Squared coefficient of variation
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Transcript of Squared coefficient of variation
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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)
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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 μ E4
Service mechanism
H3
μ1
μ2
μ3
P1
P2
P3
μ μ μ μC4
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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
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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
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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
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yrr
Y
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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
))...()((
...
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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
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Hyper-exponential distribution
P1 + P2 + P3 +…+ Pk =1
Pdf of X?
μ1
μ2
P1
P2
Pk μk
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Hyper-exponential distribution:1st and 2nd moments
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Example: H2
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Hyper-exponential: squared coefficient of variation
C2 = Var[X]/E[X]2
C2 is greater than 1
Example: H2 , C2 > 1 ?
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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
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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
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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
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l l
lii sbaaaaasf
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* ....)1()(
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Coxian model: conclusion
Most Laplace transforms Are rational functions
=> Any distribution can be represented Exactly or approximately
By a Coxian distribution
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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
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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
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First three moments of Coxian distribution
By using
We have
Squared Coefficient of variation
=> For a Coxian k distribution =>
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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
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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
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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
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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
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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
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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
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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
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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: