1.	What is Traffic Engineering ?

Traffic Engineering is a subject where you will be able to design your equipment to your requirement. It will help you not to over or under design. So with the help of traffic engineering you will be able to satisfy the users need while maintaining the correct cost for that need.

Let’s assume that you have to design a new road from A to B. Where in a new area, people has came to reside.

Main Road

New Road

A

B

Assume that there are 50 new houses with 50 vehicles around this new road where this new road will be connected to main road at B.

If we are to design a road from A to B. It has to consider the following two factors. (i) What is the width of the road (ii) How much weight this road can carry. When deciding the width are we going to assume that all these 50 vehicles are moving in this road at the same time. No, if it is so the width of the road is going to be massive and it is uneconomical.

When designing the new road, one has to assess (i) what is the busiest period that this road has to be used by the users(people) and

(ii) During that period how many vehicles will be using that road. and

(iii) how much load that this road

need to withstand to handle the number of vehicles and

(iv) what is the minimum width

that this road need to have to allow the smooth transportation for the vehicles.

What is busy hour?

Busy hour can be identified, as a hour where maximum number of vehicles will be using this road. In order to find out the busy hour one has to find out what are the busiest days this road is utilized. Busiest day means for example, in a given week 5 working days can be assumed as the busiest days, where as Saturday and Sunday can be assumed as holidays. In identifying busiest days one should not identify an exceptional day for a year where this road to be used heavily. After identifying the busiest common days, then out of those days, it has to find out what is the busiest common hour that this road is going to be maximumly used,

Busy hour will show you, a hour where a common characteristics of maximum number of vehicles can be observed. For example assume in a working day that there are 20 vehicles are utilizing between 8 and 9 a.m. You will observe close to 20 numbers of vehicles (not 20 similar people) are utilizing this road between 8 and 9 a.m. More the number of houses more accurate results, or more constant number of vehicles can be seen. Day's can be categories as working day's or holidays. Normally working days tend to be busier than holiday

Hence busy hour means in a common busy day, what is the busiest hour that this road will be utilized. When designing a road to the busy hour, one should not think that this road will allow smooth transport of vehicles in a busy hour in an abnormal busy day. Under this kind of abnormal busy day certainly you can observe traffic jam in this road, which is going to be inevitable, and try to design this road to cater such kind of situation will lead to that this road is going to be underutilized even for normal busy day and cost for building this road will be wasted. When you are building the road, you have to construct the road for the comfortable movement of the vehicles. Then you have to assess, how much is the maximum weight of the 20 vehicles that you have observed in the busy hour. Then you have to construct the road to withstand that weight.

Similarly you will further analyse about this 20 vehicles, how many vehicles need to be run parallel, in order not to have a traffic Jam. Then only you can get the optimum width of this road as well as optimum load of that need to be cater by this road.

Human behavior

Another area of study of traffic engineering is to analyse the human behavior, and its outcomes with respect to a particular equipment or service. In order to understand this let us analyse a situation you may assume the following.

A	B

From exchange A to B, there are 3 voice circuits. This means at a given time 3 people from A exchange can have simultaneous dialogue between 3 people who are connected to B exchange.

Under this situation assume the following observation, where at 10.00am all 3 circuits from A exchange to B exchange is occupied and the calls are lasted according to the following characteristics.

1 sec

1st Circuit

3 minutes

2nd Circuit

10 minutes

3rd Circuit

10.15

New assume another call is requested by a customer at exchange A, to be connected with a customer on exchange B, at 10 `o clock. What will happen to this caller. This call will be rejected and he will get a engage tone. What is the scenario, if this caller happens to dial 1 second later from 10`o clock, He will be successful. Under this scenario, how this caller knows, that the 1st circuit is going to release 1 second past 10 `o clock. To find out this type of situation people may obtain fortune tellers or Astrologers help? Is it necessary? No.In order to analyze this kind of situation. One has to find out the characteristics of calls arrival to exchange A, similarly the characteristics of calls holding at customers between A and B exchanges. In traffic Engineering we will be studying these kinds of human characteristics so that optimum number of equipments that need to be deployed to provide satisfactory service to the customers.

Customer behavior able patterns varies from country to country. It is similar to eating habits. Similarly is telephone conversations these patterns will vary from country to country. Lets try to understand a call holding characteristics. When A customer dial B customers telephone number. B customers is in a another exchange. You can observe the following basic phases.

Lifting the telephone and observing the dial tone.

Dialing the B number.

A exchange will send B customers information to B exchange. (This is called inter exchange signaling) A customer will hear nothing or some tone indication to inform him the call is in progress.

Assuming B customer is free. A customer will improved that B customers telephone is ringing, while B exchange will make B customer telephone to ring.

B customers telephone answered.

A & B are talking.

After talking A and B, both will keep the receiver on hook to make the circuit of A & B to be released.

So in a telephone conversation without your knowledge those 7 phases will be carried out by customers.

Now lets analyze the phase 6 the talking in detail. Since this is the basic function that is going to be achieved by the customers in telephone connections. Customers may visualize the following 3 major functions in a telephone conversation.

Telephone network will connect A and B telephones.

A and B persons are talking using A and B telephones.

After the conversation A & B telephones.

Will be released by the telephone network. If we further analyze the above phase with respect to a fixed telephone. (fixe telephone will be used by 4 family or a group of people is a office). You can subdivide into many phases.

Identifying each other since the normal telephones doesn’t convey

pictures first, it is essential to identify the caller & called before start of communication) (Please note that the technologies such as ISDN will allow you to have pictures to be transmitted) since this telephone is used by many people. (The latest technologies will provide to have differential running in a given telephone to differential the user)

Then commence talking even in talking phase there can be many protocols

greetings (like how are you) or certain countries (how are you how’s are family etc)

Then only actual subject will be discussed

While the actual discussion of the subject to understand properly, questions or repeating the same will be enforced. This is error correction.

A + B will end the conversations by special greetings.(Good Bye?)

It will be interesting to note that sometimes the will be interesting to note that some times the conversation may not end with Good Bye! But with other expressions, if the conversation is not upto the satisfaction of both A & B.

So you will see talking is the call holding characteristics. May very from country to country. So in traffic engineering special attention is made for human behavioral patterns, so that a proper service to be given at optimum equipment, hence for the customer service, the cost will be minimum.

(i)	What is the probability of using this equipment in the busy hour.

(ii)	What is the maximum traffic that has to be carried during the busy hour.

2.	Basic Traffic Measurement.

In telecommunication the measurement of traffic is called Erlang. This is in respect of the Denish Mathematician who introduce traffic theory Anger Krurp Erlang. Erlang unit is a measure of occupancy in a given time. For example if we observe a circuit between 0 and T hours, we see X1, X2, X3 periods were engaged during 0-T.

X1 X2 X3

0	T

X1 + X2 + X3

The traffic carried in this circuit is	T

3

 Xi

i=1

E (1) =

T

The maximum traffic that can carry is [since one circuit during T experiment time, the maximum occupancy will be T]

E(1)M = T / T = 1

Max

 Xi = T

i=1

Consider three circuits, and assume that we observed occupancies as follows.

X1 X2

X3

X4 X5

0 T

The traffic carried in this (3) circuits.

X1+X2+X3+X4+X5

E(3) =

T

5 Xi

= (

i=1 T

The maximum traffic these (3) circuits, will carry.

E(3)M = 3T / T = 3

Max

 Xi = 3T

i=1

Hence given a resource, the maximum traffic that can carry will be 1 Erlang. If

there N resource the maximum traffic that can carry will be N erlang.

3.	Measurement of Traffic (Scanning)

Now we know the basic measurement of traffic intensity. In practice how to find the traffic intensities? If there are less equipment or less circuit one can employ people with stop watches to calculate the traffic. But when the equipment and circuits are more the method, is more cumbersome. Hence a practical method of calculating traffic loads has to be achieved. This is done with the use of scanning. Consider the following case.

1st 2nd 3rd 4th 5th (scans)

4 mts 2 mts

8 mts

6 mts 1 mts

0 3 6 9 12 15 minutes

(2) (3) (1) (2) (1)( Total Number of busy scans.

CASE 1

Here you will see that there are three circuits, experiment time is 15 minutes. In order to understand the concept, the actual occupancy times are given. But assume this time we are not watching these three lines continuously.

If the number of equipment (Trunks) are small, one can employ people with stop watches to measure traffic, according to the theory, the traffic carried by these three circuits are

4 + 2 + 8 + 6 + 1

= 1.4 Erlangs

15

If the number of equipment (trunks) are small, one can deploy people with stop watches to measure traffic, according to the theory that we learned. But when the number of trunks are more, deploying personnel with stop watches may not practically feasible.

Case 1

Instead of observing 100% a trunks that we will watch every 3 mts. and observe the total number of busy circuits. The number of busy circuits are as follows.

Time	3	6	9	12	15
No of busy

circuits	1

2	2

3	3

1	1

2	1

This can be assumed as follows. The principle behind assumption is that if at the 1st scan, (3 minutes) if a circuit is busy means, it will assume that this circuit is busy from 0 to 3 minutes. If at the 5th scan or 15th minute if there is a busy scan mean’s it will assume from 12th to 15th minutes this circuit is engaged. Of course in this method we will be introducing errors. Obviously if the scanning period is less the error introduced will be less.

6 mts 3 mts

9 mts

6 mts 3 mts

0 3 6 9 12 15 m

(2) (3) (1) (2) (5)

CASE 2

If we assume the circuits occupancies as given above, traffic

obtained from the scanning method.

= 6 + 3 + 9 + 6 + 3 27

=

15 15

= 1.8 Erlang.

The answer obtained from scanning method will give rise to 0.4 Erlang difference to the actual value. If scanning period is small the traffic obtained from scanning will approach to the actual value.

The calculation of traffic by way of scanning will give rise to the following

equation.

Traffic Carried =	Total No.of busy Scans(=Y )

Total No.of Scans (=Z )

Hence Traffic carried = Y X Where X is the scanning period

(Z)X

Traffic by scanning, can be summarized as follows.

Traffic =	Total Number of busy scans

Total No. of scans

Case 2

In this exercise, analysis will be made to find out the errors that will be introduced in calculating traffic intensities by scanning method. Also it will be analyzed the important characteristics of traffic unit, i.e it is a intensity.

Consider two 15 minutes periods.

Actual

4 mts 6 mts 8

7

8 4 3 8

8

5 1/2 + 7 1/2 2

10.03 10.06 10.09 10.12 10.18 10.21 10.24 10.27

10.00 10.15 10.30 - Time

2 2 3 2 2 3 2 2 2 1 - Total Number of busy SCANS

a.	What is the actual traffic carried from 10.00 to 10.15.

T 4+6+8+4+5+0.5 27	27.5

A = = = 1.85 Er. = = 1.83 Er

0-15 15 15 15

at is the carried traffic according to scanning

T 2+2+3+2+2 11

SA = = = 2.2 Er.

0-15 5 5

c.	What is the actual traffic carried from 10.15 to 10.30.

T 2+2+3+2+2 28.5

A = = = 1.9 Er.

15-30 15 15

d.	What is the carried traffic according to scanning

T 3+2+2+2+1 10

SA = = = 2 Er.

15-30 5 5

Now you can compare how much the results are deviating from the actual	to the scanning (experiment) values. Now lets analyze traffic for all 30	minutes.

e.	What is the actual traffic carried for these 3 circuits from 10.00 to 10.30.

Is it the addition of TSA + TSA? (Scanning)

0-15 15-30

If it is so the answer is = 2.2 + 2 = 4.2 Er

What is the maximum traffic that these 3 circuits can carry? 3E.

Hence 4.2E is not the correct answer. If we calculate the carried traffic	from scanning from 10.00 to 10.30.

2+2+3+2+2+3+2+2+2+1 21

= = = 2.1 Er.

10 10

Now to get at the correct answer from

T T 2.2 + 2

A 0-15 and A 15-30 = = 2.1 Er.

2

Hence traffic is an intensity. When you are analyzing a circuit group to get the total traffic, you cannot add to get the total traffic, If the experiencing periods are different.

TRAFFIC is an INTENSITY. TRAFFIC can be added if the

Experimenting periods are same otherwise it cannot be added.

EXAMPLE (a) :

Traffic profile of a day in a circuit group is measured such that the traffic intensity is obtained for every fifteen from 00:00 hours to 24:00 hours for the day. The values are tabulated such that the first fifteen minutes of the day (that is 00:00 to 00:15) is taken as the 0th time slot. Let the traffic intensity measured for the nth slot of fifteen minutes follows the model a0 + a1n + a2n2 + a3n3 + a4n4. Prove that the hourly variation of the traffic intensity taken at every fifteen minutes of time is b0 + b1n + b2n2 + b3n3 + b4n4 where the following relationships hold:

b0 = a0 + 1.5a1 + 3.5a2 + 9a3 + 24.5a4

b1 = a1 + 3a2 + 10.5a3 + 36a4

b2 = a2 + 4.5a3 + 21a4

b3 = a3 + 6a4

b4 = a4

ANSWER (a)

1 HR 24th Hr

15 minutes

Slots n Slot

Traffic Intensity is given by

Tn = a0 + a1 n + a2 n2 + a3 n3 + a4n4

This traffic intensity has been measured for every 15 minutes.

If we want to calculate the hourly intensity, we have to add consecutive

15 minutes of traffic intensities and divide by 4

i.e Hourly Traffic Intensity.

= Tn + Tn+1 + Tn+2 + Tn+3	equation (1)

4

Tn = a0 + a1 n + a2 n2	+ a3 n3	+ a4 n4

Tn+1= a0 + a1 (n+1) + a2 (n+1)2 + a3 (n+1)3 + a4 (n+1)4

Tn+2= a0 + a1 (n+2) + a2 (n+2)2 + a3 (n+2)3 + a4 (n+2)4

Tn+3= a0 + a1 (n+3) + a2 (n+3)2 + a3 (n+2)3 + a4 (n+2)4

Hourly Traffic intensity has been given as

b0 + b1 n + b2 n2 + b3 n3 + b4 n4 equation (2)

Where as from equation 1 = equation (2)

Simplifying, and equating the constant term, or n or n to the powers, we

can obtain the relationship, b0 , b1 , b2 , b3 , b4 as given in the example.

4.	Traffic Profiles

Traffic Profiles play an important role in analyzing the exact utilization of a circuit group. Normally traffic profiles are drawn by calculation of traffic intensity per hour. That means Y-axis is Erlang and X-axis is hours.

The following are the readings obtained in a busy hour from one local exchange to an international exchange in Sri Lanka.

X = Time

Y = Sum of the number of circuits busy during scanning.

X	9.00	9.15	9.30	9.45	10.00	10.15	10.30	10.45	11.00	11.15
Y	304	279	324	367	338	320	302	292	271	398

X	11.30	11.45	12.00	12.15	12.30	12.45	13.00	13.15	13.30	13.45
Y	438	394	420	413	394	372	307	252	272	272

If the number of scans per hour is 100, draw a traffic profile from 09.00 to	13.30hrs.

ANSWER :

1.	Since observations have been made avery 15 minutes. The number of	scans for each observation is 100/4 = 25.

2.	Average holding time = 15x60/25 = 36 sec.

3.	Traffic = (Y1 x 36 / (15 x 60) = Y1 / 25 Erls.

TRAFFIC PROFILE OF THE INTERNATIONAL ROUTE.

18 .

17 .

16 .

15 .

14 .

13 .

12 .

11 .

10

09.00 09.30 10.00 10.30 11.00 11.30 12.00 12.30 13.00 13.30

From Traffic profiles one can design the tariffs to suit for maximum utilization of resources. In telecom maximum utilization of resources through out day will allow operator to maximize profits.

Hence by studying the traffic profiles one can introduce reduced rates in a particular direction in order to utilize the circuits to a maximum efficiency. Also exchange maintenance personnel should have an idea about traffic profile in each circuit group directions, so that when there is a maintenance problem on a certain circuit group, he should be aware as to how he should direct that traffic, in another circuit group, another direction to reach to the same destination.

5.	Relationship with Erlang and (BHCA)busy our call attempts.

What is busy hour call attempt ?

In Telecommunication BHCA is a very vital parameter to design common controlled equipments. In order to understand BHCA, first one has to understand what is call attempt. Call attempt means an attempt made to seize an equipment. In Telecom, a subscriber will lift the receiver, and dial the required number to be connected. However this attempt can end in either ringing no Answer customer engage or before reach to the customer equipments are busy.

Busy hour call attempt

X1 X2

X3

X4 X5

0 T

5

	 Xi

i=1

E (3) =	X1+X2+X3+X4+X5 =

T	T

5

 Xi	5

i=1

=	x

5	T

5

( Xi

= (=1	= Average Holding Time

5

5 = No of calls attempts

CA

= Average Holding Time x

Time

Normally traffic measurement is taken at the busy hour (statistical equilibrium).

Hence

Erlang = Average Holding Time x BHCA

QUESTION :	Calculate the G.O.S of a switching center which has busy hour

offered traffic of 200 Erlang when the average call duration is 3 minutes and 40 calls are lost during this hour.

ANSWER :	Erlang = Av. holding time x BHCA

Here offered traffic is 200 Erlang. To have 3 minutes as average holding time, let us calculate the number of call attempts using the above equation.

= 3 x BHCA

60

BHCA = 200 x 60 = 4000

3

Out of this 4000 call attempts.

40 calls are lost.

Hence Congestion	= 40 = .05

4000

as a percentage	= 40 x 100 = 1%

4000

Hence this switching system work at 1% G.O.S

5.	EXAMPLE :

Average holding time is linearly proportional to Erlangs. Hence one can visualize the loading of a common control equipment, with the average holding time that it will take to process the calls that it will be received. The following examples are given of the initial stages of digital switching, how a multi register (a common controlled equipment used to process telephone calls) is occupying to process calls. Here in these examples two telephone exchanges in the capital city of Sri Lanka Colombo was taken, where the city exchange was having a business customer environment and Maradana was having a residential customer environment. In 1994 Sri Lanka was having less than 1% telephone penetration, hence the call completion ratio was less than 20% resulting many unsuccessful calls. This has given rise to overload the common control equipment due to repeated call attempts.

In a telephone exchanges at Colombo - Sri Lanka the following were observed,	during the busy hour.

a.	At City

5MR	=	5 call processing processors

Each MR	=	510 registers

Busy Hour =	10.30 - 11.30 Hrs.

Date =	17-01-1994 (Monday)

Observations =	BHCA = 276,692

Traffic to the MR = 1291.94

Average Holding Time of a call processing

1291.94

Processor =	x 3600

276,692

= 16.8 Seconds.

=========

b.	At Maradana

5MR	=	Call processing processors.

Each MR	=	510

Busy Hour	=	10.30 - 11.30 Hr.

Date	=	12-01-1994

Observations	=	BHCA = 166018

Traffic to the MR	=	770.02

Average Holding Time	=	14.5 Sec. Or 16.7 Sec.

a.	At City

5MR	=	New OCB configurations R-24

Each MR	=	1530 4MR registration

Busy Hour =	10 - 11 Hrs.

Date =	30-09-2000

Observations =	BHCA = 141353

Traffic to the MR = 562.72

Average Holding Time of a call processing

562.72

Processor =	x 3600

141353

= 14.3 Seconds.

=========

b.	At Maradana

5MR	=	New OCB configuration R24.

Each MR	=	1020 4 MR registration

Busy Hour	=	10 - 11 Hrs.

Date	=	30-09-2002

Observations	=	BHCA = 114013

Traffic to the MR	=	457.33

Average Holding Time	=	14.4 Sec.

Another observations, in the year 1996 is as follows.

The two exchanges in generally remains unchanged, however more telephones in	these areas have been given. Load to City exchange has been decreased, while in	Maradana it is increased.

a.	At City

5 MR	= 5 call process

Each MR	= 510 Registers

Busy Hour	= 10.30 - 11.30 Hrs.

Date	= 26.03.96

Observations	= 229537

Traffic to the MR	= 968.9

Average Holding time of a call processing

968.9

Processor =	x 3600 = 15.19 sec.

229537

Traffic to MR=	770.02

Average Holding Time	=	14.5 Sec. Or 16.7 Sec.

b.	At City

5MR	=	5 call processing processors

Each MR	=	510 registers

Busy Hour =	10.30 - 11.30 Hrs.

Date =	26-03-1996 (Monday)

Observations =	BHCA = 251,687

Traffic to the MR = 1030.01

Average Holding Time of a call processing

1030.01

Processor =	x 3600

251,687

= 14.73 Seconds.

=========

At Maradana

5MR	=	Call processing processors.

Each MR	=	510

Busy Hour	=	10.30 - 11.30 Hr.

Date	=	26-03-1995

Observations	=	BHCA = 166018

Traffic to the MR	=	770.02

Average Holding Time	=	14.5 Sec. Or 16.7 Sec.

Average Holding Time	Maradana	City	Maradana	City

Date

Exchange Traffic

Average Holding Time

26 / 3 / 96

2803.9 E

2803.9

x 3600

229537

= 44.0 Sec.

12 / 12 / 95

3052.54 E

3052

x 3600

251687

= 43.7 Sec.

30 / 09 /2002

1720.07

1720.07

x 3600

114013

54.4 Sec.

30 /09 /2000

2069.52

1720.07

x 3600

141353

52.7 Sec.
Holding time in MR

No.of MR.

Total No.of registers

No. of registers used for speech

Traffic in MR

Holding Time

5

2560

2100

968.9

968.9

x 3600

229537

= 15.19 Sec.

5

2560

2100

1030.01

1030.01

x 3600

251687

= 14.73 Sec.

4

1020

4080

457.33

457.33

x 3600

114015

14.3 Sec.

4

1530

6120

562.72

562.72

x3600

141353

= 14.3 Sec.

QUALITY FACTORS IN TELECOM.

a.	Congestion and Grade of service.

The basic function of a switch is to receive the dial number and analyze the number then transmit the required numbers to be analyzed to the forward next exchange. Similarly the next exchange carryout the similar functions until it reaches the terminating exchange. Exchange to carryout these functions utilizes the following basic elements.

Switching element

Control element

Peripherals

Software

In strict sense, the control element will analyze the dialed number, and will be responsible to extend voice path of the customer via switching element and via a part of peripherals namely trunks to the required destination under this scenario.

Congestion can be visualized as the ratio of calls rejected due to insufficient resources. For example a call will be passing through switching network to Transmission (Trunk) to seize a circuit. A call can be rejected due to

Call Switch Trans Seize

rejected due to inadequate

rejected Number of Trunks

due to inadequate

switching

1.	Rejected due to inadequate switching may be either a link congestion in

switching or a fault in the switching module or all the out puts of the	switching Network is busy.

2.	Rejected due to all trunks busy or a fault in the transmission equipment.

In general, assumption is made that there is no switching congestion. i.e adequate No. of outlets are available to be connected to transmission, without any link congestion. Hence it is assumed that a call will be presented to Transmission or set of Trunks.

Call = N

Trunks

Rejections = R

Congestion due to inadequate Number of Trunks = R

N

If there are 25’ calls are offered and 5 call are rejected,

5

The congestion will be = = 0.2

25

Congestion can be explained in two ways.

i.e call congestion as explained earlier or time congestion.

New examples?

The time congestion is the cumulative time where all resources (in this case trunks) are busy. Under where all traffic distribution time congestion and call congestions are equal. If we assume random traffic condition in the above case time congestion =0.2. Hence in the busy hour 3600 x 0.2 sec.(= 720 sec.) all trunks were busy. In this scenario we have to emphasize that 720 seconds are not consecutive.

i.e out of 100 calls 20 calls will be rejected and 12 minuets (cumulative) all trunks	will be busy.

Now if this system is to be designed so that 10 calls out of 100 calls to be

rejected, obviously the number of trunks that has to be employed is more than

present. In this case we set the value congestion to 0.1.

The set value of congestion is named as grade of service.

G.O.S = Maximum congestion value set. by the user

Hence basic quality factor set in Telecommunication network is grade of service,

Basic Quality factors.

Quality is defined as expectation to be equal to the experience. Let us analyse what are the customers expectation from a Telephone operator. There are 3 major parameters.

Those who does not have telephone service has to be provided with telephones with quickest time.

Those who are having telephone service the following will be the expectations of the customers.

Telephone should not go out of service at least. Once in many years or uninterrupted telephone service.

They should get the correct bills

Call completion ratio has to be high.

c.	Answer to bid, answer to seizure Ratios

Another quality factor defined in telecom network is answer to bid, answer to seizure ratios.

Bid is defined as an attempt to establish a conversation. If a person

(i) Lifting a telephone is Bid or

(ii) a person lifting a telephone and partially dialing and abandoning a call also a

Bid, or

(iii) a person lifting the telephone, dialing a complete Number, and get engage

tone also a Bid,

a person lifting the telephone, dialing a complete Number, and get Ring

back tone but no answer also a Bid or

a person lifting the telephone, dialing a complete Number, but get answer

also a Bid.

A successful Bid will result in a seizure (or rejection) A seizure will result in answer (No answer, engaged tone or to any recording)

For successful calls

Bids > Seizures > Answers

Bids	=	No of calls seized the circuits + No of calls rejected due to all

circuits are busy (Assume that there are no switching congestion)

Seizures =	No of calls Answered + No of calls unanswered + No. of calls to	busy subscribers + No. of calls to recorded announcements.

In this section we will analyze the above (iii), the quality factors pertaining to customers, as well as network operators.

ANS	ANS

1 >	>

Seizure Bids

EXAMPLE 01:

Assume that there are 25 Bids (attempts) to a particular switching system for these 25 Bids the observations are as follows.

10 calls (Bids) answered, 10 calls (Bids) seized / busy tone 5 calls (Bids) No. seize at all.

Assume 10 calls seized for the answered (all) calculate the quality factors.

(i) G.O.S

(ii) A S R

(iii) A B R

5 1

Congestion experienced =	= = 0.2

25 5

5 1

A S R =	= = 0.2

25 5

5 1

A B R =	= = 0.2

25 5

Answer to bid is defined as the overall quality a subscriber can experience in a given Telecom Network. This value will be different from country to country depending upon the socio-economic development of a particular country. Socio-economic development will certainly influence the telephone penetration of a country. Sri Lanka major incumbent operator Sri Lanka Telecom has spent around us $ 400m to develop its network, which has resulted in achieving the A/B ratio from 20% to 45% . Certainly this will be influenced from the network rollout done from 1997 to 2002. Hence more the telephone penetration more A/B ratio can be achieved. This is due to the fact that less number of calls will be ended up with busy customers. A/B for Local, National, calls are in general same, whereas A/B ratio for International is in general high. (This is due to the fact, still 75% telephones in the world belongs to 9 countries where the telephone penetration is very high, In Sri Lanka A = 0.30 up to National calls and

B

For International A=0.6

B

AVERAGE CCR & NEW CONNECTIONS FOR 1994- 2002

NO:	YEAR	AVG CCR	NC

1	1994	26.69	25322
2	1995	30.98	24556
3	1996	31.89	50170
4	1997	34.32	72453
5	1998	34.73	143075
6	1999	36.95	133709
7	2000	42.68	90647
8	2001	44.87	78447
9	2002	44.66	43222

c.	Holding time & Conversation Time

Another important quality factors that can be defined in telecommunication

networks are as follows.

(a)	Holding time- average period a equipment or trunk is held irrespective	whether the call is successful or not. (Successful means, the call is

answered only).

(b)	Conversation time : Average period a equipment or a trunk is held for a	successful call (call that have being answered only).

The above two factors will be very important to design tariffs for customers (A.C.T.) as well as to reduce cost in the network(AHT). Normally operators tend to design their tariff structure to attract customers to its network.

EXAMPL 2:

For the example 1, assume the following average conversation time is 3 minutes.

Average seizure time & signaling before getting busy tone is 0.5 sec.

What is the average holding time ?

A.H.T	= 3 x 60 x 10 + 0.5 x 10 = 10 x (3 x 60) (Approximately)

20 20

A.H.T = ASR x ACT

Bids Seizures

Rejections

If Rejections = 0

A =	A

B S

In this exercise, one has to assume that that are no calls where calls will not be

observing telephones to ring with resulting no answer.

If the G.O.S = O or Rejections are Zero

Then AHT = ACT x Ans

Bids

d.	The relationship between A/B, A/S, and congestion

A	A	S

= x

B S B

S = B - R

A	A B-R A

= = 1 - Congestion

B S B S

ANS ANS

= 1- Congestion

BIDS SEIZ

EXAMPLE 1:

In a circuit group that is going from exchange X to exchange y the following are observed in the busy hour in a normal working day. i.e. On 26.10.1992 between 11.00 to 12.00 a.m.

X Y

Fo = Maximum traffic carried = 67.10E

FR = Number of rejections due to route busy = 4747

Fp = Number of seizures	= 6639

FE = Number of answers	= 1534

FQ = Equipped circuits = 345; Fs = No of circuits in operations.

From the above observations we can have the following conclusions.

Primary Conclusions

(i)	To carry 67.10 Er, we need atleast 96 circuits (If we assume 0.7E/circuit).

(ii)	Though 345 circuits are equipped only 68 circuits are operational.

(iii)	When compared the above (i) and (ii) it will be noticed that the

operational number of circuits are not enough to carry the traffic. Hence

can expect calls to be rejected about 4747 calls are rejected due to this. i.e.

(ATC 3 subscriber’s dialing ATC 3 subscribers).

(iv)	A 1534

= (From X to Y ) = = 0.23

S 6639

(v)	A 1534

= (From X to Y ) =

B 6639 + 4747

1534

=	= 0.135

11386

A A

(vi)	Since there are rejections (

B S

Secondary Conclusions

(i)	Average Holding Time	= Carried Traffic = 67.10

No of seizures 6639

= 36.4 Seconds

(ii)	Average Conversation Time	= 36.4 = 158 Seconds

0.23

EXAMPLE 2

Another Route	A	B

Observations

Fo = 44.23	Fe = 757	FS = 458-92

Fp = 5564	FR = 0	FQ = 474

Primary Conclusions

(i)	To carry 44.23E, we need at least 64 circuits.

(ii)	Though 474 circuits are equipped only 458 circuits were operational.

(iii)	When compared above (i) and (ii) and it will be noticed that circuits are	over provided from ATC 3 to ATC 5. Hence no rejections of calls due to	inadequate circuits.

(iv)	A	757

= (From ATC 3 to ATC 5) = = 0.13

S	5564

(v)	A	=	A

B	S

Secondary Conclusions

(i)	Average Holding Time = 44.23 x 3600 = 28.6 Sec.

5564

(ii)	Average Conversation Time = 28.6 = 220 Sec.

0.13

Mean Holding Time

EXAMPLE (a)

The mean time of a successful call (conversation time) for a population of subscribers in a city is estimated as 120 seconds. A master exchange is to be installed in one geographic end of the city to serve the population of that locality and it serves as an end office. It is estimated that 10% of the calls of that exchange is intra-office and the probability of failure of such calls is negligible. The rest of the calls could have a probability of failure of 0.05. Assuming that the subscribers shall not repetitively dial in the event a call fails, what is the mean holding time of a call in this exchange?

ANSWER - (a)

0.9N

N Call

Exchange

The prob.of failure of interoffice calls (0.1 N) is negligible.

Hence intra office traffic = 0.1 N x 120 Er.

3600

0.9N calls will encounter a failure of 5

100

Hence out of 0.9 N, the successful calls will be

0.9N x 95 = (0.9 x 0.95)N

100

= .855 N.

Hence the traffic carried from the rest of the calls

(other than that of interoffice) = 0.855N x 120

3600

Total traffic generated for calls in this exchange

= N x A.H.T. (Average holding time)

= 0.1N 120 + 0.855N x 120

3600

Hence average holding time = 0.1 x 120 + 0.855 x 120 (Hr)

3600

= 12 + 106.6 (Seconds)

= 118.6 seconds.

EXAMPLE (b)

If the limitation in traffic handling capacity of this exchange in Example (a) is 6000 Erlangs and calls per subscriber in the busy hour is 4, how many subscribers could be served by this exchange?

ANSWER - (b)

If the total number of customers in this exchange is C

Busy hours call attempts = 4C

Erlang = A.H.T x BHCA

6000 = 118.6 x 4C

3600

C = 6000 . x 3000

4x118.6

= 4553.1

EXAMPLE (c )

If there is a further limitation of the busy hour call attempts being limited to 250,000 calls in the above example (a) and example (b), can the above number of subscribers be served?

ANSWER (c)

Normally Telephone exchanges will have limitation on to the Busy Hour Call attempts. From 45531 customers, we can expect

45531 x 4 = 182,224 Busy Hour Call attempts

Hence this exchange can cater for the total population of the

above subscribers.

e.	Overall Quality factors.

When A dials B, subs can be a person (Telephone) out of 2 billion telephones in the world, and dialing can be categorized to one of the following.

(i)	Local call (stay within the same exchange)

(ii)	Regional call (stay within the region)

(iii)	National call (stay within a country outside the caller districts)

(iv)	International call (stay within the world outside its own country)

(v)	Special service (dial operator or programmed call etc)

Hence when A dial B, this call can be routed via 4 exchanges as shown in the	following figure.

( (

A Ex=1 Ex=2 Ex=3 Ex=4

Seizure 1 Seizure 2

Reject Reject

( Call

Seizure 3 Seizure 4

Reject Reject

At every exchange namely Ex.1 to Ex.3 the call will be successful if it goes to seizure stage, finally if the called customer answered only it will be taken as a successful call.

Unsuccessfulness can be achieved either customer was busy or Ex.3 to Ex.1 if it resulted in rejection. If we want to measure the exchange. The following transactions can be observed at each switching Node. Assuming that there are no switching congestion at any exchanges

Ex2 Seize Ex3 Seize Ex4 ANS

Seize

A’s call B

Reject reject reject reject

ASR 1 - ASR 2 ==> congestion of E x 2 (due to inadequate Trunks)

ASR 1 - ASR 2 ==> congestion of E x 3 (due to inadequate Trunks)

Exercise :

X Y Z J

( (

A Ex=1 Ex=2 Ex=3 Ex=4

Assume

(i) Internal Blocking of all exchanges are zero

(ii) G.O.S. of the links

XY = 0.02

YZ = 0.05

ZJ = 0.05

(iii) A/S = 0.95 at the input to exchange J

(iv) Find out the A/B ratio of exchange X and A/S of the link XY

A A

? ( Z - = 0.90

SY-Z S

Prob of call being successful from Z is 0.95 from exchange Z

Prob of call being successful on Z = J link [1-GOS ZJ] = 0.95

Prob of call being successful from link Z-J and from

exchange Z = 0.95 x 0.95 = 1- 5 1- 5

100 100

Neglecting higher order terms = 0.90

Hence A	= 0.90

SY-Z

A A

? ( Y	-	= 0.90

SY-Z SY-Z

Prob of call being successful from link (Y-Z) and

Exchange Y = 0.90 x 0.95 (since GOSY-Z 0.05)

1- 10 1- 5 1 - 15

=	100	100 100

(neglecting higher order terms)

=	85

100

Hence A	= 0.85

SX-Y

A A

? ( Y	-	= 0.90

SY-Z SY-Z

Prob of call being successful from link(X-Y) and exchange

Y = 0.85 x .98 (since GOSX-Y = 0.02)

1 - 15 1 - 2

100 100

=	1 - 17 = 83

100 100

Overall Grade of Service

In the above example A will reach B via 4 telephone exchanges at each exchange this call can encounter congestion. What is the overall congestion that this call can encounter ? or overall G.O.S ?

Overall G.O.S = X + Y + Z

Ex=1 Ex=2 Ex=3 Ex=4

X Y Z

Let’s assume the grade of service of the respective circuits are X,Y, Z

respectively.

The prob of success for a call from Ex 1 to Ex 2 is = (1-X).

Similarly from Ex 2 to Ex 3 = (1-Y)

Similarly from Ex 3 to Ex 4 = (1-Z)

This is only considering circuit bundles (assuming success at the switching	network). The probability of success from Ex 1 to Ex 4 is.

=	(1-x) (1-y) (1-z) = [1-x-y + xy] [1-z]

=	-2 + x2 - y2 - 2xy + 1 - x - y + xy

=	1 - [x + y + 2] + 2x + 2y + xy - 2xy

Considering only low order terms, we can neglect 2x / 2y / 2xy / xy terms.

= 1 - [x + y + z]

Net G.O.S = x + y + z

EXAMPLE :

In the above example what is the overall G.O.S.

= .02 + 0.05 + 0.05 = 0.12

In a given telephone network it is important to set an overall grade of service figure in general it is set to 0.01. But most of the administrational this value is not achieved. The following example shows the importance of setting overall grade of service.

(1)	i	What do you understand by the following terms as applicable to a calls	processed through a bundle of circuits.

a.	Congestion

b.	Answer Bid Ratio

c.	Answer Seizure Ratio

ii	Why is an upper limit for congestion specified for calls processed? What

is the human characteristic of subscriber to load the exchanges under

values of congestion?

A B C

( (

x Y

iii	A B C

X X

Above is a diagram showing calls from subscriber exchange A routed to subscriber in exchange C via exchange B. The congestion in routes connecting A to B is X and B to C is Y. If X and Y are small and second order terms are negligible prove that overall congestion for calls to C from A is X + Y. (Assume no switching congestion at A, B and C)

iv	Following gives the pictorial representation of network hierarchy.

Assuming that always priority for routing is from X to Y sides.

X Y

> Tertiary

> Secondary

> Primary

> Local

( Sub ( Sub

A	B

If the overall congestion for subscriber A to subscriber B is E, specify values of your choice for congestion in each type of route specified.

ANSWER :

i.	X	Y

From exchange X to exchange Y the following details are available.

Co = Offered number of calls

CR = Rejected calls due to all circuits busy from exchange X to exchange Y.

CS = Total number of seizures on the circuit group from exchange X to

exchange Y.

CA = Total number of calls answered from exchange X to exchange Y.

Congestion =	CR

Co

ANS	=	CA

Bid	Co

ANS	=	CA

Seizure	CS

Hence a justifiable congestion value has to be decided in order to obtain the

profit. In general one should understand that less quality leads to more cost to be incurred. It has been an experience if the congestion exceed 0.05 tends to increase redialing. Hence it is set to 0.01,5 times less than the highest congested value.

To achieve a lower value of congestion the cost will be high. Under high value of congestion, subscribers tend to load, the exchange by redialing. The time for redialing can be extended by routing to a recorded announcement, so the dynamic load control of the exchange can be achieved.

iii. The prob. a call being successful from A to B = (1-X)

The prob. a call being successful from B to C = (1-Y)

Prob. a call being successful from A to C = (1-X) (1-Y) = 1-(X-Y)+XY

XY term can be neglected.

Prob. a call being successful from A to C = 1 - (X+Y)

Hence the total congestion between A and C = X+Y

iv. The worst Route to take sub X to sub Y is as follows.

Tertiary

Secondary

Primary

Local

A B

7 trunks involved. If the congestion from X to Y is E. Then each trunk should	have a congestion of E/7.

Accordingly the congestion on the other routes can be determined and will be as	follows.

E/7

E/7	2E/7 E/7

E/7 E/7

3E/7

E/7	4E/7 E/7

5E

7

X X

TRAFFIC ENGINEERING

Types of switching system : loss or delay system loss calls arriving when all the circuits in a wanted groups are occupied, are rejected and busy signal is fed back to the calling party. Such calls must be dialed again even if a circuit becomes free immediately after the call has been connected to busy tone.

Delay or waiting : calls arriving when all the circuits in a wanted group are busy, wait until a circuit become idle, when this happens the waiting call is automatically taken out of storage and connected to a free circuit waiting calls can be service in order of arrival, in random order, or in any other descried sequence. Loss system mainly employed in complete switching system.

Access to expensive but short holding time common control devices such as multi registers and signal receivers is normally provided on waiting systems. Hence ensuring better circuit utilization.

2.	Traffic Theory

1 sec

5 sec

10 sec

0 T

The above shows the occupancies of 3 circuits going from one exchange to another exchange , the observation were made during a time between 0 to T sec. Imagine a 4th calls at t=0 sec, it will get engage tone. However if the 4th caller due to some reasons, has delayed to dial this call by 1 second, he will be successful as well the function of the telecom is achieved. Hence to study the traffic theory , study about human behaviors on the following aspect is important.

i Call originating characteristics.

ii	Conversation time characteristics or call holding time characteristics.

7.	Call holding characteristics

Call holding time characteristics will have one of the following features.

Fixed holding time characteristics

Variable holding time like negative exponential characteristics.

Normally common control equipment will follow the above (i).

In the telephone network, the characteristics of the holding time of subscribers is assumed to be following the above ( ii ).

Negative exponential characteristics.

((t) = e/h –t/h

((t) = Probability density function

h = average holding time.

If we want to find out, the probability of occupying a circuit more than t sec is given by.

P[t > T] = ((t) = [ 1 e –t / h ] = e –t / h

h

In addition to the above (1) & (2) , the following factor also important to be analyzed to study traffic.

i Busy hours readings.

Statistical equilibrium is achieved at the busy hour. i.e . The busy hour characteristics appears to be same for every day. Hence the characteristics are time independent.

For example at working days, the characteristics of BHCA Erlangs pertaining to a exchange is more or less similar. However this document mean that there is any major disturbance in that area even in a working day, the characteristics may drastically differ.

8.	Erlang B Model

Erlang has delivered a statistical model to be used for telephone applications. It assumed the following conditions.

Provisional distribution for the call arrivals

Negative exponential distribution for the call holding.

Observations are made when the statistical equilibrium achieved.

Lost call never return

Patient subscriber characteristics i.e. when subscriber receives engaged tone, it will not dial. Erlang has defined the following model2.

P[k, n] = n

(	[Ai /i]

Where A is offered Traffic.

Probability of occupying k circuits out of n circuit group is given by the above formulae. (Where n>k)

When k = n, i.e. probability of all circuits busy is called as Erlang loss formulae.

AK / K

B =

N

 A i

i=1

i

B is called the grade of service or set value for congestion Erlang B is having random characteristics.

Variance = 1

Mean

Hence call congestion = Time congestion.

Call congestion defines the percentage of calls rejected due to all circuits are busy. Time congestion defines the cumulative sum of times, where all circuits are busy.

G.O.S. : Fundamental design parameter.

For loss system G.O.S. is defined as, not being able to set up connection to a chosen route or destination because of an insufficient number of circuits or control or signaling devices.

It can refer to a individual group of circuit or to a complete connection from an originating to a terminating subscriber in a given network.. It is usually expressed as a blocking probability or a percentage of calls permitted fail.

Time Congestion	:	Proportion of time during which congestion exist.

Call Congestition	:	Probability a call will encounter congestion.

EXAMPLE 1:

Assume 1Er of Traffic is offered to a full availability group of trunks. The total No. of Trunks are 4.

(i) What are the loss probabilities of occupying 0,1,2,3,4 trunks?

(ii)	What is the carried Traffic by this trunk group.

1Er

T Lost

Loss probability of occupying 0 ccts.

1 0 / 0

P[0,4] =

4

( 1

i =0

4

(

i=0 = 1 = 1 + 1 + 1 1 1

2 3 4

1 1 1 48+12+4+1 65

= 2 + + + = =

2 6 24 24 24

P[0,4] = 24

65

P[1,4] = 24

65

P[2,4] = 12

65

P[3,4] = 4

65

P[4,4] = 1

65

All loss probabilities = 1

i.e P[0,4] + P[1,4] + P[2,4] + P[3,4]+ P[4,4] = 1

Loss probability or G.O.S

= 1

65

Offered Traffic = Carried traffic + Lost Traffic

1Er = Carried traffic + 1 x 1

65

Lost Traffic = Offered Traffic x [G.O.S or Loss prob of occupying all Trunks]

Carried Traffic = 64 Erlangs.

65

Hence = 64 Erlangs will be carried by this trunk group

65

= 1 Erlangs will be lost

65

2. If the same exercise to be followed, with only change of offered traffic

from 1Er to 2Er

4

(

i =0 A = 1 + 2 + 22 + 23 + 24

2 3 4

= 5 + 4 + 2 = 7

3 3

P[0,4] = 2 0 / 0 = 1

7 7

P[1,4] = 2 1 / 1 = 2

7 7

P[2,4] = 2 2 / 2 = 2

7 7

P[3,4] = 2 3 / 3 = 4

7 21

P[4,4] = 2 4 / 4 2

7 21

All probabilities

= 1 + 2 + 2 + 4 + 2 = 3+6+6+4+2 = 1

7 7 7 21 21 21

Offered Traffic = Carried Traffic + Lost Traffic

2Er = Carried Traffic + 2. 2Er

21

Carried Traffic = 2 - 4 38 Er.

21 21

Lost Traffic = 4 Er.

21

A = 1 Er A = 2 Er

P[0,4]	24/65 1/7

P[1,4]	24/65 2/7

P[2,4]	12/65 2/7

P[3,4]	4/65 4/21

P[4,4]	1/65 2/21

Carried Traffic	64/65 Er 38/21 Er

Lost Traffic 1/65 Er 4/21 Er

Visualizing the Characteristics of Erlang B

In Erlang B model, the basic assumptions are,

(i) Call arrive according to poisoning distribution

Call holding according to negative exponential distribution

Other assumptions will be discussed later.

Assume 2Er of offered traffic is allowed to a full availability circuit group of 3 trunks.

2 Erl

Erlang B

An / n

B =

n

( A

(=0

(

Here A = 2, n=3. What is B?

23 / 3

B =

n

( A

(=0

(

Hence the loss probability of occupying all 3 circuits when the offered traffic is

2 Erl is 0.2.

.. . Loss traffic	= AB = 0.4Erl

Carried traffic	= 2 – 0.4 Erl = 1.6 Erl

Observation :	(1)	In the normal circumstances, we should have through

Nothing is lost as the offered traffic is 2 Erl, and there are 3 Trunks. But mainly due to the call arrival and holding characteristics, we observed 1.6 Erl will be carried.

(2) B=0.2,means loss probability is equal to 0.2.Hence if there

are 100 calls arrived to this circuit group in the busy hour (or any experimented time), 20 calls have been received busy tone (or unsuccessful seizures). Also in the busy hour, it should have been noticed that a total of 3600 x 0.2 = 720 sec. All these three circuits have been engaged. (or in a experimenting period (T), 0.2 x T, will be engaged by all 3 circuits.

EXAMPLE 02

A traffic load of 2 Erl is offered to a full availability groups of 4 trunks. The average call duration is 2 Minutes.

(i) What is the average number of calls offered per hour ?

(ii) What is the probability that no calls are offered during specified period of 2 minutes?

(iii) What is the proportion of lost traffic?.

(iv) If the trunks are always tested in the same order,(sequential hunting) how much

traffic is carried by each trunk ?

(i) Av.h.T = 2M

2 = 2 x CA CA = 60

60 Hr Hr

(ii) calls offered during t time

P [n1 t] = (At)n eAt

n

P [x1 ] = x e x

x

P [n1 t] = (At)n e t

n

O calls offered during 2 minutes

-60 2

60

60 2

60 .e

P [0, t] =

L

= 2 . e-2

= 0.1352 x 2

= 0.27

====

Note 1 :

Here in this exercise, you will notice that P[R,N], the loss probability of occupying K circuits out of N circuit group given, where R circuits are randomly selected out of N circuit group. (it is not 1 to K numbering out of N circuits) Hence to find out traffic loss to K+1 circuit is not meaningful in contrary, if the circuits are always selected from circuit No 1 to K, then in order to find the traffic carried by each circuit, the above method can be applied, for example traffic carried by the 2nd circuit is No

TA[P(1,4)-P[(2,4)]

Note 2 :

Erlan 9B loss formulae is very useful to calculate the loss traffic, as well as the congestion.

(iii) Proportion of lost traffic.

24 / 4

B =

4

( Ai

i=0 2

4 AI 20 21 22 23 24

( = + + + + +

2=0 i 0 1 2 3 4

= 1 + 2 + 2 + 4 + 2

3 3

= 7

B = 2 X 1 = 2

3 7 21

Last traffic = 0.1

====

= 0.19E

21 /L1 2 2 2

(iv) P [1, 1] = = = =

Ai 20 + 2i 1+2 3

i i

Lost troffic = 4

3

Carried troffic = 2 - 4 = 2E = 67E

3 3

2nd Trunk

22 / 2 x2

Lost traffic from 2nd trunk =

20 + 21 + 22

1 2

= 2 x 2 = 4

5 5

Carried traffic in the 2nd Trunk

= 4 - 4 = 20-12 = 8

3 5 15 15

= 0.53E

=====

Lost traffic in the 3rd Trunk

23 / 3

= x 2

20 21 + 22 + 23

1 2 3

= 2.2.2 1+2+2 2.2.2

3.2.1 3.2.1

= 4 5 + 4 = 4 x 3 x2

3 4 3 19

= 4 x 2

19

= 8

19

Carried Traffic in the 3rd Trunk = 4 - 8

5 19

= 76 - 40

5 x 19

= 36

5 x 19

= 0.379E

=========

Lost traffic in the 4th Trunk = 24 x 2

4 / 20 + 21 + ..... + 23

1 3

= 4

21

Carried traffic in the 4th Trunk = 8 - 4

19 21

= 168 - 76 = 0.23

19 x 21

EXAMPLE 03 :

(1a)	Write down the Erlang B formula for Congestion B(n) for n trunks where A

erlangs is the offered traffic intensity.

(1b)	Divide B(n+1) by B(n) and establish the following :

V(n+1) = 1 + (n+1) *V(n)/A where V(n) = 1/B(n).

(1c)	Using the property that B(0) = 1 for whatever value of A, find the number of

trunks that can carry values of offered traffic in the given table below with a

loss probability of 1%

Offered Traffic	No of Trunk Required

7

8

9

10

(1d)	At an end office exchange the number of outgoing trunks to the main exchange

is the Traffic Offered on this group is 0.04 Erl per subscriber. The current number of subscribers is 100. It is expected to double the number of subscribers. Determine the increase of the number of trunks on the outgoing route for 1% loss probability.

ANSWER :

Ak / Z !k

B [k,n] =

n

 A i/ !i

i = 0

when k = n;

An / Z !n

B [n] =

n

 A i/ !i

i = 0

An 1

1(b)	B (N) =

L n n

 A i/ !i

i = 0

AN+1 1

B(N+1) =

N+1 N+1

( Ai/!i

i=0

N+1

( Ai / Li

B (N) = (N+1) i=0

B(N+1) A N+1

(	Ai / Li

i=0

N+1

( Ai / Li

V(N+1) (N+1) i=0

V(N) A N+1

( Ai / Li

i=0

N+1

( Ai / Li

N(N+1) = V (N) (N+1) 1+ AN+1 i=0

A N+1 N

( Ai/ Li

i=0

N(N+1) = V (N) (N+1) + V(N).AN 1

A N N

( Ai/ Li

i=0

V (N+1) =	V(N) (N+1) +1

A

1(c)	V (N+1) = 1 + (N+1). V(N)

A

B (N) = 1 when N=0

B (N) = 01.; V (N) = 100

V (0) = 1

V (0) = 1

V (1) = 1 + 1 = 1.2 A = 5

A

V (2) = 1 + 2 V (1)

A

V (3) = 1 + 3 V (2)

A

V [9] = 100

V [10] > 100

Check at which value of N.

V [n] > 100 } take that Value as N

Hence the answer is as follows

Offered Traffic	Trunk

7

8

9

10

13

14

16

17

1(d)	Total traffic generated to this circuit group = 0.04 x 100 = 4 Erlangs

For 4 Erlangs offered traffic

there are 10 Trunks

when the subscribers are doubled. The offered traffic for this circuit group is Erlangs for 8 Erlangs from 1(c), we need 14 circuits at 0.01 GOS.

Hence 4 Trunks to be increased when subscribers are doubled.

In Erlang B equation

B (n) = AN /LN A is the offered Traffic

N N total Number of circuits

( Ai / L i B the congestion value

i=0

Utilizing Erlang B model for practical purposes – Erlang B Charts.

In Erlang B formulae, one can find out the congestion that can be expected, if one knows the offered traffic and the number of circuits, this offered traffic is involved.

However when number of circuits (N) increased the calculation will become cumbersome.

The important utilization of erlang B will be to design the number of circuits, having known the offered traffic, to the expectation of the telephone operator having defined the congestion. Here in this case.

B,A, will be given one has to find out N? This is also tedious.

Hence if summarize the utilization of erlang B formulae one can encounter practical difficulties as follows.

If A and N is given B can be calculated. However when A is large and N is large the calculation will become tedious.

If B is given, A is given to calculate N, it is very tedious.

(3)	If B is given, N is given to calculate A also difficult.

Hence erlang B charts are developed so, that, depending upon the circumstances the required values of B, A, N can be calculated, provided two of the values being given. The following examples illustrates the application while the first example will explain how to make these charts.

EXAMPLE 4 :

Erlang B charts are widely find application in dimensioning Telecom Network. Generally an iterative method can be applied to find the no of equipment or circuits for a given offered traffic at a given G.O.S this is illustrated in the above Q.2. In order to make it simplified Erlang B charts were made to find circuit quantities for a given offered traffic at a given grade of service. These charts are annexed here.

EXAMPLE 5 :

a/ If A = 100, B = 0.01 b/ If A = 15, B = 05

N ? N ?

ANS N = ANS.

b/ If N = 300, G.O.S = 0.01 c/ If A = 400, B = 0.01

A ? N ?

ANS A = ANS = N.

ANSWER :

From the table, E15 (10) = 0.0365

Carried traffic = A[1-EN(A)] = 10(1-0.0365) = 9.635 Erlangs

From the table, E19(12)>0.0100, and E(12)<0.0100, so 20 circuits are required.

From the table, E16(8) = 0.0045, and E16(9) = 0.0111.

By interpolation, the required offered traffic is : Here we will make an assumption between E16(8) to E16(9), offered traffic and grade of service to vary linearly.

A = 8 + 0.0100 – 0.0045 = 8.8 Erlangs

0.0111 – 0.0045

The carried traffic is :

A carried = 8.8 (1-0.0100) = 8.7 Erlangs.

Note : In calculating No. of circuits required the interpolation is not required as

the circuits are integers. However for offered traffic the actual values to be obtained.

Question :

1.	Explain the meaning of Direct outward Dialing and Direct inward Dialing to

a PABX. Describe the signaling protocols, that can be used. In which instances the DOD facility is given to a PABX and if you are public network operator, what are precautions that you suggest to avoid any malpractices.

2.	A PABX with 200 extensions were having 8 Bothway Trunks with DOD + DID with the public Network. What is the blocking probability if each extension will receive 2 calls, and transmit 4 calls, per eight hours working day through (DID & DOD) trunks. (Assume average duration of any call is 2 minutes). Calculate the average number of trunks occupied. Assume that blocked calls are lost. Figure shows the probability of call blocking for a given amount of offered traffic for various Number of Trunk lines. How many Trunks are required to improve the grade of service to better than 2%.

Answer :

In normal practice a PABX will be given with DID facilities. (Direct Inward Dialing) any calls that has to be dialed out from the extensions, the PABX will access through the normal subscriber line normal protocols. Adopted is CCITT R2 signaling and CCITT No.7, though CCITT No.7 is yet to be implemented.

Under special circumstances DOD facility will be given to a PABX. (Normally another cellular operator). The protocol used is R2, and No.7, R2 is more widely used where No.7 is gradually implementing. If R2 signaling is used, the following has to be monitored since in R2 signaling the terminating exchange will control the signaling.

1. Undue loading of signaling units of public Network causing long processing times

introduced due to the following.

i.	Long processing times before sending an acknowledgment by PABX.

ii Due to poor Transmission line in between PABX and the public Network.

For terminating call to PABX, requesting callers number PABX.

For originating numbers, specially for intern calls, sending faults A number from the PABX.

8

Total traffic received	=	2/60 X 2/8 X 200	=	5/3 Er

Total traffic Transmitted	=	2/60 X 4/8 X 200	=	10/3 Er

Total B/W traffic	=	5/3 + 10/3	=	5 Er

Congestion from the chart	=	0.05

Average number of trunks occupied	= 5

Trunks required to improve the G.S.O to 3% will be = 10

Hence need increase by 2 Trunks.

Explain the Basic message of CCITT No.7 signaling, with relation to each layer. Explain the importance of ISUP, in the latest telephone network with relation to telephone user part.

A call (subscriber X to Subscriber Y) is taken from Ex.A to Ex.B. The CCITT No.7 signaling is transmitted via exchange C. There are no signalling links between exchange A and exchange B.

SP SP

voice

( (

X Y

SLK

Signaling SLK

STP

But from exchange A to C and B to C from the signaling links are available. C switching Node will act as a signaling transfer point. From exchange A to Exchange B. The Quasi associated mode of signaling is enforced via signaling transfer point C.

a. What are the basic signaling messages transferred from A to B to A, (only important

parameters of each messages to be explained)

If X and Y has a successful call and Y has released the call first

Explain how the messages are routed via C to from A and B.

Explain how a message is evolved to a non-homogeneous Network application.

A message to be transferred in a non homogeneous Network from station A to far away station B. What are the methods you suggest to use for error control for this message. (assume any type of messages that can be used for various applications).

CCITT No.7 layer 2 (link control) will be used for error control. Explain clearly the layer 2 functions.

Ex A & Ex B are linked with number of PCM systems and is expected to be working on associated are mode of signaling.

The calls that are passing through this Y PCM’s are calls from Ex.X and Ex.Y and will have the following characteristics.

Average holding time 100 secs.

5 messages are used in each direction to establish a call.

Each message consist of 256 bits.

1

PABX

A

B

C

EX

EX

2

6

4

1

3

7

5

At EX1

EX2

EX3

EX4

0 1 2 3 nth

0.1N

Answer :

DUSHYANTHI MENDIS

2003-03-27

SLT

Digital Equipment Corporation