Erlang, Hyper-exponential, and Coxian distributions
description
Transcript of Erlang, Hyper-exponential, and Coxian distributions
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
H3
μ1
μ2
μ3
P1
P2
P3
μ μ μ μC4
2
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
3
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
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
)!1(
.)...()(
..1
r
eyrryf
yrr
Y
5
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
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
))...()((
...)(
21
21*
r
rY ssssf
7
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
8
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
9
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
10
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
11
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
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
korderPolynomial
korderPolynomial
_
_
k
i
i
l l
lii sbaaaaasf
1 112100
* ....)(
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
b=1-a
μ1 μ2
15
C2 : first three moments
32
31
3212121
31
3
22
21
22121
21
2
21
)(6)(126][
)(22.2][
1][
aabXE
aabXE
aXE
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
korderPolynomial
korderPolynomial
_
_
a
b=1-a
μ1 μ2
17
3 moments method 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
18
3 moments method: solution The following expressions will be obtained
However, The following condition has to hold: X2 – 4Y >= 0
=> 3 < 2m1m3
=> the 3 moments method applies to the case where c2 > 1 for the original distribution
1
2
13
1
22
1
21
12
2
1
111
2
2
1;
46
36
;2
4
)1.(
m
Ym
mX
mmm
mmm
Y
XYXX
ma
22m
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
21
21
1
2
.
1;
2.2
1
cmm
ca
20
M/C2/1 queue
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
μ μ C2a
b
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
0,1
1,1
2,1
.
.
.
0,2
1,2
2,2
.
.
.
λ
λ
λ
λ
λ
μ2
μ2
μ2
bμ1
bμ1
bμ1
aμ1
aμ1
aμ1
2nd column: states of the systemwhere server busy in phase 2
1st column:States wheresystem in phase 1
23
M/C2/1: steady state equations Steady state equations
1st set of equations (1st column)
'
]'_[]'_[
]____[]_[
S
SStransitionSstateP
SofoutratetransitionSstateP
1,12,121,111,1
01222211111
00122111011
02201100
....).(
.
.
....).(
....).(
....
nnnn PPPbP
PPPbP
PPPbP
PPbP
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
2,11,12,2
12211222
02111122
011022
...).(
.
.
...).(
...).(
..).(
nnn PPaP
PPaP
PPaP
PaP
2,11,1
12112
02011
000
.
.
nnn PPP
PPP
PPP
PP
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
]__1Pr[...)1(.)( 110110
11 busystagestPPgPzzgi
ii
]__2Pr[...)1(.)( 120220
22 busystagendPPgPzzgi
ii
26
Expressions for the generating functions Using the 2 set of balance equations, we get
(1), (2), and (3) =>
02201100
21122
10002220111
11
....
)(..)(..)()(
)(...])([1.])([
1.
)()(
PPbP
zgzzgazg
zgzPPzgz
Pzgz
b
zg
(1)
(2)
(3)
22
11
0022121
221
211
12
12
;,
.1)(
))1(1()(
)(.)1(
.)(
where
Pbzz
zzg
zgz
azg
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))
212100
002212121
111
1.1.1
.1
)1(
aaP
Pb
g
22121212
0022
2100
)(1
).1()(
)(.)(.)(
zzb
Pzbbzg
zgzzgzPzg
28
Pn : recursive scheme
The general expression of probability Pn
2
212
2
21211
2
21
2211
11
0
110
1
;1
;1
1
,
,...2,1),(
bS
bS
b
bT
ASASA
SA
A
where
nATAPP
nnn
nnn