Traffic [Compatibility Mode]

8/18/2019 Traffic [Compatibility Mode]

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Traffic Engineering


2/78

Traffic Engineering Traffic Engineering • One billion+ terminals in voice network alone

– Plus data, video, fax, finance, etc.

• Imagine all users want service simultaneously…its not evennearly possible (despite our common intuition)

– In ractice the actual amount of e ui ment rovisioned is vastl less than would support all users simultaneously

• And yet, by and large, we get the impression of phone and data

• How is this possible?

Traffic theor !!


3/78

Traffic EngineeringTraffic Engineering –– Trade Trade- -offs offs • Design number of transmission paths, or radio channels?

– How many required normally?

– What if there is an overload?

• Design switching and routing mechanisms

–

– E.g.

• High-usage trunk groups

• Overflow trunk rou s

• Where should traffic flows be combined or kept separate?

• Design network topology

– – Number and sizing of transmission systems and locations

– Survivability


4/78

Poisson Distribution Poisson Distribution • Poisson, is a discrete probability distribution that expresses the

probability of a given number of events occurring in a fixedn erva o me an or space ese even s occur w a nown

average rate and independently of the time since the last event.

• A discrete random variable X is said to have a Poissondistribution with parameter λ > 0, if for k = 0, 1, 2, ... theprobability mass function of X is given by:

• Where e is the base of the natural lo arithm e = 2.71828... k!

is the factorial of k.

• The positive real number λ is equal to the expected value of X.


5/78

Exponential Distribution Exponential Distribution • In probability theory and statistics, the exponential distribution

(a.k.a. negative exponential distribution) is the probabilitys r u on a escr es e me e ween even s n a o sson

process, i.e. a process in which events occur continuously andindependently at a constant average rate. It is the continuous

,property of being memoryless.

• The probability density function (pdf) of an exponentialdistribution is

• ,rate parameter


6/78

binomial distribution binomial distribution • In probability theory and statistics, the binomial distribution is

the discrete probability distribution of the number of successesn a sequence o n n epen en yes no exper men s, eac o

which yields success with probability p. Such a success/failureexperiment is also called a Bernoulli experiment or Bernoulli

= ,distribution.

• The binomial distribution is frequently used to model thenumber of successes in a sample of size n drawn withreplacement from a population of size N. If the sampling iscarried out without replacement, the draws are not independent

distribution, not a binomial one. However, for N much largerthan n, the binomial distribution is a good approximation, and

.


7/78

Probability mass function Probability mass function • In general, if the random variable X follows the binomial distribution

with parameters n and p, we write X ~ B(n, p). The probability of

mass function:

• for k = 0, 1, 2, ..., n , where

• is the binomial coefficient hence the name of the distribution.

• The formula can be understood as follows: we want k successes(pk ) and n − k failures (1 − p)n − k . However, the k successes canoccur an where amon the n trials and there are different wa sof distributing k successes in a sequence of n trials.


8/78

Characterization of Telephone Traffic Characterization of Telephone Traffic •• Calling RateCalling Rate () – also called arrival rate, or attempts rate, etc.

– Average number of calls initiated per unit time (e.g. attempts perhour)

– Each call arrival is independent of other calls (we assume)

– Call attempt arrivals are random in time

– Until otherwise, we assume a “large” calling group or source pool

αγ If receive calls from a terminal in time TT:

If receive calls from mm terminals in time T:

T

αγ g

Tm

αγ

calling rate


9/78

Characterization of Telephone Traffic (2) Characterization of Telephone Traffic (2)

• Calling rate assumption:

– Number of calls in time T is x

– In our case

...2,1,0!

)(

x x

x p

T

• me e ween ca s s -ve exponen a y s r u e :

e mean

• Class Question: What do these observations about telephone trafficimply about the nature of the traffic sources?


10/78

- -ve Exponential Holding Times ve Exponential Holding Times

•• Implies the “MemoryImplies the “Memory--less” propertyless” property – Prob. a call last another minute is independent of how long the call has

alread lasted! Call “for ets” that it has alread survived to time T

t T PT T t T T P 11• Proof:

1

1111

T T P

T T t T T PT T t T T P

1

1

T T P

t T T P

hT

ht T

e

e/

/)(

1

1

ht /

Recall:

hT

ht hT

e

ee/

//

1

1

ht e

/ t T P


11/78

Characterization of Telephone Traffic (3) Characterization of Telephone Traffic (3)

•• Holding TimeHolding Time (hh)

– Mean length of time a call lasts

– Probability of lasting time t or more is also –ve exponential in

nature:

0)( / t et T P ht

–

00)( t t T P

above

– As non-voice “calls” begin to dominate, more and more calls have aconstant holding time characteristic

•• Departure RateDeparture Rate ():

h

1


12/78

Some Real Holding Time Data Some Real Holding Time Data


13/78

TrafficTraffic Volume (V) Volume (V)

= # calls in time period T

= mean o ng me

V = volume of calls in time period T

• In N. America this is historically usually expressed in terms of centum call seconds “ccsccs”:

– Hundred call seconds

“cc” “cc” “ss” – 1 ccs is volume of traffic equal to:

– r u u y r , r – two circuits busy for 50 seconds, or

– 100 circuits busy for one second, etc.


14/78

TrafficTraffic Intensity (A) Intensity (A)

• Also called “traffic flowtraffic flow” or simply “traffictraffic”.

====

h = mean holding time

= time period of observations

A

h







= calling rate = calling rate

= departure rate

T

eca :

h

1

eca :

V h

eca :

= calling rate

= departure rate

• Units:

– “ccs/hourccs/hour”, or

V = call volume

–

“ErlangErlang” unit


15/78

TheThe Erlang Erlang

• Dimensionless unit of traffic intensity

• -

• Usually denoted by symbol EE.

• 1 Erlang is equivalent to traffic intensity that keeps:

– one circuit busy 100% of the time, or

– two circuits busy 50% of the time, or

– four circuits busy 25% of the time, etc.

• 26 Erlangs is equivalent to traffic intensity that keeps :

– 26 circuits busy 100% of the time, or

– ,

– 104 circuits busy 25% of the time, etc.


16/78

ClassClass

• Could 4 E be produced as a traffic intensity by:

– 16 sources? (What is the utilization?)

– 4 sources (same)

– 1 source?

• What is special about the traffic intensity if it pertains to onesource or terminal only?


17/78

Erlang (2) Erlang (2)

• How does the ErlangErlang unit correspond to ccsccs?

0.027 E 1 ccs hour 1 hour × 60 min hr × 60 sec min

× 60 min hr × 60 sec min

1 E 36 ccs hour 1 hour × 60 min hr × 60 sec min

× 60 min hr × 60 sec min

•generated by that terminal in Erlangs, or

• Average number of circuits in a group busy at any time

• Typical usages:

– residence phone -> 0.02 E

– - .

– interoffice trunk -> 0.70 E


18/78

Traffic Offered, Carried, and Lost Traffic Offered, Carried, and Lost

•• Offered TrafficOffered Traffic (TTOO ) equivalent to Traffic Intensity ( A A)

– Takes into account all attempted calls, whether blocked or not,an uses t eir expecte o ing times

• Also Carried TrafficCarried Traffic (TTCC ) and Lost TrafficLost Traffic (TTLL )

,in other words, 0.1 E per source) and dedicated servicededicated service:

each terminal has an

outgoing trunk

(i.e. terminal:trunk ratio = 1:1)

O = A = 150 x 0.10 E = 15.0 ETC = 150 x 0.10 E = 15.0 E

L =


19/78

Traffic Offered, Carried, and Lost (2) Traffic Offered, Carried, and Lost (2)

• A = TO = TC + TLLost

Intensity Offered

Traffic

Carried

Traffic

Traffic

L = O .

= P(B) x TO = P(B) x A

•• Circuit UtilizationCircuit Utilization () - also called Circuit EfficiencyCircuit Efficiency

– proportion of time a circuit is busy, or

– average proportion of time each circuit in a group is busy

CT


20/78

Grade of Service (gos) Grade of Service (gos)

• In general, the term used for some traffic design objective

•

• In systems where blocked calls are cleared, usually use:

O L C

( )T T + T

P Bgos

•

– in busy hour, range from 0.2% to 5% for local calls, however

– generally no more that 1%

–

• In systems with queuing, gos often defined as the probability ofe ay excee ng a spec c eng o me


21/78

Grade of Service Related Terms Grade of Service Related Terms

•• Busy HourBusy Hour

– One hour period during which traffic volume or call attempts is theig est overa uring any given time perio

•• Peak (or Daily) Busy HourPeak (or Daily) Busy Hour

– Bus hour for each da usuall varies from da to da

•• Busy SeasonBusy Season

– 3 months (not consecutive) with highest average daily busy hour•• High Day Busy Hour (HDBH)High Day Busy Hour (HDBH)

– One hour period during busy season with the highest load


22/78


23/78

Hourly Traffic Variations Hourly Traffic Variations


24/78

Daily Traffic Variations Daily Traffic Variations


25/78

Seasonal Traffic Variations Seasonal Traffic Variations


26/78

Seasonal Traffic Variations (2) Seasonal Traffic Variations (2)


27/78

Typical Call Attempts Breakdown Typical Call Attempts Breakdown

• Calls Completed - 70.7%

• -

• Called Party Busy - 10.1%

• Call Abandoned - 2.6%

• Dialing Error - 1.6%

• Number Changed or Disconnected - 0.4%

• Blockage or Failure - 1.9%


28/78

3 Types of Blocking Models 3 Types of Blocking Models

• Blocked Calls Cleared (BCCBCC)

– Blocked calls leave system and do not return

– Good approximation for calls in 1st choice trunk group

• Blocked Calls Held (BCHBCH)

–have normally stayed for

– If a server frees up, the call picks up in the middle and continues

– Not a ood model of real world behaviour mathematicalapproximation only)

– Tries to approximate call reattempt efforts

– Blocked calls enter a queue until a server is available

– When a server becomes available, the call’s holding time begins


29/78

Blocked Calls Cleared (BCC) Blocked Calls Cleared (BCC)

10 minutes

2 sources

Source #1

Offered Traffic 1 3

Total Traffic Offered:

Source #2

Offered Traffic 2 4

TO = 0.4 E + 0.3 E

TO = 0.7 E

Only one server 1st call arrives and is served

2nd call arrives but

Carried 1

22nd call is cleared

1

3rd call arrives and is served

3 4

Total Traffic Carried:4th call arrives and is served

TC = 0.5 E


30/78

Blocked Calls Held (BCH) Blocked Calls Held (BCH)

10 minutes

2 sources

Source #1

Offered Traffic 1 3


Source #2

Offered Traffic 2 4

TO = 0.4 E + 0.3 E

TO = 0.7 E

Only one server1st call arrives and is served

2nd call arrives but server busy

Carried 1 21 2 3 4

2nd call is served

3rd call arrives and is servedTotal Traffic Carried:

2nd call is held until server free

4th call arrives and is served

TC = 0.6 E


31/78

Blocked Calls Wait (BCW) Blocked Calls Wait (BCW)

10 minutes

2 sources

Source #1

Offered Traffic 1 3


Source #2

Offered Traffic 2 4

TO = 0.4 E + 0.3 E

TO = 0.7 E

Only one server

ca arr ves an s serve

2nd call arrives but server busy

2nd call waits until server free

Carried 1 2 2nd call served1 2

3rd call arrives, waits, and

is served

3 4

Total Traffic Carried:4 call arrives, waits, and

is served

TC = 0.7 E


32/78

Blocking Probabilities Blocking Probabilities

• System must be in a Steady StateSteady State

– Also called state of statistical equilibrium

– – Arrival Rate Arrival Rate of new calls equals Departure RateDeparture Rate of

disconnecting calls – Why?

• ca s arr ve aster t an t ey epart

• If calls depart faster than they arrive?


33/78

Binomial Distribution Model Binomial Distribution Model

• Assumptions:

– – mm sources

– – A A Erlangs of offered traffic

• per source: TO = A/m• probability that a specific source is busy: P(B) = A/m

• Can use Binomial Distribution to give the probability that acertain number (k k ) of those m sources is busy:

k mk A Am

k P

1)(k mk

A Am

1

!


34/78

Binomial Distribution Model (2) Binomial Distribution Model (2)

• What does it mean if we only have N serversN servers (N


35/78

Binomial Distribution Model (3) Binomial Distribution Model (3)

• What does it mean if k>N?

– Impossible to have more sources busy than servers to serve them

– Doesn’t accurately represent reality

• In reality, P(k>N) = 0

– In this model, we still assign P(k>N) = A/m

– Acts as good model of real behaviour

• Some people call back, some don’t

• Which type of blocking model is the Binomial Distribution? – Blocked Calls Held (BCH)


36/78

Time CongestionsTime Congestions vs.vs. Call Congestion Call Congestion

• Time Congestion

– Proportion of time a system is congested (all servers busy)

– Probability of blocking from point of view of servers

• Call Congestion

–

– Probability of blocking from point of view of calls

• Why/How are they different?

Time Congestion:

)()( N k P BP

Call Congestion:

)()( N k P BP

Probability that all

servers are busy.

Probability that there are

more sources wanting service

an ere are servers.


37/78

Poisson Traffic Model Poisson Traffic Model

• Poisson approximates Binomial with large mlarge m and small A/msmall A/m

k

!)(

k

ek P = Mean # of

Busy Sources

Note: )(lim BinomialPoissonm

– Mean number of busy sources

– = A

!)(

k Aek P

k A


38/78

Poisson Traffic Model (2) Poisson Traffic Model (2)

• Now we can calculate probability of blocking:

)()( N k P BP )(...)1()( P N P N P

Remember:

k A Ae

k A

!)(

k

ek P

N k k !

N k

ek !

k A

k

ek

0 !1

Example:

, “P” = Poisson

“A” = Offered Traffic

,

PoissonPoisson P(B) with 10 E10 E

“N” = # Servers


39/78

Traffic Tables Traffic Tables

• Consider a 1% chance of blocking in a system with N=10 trunks

– How much offered traffic can the system handle?

A

k

k

k

A

k

ek

Aek

A

9

010 !1

!01.0

• How do we calculate A?

– Very carefully, or – Use tra ic ta es


40/78

Traffic Tables (2) Traffic Tables (2) P(B)=P(N,A)P(B)=P(N,A)

NN

A A


41/78

Traffic Tables (3) Traffic Tables (3) P(N,A)=0.01P(N,A)=0.01

N=10N=10

A=4.14 E A=4.14 E

If system with N = 10 trunksIf system with N = 10 trunks

has P(B) = 0.01:has P(B) = 0.01:

Offered traffic (A) = 4.14 EOffered traffic (A) = 4.14 E


42/78

Poisson Traffic Tables Poisson Traffic Tables P(N,A)=0.01P(N,A)=0.01

N=10N=10

A=4.14 E A=4.14 E

If system with N = 10 trunksIf system with N = 10 trunks

has P(B) = 0.01:has P(B) = 0.01:

Offered traffic (A) = 4.14 EOffered traffic (A) = 4.14 E


43/78

Efficiency of Large Groups Efficiency of Large Groups

• What if there are N = 100 trunks?

– Will they serve A = 10 x 4.14 E = 41.4 E with same P(B) = 1%?

– No!

– Traffic tables will show that A = 78.2 E!

•

– Called efficiency of large groupsefficiency of large groups:

14.4 A , . .10 N

For N = 100, A = 78.2 E efficiency %2.782.78

A

The larger the trunk group, the greater the efficiency


44/78

TrafCalc Software TrafCalc Software

• What if we need to calculate P(N,A) and not in traffic table?

– – TrafCalcTrafCalc: Custom-designed software

• Calculates P(B) or A, or

• Creates custom traffic tables


45/78

TrafCalc Software (2) TrafCalc Software (2)

• How do we calculate P(32,20)?


46/78

TrafCalc Software (3) TrafCalc Software (3)

• How do we calculate A for which P(32,A) = 0.01?


47/78

Erlang B Model Erlang B Model

• More sophisticated model than Binomial or Poisson

•

• Good for calls that can reroute to alternate route if blocked• No approximation for reattempts if alternate route blocked too

• Derived using birthbirth--death processdeath process

– See selected pages from Leonard Kleinrock, Queueing Systems

, ,


48/78

Erlang B Birth Erlang B Birth- -Death Process Death Process

• Consider infinitesimally small time tt during which only onearrival or departure (or none) may occur

• Let be the arrival rate from an infinite pool or sources

• Let = 1/h= 1/h be the departure rate per call

– o e: ca s n sys em, epar ure ra e s

• Steady State Diagram:

Blockage

0 1 2 N-1 N……

2 N(N-1)3

Immediate Service


49/78

Erlang B Birth Erlang B Birth- -Death Process (2) Death Process (2)

• Steady State (statistical equilibrium)

– Rate of arrival is the same as rate of departure

– Average rate a system enters a given state is equal to the average

rate at which the system leaves that stateProbability of moving

from state 1 to state 2? PP11

0 1 2 N-1 N……P0 P1 P2 PN-1 PN

2 N(N-1)3

from state 2 to state 1? 22PP22


50/78


0 1 2 N-1 N……

P0 P1 P2 PN-1 PN

• Set up balance equations: 2 N(N-1)3

P P P P 1 0P P

1 1 2 02P P P P

2 2 3 12 3P P P P

1 22P P

2 33P P 2 1

2P P

2

0

2

P

3 3 4 23 4P P P P P k P

3 23

P P

3

0

6

P

1 1 2( 1) N N N N N P P N P P

0

!

k

k

PP

k

1 N N 1 N N


51/78


Rule of Total Probability:

i 10

!

k

k

PP

k

Recall:

0

1i

i

P

00

!i

i

0

0

1

!

i N

i i

1

k

A h

Recall:

k

!

1

!

k i N

k P

i

0

!

!

i N k

i

k P A

i

For blocking, must be in state k = N:

N A

( ) ( , ) N

P B B N A P

“B” = Erlang B

” 0 !

i N

i

A

i

“A” = Offered Traffic


52/78

Erlang B Traffic Table Erlang B Traffic Table

Example: In a BCC system withm= sources, we can accept a

B(N,A)=0.001B(N,A)=0.001

0.1% chance of blocking in the

nominal case of 40E offered traffic.However, in the extreme case of aB(N,A)=0.005B(N,A)=0.005

,0.5% chance of blocking.

How many outgoing trunks do we

==N=59N=59

Nominal design: 59 trunks

A A48 E48 EOverload design: 64 trunks

N=64N=64

Requirement: 64 trunks

l (2)l (2)


53/78

Example (2) Example (2) P(N,A)=0.01P(N,A)=0.01

N=32N=32

A=20.3 E A=20.3 E

P(N A) & B(N A)P(N A) & B(N A) Hi h Bl kiHi h Bl ki


54/78

P(N,A) & B(N,A)P(N,A) & B(N,A) -- High Blocking High Blocking

• We recognize that Poisson and Erlang B models are onlyapproximations but which is better?

– Compare them using a 4-trunk group offered A=10E

Erlang BErlang B PoissonPoisson

, .

(1 ( ))C T A P B 10 (1 0.64666)

, .

(1 ( ))C T A P B 10 (1 0.98966)

.C

3.5330.88

.C

0.1030.026

How can 4 trunks handle 10E offeredHow can 4 trunks handle 10E offered

traffic and be busy only 2.6% of the time?traffic and be busy only 2.6% of the time?

P(N A) & B(N A)P(N A) & B(N A) Hi h Bl ki (2)Hi h Bl ki (2)


55/78

P(N,A) & B(N,A)P(N,A) & B(N,A) -- High Blocking (2) High Blocking (2)

• Obviously, the Poisson result is so far off that it is almostmeaningless as an approximation of the example.

– 4 servers offered enough traffic to keep 10 servers busy full time

(10E) should result in much higher utilization.• Erlang B result is more believable.

– All 4 trunks are busy most of the time.

• What if we extend the exercise by increasing A?

–

– Poisson result goes to 0E carried

• Illustrates the failure of the Poisson model as valid for situationswith high blocking

– Poisson only good approximation when low blocking

– Use Erlang B if high blocking

E tE t Di t ib ti M d lDi t ib ti M d l


56/78

Engset Engset Distribution Model Distribution Model

• BCC model with small number of sources (m > N)

=

= mean arrival rate of a single source = arrival rate if in the system is state k

k k == (m(m--k)k)

m (m-1) (m-2) [M-(N-2)] [m-(N-1)]

0 1 2 N-1 N……P0 P1 P2 PN-1 PN

2 N(N-1)3

Immediate Service

EngsetEngset Distribution ModelDistribution Model


57/78

Engset Engset Distribution Model Distribution Model

• Engset's equation is similar to the Erlang-B formula; however itcontains one major difference: Erlang's equation assumes ann n e source o ca s, y e ng a o sson arr va process, w e

Engset specifies a finite number of callers. Thus Engset'sequation should be used when the source population is small, .

• In practice, like Erlang's equations, Engset's formula requires

recursion to solve for the blocking or congestion probability.here are several recursions that could be used. One way todetermine this probability, one first determines an initialestimate. This initial estimate is substituted into the equation

.

calculation is then substituted back into the equation, resultingin a new answer which is again substituted. This iterative

answer.

Engset Traffic Model (2)Engset Traffic Model (2)


58/78

Engset Traffic Model (2) Engset Traffic Model (2)

• Balance equations give:

0

!

!( )!k

mP P

k m k

and 0

0

i N

i

P

mi

therefore:k

k i N

m

k P

but can show that: A

m A

0i i

N m A

( ) ( )P B P k N ( , , ) i N m A N E m N A m A

0i m “E” = Engset

Engset Traffic TableEngset Traffic Table


59/78

Engset Traffic Table Engset Traffic Table M = 30 sourcesM = 30 sources# trunks (N)# trunks (N)

Traffic offered (A)Traffic offered (A)

P(B)=E(m,N,A)P(B)=E(m,N,A) N=10N=10

A A==4.8 E4.8 E

0.16 Erlangs to a concentrator witha goal of less than 1% blocking.

How man out oin trunks do we

< .< .

need? A = 30 x 0.16 = 4.8 E

==

Check m < 10 x N?

Erlang C Distribution ModelErlang C Distribution Model


60/78

Erlang C Distribution Model Erlang C Distribution Model

• BCW model with infinite sourcesinfinite sources (m) and infinite queue lengthinfinite queue length

=

= mean departure rate per call Blockage

0 1 2 N Q1 Q2…… ……P0 P1 P2 PN PQ1 PQ2

2 NNNN3

Immediate Service

Erlang C Distribution Model (2)Erlang C Distribution Model (2)


61/78

Erlang C Distribution Model (2) Erlang C Distribution Model (2)

• Balance equations give:

0 ,

!

k

k

A PP k N

k

and 0 ,!

k

k k N

A PP k N

N N and

0 1

0! !

N i N

i

P A N A

N N A i

• But P(B) = P(k N):

0

k A P

P B

0k

P A

k N

A A

but can show that:k

A N

!k N N N

!

N k N N N N

0! k N N 0k

N N A

N

N A N

0( )!

P B P

N N A

1!( , ) N i N N N AC N A N A C” = Erlang C

Erlang C Traffic TablesErlang C Traffic Tables


62/78

Erlang C Traffic Tables Erlang C Traffic Tables N=18N=18

# trunks (N)# trunks (N) P(B)=C(N,A)P(B)=C(N,A)

Traffic offered ATraffic offered A

Example:

== C(18,7)=0.0004C(18,7)=0.0004

W at is t e pro a i ity o oc ing

in an Erlang C system with 18servers offered 7 Erlangs of traffic?

Delay in Erlang CDelay in Erlang C


63/78

Delay in Erlang C Delay in Erlang C

• Expected number of calls in the queue?

k A k N

A A

k

k N

0!k N k N N N

0!

k

k N N

0

N P A A N

( , ) A C N A

h

! N N A N A N A ,

N A

Mean #Calls Delayed h T

Recall:

ean e ay over a s =Arrival Rate of Calls

, N A

T

= h

N A

Also: hT

,

Comparison of Traffic ModelsComparison of Traffic Models


64/78

Comparison of Traffic Models Comparison of Traffic Models

Erlan C BCWErlan C BCW sourcessources

Poisson (BCH,Poisson (BCH, sources)sources)

Erlang B (BCC,Erlang B (BCC, sources)sources)

P(B)

B nom a BCH, m sourcesB nom a BCH, m sources

Engset (BCC, m sources)Engset (BCC, m sources)

Offered Traffic (A)

Efficiency of Large GroupsEfficiency of Large Groups


65/78

Efficiency of Large Groups Efficiency of Large Groups

• Already seen that for same P(B), increasing servers results inmore than proportional increase in traffic carried

example 1: (10, 4.14) 0.01P and (100,78.2) 0.01P

examp e : (32, 20.3) 0.01P (33, 20.1) 0.005P and

example 3: (8, 2.05) 0.001 B (80,57.8) 0.001 B and

• What does this mean?

– If it’s possible to collect together several diverse sources, you can

• provide better gos at same cost, or

• provide same gos at cheaper cost

Efficiency of Large Groups (2)Efficiency of Large Groups (2)


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Efficiency of Large Groups (2) Efficiency of Large Groups (2)

• Two trunk groups offered 5 Erlangs each, and B(N,A)=0.002

N1=?5 EHow many trunks total?

From traffic tables, find B(13,5) 0.002

N1=13

N2=?5 E N2=13Ntotal = 13 + 13 = 26 trunks

Trunk efficiency?

T 10 1 0.002

N

.26

38.4% utilization38.4% utilization



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• One trunk group offered 10 Erlangs, and B(N,A)=0.002

N=?10 E

From traffic tables, find B(20,10) 0.002N=20

N = 20 trun s

Trunk efficiency?

C T

N

10(1 0.002)0.499

20

49.9% utilization49.9% utilization

For same gos, we can save 6 trunks!



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B=0.1B=0.1

B=0.01B=0.01

A B=0.1B=0.1B=0.01B=0.01

B=0.001B=0.001 B=0.001B=0.001

N N

Sensitivity to Overload Sensitivity to Overload


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yy

• Consider 2 cases:

= =

B(10,4.5) 0.01, so can carry 4.5 E

What if 20% overload (5.4 E)? B(10,5.4) 0.03

3 times P(B) with 20% overload

Case 1: N = 30 and B(N,A) = 0.01

B(30,20.3) 0.01, so can carry 20.3 E

What if 20% overload (24.5 E)? B(30,24.5) 0.08

8 times P(B) with 20% overload!

“Trunk Group Splintering” “Trunk Group Splintering”

• if high possibility of overloads, small groups may be better

Incremental Traffic Carried by N Incremental Traffic Carried by N th th Trunk Trunk


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yy

• If a trunk group is of size N-1, how much extra traffic can itcarry if you add one extra trunk?

– Before, can carry: TC1 = A x [1-(B(N-1,A)]

– After, can carry: TC2 = A x [1-(B(N,A)]

2 1 N C C T T 1 ( , ) 1 ( 1, ) B N A B N A

( 1, ) ( , ) A B N A B N A

• What does this mean?

– – Random HuntingRandom Hunting: Increase in trunk group’s total carried traffic

, N for very low blocking

after adding an Nth trunk

– – Sequential HuntingSequential Hunting: Actual traffic carried by the Nth trunk in thegroup

Incremental Traffic Carried by N Incremental Traffic Carried by N th th Trunk (3) Trunk (3)


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A N

,

N

Example Example


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pp

• Individual trunks are only economic if they can carry 0.4 E ormore. A trunk group of size N=10 is offered 6 E. Will all 10run s e econom ca

( 1, ) ( , ) N A B N A B N A

10 6 (9,6) (10,6) A B B

6 0.07514 0.04314

0.192 E 0.4 E

At least the 10At least the 10thth trunk is nottrunk is not economicaleconomical

Binomial Distribution Binomial Distribution


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• In probability theory and statistics, the binomial distribution is the discrete

probability distribution of the number of successes in a sequence of n independent

es/no ex eriments each of which ields success with robabilit . Such a

success/failure experiment is also called a Bernoulli experiment or Bernoulli trial.

In fact, when n = 1, the binomial distribution is a Bernoulli distribution. The binomial distribution is the basis for the popular binomial test of statistical

significance. A binomial distribution should not be confused with a bimodal

distribution.

• An elementary example is this: Roll a standard die ten times and count the number

of sixes. The distribution of this random number is a binomial distribution with n

= 10 and p = 1/6.

• As another example, assume 5% of a very large population to be green-eyed. You

ick 100 eo le randoml . The number of reen-e ed eo le ou ick is a

random variable X which follows a binomial distribution with n = 100 and p =

0.05.

Probability mass function Probability mass function


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• In general, if the random variable K follows the binomialdistribution with parameters n and p , we write K ~ B(n , p ). Thepro a y o ge ng exac y successes n n r a s s g ven ythe probability mass function:

• for k = 0, 1, 2, ..., n and where

• is the binomial coefficient (hence the name of the distribution)" " , .

understood as follows: we want k successes ( p k ) and n − k failures (1 − p )n − k . However, the k successes can occur

, ,ways of distributing k successes in a sequence of n trials.

Poisson approximation Poisson approximation


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• The binomial distribution converges towards the Poissondistribution as the number of trials goes to infinity while thepro uc np rema ns xe .

• Therefore the Poisson distribution with parameter λ = np can beused as an approximation to B(n , p ) of the binomial distributionif n is sufficiently large and p is sufficiently small.

• According to two rules of thumb, this approximation is good if n

. ,

Poisson distribution Poisson distribution


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• In probability theory and statistics, the Poisson distribution isa discrete probability distribution that expresses the probabilityo a num er o even s occurr ng n a xe per o o methese events occur with a known average rate andindependently of the time since the last event.

• The Poisson distribution can also be used for the number of events in other specified intervals such as distance, area orvolume.



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• The distribution focuses on certain random variables N thatcount, among other things, a number of discrete occurrencessome mes ca e arr va s a a e p ace ur ng a me-

interval of given length. If the expected number of occurrencesin this interval is λ, then the probability that there are exactly k

- , = , , , ...equal to



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where

• =

• k is the number of occurrences of an event - the probability ofwhich is given by the function

• k ! is the factorial of k

• λ is a positive real number, equal to the expected number of

. ,the events occur on average 4 times per minute, and you areinterested in the number of events occurring in a 10 minuteinterval ou would use as model a Poisson distribution withλ = 10*4 = 40.

• As a function of k , this is the probability mass function. The

binomial distribution.

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