Erlang, Hyper-exponential, and Coxian distributions

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Erlang, Hyper-exponential, and Coxian distributions. Mixture of exponentials Combines a different # of exponential distributions Erlang Hyper-exponential Coxian. μ. μ. μ. E 4. μ. Service mechanism. μ 1. P 1. μ 2. H 3. P 2. P 3. μ 3. μ. μ. μ. μ. C 4. - PowerPoint PPT Presentation

Transcript of Erlang, Hyper-exponential, and Coxian distributions

  • *Erlang, Hyper-exponential, and Coxian distributionsMixture of exponentialsCombines a different # of exponential distributionsErlang


    CoxianE4Service mechanismH3123P1P2P3C4

  • *Erlang distribution: analysisMean service timeE[Y] = E[X1] + E[X2] =1/2 + 1/2 = 1/

    VarianceVar[Y] = Var[X1] + Var[X2] = 1/42 v + 1/42 = 1/221/21/2E2

  • *Squared coefficient of variation: analysisX is a constantX = d => E[X] = d, Var[X] = 0 => C2 =0

    X is an exponential r.v.E[X]=1/; Var[X] = 1/2 => C2 = 1

    X has an Erlang r distributionE[X] = 1/, Var[X] = 1/r2 => C2 = 1/r

    fX *(s) = [r/(s+r)]r C20constant1exponentialHypo-exponentialErlang

  • *Probability density function of Erlang rLet Y have an Erlang r distribution

    r = 1 Y is an exponential random variable

    r is very large The larger the r => the closer the C2 to 0

    Er tends to infintiy => Y behaves like a constant

    E5 is a good enough approximation

  • *Generalized Erlang ErClassical Erlang rE[Y] = r/

    Var[Y] = r/2

    Generalized Erlang rPhases dont have same


  • *Generalized Erlang Er: analysisIf the Laplace transform of a r.v. YHas this particular structure

    Y can be exactly represented by An Erlang Er

    Where the service rates of the r phase Are minus the root of the polynomials

  • *Hyper-exponential distributionP1 + P2 + P3 ++ Pk =1

    Pdf of X? 12P1P2Pkk..X

  • *Hyper-exponential distribution:1st and 2nd momentsExample: H2

  • *Hyper-exponential: squared coefficient of variationC2 = Var[X]/E[X]2C2 is greater than 1

    Example: H2 , C2 > 1 ?

  • *Coxian model: main ideaIdeaInstead of forcing the customer to get r exponential distributions in an Er model

    The customer will have the choice to get 1, 2, , r services

    ExampleC2 : when customer completes the first phaseHe will move on to 2nd phase with probability a Or he will depart with probability b (where a+b=1)


  • *Coxian model1234a1b1b2b3a2a311b1a1 b22123a1 a2 b31234a1 a2 a3

  • *Coxian distribution: Laplace transformLaplace transform of CkIs a fraction of 2 polynomialsThe denominator of order k and the other of order < k

    ImplicationA Laplace transform that has this structureCan be represented by a Coxian distributionWhere the order k = # phases, Roots of denominator = service rate at each phase

  • *Coxian model: conclusionMost Laplace transforms Are rational polynomials

    => Any distribution can be represented Exactly or approximately

    By a Coxian distribution

  • *Coxian model: dimensionality problemA Coxian model can grow too bigAnd may have as such a large # of phases

    To cope with such a limitation Any Laplace transform can be approximated by a c

    To obtain the unknowns (a, 1, 2)Calculate the first 3 moments based on Laplace transform

    And match these against those of the C2


  • *C2 : first three moments

  • *Rational polynomials approximated by Coxian-2 (C2)The Laplace transforms of most pdfsHave the following shape

    Can be exactly represented by A Coxian k (Ck) distribution The # stages = the order of the denominatorService rates given by the roots of the denominator

    If k is very large => dimensionality problemSolution: collapse the Ck into a C2


  • *3 moments methodThe first way of obtaining the 3 unknowns (1 2 a)3 moments methodLet m1, m2, m3 be the first three moments of the distribution which we want to approximate by a C2

    The first 3 moments of a C2 are given by

    by equating m1=E(X), m2=E(X2), and m3 = E(X3), you get

  • *3 moments method: solutionThe following expressions will be obtained

    However,The following condition has to hold: X2 4Y >= 0=> 3 < 2m1m3

    => the 3 moments method applies to the case where c2 > 1 for the original distribution

  • *Two-moment fitIf the previous condition does not holdYou can use the following two-moment fit

    General ruleUse the 3 moments method

    If it doesnt do => use the 2 moment fit approximation

  • *M/C2/1 queueWhat is the state of the system?How many variables are needed to Describe the state of this queue?

    A 2-dimensional state (n1 ,n2) is neededn1 represents the number of customers in the queue

    n2 is the state of the server 0 => server is idle; 1 => the server is busy in phase 1

    2 => server is busy in phase 2


  • *M/C2/1 queue: analysisTo analyze this queue, one must go thru Rate diagram that depicts state transition

    Steady state equationsDerive the balance equations

    Based on these equationsObtain the generating functions

    Using a recursive schemeSolve the M/C2/1 queue and

    Determine P(n) = Pn = prob of having n customers in system

  • *M/C2/1: state diagram(n1 , n2)Where n1 is the # customers in the queueExcluding the one in service

    n2 is the state of the server (0:idle, 1:phase1, 2:phase2)0,00,11,12,1...0,21,22,2...

  • *M/C2/1: steady state equationsSteady state equations

    1st set of equations (1st column)

  • *Balance equations2nd set of steady state equations (2nd column)

    We are interested in finding PnProb of having n customers in the system

    that can be obtained based on Pn1,n2

  • *Generating functionsLet us define generating function g1 (z) involving all probabilities Pi1Where i customers are in the queue and 1st phase is busy

    generating function g2 (z) based on {P02, P12, P22,}Where the server is busy in phase 2

    generating function g(z) based on PnThe probability of having n customers in the system

  • *Expressions for the generating functionsUsing the 2 set of balance equations, we get

    (1), (2), and (3) =>(1)(2)(3)

  • *Finding P00g1 (1) = P01 + P11 + P21 +=prob stage 1 is busyThis is equal to traffic intensity for stage 1 => 1

    g (z) = f(g1 (z), g2 (z))

  • *Pn : recursive schemeThe general expression of probability Pn

    *The reason people was studying these distributions was to introduce more realistic representation of real life service times. *The following condition has to hold: 3 < 2m1m3 . Note that the above method of moment applies to the case where the c2 of the original distribution (which we approximate by a C2) is greater than 1.*We will start by construct the flow balance equations. In order to do this, first we need to draw the state diagram that represents all possible transitions from one state to another. Starting at (0,0), the system goes to (0,1) because the arriving customer goes directly into service. Other arrivals will lead to queuing up customers. What happens when you have a service completion? From (0,1) with some probability the system proceeds to (0,0) and with another probability it goes to (0,0). The state diagram consists of 2 columns. The first column involves states of the system where the server is busy in phase 1. The other column includes all the states of the system where the server is busy in phase 2. In order to obtain the expressions for these probabilities, first we need to come up with their corresponding moment generating functions g1(z) and g2(z).