Network Planning VITMM215

Markosz Maliosz PhD

10/05/2016

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Outline

Telephone network dimensioning

– Traffic modeling

– Erlang formulas

– Exercises

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Telephone Network

Circuit switching

Each voice channel is identical

For each call one channel is allocated

A call is accepted if at least one channel is idle

Goal: network dimensioning

Question to answer: How many circuits are required to satisfy subscribers’ needs?

Input: traffic statistics – subscribers’ behavior: when, how often are calls arriving?

how long are the call durations?

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Arrival Process

In our case: telephone calls arriving to a switching system

described as stochastic point process

we consider simple point processes, i.e. we exclude multiple arrivals

the ith call arrives at time Ti

N(t): the cumulative number of calls in the half-open interval [0; t[

N(t) is a random variable with continuous time parameter and discrete space

4 t

N(t)

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Arrivals and Departures

N(t) to be the cumulative number of arrivals up to time t

D(t) to be the cumulative number of departures up to time t

L(t) = N(t) - D(t) is the number of calls at time t

N(t)

D(t)

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Equations

Average arrival rate: λ(t) = N(t)/t

F(t) = area of shaded region from 0 to t in the figure = total service time for all customers

= carried traffic volume

Average holding time: W(t) = F(t)/N(t)

Average number of calls: L(t) = F(t)/t

= W(t)N(t)/t = W(t)λ(t)

N(t)

D(t)

N(t)

D(t)

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Traffic Volume

Volume of the traffic: the amount of traffic carried during a given period of time

Traffic volume in a period divided by the length of the period is the average traffic intensity in that period = average number of calls

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Traffic Variations

Traffic fluctuates over several time scales – Trend (>year)

Overall traffic growth: number of users, changes in usage Predictions as a basis for planning

– Seasonal variations (months) – Weekly variations (day) – Daily profile (hours) – Random fluctuations (seconds – minutes)

In the number of independent active users: stochastic process

Except the last one, the variations follow a given profile, around which the traffic randomly fluctuates

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Traffic Variations

Source: A. Myskja, An introduction to teletraffic, 1995.

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Busy Hour

It is not practical to dimension a network for the largest traffic peak

describe the peak load, where singular peaks are averaged out

Busy Hour = The period of duration of one hour where the volume

of traffic is the greatest.

Operator’s intention: spreading the traffic

– By service tariffs

busy hour period is the most expensive

less important calls are started outside of the busy hour, and typically last

longer

Recommendations define how to measure the busy hour traffic

– There are several definitions (ITU E.600, E.500)

– An operator may choose the most appropriate one

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Busy Hour Measurements 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

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

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Traffic Model

Average arrival rate: λ(t) – depends on time, however it has a very strong deterministic component according to the profiles

In the busy hour period the average arrival time is considered stationary: λ, and the arrival process is considered as a Poisson process with intensity λ – Time homogenity

– Independence The future evolution of the process only depends upon the actual

state.

Independent of the user(!) – modeling all users in the same way

The average holding time (W(t)) is also considered to be stationary, and exponentially distributed with intensity μ

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Traffic Model

N(t) – Poisson process:

– in time interval (t,t + τ] the number of calls follows a Poisson distribution with parameter λτ

– Expected number of calls = λτ

– λ = arrival intensity [1/hour]

W(t) = W – exp. distribution

– Expected value = 1/μ = h

– h – average holding time [min] (!)

f(x; μ) = μe-μx

Poisson

Exp.

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Traffic Intensity

A – traffic intensity – A = λ * h – A [1], often written as Erl (Erlang)

Example: individual subscriber – λ = 3 [1/hour] – h = 3 [min] = 0.05 [hour] – A = 3 [1/hour]* 0.05 [hour] = 0.15 [Erl]

Example: 10 000 line switch – λ = 20 000 [1/hour] – h = 3 [min] = 0.05 [hour] – A = 20 000 [1/hour]* 0.05 [hour] = 1000 [Erl]

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Typical Traffic Intensities

Typical traffic intensities per a single source are (fraction of time they are being used)

– private subscriber 0.01 - 0.04 Erlang

– business subscriber 0.03 - 0.06 Erlang

– mobile phone 0.03 Erlang

– PBX (Private Branch Exchange) 0.1 - 0.6 Erlang

– coin operated phone 0.07 Erlang

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Traffic Modeling

Agner Krarup Erlang (1878 – 1929) – Danish mathematician, statistician and engineer

Conditions: – n identical channels

– Blocked Calls are Cleared (BCC)

– The arrival process is a Poisson process with intensity λ

– The holding times are exponentially distributed with intensity μ (corresponding to a mean value 1/μ)

The traffic process then becomes a pure birth and death process, a simple Markov process – A= λ/μ

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Infinite number of channels

State diagram – nr. of busy channels:

If the system is in statistical equilibrium, then the system will be in state [i] the proportion of time p(i), where p(i) is the probability of observing the system in state [i] at a random point of time, i.e. a time average

When the process is in state [i] it will jump to state [i+1] λ times per time unit and to state [i-1] iμ times per time unit

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Infinite number of channels

In equilibrium state

– Node equations:

– Cut equations:

Normalization restriction:

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Infinite number of channels

Derivation of cut equations:

A= λ/μ

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Infinite number of channels Using the normalization constraint:

State probabilities:

Carried traffic = offered traffic = A

No congestion, no traffic loss

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Limited number of channels

State diagram:

Normalization condition becomes:

State probabilities:

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Erlang B formula

Time congestion: – The probability that all n channels are busy at a random point of

time is equal to the portion of time all channels are busy (time average)

Call congestion: – The probability that a random call attempt will be lost is equal to

the proportion of call attempts blocked. If we consider one time unit, we find by summation over all possible states:

Carried traffic =

Lost traffic =

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Erlang B formula

Conditions for applicability:

– Gives good results if number of subscribers is much greater, than the number of channels (around 10x)

– Subscribers initiate calls independently from each other (not applicable e.g. if a TV advertisement presents a phone number and many people call it)

– The only reason for blocking is if all channels are busy

– Blocked Calls are Cleared, no waiting queue

– Subscribers do not repeat call attempt, if call was blocked

– The channel is occupied only by the particular subscribers, no resource sharing

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Erlang B formula

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Erlang B formula

Example: 3 employees in an office, each of them calls 3 times in an hour with 3 minutes talking time.

Question: How many channels are needed for max. 5% blocking? (1? 2? 3??)

Answer: – λ = 3*3 [1/hour]

– h = 3 [min] = 0.05 [hour]

– A = 3*3 [1/hour]* 0.05 [hour] = 0.45 [Erl] E(1)=31%

E(2)=6.5% (not enough!)

( E(3)=1%, in reality: E(3)=0)

– 3 channels are needed

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Erlang B formula

E.g. 1000 subscriber and n channels:

– λ = 1000*3 [1/hour]

– h = 3 [min]

– A = 1000*3 [1/hour]* 0.05 [hour] = 150 [Erl]

– E(n):

If the number of subscribers are large, the required number of channels (n) for a satisfactory blocking ratio converges to A

n 100 150 155 160 200

E(n) 34% 6,2% 4,3% 2,8% 0,0015%

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Erlang B formula

If A and achievable blocking is given, how to calculate n?

By probing

Recursive method:

– Def.: In(A) = 1 / En(A)

– I0(A) = 1, that is with 0 channel the blocking = 1

– In(A) = In-1(A) * n / A + 1

– E.g. if the goal is: En(A) = 1 / In(A) < 0.05

– In(A) > 1/0.05 = 20

n

i

i

n

n

i

A

n

A

AE

0 !

!)(

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Extended Erlang B

Extended Erlang B: a certain percentage of blocked calls are reattempted – Iterative calculation with extra parameter, the Recall

Factor: Rf

– A0:initial traffic intensity

1. Calculate En(A0) with Erlang B

2. Nr. of blocked calls: B = A0 En(A0)

3. Nr. of recalls: R = Rf B

4. New offered traffic: A1 = A0+ R

5. Return to step 1 and iterate until value of A is stabilized

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Erlang C formula

Erlang C: blocked calls remain in the system (waiting in a queue), until they get served

– E.g. call centers

Probability of waiting:

1

0 )(!!

)(!),(

n

i

ni

n

w

Ann

nA

i

A

Ann

nA

AnP

Exercises

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Problems What is the blocking probability if traffic intensity is 2 Erl and 5 lines are

available?

Analyze the following diagram!

– Examine the network utilization if the number of available channels is low!

– Examine the network utilization depending on the blocking ratio!

0%

10%

20%

30%

40%

50%

60%

70%

80%

90%

100%

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

A/n

n

1% blocking

10% blocking

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Problems

How many lines are required for 100 subscribers, when they individually generate 0.04 Erl traffic intensity? – if the allowed blocking ratio is 20%?

– if the allowed blocking ratio is 1%?

20 employees work in an office with 2 lines. What is the blocking ratio if employees call with 0.1 Erl intensity?

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Problems

10 employees work in an office with 3 lines. What is the blocking ratio if employees initiate once a 15 min long call in the busy hour?

– Average utilization of lines? – Blocking ratio? – Is it a well dimensioned system?

A subscriber generates 0.1 Erl traffic intensity. How many lines are required, if the blocking requirement is 1% and the number of subscribers are – 10? (5) – 100? (18) – 1 000? (117) – 4 000? (426) – 10 000? (1029)