1. 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

2. 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

3. 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

4. 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

5. 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

6. 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

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

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

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

10. 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

11. Coxian model μ1 μ2 μ3 μ4 a1 a2 a3 b1 b2 b3 μ1 b1 a1 b2 μ1 μ2 a1 a2 b3

12. 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

13. Coxian model: conclusion
Most Laplace transforms
Are rational polynomials
=> Any distribution can be represented
Exactly or approximately
By a Coxian distribution

14. 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 c
To obtain the unknowns (a, μ1, μ2)
Calculate the first 3 moments based on Laplace transform
And match these against those of the C2 a μ1 μ2
b=1-a

15. C2 : first three moments

16. Rational polynomials approximated by Coxian-2 (C2)
The Laplace transforms of most pdfs
Have the following shape
Can be exactly represented by
A Coxian k (Ck) distribution
The # stages = the order of the denominator
Service rates given by the roots of the denominator
If k is very large => dimensionality problem
Solution: collapse the Ck into a C2 a μ1 μ2
b=1-a

17. 3 moments method The first way of obtaining the 3 unknowns (μ1 μ2 a)
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

18. 3 moments method: solution
The following expressions will be obtained
However,
The following condition has to hold: X2 – 4Y >= 0
=> < 2m1m3
=> the 3 moments method applies to the case where
c2 > 1 for the original distribution
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.

19. Two-moment fit 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

20. M/C2/1 queue What is the state of the system?μ μ C2 a b
What 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 needed
n1 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

21. M/C2/1 queue: analysis To analyze this queue, one must go thru
Rate diagram that depicts state transition
Steady state equations
Derive the balance equations
Based on these equations
Obtain the generating functions
Using a recursive scheme
Solve the M/C2/1 queue and
Determine P(n) = Pn = prob of having n customers in system

22. M/C2/1: state diagram (n1 , n2)
Where n1 is the # customers in the queue
Excluding the one in service
n2 is the state of the server (0:idle, 1:phase1, 2:phase2)
0,0 μ2 λ
bμ1
aμ1
2nd column: states of the system
where server busy in phase 2
1st column:
States where
system in
phase 1
0,1
0,2
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).
1,1
1,2
2,1
2,2 . .

23. M/C2/1: steady state equations
1st set of equations (1st column)

24. Balance equations 2nd set of steady state equations (2nd column)
We are interested in finding Pn
Prob of having n customers
in the system
that can be obtained based on Pn1,n2

25. Generating functions Let us define
generating function g1 (z) involving all probabilities Pi1
Where 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 Pn
The probability of having n customers in the system

26. Expressions for the generating functions
Using the 2 set of balance equations, we get
(1), (2), and (3) =>
(1)
(2)
(3)

27. Finding P00 g1 (1) = P01 + P11 + P21 +…=prob stage 1 is busy
This is equal to traffic intensity for stage 1 => ρ1
g (z) = f(g1 (z), g2 (z))

28. Pn : recursive scheme
The general expression of probability Pn