1

Squared coefficient of variation

The squared coefficient of variation Gives you an insight to the dynamics of a r.v. X

Tells you how bursty your source is C2 get larger as the traffic becomes more bursty

For voice traffic for example, C2 =18

Poisson arrivals C2 =1 (not bursty)

22

][

][

XE

XVarC

1/1

/1

][

][2

2

22

XE

XVarC

2

Erlang, Hyper-exponential, and Coxian distributions

Mixture of exponentials Combines a different # of exponential distributions

Erlang

Hyper-exponential

Coxian

μ μ μ μ E4

Service mechanism

H3

μ1

μ2

μ3

P1

P2

P3

μ μ μ μC4

3

Erlang distribution: analysis

Mean service time E[Y] = E[X1] + E[X2] =1/2μ + 1/2μ = 1/μ

Variance Var[Y] = Var[X1] + Var[X2] = 1/4μ2 v + 1/4μ2 = 1/2μ2

1/2μ 1/2μ

E2

2

*

**

22

21

2

2)(

.2

.2)()(

..2)(;..2)(

21

2

2

1

1

ssf

ssfsf

exfexf

Y

XX

xX

xX

4

Squared coefficient of variation: analysis

X is a constant X = d => E[X] = d, Var[X] = 0 => C2 =0

X is an exponential r.v. E[X]=1/μ; Var[X] = 1/μ2 => C2 = 1

X has an Erlang r distribution E[X] = 1/μ, Var[X] = 1/rμ2 => C2 = 1/r

fX *(s) = [rμ/(s+rμ)]r

C2

0

constant

1

exponential

Hypo-exponential

Erlang

5

Probability density function of Erlang r

Let 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

)!1(

.)...()(

..1

r

eyrryf

yrr

Y

6

Generalized Erlang Er

Classical Erlang r E[Y] = r/μ

Var[Y] = r/μ2

Generalized Erlang r Phases don’t have same μ

rμ rμ … rμ

Y

μ1 μ2… μr

Y

))...()((

...

.....)(

21

21

2

2

1

1*

r

r

r

rY

sss

ssssf

7

Generalized Erlang Er: analysis

If the Laplace transform of a r.v. Y Has 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

))...()((

...)(

21

21*

r

rY ssssf

8

Hyper-exponential distribution

P1 + P2 + P3 +…+ Pk =1

Pdf of X?

μ1

μ2

P1

P2

Pk μk

.

.

X

k

i

xii

XkXX

i

k

eP

xfPxfPxf

1

.

1

..

)(....)(.)(1

9

Hyper-exponential distribution:1st and 2nd moments

22

12

2

1

1

*

][][][

2.][

][

.)(

XEXEXVar

PXE

PXE

sPsf

k

i i

i

k

i i

i

k

i i

iiX

Example: H2

2

2

2

1

122

221

1

22

221

12

2

2

1

1

2.

2.][

2.

2.][

][

PPPPXVar

PPXE

PPXE

10

Hyper-exponential: squared coefficient of variation

C2 = Var[X]/E[X]2

C2 is greater than 1

Example: H2 , C2 > 1 ?

0)11

(211

0.

..2)1()1(

0.

..2

11)//(

/2/2

2

212122

21

21

2122

2221

11

21

2121

22

21

21

22

221

1

22211

222

2112

PPPPPP

PPPPPP

PP

PPC

11

Coxian model: main idea

Idea Instead 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

Example C2 : when customer completes the first phase

He will move on to 2nd phase with probability a Or he will depart with probability b (where a+b=1)

a

b

12

Coxian model

μ1 μ2 μ3 μ4a1

b1 b2 b3

a2 a3

μ1

μ1

b1

a1 b2 μ2

μ1 μ2 μ3

a1 a2 b3

μ1 μ2 μ3 μ4

a1 a2 a3

13

Coxian distribution: Laplace transform

Laplace transform of Ck

Is a fraction of 2 polynomials The denominator of order k and the other of order < k

Implication A Laplace transform that has this structure

Can be represented by a Coxian distribution Where the order k = # phases, Roots of denominator = service rate at each phase

korderPolynomial

korderPolynomial

_

_

k

i

i

l l

lii sbaaaaasf

1 112100

* ....)1()(

14

Coxian model: conclusion

Most Laplace transforms Are rational functions

=> Any distribution can be represented Exactly or approximately

By a Coxian distribution

15

Coxian model: dimensionality problem

A Coxian model can grow too big And may have as such a large # of phases

To cope with such a limitation Any Laplace transform can be approximated by a Coxian 2

The unknowns (a, μ1, μ2) can be obtained by Calculating the first 3 moments based on Laplace transform

And then matching these against those of the C2

a

b=1-a

μ1 μ2

16

Some Properties of Coxian distribution

Let X be a random variable Following the Coxian distribution

With parameters µ1,µ2, and a

PDF of this distribution

Laplace transform

17

First three moments of Coxian distribution

By using

We have

Squared Coefficient of variation

=> For a Coxian k distribution =>

18

Approximating a general distribution

Case I: c2 > 1 Approximation by a Coxian 2 distribution

Method of moments Maximum likelihood estimation Minimum distance estimation

Case II: c2 < 1 Approximation by a generalized Erlang distribution

19

Method of moments The first way of obtaining the 3 unknowns (μ1 μ2 a)

3 moments method Let 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

32

31

3212121

31

3

22

21

22121

21

2

21

)(6)(126][

)(22.2][

1][

aabXE

aabXE

aXE

20

3 moments method The following expressions will be obtained

Let X=µ1+µ2 and Y=µ1µ2, solving for X and Y and

=> and

However, The following condition

has to hold: X2 – 4Y >= 0 =>

Otherwise, resort to a two moment approximation

21

Two-moment approximation

If the previous condition does not hold You can use the following two-moment fit

General rule Use the 3 moments method

If it doesn’t do => use the 2 moment fit approximation

21

21

1

2

.

1,

2.2

1

cmm

ca

22

Maximum likelihood estimation A sample of observations

from arbitrary distribution is needed Let sample of size N be x1, x2, …, xN

Likelihood of sample is:

Where is the pdf of fitted Coxian distribution

Product to summation transformation

Maximum likelihood estimate: Maximize L subject to

and

23

Minimum distance estimation

Objective Minimize distance

Between fitted distribution and observed one

=> a sample observation of size N x1, x2, …, xN

is needed

Computing formula for the distance , where

X(i) is the ith order statistic and Fθ(x) is the fitted C2

A solution is obtained by minimizing the distance No closed form solution exists

=> a non-linear optimization algorithm can be used

24

Concluding remarks on case I

If exact moments of arbitrary distribution are known

Then the method of moments should be used

Otherwise, Maximum likelihood or minimum distance estimations

Give better results

25

Case II: c2 < 1

Generalized Erlang k can be used To approximate the arbitrary distribution In this case

Service ends with probability 1-a or Continues through remaining k-1 phases with probability a

The number of stages k should be such that

Once k is fixed, parameters can be obtained as: