matrix-geometric analysis and its applicationscslui/csc5420/matrix-geometric_elegant.pdf ·...

OutlineIntroduction

Matrix-Geometric in ActionGeneral Matrix-Geometric Solution

Application of Matrix-GeometricProperties of Solutions

Computational Properties of R

Matrix-Geometric Analysis and ItsApplications

John C.S. Lui

Department of Computer Science & EngineeeringThe Chinese University of Hong Kong

Email: [email protected]

Summer Course at Tsinghua, 2005.

John C.S. Lui Matrix-Geometric Analysis and Its Applications

OutlineIntroduction

IntroductionWhy we need the Matrix-Geometric Technique?

Matrix-Geometric in ActionKey Idea

General Matrix-Geometric SolutionGeneral Concept

Application of Matrix-GeometricPerformance Analysis of Multiprocessing System

Properties of SolutionsProperties

Computational Properties of RAlgorithm for solving R

OutlineIntroduction

Why we need the Matrix-Geometric Technique?

OutlineIntroduction

Why we need the Matrix-Geometric Technique?Matrix-Geometric in Action

Key IdeaGeneral Matrix-Geometric Solution

General ConceptApplication of Matrix-Geometric

Performance Analysis of Multiprocessing SystemProperties of Solutions

PropertiesComputational Properties of R

Algorithm for solving R

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Motivation

� Closed-form solution is hard to obtain.� Need to seek efficient, numerical stable solutions.� Can be viewed as a generalization of conventionalqueueing analysis.� A Special way to solve a Markov Chain.

OutlineIntroduction

Motivation

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Motivation

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Motivation

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Key Idea

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Key Idea

Key Ideas� It is a technique to solve stationary state probability forvector state Markov processes. Two parts:

1. Boundary set2. Repetitive set� Example: a modified M

�M�1 �� if the system is empty,

else � . Customers require two exponential stages ofservice, � 1 � and � 2

S �� i � s �� i � 0 and it is the no. of customer in the queue, �s is the current stage of service, s �� 1 � 2 �� s � 0 if no customer in the system� Well, let us proceed to specify the state transition diagram,

then the Q matrix.John C.S. Lui Matrix-Geometric Analysis and Its Applications

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Key Idea

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Key Idea

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Key Idea

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Key Idea

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Key Idea

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Key Idea

λ λ λ λ

µ1 µ1 µ1 µ1µ2 µ2

..........

Let ai �� i , i � 1 � 2. Arrange states lexicographically,� 0 � 0 �� 0 � 1 �� 0 � 2 �� 1 � 1 �� 1 � 2 �� .John C.S. Lui Matrix-Geometric Analysis and Its Applications

OutlineIntroduction

Key Idea

The transition rate matrix Q is:

Q �� 0 0 0 0 0 0 0 � � 0 � a1 � 1 � 0 0 0 0 0 � � � 2 0 � a2 0 � 0 0 0 0 � � 0 0 0 � a1 � 1 � 0 0 0 � � 0 � 2 0 0 � a2 0 � 0 0 � � 0 0 0 0 0 � a1 � 1 � 0 � � 0 0 0 � 2 0 0 � a2 0 � � � ...

......

...... � �

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Key Idea

Let us re-write the Q in matrix form:

A0 �"! � 00 �$#&% A1 �"! � a1 � 1

0 � a2 #&% A2 �"! 0 0� 2 0 #B00 ')(*,+.-0/ -0/ 0

0 + a1 1 11 2 0 + a2

2354B01 ')(* 0 0- 0

0 -2364

B10 '87 0 0 00 1 2 0 9

Q � ��B00 B01 0 0 0 � � B10 A1 A0 0 0 � � 0 A2 A1 A0 0 � � 0 0 A2 A1 A0 � � ...

......

...... � �

OutlineIntroduction

Key Idea

Let us solve it. For the repetitive portion:j ; 1A0 � : jA1 � : j < 1A2 � 0 j � 2 � 3 �� (1)

This is similar to the solution of M�M�1. Therefore, : j is a

function only of the transition rates between states with j � 1queued customers and states with j queued customers.:

j � :j ; 1R j � 2 � 3 ��

or :j � :

1Rj ; 1 j � 2 � 3 �� (2)

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Key Idea

Putting (2) into (1), we have::1Rj ; 2A0 � : 1R j ; 1A1 � : 1RjA2 � 0 j � 2 � 3 ��

Since it is true for j � 2 � 3 �� , substitute j � 2, we have:

A0 � RA1 � R2A2 � 0

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Key Idea

For the initial portion: :0B00 � : 1B10 � 0:

0B01 � : 1A1 � : 2A2 � 0

or = :0 � : 1 > ! B00 B01

B10 A1 � RA2 # � 0

We also need:

1 � :0e � : 1 ?@

Rj ; 1e � : 0e � : 1 � I � R ; 1 e= :0 � : 1 > ! e B �00 B01� I � R ; 1e B �10 A1 � RA2 # � = 1 � 0 >

where M � is M with first column being eliminated.John C.S. Lui Matrix-Geometric Analysis and Its Applications

OutlineIntroduction

Key Idea

BNq � E[queued customers]� ?@

j : je � ?@j A 1

� j : 1Rj ; 1e � : 1 � I � R ; 2 e

Note: S � ?@j A 1

R j ; 1 � I � R � R2 �C � � SR � R � R2 � R3 �C � �

S � I � R D� I

S � I � I � R ; 1 �E� I � R ; 1

This is true only when the spectral radius of R is less than unity.John C.S. Lui Matrix-Geometric Analysis and Its Applications

OutlineIntroduction

General Concept

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General Concept

Q ��

B00 B01 0 0 0 � � B10 B11 A0 0 0 � � B20 B21 A1 A0 0 � � B30 B31 A2 A1 A0 � � B40 B41 A3 A2 A1 � �

......

We index the state by � i � j where i is the level, i � 0 and j is thestate within the level, 0 F j F m � 1 for i � 1.

OutlineIntroduction

General Concept

For the repetitive portion,?@k A 0

:j ; 1 < k Ak � 0 j � 2 � 3 �� (3):

j � :1R j ; 1 j � 2 � 3 �� (4)

putting (4) to (3), we have:?@k A 0

Rk Ak � 0

For the boundary states, we have:= :0 � : 1 > ! B00 B01G ?k A 1 Rk ; 1Bk0

G ?k A 1 Rk ; 1Bk1 #John C.S. Lui Matrix-Geometric Analysis and Its Applications

OutlineIntroduction

General Concept

Procedure (continue:)

Using the same normalization, we have= :0 � : 1 >IH e B �00 B01� I � R ; 1 e J G ?k A 1 Rk ; 1Bk0 K � G ?k A 1 Rk ; 1Bk1 L � = 1 � 0 >

Therefore, it boils down to

1. Solving R.

2. Solving the initial portion of the Markov process.

OutlineIntroduction

General Concept

Procedure (continue:)

Using the same normalization, we have= :0 � : 1 >IH e B �00 B01� I � R ; 1 e J G ?k A 1 Rk ; 1Bk0 K � G ?k A 1 Rk ; 1Bk1 L � = 1 � 0 >

Therefore, it boils down to

1. Solving R.

2. Solving the initial portion of the Markov process.

OutlineIntroduction

Performance Analysis of Multiprocessing System

OutlineIntroduction

Multiprocessing System

We have a multiprocessing system in which� K homogeneous processors.� Each processor is subjected to failure with rate M .� A single repair facility with repair rate N .� Jobs arrive at a Poisson rate � .� Whenever there is no processor available, all jobs are lost.

OutlineIntroduction

Markov Model

0,K 1,K 2,K

0,3 1,3 2,3

0,2 1,2 2,2

0,1 1,1 2,1

3µ µ3

2µ µ2

ααα γK γK γK

γ3 γ3 γ3

γ2 γ2 γ2

γγ γ

.................

OutlineIntroduction

Define bi ��O� i MP�QN for i � 1 � 2 ��R� K . We have:

B00 � = � N > % B01 � = N.� 0 �� S� 0 > % Bj0 � �� M0...0

TVUUUW % B0 X j � 0 j � 2 � 3 ��R�B1 X 1 �

�� b1 N 0 0 � � 0 0 02 M � b2 N 0 � � 0 0 00 3 M � b3 N � � 0 0 0...

......

...0 0 0 0 � � Y� K � 1 ZM � bK < 1 N0 0 0 0 � � 0 K M � bK

TVUUUUUUUWJohn C.S. Lui Matrix-Geometric Analysis and Its Applications

OutlineIntroduction

� The matrices of the repeating portion of the process are:

A0 �[� I % A1 � B1 X 1 % A2 �� 0 � � 0 00 2 � � � 0 0...

... � � ......

0 0 � � Y� K � 1 \� 00 0 � � 0 K �

TVUUUUUW� This can be solved numerically rather than using thetransform method.

OutlineIntroduction

� The matrices of the repeating portion of the process are:

A0 �[� I % A1 � B1 X 1 % A2 �� 0 � � 0 00 2 � � � 0 0...

... � � ......

0 0 � � Y� K � 1 \� 00 0 � � 0 K �

TVUUUUUW� This can be solved numerically rather than using thetransform method.

OutlineIntroduction

Properties

OutlineIntroduction

Properties� Informally, the stability of a process depends on the drift ofthe process for states in the repetitive portion.� For example, M

�M�1, the expected drift toward higher

states is �^]1. The expected drift toward lower states is�_� � 1 _� � � . The drift of the process is � � � . Process isstable if the total expected drift is NEGATIVE, or �a`Q� inour case.� Now suppose the process can go up by 1 and go down byat most K steps. Let the rate for l steps be r � l ,l � � K � � K � 1 ��R� 0 � 1.

r � 1 b� K@l A 1

� � l r � � l .c r � 1 d` K@l A 1

lr � l John C.S. Lui Matrix-Geometric Analysis and Its Applications

OutlineIntroduction

r � 1 b� K@l A 1

� � l r � � l .c r � 1 d` K@l A 1

OutlineIntroduction

r � 1 b� K@l A 1

� � l r � � l .c r � 1 d` K@l A 1

OutlineIntroduction

Properties� Analogous to the scalar case, we can think of the drift ofthe process in terms of levels. Assume that for therepetitive portion, we have m states, a transition from leveli , i efe 0, to level i � k , 1 F k F K� k

m@l A 1

Ak < 1 � j � l where Ak < 1 � j � l is the transition from state j in level i tostate l in level i � k .� Let fj , 0 F j F m � 1 be the probability that the process isin inter-level j of the repeating portion of the process oflevel i efe 0. The average drift from level i to level i � k is� k

m ; 1@j A 0

fjm ; 1@l A 0

Ak < 1 � j � l John C.S. Lui Matrix-Geometric Analysis and Its Applications

OutlineIntroduction

Properties� Analogous to the scalar case, we can think of the drift ofthe process in terms of levels. Assume that for therepetitive portion, we have m states, a transition from leveli , i efe 0, to level i � k , 1 F k F K� k

m@l A 1

Ak < 1 � j � l where Ak < 1 � j � l is the transition from state j in level i tostate l in level i � k .� Let fj , 0 F j F m � 1 be the probability that the process isin inter-level j of the repeating portion of the process oflevel i efe 0. The average drift from level i to level i � k is� k

m ; 1@j A 0

fjm ; 1@l A 0

Ak < 1 � j � l John C.S. Lui Matrix-Geometric Analysis and Its Applications

OutlineIntroduction

Properties

� To get the total drift, we sum the previous equation for all k ,0 F k F K � 1.� But what is fj? Let us define A � G K < 1

l A 0 Al , we havef �E� f0 � f1 ��g� fm ; 1 . Therefore:

f A � 0 h f e � 1� The stability condition is:

f A0e ` K < 1@k A 2

� k � 1 f Ak e

OutlineIntroduction

Properties

f A0e ` K < 1@k A 2

� k � 1 f Ak e

OutlineIntroduction

Properties

f A0e ` K < 1@k A 2

� k � 1 f Ak e

OutlineIntroduction

Algorithm for solving R� It is an iterative method. Let

R � 0 i� 0

R � n � 1 i� � ?@l A 0 X l jA 1

R l � n AlA; 11 n � 0 � 1 � 2 �� The iterative process halts whenever entries in R � n � 1

and R � n differ in absolute value by less than a givenconstant.� The sequence R � n �� are entry-wise non-decreasing andconverge monotonically to a non-negative matrix R.� the number of iteration needed for convergence increasesas the spectral radius of R increases. This is similar to thescalar case where klc 1. As the system utilizationincreases, it becomes computationally more difficult to getR. John C.S. Lui Matrix-Geometric Analysis and Its Applications

OutlineIntroduction

R � 0 i� 0

R � n � 1 i� � ?@l A 0 X l jA 1

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R � 0 i� 0

R � n � 1 i� � ?@l A 0 X l jA 1

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R � 0 i� 0

R � n � 1 i� � ?@l A 0 X l jA 1

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Replicated Database

� Poisson arrival with rate � .� Probability it is a read request: r .� A read request can be served by any server.� A write request has to be served by BOTH servers.� What is the proper state space?

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Replicated Database

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Replicated Database

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Replicated Database

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Replicated Database

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The state space S �E� i � j where i is the number of queuedcustomers and j is the number of replications that are involvedin service. So i � 0 and j � 0 � 1 � 2.

λ(1-r) λ

λ λ λ

µ2 r µ2 r µ2 r

λ(1-r)

µ2 (1-r) µ2 (1-r)

...........

OutlineIntroduction

Q ' (mmmmmmmmm*+.- - r -on 1 + r p 0 0 0 q�q�q1 +rns-ut 1 p - r -vn 1 + r p 0 0 q�q�q0 2 1 +rns-wt 2 1 p 0 - 0 q�q�q0 0 1 +rnx-yt 1 p 0 - q�q�q0 0 2 1 r 2 1 n 1 + r p +rnx-ut 2 1 p 0 q�q�q0 0 0 0 1 +rns-wt 1 pzq�q�q...

......

... q�q�qB00 � �� r �b� 1 � r � � �\��{�| � r

0 2 � � �\�O� 2 �| TW % B01 � �� 0 0�b� 1 � r 00 � TW

B10 �}! 0 0 �0 0 2 � r #&% B11 � A1 �"! � �\�O�{�| 0

2 �_� 1 � r � �\�O� 2 �| #&%John C.S. Lui Matrix-Geometric Analysis and Its Applications

OutlineIntroduction

A0 �"! � 00 �$#&% A2 �"! 0 �

0 2 � r #To determine the stability:

A �}! � � �2 �_� 1 � r � 2 �_� 1 � r l# � A0 � A1 � A2

f1 � 2 � 1 � r 3 � 2r % f2 � 1

3 � 2r

f A0e ` K < 1@k A 2

� k � 1 f Ak e c~��` 2 �3 � 2r

matrix-geometric analysis and its applicationscslui/csc5420/matrix-geometric_elegant.pdf ·...

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