Network Planning VITMM215 Markosz Maliosz PhD 10/05/2016 Outline - - PowerPoint PPT Presentation

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-

• pen interval [0; t[

 N(t) is a random variable with continuous time

parameter and discrete space

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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

• f 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

• f 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 A E ! ! ) (

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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

) ( ! ! ) ( ! ) , (

n i n i n w

A n n n A i A A n n n A A n P

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)