Basic teletraffic concepts
An intuitive approach
(theory will come next)
Focus on “calls”
Giuseppe Bianchi

1 user making phone calls
TRAFFIC is a “stochastic process”
BUSY 1
IDLE 0 time
 How to characterize this process?
 statistical distribution of the “BUSY” period
 statistical distribution of the “IDLE” period
 statistical characterization of the process “memory”

E.g. at a given time, does the probability that a user starts a call result different depending on what happened in the past?
Giuseppe Bianchi

Traffic characterization suitable for traffic engineering traffic amount intensity

A i
 lim
 t
 
 average number
 of calls of busy time
 t per min in
 t

  average call duration

 probabilit y that, at a random time t, user is in BUSY state
 min


 mean process value
All equivalent (if stationary process)
Giuseppe Bianchi

Traffic Intensity: example
 User makes in average 1 call every hour
 Each call lasts in average 120 s
 Traffic intensity =
 120 sec / 3600 sec = 2 min / 60 min = 1/30
 Probability that a user is busy
 User busy 2 min out of 60 = 1/30 adimensional
Giuseppe Bianchi

U1
Traffic generated by more than one users
Traffic intensity
(adimensional, measured in Erlangs):
U2
A
 i
4 

1
A i

4 A i
U3
U4
P
 k active calls
E
 active calls






4 k

 A i k

1

A i

4
 k
4

A i

A
TOT
Giuseppe Bianchi

example
 5 users
 Each user makes an average of 3 calls per hour
 Each call, in average, lasts for 4 minutes
A i


3  calls hour



4
60
 hours
  
5
A

5

1    
5
Meaning: in average, there is 1 active call; but the actual number of active calls varies from 0 (no active user) to 5 (all users active), with given probability
Giuseppe Bianchi number of active users probability
0 0,327680
1
2
3
4
5
0,409600
0,204800
0,051200
0,006400
0,000320

Second example
 30 users
 Each user makes an average of 1 calls per hour
 Each call, in average, lasts for 4 minutes
A

30
 
1
4

2 Erlangs
60
SOME NOTES:
-In average, 2 active calls (intensity A);
-Frequently, we find up to 4 or 5 calls;
-Prob(n.calls>8) = 0.01%
-More than 11 calls only once over 1M
TRAFFIC ENGINEERING: how many channels to reserve for these users!
Giuseppe Bianchi n. active users binom
0
1
2
3 probab cumulat
1 1,3E-01 0,126213
30 2,7E-01 0,396669
435 2,8E-01 0,676784
4060 1,9E-01 0,863527
7
8
9
4
5
6
27405 9,0E-02 0,953564
142506 3,3E-02 0,987006
593775 1,0E-02 0,996960
2035800 2,4E-03 0,999397
5852925 5,0E-04 0,999898
14307150 8,7E-05 0,999985
16
17
18
19
20
21
10
11
12
13
14
15
30045015 1,3E-05 0,999998
54627300 1,7E-06 1,000000
86493225 1,9E-07 1,000000
119759850 1,9E-08 1,000000
145422675 1,7E-09 1,000000
155117520 1,3E-10 1,000000
145422675 8,4E-12 1,000000
119759850 5,0E-13 1,000000
86493225 2,6E-14 1,000000
54627300 1,2E-15 1,000000
30045015 4,5E-17 1,000000
14307150 1,5E-18 1,000000
22
23
24
25
26
27
28
29
30
5852925 4,5E-20 1,000000
2035800 1,1E-21 1,000000
593775 2,3E-23 1,000000
142506 4,0E-25 1,000000
27405 5,5E-27 1,000000
4060 5,8E-29 1,000000
435 4,4E-31 1,000000
30 2,2E-33 1,000000
1 5,2E-36 1,000000

A note on binomial coefficient computation


60
12



60 !
12 !
48 !

1 .
39936 e

12 but 60 !

8 .
32099 e

81 ( overflow problems!
!
)


60
12


 exp

 log


60
12




 exp
 log
 
 log
 
 log
  

 exp

 i
60 

1 log
 i
12 

1 log
 i
48 

1 log


(no overflow!
!
before exp...)


60
12

 A i
12

1

A i

48 
 exp

 i
60 

1 log
 i
12 

1 log
(no overflow!
!
never!
)
 i
48 

1 log

12 log
  i

48 log

1

A i



Giuseppe Bianchi

Infinite Users
P
Assume M users, generating an overall traffic intensity A
(i.e. each user generates traffic at intensity A i
We have just found that
=A/M).
   
 


 M
P k active calls, M users




M k 


A i k
1

A i
M
 k 
M
M
 k
!
!
k !



A
M


 k


A
M


 k


A
M
Let M  infinity, while maintaining the same overall traffic intensity A
 k active calls,
 users


M lim
 

M
M

!
k

!

1 k !

A k
M k

A
M
M
 
1
A
M
 k


A k k !
 lim
M
 
M

M

1
 
M
M k
 k

1






1
A
M

M
A




A
 
1
A
M
 k
 e

A
A k k !
Giuseppe Bianchi

P k
Poisson Distribution
30%
25%
20%
A=2 erl poisson binomial (M=30)
A=10 erl
15%
10%
5%
0%
0 2 4 6 8 10 12 14 16 18 20 22
 e

A
A k k !
Very good matching with Binomial
(when M large with respect to A)
Much simpler to use than Binomial
(no annoying queueing theory complications)
Giuseppe Bianchi

Limited number of channels
THE most important problem in circuit switching
 The number of channels
C is less than the number of users M (eventually infinite)
U2
U3
 Some offered calls will be
“blocked”
 What is the blocking probability?
U4
 We have an expression for
P[k offered calls]
 We must find an expression for
P[k accepted calls]
 As:
P [ block ]

P

C accepted calls

TOT
Giuseppe Bianchi
X
X
No. carried calls versus t
No. offered calls versus t

Channel utilization probability offered traffic: 2 erl - C=3
 C channels available
 Assumptions:
 Poisson distribution (infin. users)
 Blocked calls cleared
 It can be proven (from
Queueing theory) that:
35%
30%
25%
20%
15%
P [ k calls

C
P
 in the system, k offered
P
 i offered calls calls
 k
 i


0

(0, C) ]

10%
5%
0%
0 1 2 3 4 offered calls accepted calls
5 6
(very simple result!)
 Hence:
P [ system full ]

P [ C accepted calls ]

P

C offered calls
 i
C 

0
P
 i offered calls

Giuseppe Bianchi
7 8

Blocking probability: Erlang-B
 Fundamental formula for telephone networks planning
 A o
=offered traffic in Erlangs
 Efficient recursive computation available
E
1 , C
  o

C
A o

E
1 , C
A o

1
E
1 , C
 

1
A o
  o
A o
C
 block
 j
C 

0
C !
A o j j !

E
1 , C
  o
100,00%
10,00%
1,00%
C=1,2,3,4,5,6,7
0,10%
0,01%
0 1 2 3 offered load (erlangs)
4 5
Giuseppe Bianchi

NOTE: finite users
 Erlang-B obtained for the infinite users case
 It is easy (from queueing theory) to obtain an explicit blocking formula for the finite users case:
 Erlang-B can be re-obtained as limit case
 M  infinity
 A i
 0
 M·A i
 A o
 Erlang-B is a very good approximation as long as:
 A/M small (e.g. <0.2)
 ENGSET FORMULA:
 block

A
C i

M

C
1

 k
C 

0
A i k

M
 i
1


A i

A o
M
Giuseppe Bianchi
 In any case, Erlang-B is a conservative formula
 yields higher blocking probability
 Good feature for planning

Capacity planning
 Target: support users with a given Grade Of
Service (GOS)
 GOS expressed in terms of upper-bound for the blocking probability

GOS example: subscribers should find a line available in the
99% of the cases, i.e. they should be blocked in no more than
1% of the attempts
 Given:

C channels
 Offered load A o
 Target GOS B target
 C obtained from numerical inversion of B target

E
1 , C
  o
Giuseppe Bianchi

Channel usage efficiency
Offered load (erl)
A o
C channels
Carried load (erl)
A c

A o

1

B

A o
B efficiency :
 
A c
C
Blocked traffic

A o

1

E
1 , C
C
  o


A o if small blocking
C
Fundamental property: for same GOS, efficiency increases as C grows!! (trunking gain)
Giuseppe Bianchi

example
100,0%
10,0%
1,0%
A = 40 erl
A = 60 erl
A = 80 erl
A = 100 erl
0,1%
0 20 40
GOS = 1% maximum blocking.
Resulting system dimensioning and efficiency:
Giuseppe Bianchi
60 capacity C
80 100
40 erl C >= 53
60 erl C >= 75
80 erl C >= 96
100 erl C >= 117
 = 74.9%
 = 79.3%
 = 82.6%
 = 84.6%
120

Erlang B calculation - tables
Giuseppe Bianchi