∑ AK/K! K=0

N

(3)

The grade of service of loss system with N full availability trunks , offered A erlang of traffic is given by E 1.N (A) . From equation 3 = AN-1/( N-1 )!

∑ AK/K!

K=0

N-1E 1.N -1

Lost Call System

∑N

K=0

AK/ K! = A N-1/ (N-1)!

E 1.N-1 (A)

+ AN / N!

Substituting in equation (3 )

E 1.N (A) = A E 1.N -1 (A)

N + AE1.N-1 (A)

Lost Call SystemExample

A group of 5 trunks is offered 2 erlang of traffic , find. 1- The grade of service

2- The probability that only one trunk busy

3- The probability that only one trunk free

4- The probability that at least one trunk free

Solution

1- From equation (3) B = E 1.N (A) = 25 / 5!

1 + 2/1 + 4/2 + 8/6+ 16/24 + 32/120

= 0.2667/ 7.2667 = 0.037

2 - From equation (2)P (x) = Ax/ x!

∑ AK/ K!N

K=0

Lost Call System P (1) = 21/1!

7.2667 = 2 / 7.2667

3- p (4) = 24/4!

7.2667= 16/24

7.2667= 0.0917

4- P (x < 5) = 1 – P(5) = 1-B = 1 – 0.037 = 0.963

= 0.275

Queuing Systems• Erlang determined the probability of encountering delay when

traffic A is offered to a queuing system with N trunks as shown in fig below.

• Erlang’s solution depend upon following assumptions:

1. Pure chance traffic

2. Statistical equilibrium

3. Full availability

4. Calls which encounters congestion enter a queue and are stored there till a trunk become free.

Traffic offered A Erlangs

QueueN

Trunks

Fig . Queuing System

Queuing Systems• The assumption of pure chance traffic implies that call

arrival and termination are independent random event. If call arrivals are independent random events, their occurrence is not affected by previous calls. This traffic is therefore sometimes called memory less traffic .

• Statistical equilibrium implies that the probabilities do not change.

• Full availability means that every call that arrives can be connected to any outgoing trunk which is free.

• If A ≥ N calls are entering the system at a greater rate than they leave. As a result ,the length of the queue must continually increase toward infinity.

Queuing Systems• Let x be total number of calls entering the system. When

x < N , then x calls are being served and there is no delay . When x > N , then all the trunks are busy and incoming calls encounter delay , there are N calls being served and x- N calls in queue.

• If x ≤ N :

There is no queue and there is behavior of system is same as that of a lost call system in the absence of congestion.

Ax

x! P ( 0 ) (4.9)For 0 < x < N

• If x ≥ N :The probability of call arrival in a very short period of time , δt , from equation on next slide :

P (x) =

Queuing Systems

P(a) = A δt /hwhere h is mean service time

• The probability of transition from x-1 to x calls in the system during δt is given by :P (x-1 → x ) = P (x-1) A δt /h Since all trunks are busy , only N calls are being served can terminate P (e) = N δt /h and the probability of a transition from x to x-1 is given by P ( x → x-1 ) = P(x) P(e) = P (x) δt /h

• For statistical equilibrium , P (x-1 → x) = P (x→ x-1). Therefore P (x) N δt /h = P (x) A δt /h

and

NP (x-1)

(4.10)P(x) = A

Queuing Systems

P (N) = AN

N!P (0) From equation 4.7

P (N+1) = A

NP (N) =

A N+1

N. N!P (0)

P (N+2) =A

NP (N+1) =

A N+2

N2. N!P (0)

For x ≥ N :

P (x) = Ax

Nx-N. N!

P (0) = NN

N!

A

N

x

P(0) (4.11)

If there is no limit to the possible length to the queue , then x can have any value between zero and infinity.

∞∑ P (x) = 1X=0

Queuing Systems

• Thus from equation (4.7 ) and (4.11) :

1

P (0)=

N -1

∑ Ax

X=0 xỊ+

NN

NỊA

N

N ∞∑K=0

k(4.12)

Where k = x – N , since A/N ≤ 1 then :

∞∑K=0

A

N

k= -1

AN

-1

1

P (0)=

N -1

∑ Ax

X=0 xỊ+

AN

NỊ1

AN

-1

P(0) = NAN

NỊ (N –A )+ N -1

∑X = 0

AX

XỊ

-1(4.13)

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Traffic 2 by Mushtaq 12042010