IEEE ICCA 2010 – Xiamen, June 11, 2010

On Closed Form Solutions for Equilibrium Probabilities in the Closed Lu-Kumar Network

under Various Buffer Priority Policies

Seunghwan Jung and James R. MorrisonKAIST, Department of Industrial and Systems Engineering

IEEE ICCA 2010 Xiamen, ChinaJune 11, 2010

IEEE ICCA 2010 – Xiamen, June 11, 2010 - 2

Presentation Overview

IntroductionSystem Description Equilibrium Probabilities Under the LBFS PolicyEquilibrium Probabilities Under the FBFS PolicyConclusion

IEEE ICCA 2010 – Xiamen, June 11, 2010 - 3

Introduction

Server 1

Server 2Custome

rs arrive

Customers arriveCustom

ers exitCustomers exit

< Jackson network >

• Jackson network is one of the rare class of network that possess closed form equilibrium probability distributions.

IEEE ICCA 2010 – Xiamen, June 11, 2010 - 4

Introduction

< General reentrant network [1] >

• Except for some classes of networks, few networks possess closed form equilibrium probability distributions.

[1] James R. Morrison, “Implementation of a Fluctuation Smoothing Production Control Policy in IBM’s 200mm Wafer Fab”, Euro-pean Control Conference, pp. 7732-7737, 2005.

IEEE ICCA 2010 – Xiamen, June 11, 2010 - 5

Introduction

< Closed Lu-Kumar network >

• Obtain closed form equilibrium probabilities.• Allows complete characterization of the steady state behavior.

IEEE ICCA 2010 – Xiamen, June 11, 2010 - 6

System Description: Network Model

Two stations : σ1 and σ2

Buffers : b1, b2 , b3 , b4

Service time for a customer in buffer bi : exponential with rate μi

N trapped customers circulate within the network

A closed reentrant queueing network

IEEE ICCA 2010 – Xiamen, June 11, 2010 - 7

System Description: Last Buffer First Served

Non-idling , preemptiveGives priority b1 over b4 and b3 over b2

A closed reentrant queueing network

IEEE ICCA 2010 – Xiamen, June 11, 2010 - 8

System Description: First Buffer First Served

Non-idling , preemptiveGives priority b4 over b1 and b2 over b3

A closed reentrant queueing network

IEEE ICCA 2010 – Xiamen, June 11, 2010 - 9

Equilibrium Probabilities under LBFS

System state at time t : S(t)={w(t),x(t),y(t),z(t)} w(t),x(t),y(t),z(t) : Number of customers in

buffers b1, b2, b3, b4 at time t Uniformization : Get Discrete time Markov chain Steady state probability of state S : Πs

A closed reentrant queueing network

Transition diagram under LBFS

1 N-1

00

IEEE ICCA 2010 – Xiamen, June 11, 2010 - 10

Equilibrium Probabilities under LBFS

Transition diagram under LBFS

• To find equilibrium probability : Balance equations Π=ΠP•“Flow in” = “Flow out”

So, assuming that we know ,we can obtain .

),0,0,0( N

)1,1,0,0( N

)1,1,0,0(3),0,0,0(4 NN

)1,0,1,0()1,1,0,0()( 243 NN

So we can express in terms of Recursively, we can express whole steady state

probabilities in terms of initial condition .

)1,0,1,0( N ),0,0,0( N

),0,0,0( N

IEEE ICCA 2010 – Xiamen, June 11, 2010 - 11

Equilibrium Probabilities under LBFS

1,1,,031,0,1,02

2,1,,111,0,,10

:][and:][

,:][,:][

nNnnNn

nNnnNn

nXnX

nXnX

]0[0X

]1[0X

]1[0 NX

To specify our main idea, we redefine the state as below :

1]-[N0X11]-[N2X2

1]-[N2X22]-[N1X11]-[N3X3

2]-[N2X41]-[N0)X21(

21[n]2X21]-[n1X1[n]3)X43(

20]1[n3X3[n]0X1[n]2)X42(

20[n]3X41][n0X2[n]1)X31(

211]-[n2X4[n]1X3[n]0)X21(

[0]2X2[0]3)X43(

[-1]2X4[0]1X3[0]0X1

[0]3X3[-1]2X4

N-n,

N-n,

N-n,

N-n,

IEEE ICCA 2010 – Xiamen, June 11, 2010 - 12

Equilibrium Probabilities under LBFS Overall steps for obtaining closed form solutions

Step 1: We make the equation involving only one type of signal by combining given equations

Step 2: Taking z-transform and inverting it give a closed form solution for the signal

Step 3: Plugging the closed form solution into the other balance equations gives closed form solutions for them

50,][]1[]2[]3[]4[ 00000

NnnXDnXCnXBnXAnX

,])0[]0[(])0[]0[(])0[]0[(]0[

)(234

36052

34033

32014

00^

DCzBzAzz

zxcxczxcxczxcxczxzX

10]0[]2[]0[]1[][ 3

3

10

3

10

NnXpXpnXi

nii

i

nii

IEEE ICCA 2010 – Xiamen, June 11, 2010 - 13

Equilibrium Probabilities under LBFS Overall steps for obtaining closed form solutions (continued)

Step 4: Using the balance equations, all Xk[n] are expressed in terms of X0[0]

Step 5: Summing all probabilities and setting them equal to 1 to get X0[0]

IEEE ICCA 2010 – Xiamen, June 11, 2010 - 14

Equilibrium Probabilities under FBFS

A closed reentrant queueing network

Transition diagram under FBFS

System state at time t : S(t)={w(t),x(t),y(t),z(t)} w(t),x(t),y(t),z(t) : Number of customers in

buffers b1, b2, b3, b4 at time t Uniformization : Get Discrete time Markov chain Steady state probability of state S : Πs

IEEE ICCA 2010 – Xiamen, June 11, 2010 - 15

Equilibrium Probabilities under FBFS

Transition diagram under FBFS

)0,1,1,0(2)0,,0,0(3 NN

)0,2,2,0(2)0,1,0,1(1)0,1,1,0(2 NNN

Recursively, we can express whole steady state probabilities in terms of initial conditions.

• To find equilibrium probability : Balance equations Π=ΠP•“Flow in” = “Flow out”

Initial conditions

So, assuming that we know ,we can obtain .

)0,,0,0( N

)0,1,1,0( N

IEEE ICCA 2010 – Xiamen, June 11, 2010 - 16

Equilibrium Probabilities under FBFS

0,,,,,0, :][:][ knnkNknnkNkk nYandnX

To specify our main idea, we redefine the state as below :

]1[][20

],[]2[]1[20],[]1[)(

20,21],2[][]1[)(

]1[]0[20],1[]1[]0[)(

]1[]0[20,21],2[][]1[)(

20],[]1[)(20

],[]2[]1[]1[][

0102

21112

01021

12121

141

422131

1203

14343

03043

41314

0304

NYNYNn

nNYnNYnNYNnnYnY

kNnNknYnYnY

XXNnXYX

YXkNnNknXnXnX

NnnXnXNn

nNXnNXnNXNXNX

nnn

kkk

NN

nnNn

N

kkk

nnn

]0[0X

]1[0X

IEEE ICCA 2010 – Xiamen, June 11, 2010 - 17

Equilibrium Probabilities under FBFS Overall steps for obtaining closed form solutions

Step 1: Investigating X0[n], we obtain relationship below:

Step 2: Using relationship between Xk[m] and Xk-1[n], we obtain X1[n].

Step 3: Recursively, we can obtain

]0[][ 00 XnX n20][]1[ 00 NnnXnX

]0[]0[][ 01

11 XnXnX nn

20,21],2[][]1[ 1

kNnNknXnXnX kkk

.10,10],0[)()!(!

)!12(]0[][1

kNnNkXini

innXnXk

iik

ink

nk

][]2[]1[ 41314 nNXnNXnNX nnn

43

3

43

4

IEEE ICCA 2010 – Xiamen, June 11, 2010 - 18

Equilibrium Probabilities under FBFS

Step 4: By symmetry, we get the inverse transforms for the lower region

Step 5: Using remaining balance equations, we express all Xk[n] in terms of X0[0].(Toeplitz matrix structure)

.11,10,]0[)()!(!

)!12(]0[][1

kNnNkXini

innXnYk

iikN

inkN

nk

.)()!1()!1(

)!322(][

)!1(!)!22(

][

)!1(!)!22(

][,

]0[

]0[]1[]2[

.

.

.]3[]2[]1[

]0[]0[]0[

.

.

.]0[]0[]0[

]1[]2[...]3[]2[]1[

]1[1]1[...]4[]3[]2[]2[]1[1...]5[]4[]3[

.

.

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.]3[]4[]5[...1]1[]2[]2[]3[]4[...]1[1]1[]1[]2[]3[...]2[]1[1

12

1

1

1

1

0

1

2

3

2

1

3

1

nNnN

i

inNi

nn

nn

N

N

N

nNinNinN

inC

nnn

nB

nnn

nAwhere

X

CNBNB

BBB

XXX

XXX

NCNCCCC

ABNBNBNBAANBNBNB

NANANABBNANANAABNANANAAA

IEEE ICCA 2010 – Xiamen, June 11, 2010 - 19

Equilibrium Probabilities under FBFS

Step 5: Summing all probabilities and setting them equal to 1 to get X0[0]

Note: Not a complete closed form

.)()!1()!1(

)!322(][

)!1(!)!22(

][

)!1(!)!22(

][,

]0[

]0[]1[]2[

.

.

.]3[]2[]1[

]0[]0[]0[

.

.

.]0[]0[]0[

]1[]2[...]3[]2[]1[

]1[1]1[...]4[]3[]2[]2[]1[1...]5[]4[]3[

.

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.]3[]4[]5[...1]1[]2[]2[]3[]4[...]1[1]1[]1[]2[]3[...]2[]1[1

12

1

1

1

1

0

1

2

3

2

1

3

1

nNnN

i

inNi

nn

nn

N

N

N

nNinNinN

inC

nnn

nB

nnn

nAwhere

X

CNBNB

BBB

XXX

XXX

NCNCCCC

ABNBNBNBAANBNBNB

NANANABBNANANAABNANANAAA

IEEE ICCA 2010 – Xiamen, June 11, 2010 - 20

Concluding Remarks

LBFS : Indeed obtained a closed form solutionFBFS : Enough structure to reduce the computational complexity

• To obtain equilibrium probabilities by “Π=ΠP”, we have to inverse

(N+1)2╳(N+1)2 matrix.

Future worksAttempting to obtain a closed-form expression for the inverse of the

Toeplitz matrix from the FBFS case.Extend the structure to more general cases.