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

Network Planning VITMM215 Markosz Maliosz PhD 10/05/2016

Outline  Telephone network dimensioning – Traffic modeling – Erlang formulas – Exercises 2

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? 3 3

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 i th call arrives at time T i  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 N(t) t 4 4

Arrivals and Departures N(t) D(t)  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 5

N(t) N(t) Equations D(t) D(t)  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) 6

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 7

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 8

Traffic Variations Source: A. Myskja, An introduction to teletraffic, 1995 . 9

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 10

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 11

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 12

Traffic Model  N(t) – Poisson process: Poisson – in time interval ( t , t + τ] the number of calls follows a Poisson distribution with parameter λτ – Expected number of calls = λ τ – λ = arrival intensity [1/hour] Exp.  W(t) = W – exp. distribution – Expected value = 1/ μ = h – h – average holding time [min] (!) f(x; μ ) = μ e - μ x 13 13

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] 14 14

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 15

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 16

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] i μ times per time unit 17

Infinite number of channels  In equilibrium state – Node equations: – Cut equations:  Normalization restriction: 18

Infinite number of channels  Derivation of cut equations: A= λ / μ 19

Infinite number of channels  Using the normalization constraint:  State probabilities:  Carried traffic = offered traffic = A  No congestion, no traffic loss 20

Limited number of channels  State diagram:  Normalization condition becomes:  State probabilities: 21

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 = 22

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 23 23

Erlang B formula 24 24

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 25 25

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): n 100 150 155 160 200 E(n) 34% 6,2% 4,3% 2,8% 0,0015%  If the number of subscribers are large, the required number of channels (n) for a satisfactory blocking ratio converges to A 26 26

Erlang B formula If A and achievable blocking is given, how to calculate n? n A ! n   By probing E ( A ) n i n  A  Recursive method: i !  i 0 – Def.: I n (A) = 1 / E n (A) – I 0 (A) = 1, that is with 0 channel the blocking = 1 – I n (A) = I n-1 (A) * n / A + 1 – E.g. if the goal is: E n (A) = 1 / I n (A) < 0.05 – I n (A) > 1/0.05 = 20 27 27

Extended Erlang B Extended Erlang B: a certain percentage of  blocked calls are reattempted – Iterative calculation with extra parameter, the Recall Factor: R f – A 0 :initial traffic intensity 1. Calculate E n (A 0 ) with Erlang B 2. Nr. of blocked calls: B = A 0 E n (A 0 ) 3. Nr. of recalls: R = R f B 4. New offered traffic: A 1 = A 0 + R 5. Return to step 1 and iterate until value of A is stabilized 28