This document discusses key concepts in telecommunications network planning and traffic engineering. It covers: - Types of random processes used to model network usage patterns like call arrival rates and durations. - How traffic engineering balances factors like grade of service, resources, blocking vs. delay systems based on traffic amounts. - Key metrics like erlangs, traffic intensity, busy hour, traffic volume that are used to quantify network usage and demand. - Concepts like grade of service, blocking probability, and how they measure network performance during busy periods.

1. Department of Electrical Engineering

2. 
Covers specific types of random processes in
telecommunications
– Average connection duration
– Average number of users
– Busy time
– Service time
– Call arrival
2

3. 

Required in telecommunications network planning to ensure
that network costs are minimized without compromising the
quality of service (QoS) delivered to the user of the network.
– It is based on probability theory and can be used to analyze
mobile radio networks as well as other telecommunications
networks.
Mobile radio networks have traffic issues that do not arise in the
fixed line PSTN. A mobile handset, moving in a cell, receives a
signal with varying strength. This signal strength is subject to:
– slow fading,
– fast fading
– interference from other signals,
3

4. •
•
Traffic engineering balances the following factors based on given
amount of traffic
• Grade of Service (GOS)
• Resources (e.g. trunk channels)
Two types of systems implemented to provide voice
communications
– Blocking
• Voice or data is blocked (by a busy signal) if network
resource (e.g trunk channel) is not available.
• GOS = Blocking probability
– Delay System
• Voice or data is queued until network resource is available
• GOS = Queuing Probability and average time in queue
4

5. 


Holding Time - the length of time that a resource is
being held
(e.g the duration of a phone call)
Traffic volume - for an interval is the sum of all the
traffic holding times for that interval
Traffic intensity = traffic volume / time interval which
is a measure of demand
5

6. 

Erlangs - describe traffic intensity in terms of the
number of hours of resource time required per hour of
elapsed time
CCS( Centum Call Seconds) - measures the exact
same traffic intensity as the Erlangs but expresses it as
the number of 100 second holding times required per
hour. Traffic registers sample stations every 100 seconds
per hour to check for busies. Since there are 36 sets of
hundred seconds in an hour
36 CCS = 1 Erlangs
6

7. 

Erlangs:
Traffic intensity (named after of a Danish mathematician) is the
average number of calls simultaneously in progress over a
certain time. It is a dimensionless unit.
– Erlang
• one hour of continuous use of one channel = 1
Erlang
• 1 Erlang = 1 hour (60 minutes) of traffic
– In data communications, an 1 E = 64 kbps of data
– In telephone, 1 Erlang = 60 mins = 1 x 3600 call seconds
% of Occupancy
7

8. 
For example, if a group of user made 30 calls in one
hour, and each call had an average call duration of 5
minutes, then the number of Erlangs this represents is
worked out as follows:
Minutes of traffic in the hour = number of calls x duration
Minutes of traffic in the hour = 30 x 5
Minutes of traffic in the hour = 150
Hours of traffic in the hour = 150 / 60
Hours of traffic in the hour = 2.5
Traffic figure = 2.5 Erlangs
8

9. ◦ Call holding time is the length of time during
which a traffic source engages a traffic path or
channel. 1 – 3 minutes typical, >10 minutes
infrequent for voice.
H = average
holding time,
3 minutes.
Negative
exponential

10. 
Busy hour is that continuous 60 minutes time
span of the day during which the highest
usage occurs.

11. 
Note:
◦
◦
◦
◦

may not occur at the same time every day
weekly variation
week day /weekend variation
seasonal variation
Mathematical formulas assume the busy hour
traffic intensity is the average of an infinite
number of busy hours.

12. 
ADPH (Average Daily Peak Hour)
◦ − one determines the busiest hour separately for each day
(different time for different days), and then averages over e.g. 10
days
◦ − the resolution of the start time of the busy hour may be either a
full hour (ADPH-F) or a quarter of an hour (ADPH-Q)

TCBH (Time Consistent Busy Hour)
◦ − a period of one hour, the same for each day, which gives the
greatest average traffic over e.g. 10 days

FDMH (Fixed Daily Measurement Hour)
◦ − a predetermined, fixed measurement hour (e.g. 9.30-10.30);
the measured traffic is averaged over e.g. 10 days

aFDMH ≤ aTCBH ≤ aADPH

13. 

Traffic Density is defined as the number of
simultaneous calls at a given moment.
Traffic Intensity,A represents the average
traffic density (occupancy) during any one
hour period.
◦ Occupancy is any use of a traffic resource
regardless of whether or not a connection (call) is
completed.
◦ Occupancy is the probability of finding the trunk
busy is equal to the proportion of time for which
the trunk is busy

14. 
A Loss System is one in which a call attempt
is rejected when there is no idle resource to
serve the call. (BCC – Blocked Call Cleared)
◦ Blocked calls…

A Delay System is one in which call attempts
are held in a waiting queue until resource are
available to serve the calls.
◦ Delayed calls…

15. 


Offered traffic is the traffic intensity that
would occur if all traffic submitted to a group
of circuits could be processed.
Carried traffic is the traffic intensity actually
handled by the group.
Blocked traffic is that portion of traffic that
cannot be processed by the group of circuits
(I.e. offered traffic minus carried traffic).
◦ Blocked traffic may be rejected, retried or offered to
another group of circuits (overflow).

16. Let
N channels
Carried traffic (C)
Offered traffic (A)
◦ A be the offered traffic
◦ C be the carried traffic,
Lost traffic; (BxA )
◦ B is blocking probability
◦ Lost traffic (A- C) = B*A
◦ Probability of blocking
◦ System utilization:
1 B
A1 B
N
C
A

17. Is a measure of the average occupancy of a resource
during a specified period of time, normally a busy hour.
The traffic intensity offered by each user is:
A = μH Erlangs
where
H is the average holding time of a call
μ is the average number of call requested/hour
If there are U users and an unspecified number of channels.
The total offered traffic intensity is:
AT = UA Erlangs
Busy hours traffic: Calls/busy hours *Mean call hold time
17

18. In a trunks system of C channels and equally
Distributed traffic among the channels, the traffic
intensity per channel is:
AC = UA/C
Erlangs/channels
The traffic volume
is a measure of the total work done by a resource or
facility, normally over 24 hours
VT = A * T Erlangs-Hours
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19. ◦ Traffic Flow = (no. of calls)*(mean call holding time)
◦ Example :
If 100 calls are generated in 1 hour of 3 minutes
average duration
we have 3*100 = 300 call minutes or 300/60 = 5
call hours.

20. 

The international dimensionless unit of
telephone traffic is called the Erlang after A.
K. Erlang (1878 – 1929) a Danish scientist.
Defined as one circuit occupied for one hour.
◦ 1 Erlang = 1 Call–hour / hour

Busy hour traffic
◦ Erlangs = (Calls/busy hour)*(mean call holding
time)
 (careful with units, all times in hours)

1 Erl = 36 CCS

21. A call established at 1am between a mobile and MSC. Assuming a
continuous connection and data transfer rate at 30 kbit/s, determine the
traffic intensity if the call is terminated at 1.50am.
Solution:
Traffic intensity = (1 call)*(50 mins)*(1 hour/60 min) = 0.833 Er
Note, traffic intensity has nothing to do with the data rate, only the holding time is
taken into account.
Note:
• If the traffic intensity > 1 Erlang: The incoming call rate exceeds the
outgoing calls, thus resulting in queuing delay which will grow without
bound (if the traffic intensity stays the same).
• If the traffic intensity is < 1 Erlang, then the network can handle more
average traffic.
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22. Consider a PSTN which receives 240 calls/hr. Each call lasts an average
of 5 minutes. What is the outgoing traffic intensity to the public network.
Solution:
A = μ *H
μ = 240 calls/hr and H = 5 minutes
A = (240 calls /hr) x (5 min/call) = 1200 min/hr
Erlang cannot have any unit so
A= 1200 min/hr * (1 hour/60 minutes) = 20 Erlangs
So 20 hours of circuit talk time is required for every hour of elapsed time. An
average
of T1 voice circuits busy at any time is 20. (Or 20 hours of continuous use of
20 channels.)
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23. 
Call established at 2 am between a central computer
and a data terminal. Assuming a continuous
connection and data transferred at 34 kbit/s. What is
the traffic if the call is terminated at 2-45am?
Solution:
Traffic
= (1 call)*(45 min)*(1 hour /60 min)
= 0.75 Erlangs.
It’s nothing to do with the data rate of communication,
only the call holding time.

24. 
A group of 20 subscribers generate 50 calls with an
average holding time of 3 minutes, what is the
average traffic per subscriber?
Solution:
Traffic = (50 calls)*(3min)*(1 hour/60 min)
= 2.5 Erlangs
Average traffic per subscriber = 2.5 / 20
= 0.125 Erlangs per subscriber.
Individual (residential) calling rates are quite low and may
be expressed in milli-Erlangs, i.e. 0.125 Erlangs = 125
milli-Erlangs.

25. There are 100 subscribers with the following telephone traffic
profile: 20 make 1 call/hour for 6 minutes; 20 make 3 calls/hour
for half a minute; 60 make 1 call/hour for 1 minute.
Solution:
The traffic they generate is:
20x1x (6/60) = 2 E
20x3x(0.5/60) = 0.5 E
60x1x(1/60) = 1 E
a total traffic of 3.5 E.
On average, each subscriber generates 3.5/100 = 35 mE.
In practice on average telephone subscribers generate between
25 to 35 mE during the busiest hour

26. 

Grade of Service is a measure of the probability
that a percentage of the offered traffic will be
blocked or delayed.
◦ the ability to interconnect users
◦ the rapidity with which that connection is made
Commonly expressed as the fraction of calls or
demand that
◦ fails to receive immediate service (blocked calls)
◦ is forced to wait longer than a given time
(delayed calls)
For example, if GOS = 0.05, one call in 20 will be blocked during the
busiest hour because of insufficient

27. 
Grade of Service, B
B=
Or
B=
Lost traffic
Offered traffic
Lost call
Offered call

28. A measure of the performance of a telephone
system
GOS is a measure of the ability of a user to access a
trunked system during the busiest hour
Also an indication of the user not being able to
secure a channel during the busiest hour
Telephone networks are designed with specified
GOS, usually for the busiest hour. If a subscriber is
able to make a call during the busiest hour, he will
be able to make a call at any other time

29. 
Grade Of Service, B is used to observe and
measure how many calls are offered, carried
and lost in the system.
B=
B=

Number of calls lost
Number of calls offered
Traffic lost
Traffic offered
The lower this number, the higher the GOS.

30. The offered traffic: Volume of traffic offered to a switch that are all processed
is defined as:
Offered traffic = carried traffic + overflow
The carried traffic: The actual traffic carried by a switch.
Overflow (blocked) traffic: Portion of the traffic not processed.
Busy Hour Call Attempts (BHCA)
• Used to evaluate and plan capacity for telephone networks
• Is the number of telephone calls made at the peak hour
• The higher the BHCA, the higher the stress on the network processors.
• Not to be confused with Busy Hour Call Completion (BHCC), which
truly measures the throughput capacity of the network.
30

31.  For example, if GOS = 0.05, one call in 20 will be
blocked during the busiest hour because of
insufficient capacity

For GOS = 0.02. This means that two users of the
circuit group out of a hundred will encounter a call
refusal during the busy hour.

32. 
1200 calls are offered to a channel and 6
calls are lost. Duration of a call is 3 minutes.
Find:
a)
b)
c)
d)
e)
Offered traffic, A
Carried traffic
Lost traffic
GOS, B
Congestion time

33. A = CH/T = (1200 x 3) / 60 = 60 E
b)
Carried traffic = [(1200-6) x 3] / 60 = 59.7 E
c)
Lost traffic = (6 x 3) / 60 = 0.3 E
d)
B = Lost traffic / Offered traffic
= 6/1200
= 0.005
e) Congestion time
= B x 1 hours (second)
= 0.005 x 60 x 60
= 18 seconds
a)

34. 
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;
◦ A. The outgoing traffic
◦ B. The incoming traffic
◦ C. The total traffic

35. 


A. The outgoing traffic is 120 x 2/60 = 4E
B. The incoming traffic is 200 x 3/60 = 10E
C. Total Traffic is 4E + 10E = 14E