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1
Exponential distribution: main limitation So far, we have considered
Birth and death process and its direct application To single server queues
With the 2 following assumptions Inter-arrival time is exponential
Service time is exponential
Problem with exponential distribution It has the memoryless property (not very good)
Since, the longer you get served,
=> the more likely your service will complete
2
Exponential assumption: example Let us consider
Queue in one of the output ports of a router Is it ok to model this queue as an M/M/1 queue?
A lot of people do, but is it correct?
The implication of such a model Transmission delay of a packet is exponential?
If you are doing file transfer all packets are 1500 bytes long
The latest statistics from SANs (Storage Area Networks)
1500 bytes P1
500 bytes P2
64 bytes P3
3
Inter-arrival time: is it exponential? Back in the 80s
When the link speeds were very small It was OK to consider packet arrival to be Poisson
But as speeds picked up with the advent of SONET Because of the tremendous speed, the utilization of links
No longer regular (spaced out with exponential inter-arrival)
Instead, traffic is bursty (packets are grouped together) You go thru an ON period with back to back packets
Followed by an off period
It is better to use Interrupted Poisson distribution
4
Erlang model: mixture of exponentials Erlang
Realized the exponential distribution
was not an adequate modeling technique
But it is important to keep it as it leads to easy mathematics
Came up with another idea Mixture of exponentials represented by the letter E
Combine several exponentials to make up a service
μ Break it up into r exponentials in tandem
rμ rμ … rμ
5
Erlang model: main concept
The arriving customer Takes a sequence of exponential services
Instead of a single service time
Breaking service time into r service times Is called Er Erlang distribution with r stages
However, the r servers are equivalent to one server Only one customer is accommodated in service at a time
Example: E2
Why is it important? Arbitrary pdf can be represented by the Erlang model
1/2μ 1/2μ
6
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
7
Laplace transform
X is a continuous random variable fX (x) is the probability density function of X
Laplace transform of X
Since,
0
* ][)()( sxX
sxX eEdxxfesf
x X dxxfxgXgEYEXgY )().()]([][)(
8
Main property of Laplace transform Main property
Proof: first and second moment
][)1()(0
* nn
s
X
n
XEsfds
d
][)()0(
)(.)(..)()(
][).(.)0(
)(..))(()()(
22*2
2
2**2
2
*
*
XEdxxfxfds
d
dxxfexdxxfexds
dsf
ds
d
ds
dsf
ds
d
XEdxxfxfds
d
dxxfexdxxfeds
ddxxfe
ds
dsf
ds
d
XX
Xsx
Xsx
XX
XX
Xsx
Xsx
Xsx
X
9
Moment generating functions
Laplace transform is a special case of Moment generating function ø(t)
It is called as such because of all of the moments of X Can be obtained by successively differentiating it
Laplace transform is obtained t = -s
x
Xtx
x
tx
continuousXdxxfe
discreteXxXPe
_;)(
_];[.
]E[eø(t) tX
10
Laplace transform of an exponential distribution
Let X be a continuous random variable That follows the exponential distribution
The Laplace transform of X is given by
xX exf .)(
se
s
xsdes
dxe
dxeedxeedxxfesf
xs
xsxs
xsxxsx
x
Xsx
X
0
)(
0
)(
0
)(
00
*
.
)(
.)(.)(
11
Convolution of 2 discrete random variables
Let X1 and X2 be discrete random variables
Whose probability distributions are known P[X1 = x1] and P[X2 = x2]
Y be the sum of X1 and X2
What is the probability distribution of Y = X1 + X2?
P[Y = y]
1
][].[][ 1211x
xyXPxXPyYP
12
Convolution of 2 continuous random variables
Let X1 and X2 be 2 continuous random variables fX1 (x1) and fX2 (x2) are known
=> fX1 *(s) and fX2
*(s) are known
Let Y = X1 + X2
What is the pdf of Y?
)().()(
)()()(
)().()(
***
21
0
111
21
21
21
sfsfsf
xfxfyf
dxxyfxfyf
XXY
XXY
y
XXY
13
Application to Erlang model
X1 and X2
are exponentially distributed and independent
with mean 1/2μ each
1/2μ 1/2μ
X1 X2
Y=X1+X2
2
*
**
22
21
2
2)(
.2
.2)()(
..2)(;..2)(
21
2
2
1
1
ssf
ssfsf
exfexf
Y
XX
xX
xX
14
Erlang service time: first and second moment
The mean of Y?
The variance of Y?
Exercise at home Obtain the mean and variance of E2 from
its Laplace transform
1
][][][][ 2121 XEXEXXEYE
222
2121
2
1
4
1
4
1
][][][][
XVarXVarXXVarYVar
15
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
16
Poisson arrivals vs. bursty arrivals
Why do we care If arrival is bursty or Poisson
Bursty traffic will place havoc into your buffer
Example: router design
% loss
C2
17
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
18
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
19
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
20
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
21
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
22
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
23
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
24
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
25
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
26
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
27
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
28
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
* ....)(
29
Coxian model: conclusion
Most Laplace transforms Are rational polynomials
=> Any distribution can be represented Exactly or approximately
By a Coxian distribution
30
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
b=1-a
μ1 μ2