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Traffic Engineering
Traffic analysis
Network capacity (e.g., number of channels between exchanges, exchange sizes, number of
radio channels in a cellular network) should be increased where the bottlenecks of the
network are found.
Therefore, the utilization of the network is continuously measured and traffic demand in
the future is estimated.
Grade of Service
nch
Grade of service (GoS) i.e. availability or quality of service,
measures the service quality customers achieve. GoS
depends on the network capacity that should meet the
service demand of the customers.
In a circuit-switched service the most important factor is
whether the call is successful or blocked.
Blocking probability: blocking occurs if more than nch
Local exchange and blocking
subscribers make external calls at a time.
If more than nch subscribers make a call at a time, some of them are blocked and have to retry.
The number of external calls varies randomly. To be sure that blocking never occurs nch should
equal number of subscribers. This is too expensive because number of subscribers connected
to a local exchange is very large and on average only a few place calls at the same time.
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The Erlang: measure of Traffic Intensity
The “Erlang” as a telephone traffic measure
The Erlang unit (Erl) is defined as a unit of
telephone traffic specifying the percentage of
average use of a line or circuit (one channel); in
other words it is the ratio of time during which a
circuit is occupied and the time for which the
circuit is available to be occupied.
Example: traffic that occupies a circuit for 1 hour
during a busy hour is equal to 1 Erl.
The Danish mathematician A. K. Erlang,
founder of the Traffic Theory.
Busy-hour traffic
The typical average busy hour traffic volume
generated by one subscriber is in the range of 10
to 200 mErl. Low values are typical for residential
use and high values for business subscribers.
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Traffic Parameters
•
Definition of traffic:
– 1 Erlang: one channel utilized for all the time
– Offered Traffic: A0 [Erl]
– Consigned Traffic: AS [Erl]
PBLO·A0
Offered Traffic
(Traffico offerto), A0
Consigned Traffic
(Traffico smaltito), As
•
AS=A0·(1-PBLO)=A0·GS
•
GS = 1-PBLO = grade of service (grado di servizio)
•
Assumptions:
– Infinite population (in practice > 30 users)
– Offered Traffic is generated according to the Poisson model with interarrival
time, λ (negative exponential)
– Average service time (negative exponential), µ
Elementary Traffic generated by single user: A0=λ/µ
•
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Performance parameters
• A single exchange has a certain number of channels (nch)
• Problem: calculate the probability that one user, willing to
connect, does not find available resources
• This is called blocking probability, PBLO
• Under certain assumptions, always valid for classical telephone
exchanges and voice traffic, blocking probability can be
computed with the so called “Erlang B” formula:
n
PBLO
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A0 ch
n ch !
(n ch , A0 ) =
n
n ch A 0
∑ n = 0 n!
Legend:
A0 = total offered traffic (Erlang)
PBLO = blocking probability or fraction of
lost calls
nch = number of servants (i.e. channels)
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Exercise
• Problem:
• Determine the minimum number of channels, nch ,
necessary to support 100 users with GOS equal to
0.95 (95%). Each user makes a call on average every
half hour and the call lasts an average of 3 minutes.
Both the inter-arrival time and the service time are
distributed according to an exponentially negative law.
•
•
Solution:
Aou = user offered traffic = λu / µu = 0.1 Erl
– λu = inverse of the average inter-arrival time
between the service requests of one single user =
1/(30 min)
– µu = inverso del tempo medio di servizio = 1/(3
min)
•
Ao = offered traffic (all population) = 100·Aou = 10 Erl
•
Equation: Probability of call being accepted ≥ 0,95 ,
i.e. blocking probability ≤ 0,05 PBLO(nch,10) ≤0,05
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Erlang B (nch, A0)
A0
nch
8
0,89
1
2
0,78
0,68
3
0,57
4
5
0,48
0,39
6
0,31
7
0,24
8
0,17
9
10
0,12
0,08
11
0,05
12
0,03
13
0,01
14
15
0,009
0,005
16
0,002
17
18
0,001
10
0,91
0,82
0,73
0,65
0,56
0,48
0,41
0,34
0,27
0,21
0,16
0,12
0,08
0,06
0,04
0,02
0,01
0,007
12
0,92
0,85
0,77
0,7
0,63
0,56
0,49
0,42
0,36
0,3
0,25
0,2
0,15
0,12
0,09
0,06
0,04
0,03
PBLO(nch,10) ≤0,05 nch=15
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Signals Carried over the Network
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Power Levels of Signals and Decibels
The decibel (dB) is a logarithmic measure of relative signal levels (power
ratios) along communications systems.
P1
P2
Gain: G = P2/P1
Linear System
Attenuation: A = P1/P2
Relationships and definitions:
A = G -1
For any linear system the attenuation is always the reciprocal of the gain
G (dB) = 10 log10 G = 10 log10 P2 – 10 log10 P1 Gain in Decibel
(note P1 and P2 in same unit!, e.g., both in Watt)
A (dB) = 10 log10 A = 10 log10 P1 – 10 log10 P2 = - 10 log10 G = - G(dB)
When expressed in decibel, the attenuation is always the opposite of the gain (& vice versa)
In a linear system, if G > 1, i.e. G(dB) > 0 this linear system is called Amplificator
In a linear system, if A > 1, i.e. A(dB) > 0 this linear system is called Attenuator
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Examples: how to use dB’s
For example, a gain of 100,000,000 = 108 corresponds to the gain of 80 dB
Gain of cascaded linear systems:
amplifier
amplifier
g = g1/a2
a2
g = g1 – a2 (dB)
amplifier
attenuator
Other logaritmic measures:
Es. g1 = 5 dB, a2 = 8 dB
Therefore g = - 3dB , so a = 3 dB. The
cascade is a 3 dB attenuator (half-power)
P (dBW) = 10 log10 P/P1, when P1 = 1 Watt
P (dBm) = 10 log10 P/P1, when P1 = 1 milliWatt
Since 1 W = 1000 mW
P (dBW) = 10 log10 P/(103 mW) = -30 dB + P (dBm)
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Exercise: radio relay
Problem: determine the received radiofrequency power level.
Change power P1 in dBm:
P1 (dBm) = 10 log10(P1/1 mW) = +30 dBm
P2 (dBm) = P1 (dBm) + gT (dB) – L (dB) + gR (dB) =
+30 dBm + 30 dB –110 dB +30 dB = –20 dBm.
P2 (Watt) = 10 P2(dBW) /10 = 10 [P2(dBm) -30] /10 = 10 -5 Watt
Remember: 1 mW = 10 -3 W
1 µW = 10-6 W
1 pW = 10 -9 W
1 nW = 10 -12 W
Therefore: P2 = 0,01 mW = 10 µW = 10000 pW
T
radio relay
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Exercise: optical fiber link
Problem: determine the received optical power level.
First calculate the Total fiber attenuation , given the fiber length of 40 km and the fiber
specific attenuation of 0.5 dB/km (specific attenuation is, by definition, the attenuation
of a unit length of a fiber – or more generally cable – trunk).
We have: L (dB) = 40 km⋅ 0.5 dB/km = 20 dB
Therefore: P2 = P1 - 20 dB = 0 dBm – 20 dB = - 20 dBm
optical fiber system
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