Prof. Krishna R. Pattipati Dept. of Electrical and ...
Transcript of Prof. Krishna R. Pattipati Dept. of Electrical and ...
Copyright ©2004 by K. Pattipati
Prof. Krishna R. Pattipati
Dept. of Electrical and Computer Engineering
University of Connecticut Contact: [email protected] (860) 486-2890
Lecture 3
ECE 336
Stochastic Models for the Analysis of Computer Systems
and Communication Networks
Copyright ©2004 by K. Pattipati
2
Summary of Lecture 2
Properties of Exponential and Geometric Distributions
Properties of the Poisson Process
Steady-state solutions to DTMC and CTMC
State Transition Rate Diagrams
Applications to M/M/1,M/M/1/N,M/M/m,M/M/∞,M/M/m/m Queues
Performance measures for M/M/1, M/M/∞, and M/M/m Queues
Outline of Lecture 3
Copyright ©2004 by K. Pattipati
3
DTMC:
• Specified by
CTMC:
• Specified by
ijPi)
nX/j
nP(X
1
noffunction
pmfGeometricn
iiiii
N
ii
nT
n
T
N
jij
T
PPnPipmf
ppppPpPp
NiPnpPnp
1
0
0
)1()(:stategivenainspentstagesof# ofThe
1 also;]...[)(lim&,statesteadytheIn
,.......,1,01);()1(
Independent of n
non-homogeneous
homogeneous
iiiiiτ
T
i
T
ijii
N
jij
iiij
qt(t)hi
pQop
tQitXtppQp
oeQ(t)qtqtq
tottqitXittXPjitottqitXjttXP
i
)),(exp( : stategiven ain spent timeofdensity The
state,staedy In the
chain Markov shomogeneou invariant time)(};)(Prob{)(,
0 since0)(0)(
)()(1))(/)((;),()())(/)((
0
• Unconditional probabilities
Summary of Lecture 2
Copyright ©2004 by K. Pattipati
4
Questions:
• What are the properties of exponential density & Geometric pmf ?
• Since inter arrival times of a Poisson distribution are exponentially
distributed, what are the properties of Poisson process ?
• Is there a general way of looking at arrival and departure processes?
…. Renewal processes
• When does steady-state solutions exist for DTMC and CTMC?
• Are there simple ways of visualizing state transitions?
• How to analyze simple queues ?
Issues Addressed in Lecture 3
Copyright ©2004 by K. Pattipati
5
Properties of Exponential pdf and Geometric pmf
a) Memory-less Property:
Suppose that we have already spent in state i for r time units (k steps).
want:
)(
)(
)(
)(
),(
)|()|(
tP
e
rP
trP
rP
rtrP
rtrPrtYP
i
t
i
i
i
ii
iiii
i
)(
)(
)(
)(
),(
)|(
,...2,1)|(
nTP
PkTP
knTP
kTP
kTknTP
kTknTP
nkTnYP
i
n
ii
i
i
i
ii
ii
ii
i
ri
Yi
Ti
Yk
Unconditional pdf
CTMC
Unconditional pmf
Exponential Density and Geometric pmf -1
DTMC
Copyright ©2004 by K. Pattipati
6
Implication: Suppose denotes the time to failure of a component. Result says
that if the component has lasted until time r, the distribution of the remaining
(residual) time Y= -r is the same as the time to failure of a new component.
If the time to failure is exponentially distributed, then a used component is “as
good as new”.
b) Hazard Rate:
Hazard rate is a constant
iiP
ni
TP
ni
TP
ni
TP
ni
TP
nZ
itie
tie
i
t
i
H
t
i
h
ti
Z
1
)1(1
)(
)(1
)(;
)(1
)(
)(
Markovian property (e.g., Poisson)
Exponential density Constant hazard rate
Exponential Density and Geometric pmf -2
Copyright ©2004 by K. Pattipati
7
c) Among all the densities having the same mean, exponential density has
the largest entropy.
)(ln)(:pmf geometricfor result similar showCan
1)( so, .
11e :yields constraintMean
)(1 :sconstraintmean &ion normalizat From
)(0)1()(ln
)(1)()(ln)(
:Lagrangian Define
)(ln)()(ln
0
/
0
)1(
)1(
)1(
000
0
nPnPH
ehd
ehee
ehh
dhdhdhhL
dhhhEH
nii
Exponential Density and Geometric pmf - 3
Copyright ©2004 by K. Pattipati
8
d) Suppose have n independent random variables with parameters
, respectively. Thenm ,...,,
21
m ,...,,
21
m
iim
(τ1
21parameter with lexponentia is ),...,,min
Proof:
Exponential Density and Geometric pmf - 4
rate relative
)()|()(:
)1( )()( parameter with geometric is
)()( Then,
parameter with GeometricT where),...,,min(
:pmf geometricfor Similarly
parameter with lexponentia
)(...)()()(
00
1
1
1121
ii21
1
)(
21
1
ji
i
tt
iijiji
nm
iii
nm
iii
m
i
n
iim
im
m
ii
t
m
dteedtthtPPAside
PPnTPPPT
PPn)n)....P(Tn)P(T P(Tn) P(T
PTTTT
etPtPtPtP
ij
i
m
i
i
Copyright ©2004 by K. Pattipati
9
e) Suppose have n independent random variables with the same
parameter . Thenm ,...,,
21
)),((or is
1
ondistributimGammaErlang-mSm
iim
Proof:
Exponential Density and Geometric pmf - 5
masm
Cm
mmSE
mSE
tem
tth
ssL
ssL
mmS
mm
i
S
mm
t
m
S
m
S
01
)1()(;)(
0;)!1(
)()()(
)()(:MGF Recall
2
2
2
2
1
Aside: Suppose have different parameters, jim ;,...,,
21m ,...,,
21
0;)()(111
teths
sLm
ijj
ij
jtm
iiS
m
ii
i
S
i
mm
Application:
Pdf of the time until mth
arrival.
X(t) Poission X(t)=
max{m: Sm t} & S0= 0..
Copyright ©2004 by K. Pattipati
10
a) Merging of Poisson Processes results in a Poisson Process
Poisson Stream
With rate
X
{X
X X
X X
XX X X X
Individual
streams
Merged
stream
m
ii
1
11),( tN
22),( tN
mmtN ),(
Proof:
...2,1,0;!
)()(),(
MGFs ofproduct nconvolutio i.i.d. of sum
)(.....)()(
)1()1()(
1
1
nen
ttpeetzG
tNtNtN
t
n
n
ztzt
N
m
m
i
i
Properties of Poisson Process - 1
Copyright ©2004 by K. Pattipati
11
b) Splitting or Decomposition of a Poisson Process
)(tN
mp
1p
2p
)(11
tNp
)(22
tNp
)(tNpmm
Poisson stream
with rate
rates,th poisson wi andt independen are streams Individual
!
)(
!
)(
!!...!
...!
)()|,...,,();,...,,(
...!!...!
!)|,...,,(
)(...)()()(
1
...
21
21)(
2121
21
21
21
21
2121
21
i
m
ii
N
i
tp
NNN
m
N
m
NNtp
mm
N
m
NN
m
m
m
p
N
tpe
N
t
NNN
pppNe
NPNNNNPNNNNP
pppNNN
NNNNNP
tNtNtNtN
ii
mm
i
m
Properties of Poisson Process - 2
Splitter
Copyright ©2004 by K. Pattipati
12
c) Conditional Distribution of the Arrival Times
tsst
n
nte
eeee
ntNP
nssshnsssP
SSS
,t
SSSnntN
t
s
te
eλse
tNP
PsP
tNP
sP
tNP
tNsPtNsP
tuniform
t
nn
nt
stsssss
n
n
ii
n
t
sts
nnn
....0;!
!/)(
...
))((
),,....,,()|,....,,(
0; : timesarrival-Inter
]0[ interval in the ddistributeuniformly and
i.i.d. are ,....,, timesarrival the,)( Given that :Aside
uniform
)1)((
}t)[s,in events 0{)}[0,in event 1{
)1)((
}t)[s,in events 0),[0,in event 1{
)1)((
)1)(,()1)(|(
.],0[in is occurredevent which the
at time theofon distributi then,informatio Given this
. by time, place taken has processPoisson a ofevent one Suppose
1
)()()(
21
21
01i
21
)(
1
1
1121
Properties of Poisson Process - 3
Copyright ©2004 by K. Pattipati
13
d) Poisson pmf is a Special Case of Binomial pmf
,...2,1,0;!
)()( As
)(!
)1)...(1(
)1()(!
)1)...(1(
)1()(
!!
!)1()(
!!
!)(
lim
Let
)1()(
)(.
0
nen
tnpN
etNn
nNNN
N
tt
Nn
nNNN
N
t
N
t
nnN
Ntt
nnN
Nnp
tNp
t
tN
tp
ppn
Nnp
t
n
N
nNt
n
n
N
nNt
t
N
n
n
nN
n
n
nNn
t
nNn
Properties of Poisson Process - 4
Copyright ©2004 by K. Pattipati
14
Suppose are i.i.d. nonnegative random variables (not
necessarily exponential).
],,,2,1,0:{ ii
Let 1
n
iin
S
Digression: Renewal Processes
timerenewal-inter
1renewals of rate termlong
][
1|
)(
)()(
)(lim
)(lim)(lim)(1
)( :domain LaplaceIn
)()()()()()(
)(,
)()()()(
)()()()()()()(
)()()()]([)()())((
)()()()(}{))((
process counting ~)(
)()()(*)()(
0
0
01
0
011
0
1 1
11
011
1
1
1
E
ds
sdLds
sdLssL
tm
ssLtmsL
sL(s)L
dhtmthtmthdt
tdMtmrenewalsofRateLet
dhtMtHtM
dhtHtHdhtHtHtM
tHtHtHtXEtMtHntXP
tHtHtSPtSPStSPntXP
tX
dhththththSS
st
mst
m
t
nS
t
t
nS
n
t
S
n nSSS
SSnnnn
t
SSSnnn
n
nn
nnn
nn
nnn
Equation Renewal
Equation Renewal lFundamenta
Copyright ©2004 by K. Pattipati
15
Suppose are i.i.d. exponential random variables],,,2,1,0:{ ii
Let 1
n
iin
S
Poisson Process is a Renewal Process
,..2,1,0;!
)()()())((
!
)(1)()(
)!1(
)(),()()]([)(,
t][0,in renewals ofnumber Expected )(
)()(
)()(
1
1
00
1
nn
ettHtHntXP
ek
tdhtH
n
etnErlangthsLsLAlso
PoissonofmeanttM
tms
sL
ssLlExponentia
tn
SS
tn
k
kt
SS
tn
S
n
S
m
nn
nn
nn
Copyright ©2004 by K. Pattipati
16
When does steady-state solution to pQpPp TT 0&
We need n-step transition probabilities. Let
theorycontrolin matrix transition-state oSimilar t )()1()(
)1(
.)(
MC shomogeneou- time)(
0
010
1
0
n
N
kkjik
n
N
knnij
nmnmij
PnPnn
Pn
ikPkjPn
ijPijPn
exist?
Two states i and j communicate if
• state i communicates with itself
• if i and j communicate, then j and i communicate
• if i and j communicate, and j and k communicate, then i and k
communicate (transitive)
• communicating states define a class comm. divides states into classes
0)()'( and 0)()( ', ji
n
jiij
n
ijPnPnnn
If all states communicate (i.e., one class) Markov chain is irreducible
Steady-state Probabilities -1
Copyright ©2004 by K. Pattipati
17
First Passage Probabilities and Recurrence Times
recurrent) (Positive nonnullrecurrent
recurrent )(
:Re
1;)(1
state toreturningever Prob)(
)(1
1)()()(1)(,For
.0)0(,1)0(;)1(;0)0(:I
1;)()()()(y probabilit
withsteps remaining in theagain and )(y probabilit
with step at the first time for the reached becan State
})0(|)({
state leaving of stepsn after occurs state return to )(
1
1
1
i
in
iii
iiii
niiii
ii
iiiiiiii
ijjjijijij
jj
n
kijijjj
ij
th
ij
M
nullMnnfM
TimecurrenceMean
transientfrecurrentpersistentf
inff
zFzzzFzji
Pffconditionsnitial
nknkfnkn
rnkf
kj
iXjnXtimefirstP
ijfirstPnf
First Passage Probabilities
Null Recurrent
recurrent transient
1ii
f 1ii
f
Positive
Recurrent
(irreducible)
States
Copyright ©2004 by K. Pattipati
18
Further classification of Markov-chain states
• If the return to state j occurs only at γ,2γ,3γ,… (γ >1),
then state j is periodic
• If γ =1 for state i, then it is said to be aperiodic
Steady-state Probabilities - 2
Null Recurrent
recurrent transient
1ii
f 1ii
f
Positive
Recurrent
periodicaperiodic
States
If one state in a class is aperiodic, all states in that class are
If one state in a class is transient, all states in that class are
If one state in a class is recurrent, all states in that class are
If one state in a class is periodic, all states in that class are
Irreducible aperiodic (i.e., ergodic) Markov chains have:
ondistributi stationary unique theis 1
002)
capacity) servicerate arrival on(e.g.,distributi state-steady no 00)1
M
pipi p
ip
i
iii
i
Absorbing state: Pii=1 Mi =1 pi =1
Irreducible recurrent non-null (positive recurrent) and aperiodic
}, unless 0)({ knnii
}, 0)({ nnii
Copyright ©2004 by K. Pattipati
19
Examples:
Examples of State Classification - 1
0.5
0.5
0.3
0.3
0.4
1
1
0.5
0.5
(4) state Absorbing :3 Class
(2,3) statesRecurrent :2 Class
(0,1) statesTransient :1 Class
0 1 2 3
4
(2,3) states of classin trappedgetting ofy probabilit finite
0.2434] 0.5044 0.2522 0 [0
4 statein trappedischain 1] 0 0 0 [0
classesrecurrent twosince eseigen valuunity Two
: two theseof one becan iesprobabilit stateSteady
•
•
Copyright ©2004 by K. Pattipati
20
Examples:
odd 01
10
even 10
01
)(01
10
n
n
PnP n
starting in state 0, we return to state 0
in even # of steps
γ =2 periodic
}0)()({2/12/1
3/23/1nPnP
ii
n
ii
aperiodic Markov chain
01
011
0 1
1
State 0: absorbing
State 1: transient
We will consider irreducible, aperiodic Markov chains
Examples of State Classification - 2
]01[:yprobabilitstatesteady
Copyright ©2004 by K. Pattipati
21
Examples
• Poisson Process
0 1 2 3
n+1nn-1
210
)(t )(t
t
n
n
nnn
en
ttp
tptpp
!
)()(
)()(1
• Non-homogeneous Poisson process
n-1 n n+!
)(t)(t
t
d
nt
n
nnn
en
d
tp
tpttptp
0
)(0
1
!
)(
)(
)()()()(
ii
i
N
jij
iij
i
j
N
jij
ji
iii
N
jij
jjij
N
jjii
qtpqtpq
tptqtpqtptqp
know toneed No)()(
)()()()()(
NODE OF OUTFLOW
0
NODE INTOFLOW
0
00
State Transition Rate Diagrams - 1
Copyright ©2004 by K. Pattipati
22
• State-dependent birth process
0 21
0
1
n-1 n+1n
1n
n
t
nnn
t
n
n
nn
n
nnnnn
pdepetps
spsp
tptptp
nn
0
11
11
11
)0()()()(
)(
)()()(
• Birth-death processes: Forms the basis of all Markovian Queuing systems
Server
A(t)
Buffer space N
Assume N= for now
Q(t)=A(t)-D(t)
Infinite server
Single Server
Multiple servers
State-dependent
Any type
D(t)
State Transition Rate Diagrams - 2
Copyright ©2004 by K. Pattipati
23
occurcan departureor arrival oneonly t timeof intervalan in that assume We
tat timelength system )( t timeupto deaths total)(
t timeupto births total)(
tQ
tD
tA
n+1
n
n-1
.
.
.
.
.
.
.
.
.
t tt
)(1
totn
)()(1
)1)(1(
tot
tt
nn
nn
We assume S=1 (service demand=1)
0P
1P 2
P
1nP n
P 1nP
0
1
1n
n
1 2
n
)(1
totn
n
1n
State Transition Rate Diagrams - 3
)()()(
1);()()()()(
11000
1111
tptptp
ntptptptpnnnnnnnn
Copyright ©2004 by K. Pattipati
24
B-D processes model a wide variety of Markovian (exponential) queues
1) M/M/1 Queue M ~ Poisson arrivals or exponential inter-arrival pdf
M ~ exponential service demands
1 ~ single server
n
n0P
1P 2
P 1nP n
P 1nP
1
n
2) M/M/∞ queue Infinite server case No waiting
nn
n 0P1
P 2P 1n
P nP 1n
P
2n )1( n
Application to Simple Queues - 1
….
Copyright ©2004 by K. Pattipati
25
3) M/M/m queue m server case Model of a circuit
switch-Queued mode
1
2
m
mnm
mnnn
n
1mP
mP 1m
P
m m)1( m
0P
1P 2
P
2
At most
N
4) M/M/1/N queue Model of a single machine with a finite buffer
Nn
Nn
Nn
Nn
n
n
0
0
1
2
P
0P1
P
1NP
NP
Application to Simple Queues - 2
….
….
Copyright ©2004 by K. Pattipati
26
5) M/M/m/m modelBlocked call loss system or Blocked calls cleared (BCC)
mn
mnn
mn
n
n
0
10
1mP
mP
m
0P1
P
2
Application to Simple Queues - 3
….
Copyright ©2004 by K. Pattipati
27
6) Engset model (M>N) M/M/N/N/M queue
Nnn
MN
NnnM
n
n
1;
0;)(
Poisson
arrivals
)( nMn
nn
n calls in progress
NP
N
)1( NM
0P1
P
M
)1( M
2
Application to Simple Queues - 4
Copyright ©2004 by K. Pattipati
28
7) Machine Repairman model….forms the basis of closed networks
M/M/1/N/N queue
1
N
1
)1(
n
n1
n
n
NnnN
otherwise 0
0)(
8) M/M/1 queue with feedback …. A simple open network
0P1
P
N
)1( N
1NP
NP
Application to Simple Queues - 5
Copyright ©2004 by K. Pattipati
29
)(lim
only.solution state-steadyth content wi be will Wefunctions. Bessel
involvessolution queue), M/M/1 (i.e., , If methods. numerical viaSolution
asknown
)()()(
1);()()()()(
)()(])(1[)()(
n
1100
0
1111
1111
tpp
tptpdt
tdp
ntptptpdt
tdp
ttptptttpttp
nt
n
n
nnnnnnn
n
nnnnnnnn
equations balance global
Coming back to birth-death model
11
nnnnpp
nnnnpp
11
1nP
nP
1n
n
nnnnpp
11Local balance
n detailed Balance Equations
Solution of Birth-Death Model - 1
Copyright ©2004 by K. Pattipati
30
1
1
1
01
00
0
1
01
0
21
110
1
1
]1[1 Since
)(...
...
n
n
ii
i
nn
n
ii
i
n
n
n
n
n
n
pp
pppp
Once the distribution is known, we can obtain a variety of performance
measures
1) Throughput: n
nnpX
1
2) System (or Queue) length
1n
nnpQ
3) Utilizationn
n
nnn
pU
max
1
4) Average Response time:X
QR
Solution of Birth-Death Model - 2
Little’s law
Copyright ©2004 by K. Pattipati
31
1. M/M/1 Queue
)1( versus note pmf... Geometric Modified ,...2,1,0)1(
1or;111
intensity traffic;,
1
10
0
nnnp
Needp
pp
n
n
n
n
n
nnn
0 1 2 3 4
n
1n
p)1(
)1(2
)1(3
)1(4
Application to M|M|1 Queue - 1
Copyright ©2004 by K. Pattipati
32
11
1)1(11)1(
2
)1(
2
)1(
)1(2)()(
1)1(
)1()()(
1
)1()(
2
2
2
2
2
2
2
3
2
1
2
2
2
1
2
1
0
QC
dz
zGdQnE
zdz
zdGQnE
zzpzG
n
n
nn
z
Zz
n
n
n
• Throughput
)1)((1 1n n
n
nnpX
• Average response time)1(
1
X
QR
Higher moments of response time requires analysis… later
Application to M|M|1 Queue - 2
Copyright ©2004 by K. Pattipati
33
• Utilization
U
• Average waiting time
1
11RW
• Average waiting queue length
)11
:(1
22
QQcheckXWQ
WW
QQ , Q
Q
1
Application to M|M|1 Queue - 3
Copyright ©2004 by K. Pattipati
34
2. M/M/∞ Queue nnn ,
0lim
0
01
1
shouldit as )!1(!
. Throughput
1
,...2,1,0;!
:result The
!1;
!
1
1
11
1
1001
nU
Q
RW
R
en
en
npX
CQ
nen
p
en
ppn
ppn
p
n
W
n
n
n
n
nn
n
n
n
n
n
n
nn
nnn
POISSON DISTRIBUTION WITH MEAN
Application to M|M| Queue
Results valid for M|G| queue
Copyright ©2004 by K. Pattipati
35
1. M/M/m Queue
1
1
00
0
1
00n
0
0
1
00
1
01
1
1
!!
1p !!
, 1 From
!
0!
:Result
;.:For
1!
1
)1(1for
m
n
mn
m
n mn
mn
mn
n
mn
m
n
n
m
mn
nnn
nn
innn
m
mnp
mmnp
mnpmm
mnpn
p
mnpm
ppm
pmn
mnpn
pi
pmnpn
p
Application to M|M|m Queue -1
Copyright ©2004 by K. Pattipati
36
formula) second sErlang'or formula-C sErlang'or formuladelay s(Erlang'
;1
1!
0
m
p
m
p
mp
mpP mm
mn
mn
m
mnnQ
Application to M|M|m Queue - 2
Probability of Queuing, PQ
Average Number of Customers Waiting (not in service)
mP
Q
Ppnpp
mnnpQ
Q
W
Q
mn
n
m
n
n
m
nmnW
;1
)1()1(
p
-1
1
d
d.
! 2
m
00
00
Throughput
shouldit as ..!
2
010
1
11
mn
m
nnn
m
n
n
mnnn
nn
pppnn
pmpX
Copyright ©2004 by K. Pattipati
37
Average Waiting Time
1
1 QWPQ
W
Average Response Time ]1
1[11
QP
WR
Average Queue Length ]1
1[.
QP
RQ
Average number of busy servers
WQQ
An alternative expression for system length … a trick that will be useful
for state-dependent nodes of networks
1
)(
nn
pn
p
)()....1()(
nodedependent -State serverMulti
mmnm
mnnn
Application to M|M|m Queue - 3
1 bemust but 1 becan Note
m
Copyright ©2004 by K. Pattipati
38
Queue Length
1
11
1
11
1
1
1 )()()()( n
nm
n
nn
n
n
nn
m
np
m
np
n
np
n
npnpQ
1
1
1
1)(
)(1
)(n
m
n
pn
mnQ
mQ
or
1
1Q
for multi-server nodes
= 0 for single server nodes
1
1
1
)(
n
m
n
pnm useful for numerical computations
m ; Note: need p0, p1,…., pm-2 only
Utilization
mU
Application to M|M|m Queue - 4
Copyright ©2004 by K. Pattipati
39
Higher Moments of Queue Length
)1(
)(
)1(
)1(])[(
)1(
)41()](331[
1)(3
1
)1()1()1(
)1()21(
1)(2
11
]}[{1
)(
][
)1()11()(
2
3
3
3
2
23
2
2
2
2
1
0
0
11
011
11
1
nEQnEskewness
nEQnEk
QC
QQnEk
Qk
nEr
knE
nEr
k
pnr
kpnpnpnnE
n
n
rk
r
k
rk
r
nn
rk
rnn
k
nn
k
nn
kk
n
Higher Moments for M|M|1 Queue
Copyright ©2004 by K. Pattipati
40
Properties of Exponential and Geometric Distributions
Properties of the Poisson Process
Steady-state solutions to DTMC and CTMC
State Transition Rate Diagrams
Applications to M/M/1,M/M/1/N,M/M/m,M/M/∞,M/M/m/m Queues
Performance measures for M/M/1, M/M/∞, and M/M/m Queues
Summary of Lecture 3