queuing. problem solving for performance evaluation analysis: use mathematical model to solve...

Queuing

Problem Solving for performance evaluation Analysis: use mathematical model to

solve usually need assumption to make it easy to solve Can have full domain of solution, I.e. optimal

solution Simulation: more realistic assumption

heuristic solution Analysis used to be the main research

area, now simulation is a popular method, but…

Queuing examples Customer service

Bank, post office , number of clerks vs. waiting time , service time

Computer servers and jobs (requests) Traffic flow on freeway or computer

networks.

Why probability ? Because everything is unpredictable. We

use ‘mean’ , ‘variance’ to be the metrics.

Sampling of Problems

Resource Scheduling Demands:

Unpredictable arrivals Unpredictable service demands Everybody wants immediate

service Conflict resolution

Waiting Loss sharing

Demand Resource

Single Resource with waiting (queuing)

Assume: Arrival Each service take Service

departures

serverqueue

Arrivals

)sec(customers λTT1rate arrival seconds every

seconds S

)sec(1 customersS

ccapacity

Single Resource with waiting (queuing)

Def: IF constant inter-arrival times AND

constant service times.

IF random inter-arrival times AND random service times.

. time response Average system) in (TimeR

SRc Rc

unstablec


Every disk access is equally likely to be to any cylinder on the diskWhat is the average distance (in cylinders) of a seek ?

Expected execution time variance?

t1

branch

t2

t3

start

noexit

0.25

0.75

yes


fi = fraction time link i is down at a random time , what is the prob. That there is

Cherry GOES DOWN TWICE A WEEK. What’s the prob. You lost your term report ?

S

f1

f2

f3

f4

f5f6

f7f8

f9

f10D

Probability Random experiment.

Real world experiment.outcome not predictable.

Associated with a random experiment. Set of all possible outcomes.

Mutually exclusive and exhaustive. Events

Event occurs or does not occur for each outcome. Defines a subset of outcomes.

occurs event then events if EOiOiE |

Probability

Independent trials If outcome of any trial does not affect

the outcome of any other trial If trials are performed under identical

conditions.

Probability

Statistical regularityn independent trials.

Def: likelihood differing

N as 0 to goes sexperiment of sets twofor than

moreby differing of likelihood

occurring.event A ofy probabilit

large For

NAn

AiAn

n

)(

)(~)(

(“ 點”機率不能用此法 )

Probability Sample space set of all possible outcomes

of a random experiment Not required, but often deal with equally probable outcomes.

Suppose event A occurs iffOi1 or Oi2 or … Oij occurs

NOi

jinOjn

nOinNiOi

1)(

,,)()(,...,2,1|

outcomes. possible all ofset Let

Probability

i.e.

jjOijA ,...,2,1|

J

j NJOij

J

j nOijn

n

J

jOijn

nAnAP

J

jOijnAn

1)(

1

)(1)(

)()(

1)()(

Probability Counting Things – Combinations

Product ruleLet A1 be a set containing n1 objects, A2 be a set containing n2 objects.The number of ordered pairs

(a1, a2) s.t is n1*n2

Generalizes tor sets A1,A2,…Ar of cordinalistics n1,n2,...,nrNumber of distinct r-tuples with i-th componemt an object from Ai is n1*n2*n3*…*nr

2211AaAa and,

Probability Some fundamental cases

Ways of selecting r out of n objectsTwo dimensions:

Ordered or unordered samples. (sequence v.s. a set)

With or without replacement. (whether can repeat an object)

Ordered sampling with replacement. The number of ways of selecting r objects in

succession from a set of n objects, where each object is returned prior to the next selection is nr

Probability Ordered sampling without replacement

The number of ways of selecting r objects in succession from a set of n objects (r ≤ n) where each object selected is removed from the set prior to the next selection is, n(n-1)…(n-r+1)= n!/(n-r)!

Sampling without replacement and without regard for order

The number of ways of selection r objects from a set of n objects without replacement & without regard for order is :

nrr

nrrn

n

, !)!(

!

Probability Sampling with replacement and without

regard for order r out of n

Ans:

r

rn 1000|0|00||00000…|000

1 2 3 4 5 …How many “0”s rHow many “1”s n-1

n

rrnyrx

r

nnyx0

)( : Theorem Binomial

r

n

r

n

r

n

r

n

r

n

nn

r

n

r

n

1

1

2

0)1(

0

0

Probability Probability Axioms

Let S be a sample space of a random experiment Axioms

(A1) for any event A,

(A2) P(S) = 1

(A3)

It follows easily that (n finite)

if Ai are mutually exclusive

0)( A

)()()( BPAPBAP exclusivemutually are B & A i.e. , whenever BA

n

iin APAAAP

121 )()...(

Probability

(A3’) For any countable sequence of events A1,A2,...,An,… ， that are mutuall

y exclusive. (i.e. )

kjAA kj whenever

11)()(

nnnn

APAP

Some implications of the Axioms

1)()()()(

)(1)(

SPApApAApSAAAA

SAasAApAp

, :pf

w.r.t of complement where

0)( p

)()()()()()()(

)(:)()()()(

BApBpApBApBABpApBAp

BABABApfBApBpApBAp

Principles of inclusion & exclusion

If A1,A2,…,An are any events then

)...(1

)....()()(

)...()(

211

111

211

n-

nkjkji

njiji

n

ii

nn

ii

AAAP)(-

AAAPAAPAP

AAAPAP

Proof by induction on the number of events n (recitation)Related result in combinatoric S

Conditional Probability Probability that event A has occurred given

that B has occurred , notation: P(A|B) Def:

BSif

BSifBPSP

BSPi

ii

i

0)()(

)|(

Makes intuitive sense in terms of relative frequencies nBnn

iSn

)(

)(

)(

)(

)(

)(

)(

)(

)|()|()|(

BP

BAP

BP

SP

BP

SP

BSPBSPBAP

BASi

BAS

i

BAS BASji

i

i

i j

Generalization:

)|()|()(

)(

BACPABPAP

CBAP

Independent Events Def: Events A & B are said to be indepe

ndent iff In words, given that B occurred, given

no information about the probability that A occurred.

Note: If P(A|B)=P(A) ， that

i.e. symmetric

)()|( APBAP

)()(

)|()(

)(

)()|( BP

AP

BAPBP

AP

BAPABP

Independent Events

Alternative Def:follows from first def: but if P(A|B)=P(A)

then P(A)P(B) = P(AB)

)()()( BPAPBAP

)(

)()|(

BP

BAPBAP

Independent Events

Ex:Pi=prob. of introducing an error, assume that n msg arrives in errorP(msg arrives with error)=?Let Ei=event that link I introduces an errorP(msg arrives with error)

assume are independent. are indep. ??=> P(msg arrives with error)

LA NYP1 P2

Kansas City

)(1

)(

21

21

EEP

EEP

21 E&E 21 E&E

))(1))((1(1

)(1

21

21

EPEP

EEP

Generalized

N

iiEP

1

))(1(1

Some random properties

If

If A & B are independent And B&C are independent

A&C are independent

0)(0)(

BPAP

BA

or then

t,independen are B & A and

Ex: B={ 1, 2, 3, 4 } A={ 1, 3, 5 } C={ 2, 4, 6}

Independence of a set of events

Def: A list of n events A1,A2,…,An is said to be mutually indep. iff for each set of k (2 ≤ k ≤ n) distinct indices i1,i2,…,Ik

pairwise indep. mutually indep.)(...)()()...( 2121 ikiiikii APPAPAPAAAP

Note: duality mutually exclusive :

mutually indep :

k

iik APAAAP

121 )()...(

k

iik APAAAP

121 )()...(

Ex: N blocks in a file. n accesses, uniformly distributed. What is the prob. That the first block

is accessed at least once ?(A) P[1st block accessed] =(B) #ways to make the n accesses = Nn

#ways with 0 accesses to 1st block = (N-1)n

#ways with >0 accesses to 1st block = Nn- (N-1)n

∴

n

n)

11(1

nnn

nn

)N1

(11)N

1N(1

N1)(NN

] P[

Let Ei = event that the first block is accessed on i-th accessP[1st block accessed at least once]

nn

i

n

ii

n

ii

n

ii

n

ii

NNEP

EP

EPEP

)1

1(1)1

1(1)(1

][1

][1][

11

1

11

Bayes Rule In general:

Let B1, … , Bn be mutually exclusive and exhaustive events.

Let A be any event.

kkk

jj

jj

BPBAP

BPBAP

AP

ABPABP

)()|(

)()|(

)(

)()|(

B1B2

B4

B3S A

jiBiB

SB

j

n

jj

,1

{Bi}exhaustive

Mutually exclusive

n

kk

n

kk

n

kk

BAP

BAPBAP

SAPAP

1

11

)(

))(())((

)()(

Ex: Drawers G G G SSelect a drawer at random reach in and take a coin without looking.The coin is gold, what is the prob that drawer 1 was chosen ?

Find P(drawer1|gold)=?P(drawer1)=p(drawer2)=1/2P(gold|drawer1)=1P(gold|drawer2)=1/2

3

2

4/3

2/1

2/1*2/12/1*1

2/1*1

drawer2)drawer2)P(|P(golddrawer1)drawer1)P(|P(golddrawer1)drawer1)P(|P(gold

gold)|P(drawer1

Chap 2 Discrete Random Variable

Def: A random variable X on a sample space S is a function X: SR from S to the reals Numbers often are naturally associated

with sample space outcomes. e.g. face value on a die

But not always e.g what if the die had colors

Probability mass function

Def: Px(x)=P(X=x)is also called the density function for X

(Cumulative) Distribution function

xSX

SP

xSXSPxXPAxP

)(

)(

}))(|({][)(

txX

X

xP

tXPtF

)(

)()(

1/36

3/36

2 3 4

Ex: Bruin rot1/10000 of having to disease Medical test.Two possibilities P (positive) or N (negative)P[P|BR]=0.99P[P|NBR]=0.01

Henry gets a test result is positive.P[BR|P] =

01.0*)101(99.0*1099.0*10

]|[][]|[][]|[][

44

4

-

nBRPPnBRPBRPPBRPBRPPBRP

Particular Cases Bernoulli

k]P[yΔP

xxxxY

q-p]P[X

pΔ] P[X

nk

inn

Let

r.v Bernoulli a is each

Binomial

1 & 0 valueswith R.V. discrete ,X

21

10

1

Particular Cases Experiments with two outcomes, repeat some num

ber of this n, There are 2n possible sequences of length n for outcomes of the experiments.

Call the outcomes “success” and “failure” Letfor any single experiment

P(success)=p P(failure)=1-p=q

Consider a sequence of length n 以下看不懂

Particular Cases Consider

knkknkk

k

qpk

nqpSeqpP

P

.)(

trivals n in success k of prob.

Seq. distinctSequence of length n With k successes

All seq with k success

All have the same prob

Particular Cases Generalization

m

m

nm

nn

m

mi

nm

nn

mm

PPPnnn

n

PPP

, P, , PP, O, , OO

m

21

21

2121

121

2121

)!!!

!(

nm)n2,P(n1,

trials. n of sequence have them

iesprobabilit with Say

trial of outcomesdifferent are there Suppose

Geometric r.v Z

(An) Interpretation: Number of Bernoulli trials until a success

is obtained sample space is infinite All sequences of some number of 0’s follow

ed by a 1 or fff….f s PZ(i)=qi-1*p , I >= 1

Geometric r.v Z Distribution functio

n Fz(t)=P(Z<=t) 字被擋住了，看不到

t

z

-q

s]st t trialess in fir-p[no succ

is event]ment of th-p[complet

is eventprob of th(t)F

ti

1

1

1

event an is with Sequences

Geometric r.v Z

stnyo"" ofset space sample

trials. first in success

no were therethat given success a

before beyond trials ofnumber

r.v. lConditiona

0

0

k

kr.vΔLet w

pqq

pq

]kp[z

]kkp[z

]kp[z

]kzkkp[z]k|zkkp[zk]p[w

k-k

kk1

1

0

0

0

0000

0

0

:property Memoryless

Poisson Pmf

k

n

k

kknk

nt

nt

n

t

kkn

n

n

t

n

t

k

nPk(t)

,t)(

λt/n as n

t/n

1

1

!)!(

!1

0 in events k ofy probabilit the

trial.t independen an is t/n interval each

interval an inevent an ofy probabilit the Suppose

這邊字看不清楚

Probability Generating Function (Z transform)

Transformation: Will simplify many calculations later.

(non negative integer valued r.v.)

xzG

pkdz

zGd

G

zpzG

pkxpLet

x

k

z

kx

k

x

k

kkx

k

of ondistributi complete

:2Property

:1Property

)(

!)(

1)1(

)(

)(

0

0


An example

0 0

0 0

0

0

)()(

)()(

)(

)()()(

n j

jnj

n

n

j

n

nnz

n

j

zjnypzjxp

zjnypjxp

zpzG

jnypjxpnzppn

sum the i.e. y xz Define

n,0,1, range withr.v t independen bey Let x,


一些看不懂的

So )()()( zGzGzG yxz

Some transforms

Bernoulli r.v.x

nx

knk

x

pzpzG

qpk

nkxp

pzppzqzzG

pxp

pqxp

)1()(

)(

1)(

)1(

1)0(

10

r.v.x binomial 這邊有字看不懂

Some transforms

Poisson

)1)(()1()1(

21

21

)1(

00

2121

21)()()(

&

!)()(

!)(

zzzxxz

z

k

zkk

k

k

k

eeezGzGzG

xxz

xx

eeezk

ezkXpzG

ek

kxp

21,parameter withr.v poison indep. two be Let

? know we do How

poisson is sr.v' poisson indep. two of sum

Linear recurrence relation

2

2 2 221

21

21

)(

11

102

n

nn

n n n

nn

nn

nn

nn

nn

nn

n

nnn

zfzGDefine

zfzfzf

zfzfzf

znth

f

fnfff

to 2n sum Now

by equation multiply

condition initial Let

接下頁

Linear recurrence relation

nnn

n n

nn

nn

n

nn

n

nn

BAz

z

B

z

A

zzzG

zfzffzzzG

zGzfzGz

zfzzfz

zfzzfzzffzG

21

212

0102

20

1 0

2

2

22

2

2

1110

111

1)(

]1)[(

)(])([

)(

Review

pq

p

xXpxF

xx)(xp

xxx

RSX

X

iix

1

][)(

,2,1

success of prob Bernoulli

fun. onsdistributi

prob

valueson takes

fun. mass prob

r.v Discrete

：

：

Review

pqp

vr

qpC

kpk

kk

knknk

1

.

successfirst until trials Bernoulli of #

Gexxxxxx

n) of(out success of prob

trials Bernoulli Binomial

：

：

字看不懂

這邊的也看不懂

Z-transform Fn n=0,1,2,3… Define Property 1:

Property 2:

0

)(n

nn zfzF

fun. mass prob a repress }{1)1( fnifF

0

)(

!

1

z

k

k

k dz

zFd

kf

Z-transform

i

i

n

ek

kXp

tX

zFzFz

Xz

nix

zFzFzFyxz

zFzFyx

ki

i

ii

n

ix

i

i

YXZ

YX

!)(

)()(

,,2,1

)()()(

)()(

1

)( parameter with r.v Poisson 2.

indep.mutually are , generalize

transform-z has r.v. the then

& transforms-z withr.v indep. are & If 1.

續下頁

Z-transform

k

j

jkj

jkk

j

j

k

j

zzzY

ej

k

k

jk

e

j

e

jkxpjxpkYP

eeezF

XXY

0

)(21

2

0

1

0

)1)(()1()1(

21

21

21

2121

!

1

)!(!

)2()1(][

)(

Let

下面好像還好，看不清楚

Z-transform kleinrock Appendix I

a

zF

a

f

z

zFf

zFzffdz

zdFznf

fzFz

f

azFfa

zFfn

n

n

kn

nn

n

n

nn

)(

1

)(

)()1(

)(

))((1

)(

)(

0

1

01

Z-transform

32

2

2

)1(

)1(2

)1(

)1(

1

101

1

1,2,1,01

0

1

100

01

z

zn

z

zn

z

zn

z

AA

z

zknU

znU

zkn

kn

n

n

n

n

k

kn

n

kkn

n

Last 4 are special cases of the following

Z-transform

)(

)()(

)(

!

1

!

1

1)1(

1)1()1)((

!

1

0

1

zD

zNzF

dz

zFd

nf

en

mz

nmnmnm

z

n

n

n

z

mn

Expansion Fraction Partial

theorem) valueate(Intermedi methodpower

inversion) of (method

,

p.335

Discrete random vectors Multivariate distributions

16

6,3

16

5,2

16

5,1

)(

),,,()(

),,,(

2211

21

pmf marginal

r.v. are where

x

x

x

xp

xXxXxXpxXp

xXXXX

X

nn

nn

1 2 3

1

2

X

Y

1/4 1/4 1/4

1/16 1/16 1/81/4

3/4

),( yxpXY

續下頁

Discrete random vectors What is the condition pmf )2/( yixp

z

j

jzypjxpzZp

yxDef

0

)()()(

:

nconvolutio (discrete)

YXZ old of terms in r.v. new Defining

spmf' marginaltheir ofproduct the is pmf

joint their ifft independen are and

Continuous r.v

x

XXx

X

X

dttfxFdx

xdFxf

xFx

xxXpxF

)()()(

)(

)(,

)()(

or

fun.density y Probabilit

.continuous is r.v. continuous afor

: Function onDistributi

FX(x)

x0 x0

Limit=0

Limit=1

FX(x)

xx0 a b

字看不懂nondecreasing

Application to simulation

How to generate a r.v. with desired FX(x) 先 generate a uniformly distributed r.v. y Then x0=FX

-1(y0) got it Nonnegative r.v

)()]([][

)(

][][

000

00

00

xFxFypxxp

xFy

yypxxp

XX

X

x

y

x0

y0

這邊有字看不到

Exponential distribution Service time at a resource (server) Distribution of time between arivals

00

0)()(

01)()(

x

xe

dx

xdFxF

xexFxXpx

x

,

,

,

function ondistributi

x

F(x)

Exponential distribution

)()(1

1)( ytf

tFf x

Xyt

get a ofcomponent a of life remaining

component new a of time life

tY

X

Fx(x)

xtRemaining life distribution

)(1

)(

)(1

)()(

)(

)(1

)()(

][

][

]|[][)(

tF

ytfx

tF

ytFdyd

dy

ydFyf

tF

tFytF

tXp

ytXtp

txytXpyYpyF

XX

Xy

y

X

XX

ty

t

t

t

Exponential distribution Memoryless property

yt

ty

t

ty

X

Xx

xX

ee

e

e

e

tf

tyfyf

exf

xxp

xxxxpxxxxxp

t

)()(

0

0000

)1(1)(1

)()(

)(

)(

)()|(

Poisson distribution of exponential distribution

Poisson distribution was for the xxxxxxx of “events” in an interval of length t

where

0)(

lim)(

)(1[

)(10[

)(1[

t

tOttO

tOp

tOtp

tOtp

Δt

0ts.t of functionany is

t] inevent

t] inevent

t] inevent

length of lsubinterva eachfor

Poisson distribution of exponential distribution

(a) Exponential interarrival

(b) Poisson arrival process memoryless property exponential interarrival times

What is the “memoryless property” important ?Real answer must wait but intuitively.

)(

2

))((

)2

)(1(1

1),([

tO

t

tt

tt

etttp

in arrival

Arrival process departures

Expectation Def of expectation

X: a random variable

Moments

r.v continous

r.v discrete valueexpectedor mean

dxxxf

xpxxE i

ii

)(

)(][

dxxfx

xpxEYE

xY

ii

)()(

)()]([][

)(

Let

Expectation

2meanvariance

sequared varianceoft coefficien

deviation Standard

Variance

or

moment Second

22

222

]])[[(

)(][

xExE

pxdxxfxxEi

ix

xi

Expectation Expectation is a linear operation

Variance Xi are mutually indep.

][][

,,,

][][

1

21

i

n

ii

n

XEYE

XY

xxxn

bxaEbaxE

variablesrandom of Sum

N

i

N

ixii

x

jijjii

N

iii

N

iii

N

iii

i

i

i

xxE

xxxxxxE

xxExxEyyE

XY

1 1

22

0

1

2

2

1

2

1

2

])[(

]))((2])([

]))([(])[(])[(

indep. is if

Expectation Expectation of functions of more than one random variable nxxx ,,, 21

disance of valueexpected find towant

distance

disk afor time seek expected the

or

|-X|X

Example

xxpxx

dxdxxxfxxYE

xxY

nxnn

x

nnn

n

12

11

111

1

:

),,(),,(

,),,(),,(][

),,(

1

0 aNextrequest

currentrequest

x1x2

Conditional Distribution and Conditional Expectation

Conditional Probabilities

xXYX

xY

XYX

x

XY

xypxpyxpyp

xypxpyxp

xp

yxp

xXp

xXyYpxXyYpxyp

yxpYX

Bp

BApBAp

)|()(),()(

)|()(),(

)(

),(

][

][]|[)|(

),(&

)(

)()|(

|

|

|

Yfor pmf marginal the And

then Also

pmf with (discrete) r.v.dependent be Let

Marginal pmf


Continuous r.v.

)|()(),(

)(

),(

][

][

]|[)|(

)(

),()|(

|

|

|

xyfxfyxf

dxxf

dxdyyxf

dxxXxp

dxxXxdyyYyp

dxxXxdyyYypdyxyf

xf

yxfxyf

XYX

X

XY

XXY

yprexxxxxxl As

xxxxxxxxxIn 看不懂的字

看不懂的字

續下頁


)(

),(),()|(

)|()(),()(

||

|

xf

dttxfdttxfxyF

dxxyfxfdxyxfyf

X

y

y

XYXY

XYXY

Function onDistributi lConditiona

這邊的三個 example 字我看不太懂，先跳過了…

Conditional Expectation

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1. Explicit inversion formula -- we won’t use

2. Inspection – using table3. Partial fraction expansion

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queuing. problem solving for performance evaluation analysis: use mathematical model to solve...

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