telecommunication trafficii
TRANSCRIPT
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