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)

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 μ1 P1 μ2 H3
The reason people was studying these distributions was to introduce more realistic representation of real life service times. P2 P3 μ3 μ μ μ μ C4

3. Erlang distribution: analysis
1/2μ
1/2μ E2
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

4. Squared coefficient of variation: analysis
constant
exponential C2
Erlang 1
Hypo-exponential
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

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

6. Generalized Erlang Er Classical Erlang r Generalized 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

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

8. Hyper-exponential distributionμ1 P1 μ2 X P2 . Pk μk
P1 + P2 + P3 +…+ Pk =1
Pdf of X?

9. Hyper-exponential distribution:1st and 2nd moments
Example: H2

10. Hyper-exponential: squared coefficient of variation
C2 = Var[X]/E[X]2
C2 is greater than 1
Example: H2 , C2 > 1 ?

11. Coxian model: main 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 μ4 a1 a2 a3 b1 b2 b3 μ1 b1 a1 b2 μ1 μ2 a1 a2 b3

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

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 μ1 μ2
b=1-a

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
𝑓 𝑋 𝑥 = 1−𝑎 𝜇 1 𝑒 − 𝜇 1 𝑥 +𝑎 𝜇 1 𝜇 2 𝜇 2 − 𝜇 1 𝑒 − 𝜇 1 𝑥 + 𝜇 1 𝜇 2 𝜇 1 − 𝜇 2 𝑒 − 𝜇 2 𝑥
Laplace transform
𝑓 𝑋 ∗ 𝑠 = 1−𝑎 𝜇 1 𝑠+ 𝜇 1 +𝑎 𝜇 1 𝑠+ 𝜇 1 𝜇 2 𝑠+ 𝜇 2 = 𝜇 1 𝑠 1−𝑎 + 𝜇 1 𝜇 2 𝑠 2 + 𝜇 1 + 𝜇 2 𝑠+ 𝜇 1 𝜇 2

17. First three moments of Coxian distribution
By using
𝑑 𝑛 𝑓 𝑋 ∗ 𝑠 𝑑𝑠 𝑠=0 = −1 𝑛 𝐸[ 𝑋 𝑛 ]
We have
𝐸 𝑋 = 1 𝜇 1 + 𝑎 𝜇 2
𝐸 𝑋 2 = 2(1−𝑎) 𝜇 1 2 − 2𝑎 𝜇 1 𝜇 2 −2𝑎 𝜇 1 + 𝜇 𝜇 1 2 𝜇 2 2
𝐸 𝑋 3 = 6(1−𝑎) 𝜇 1 3 − 12𝑎 𝜇 1 𝜇 2 𝜇 1 + 𝜇 2 −6𝑎 𝜇 1 + 𝜇 𝜇 1 3 𝜇 2 3
Squared Coefficient of variation
𝑐 2 =1− 2𝑎 𝜇 1 𝜇 2 − 𝜇 1 (1−𝑎) 𝜇 2 +𝑎 𝜇 => 𝑐 2 ∈[0.5,∞)
For a Coxian k distribution => 𝑐 2 ∈[ 1 𝑘 ,∞)

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

20. 3 moments method The following expressions will be obtained However,
𝑎= 𝜇 2 𝜇 1 𝑚 1 𝜇 1 −1
Let X=µ1+µ2 and Y=µ1µ2, solving for X and Y
𝑌= 6 𝑚 1 −3 𝑚 2 𝑚 𝑚 𝑚 1 − 𝑚 3 and 𝑋= 1 𝑚 𝑚 2 𝑌 2 𝑚 1
=> 𝜇 1 = 𝑋+ 𝑋 2 −4𝑌 and 𝜇 2 =𝑋− 𝜇 1
However,
The following condition
has to hold: X2 – 4Y >= 0
=> 3 𝑚 2 2 <2 𝑚 1 𝑚 3
Otherwise, resort to a two moment approximation
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.

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

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:
𝐿 𝜃 = 𝑖=1 𝑁 𝑓 𝑋 ( 𝑥 𝑖 )
Where 𝑓 𝑋 𝑥 is the pdf of fitted Coxian distribution
Product to summation transformation
𝐿𝑜𝑔 𝐿 𝜃 = 𝑖=1 𝑁 log⁡( 𝑓 𝑋 𝑥 𝑖 )
Maximum likelihood estimate: 𝜃 = 𝜇 1 , 𝜇 2 , 𝑎
Maximize L subject to
𝜇 𝑖 ≥0, 𝑖=1,2 and 0≤𝑎≤1

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
𝑑 2 = 1 12𝑁 + 𝑖=1 𝑁 𝐹 𝜃 𝑥 𝑖 − 𝑖−0.5 𝑁 , 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
1 𝑘 ≤ 𝑐 2 ≤ 1 𝑘−1
Once k is fixed, parameters can be obtained as:
𝑎=1− 2𝑘 𝑐 2 +𝑘−2− 𝑘 2 +4−4𝑘 𝑐 2 1/2 2( 𝑐 2 +1)(𝑘−1)
𝜇= 1+ 𝑘−1 𝑎 𝑚 1