IE 332 - Chapter 5 - kaist

On Closed Form Solutions for Equilibrium Probabilities in the Closed Lu-Kumar Network under Various Buffer Priority Policies

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

IEEE ICCA 2010 Xiamen, China June 11, 2010

IEEE ICCA 2010 – Xiamen, June 11, 2010

Presentation Overview  Introduction  System Description  Equilibrium Probabilities Under the LBFS Policy

 Equilibrium Probabilities Under the FBFS Policy  Conclusion

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

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

Customers arrive

Server 1

Customers exit

Server 2

Customers arrive

< Jackson network >


Customers exit

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

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


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

< Closed Lu-Kumar network >


System Description: Network Model

A closed reentrant queueing network

 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


System Description: Last Buffer First Served


 Non-idling , preemptive  Gives priority b1 over b4 and b3 over b2


System Description: First Buffer First Served


 Non-idling , preemptive  Gives priority b4 over b1 and b2 over b3


Equilibrium Probabilities under LBFS N-1

1

0

0


 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


Transition diagram under LBFS

Equilibrium Probabilities under LBFS • To find equilibrium probability : Balance equations Π=ΠP •“Flow in” = “Flow out”  4  ( 0,0,0, N )   3   ( 0,0,1, N 1)

So, assuming that we know we can obtain  ( 0,0,1, N 1) .

 ( 0, 0, 0, N )

,

(  3   4 )  ( 0,0,1, N 1)   2   ( 0,1,0, N 1)

So we can express  ( 0,1,0, N 1) in terms of  ( 0,0,0, N ) Recursively, we can express whole steady state probabilities in terms of initial condition  ( 0,0,0, N ) . Transition diagram under LBFS


Equilibrium Probabilities under LBFS X 0 [ N  1]

To specify our main idea, we redefine the state as below : X 0 [n] : 1,n,0, N n1 , X 1[n] : 1,n,1, N n2 , X 2 [n] :  0,n1,0, N n1 and X 3 [n] :  0,n,1, N n1  4 X 2 [-1]   3  X 3 [0] 1 X 0 [0]   3  X 1 [0]   4  X 2 [-1] (  3   4 )X 3 [0]   2  X 2 [0]

X 0 [1]

X 0 [0]

( 1   2 )X 0 [n]   3  X 1 [n]   4  X 2 [n - 1],

1  n  N-2

( 1   3 )X 1 [n]   2  X 0 [n  1]   4  X 3 [n] ,

0  n  N-2

(  2   4 )X 2 [n]  1  X 0 [n]   3  X 3 [n  1],

0  n  N-2

(  3   4 )X 3 [n]  1  X 1 [n - 1]   2  X 2 [n] ,

1  n  N-2

( 1   2 )X 0 [N - 1]   4  X 2 [N - 2]

 3 X 3 [N - 1]  1  X 1 [N - 2]   2  X 2 [N - 1]  2 X 2 [N - 1]  1  X 0 [N - 1]


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 X 0 [n  4]  A  X 0 [n  3]  B  X 0 [n  2]  C  X 0 [n  1]  D  X 0 [n] , 0  n  N 5

Step 2: Taking z-transform and inverting it give a closed form solution for the signal ^ x [0]z 4  (c1x0 [0]  c2 x3[0]) z 3  (c3 x0 [0]  c4 x3[0]) z 2  (c5 x0 [0]  c6 x3[0]) z X 0 ( z)  0 , 4 3 2 z  Az  Bz  Cz  D

 3   3      X 0 [ n]    i [1]  pi n   X 0 [0]    i [2]  pi n   X 3[0]      i 1   i 1 





0  n  N 1

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


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]


Equilibrium Probabilities 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


Transition diagram under FBFS

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

Initial conditions

 3  ( 0,0, N ,0)   2   ( 0,1, N 1,0)

So, assuming that we know  ( 0,0, N ,0) , we can obtain  ( 0,1, N 1,0) .  2  (0,1, N 1,0)  1   (1,0, N 1,0)   2   (0, 2, N 2,0)

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

Transition diagram under FBFS


Equilibrium Probabilities under FBFS X 0 [1] X 0 [0]

To specify our main idea, we redefine the state as below : X k [n] :  k ,0, N  k  n, n and Yk [n] :  N  k  n, n, k ,0  4 X 0 [ N ]   3  X 0 [ N  1]  4 X n 1 [ N  n  1]   3  X n 1 [ N  n  2]   4 X n [ N  n], 0 n N 2 (  3   4 ) X 0 [n  1]   3  X 0 [n],

0 n N 2

(  3   4 ) X k [n  1]   3  X k [n]   4  X k 1 [n  2], 1  k  N  2, 0  n  N  k  2

 3 X 0 [0]   2YN 1 [1] ( 1   3 ) X n 1 [0]   2  YN  n  2 [1]   4  X n [1], 1 X N [0]   4  X N 1 [1] ( 1   2 )Yk [n  1]  1  Yk [n]   2  Yk 1 [n  2],

0 n N 2

1  k  N  2, 0  n  N  k  2 ( 1   2 )Y0 [n  1]  1  Y0 [n], 0 n N 2

 2Yn 1 [ N  n  1]  1  Yn 1 [ N  n  2]   2Yn [ N  n], 0 n N 2  2Y0 [ N ]  1  Y0 [ N  1]


Equilibrium Probabilities under FBFS  Overall steps for obtaining closed form solutions

Step 1: Investigating X0[n], we obtain relationship below: X 0 [n  1]    X 0 [n]

0 n  N 2

X 0 [n]   n X 0 [0]

 3        3 4  

Step 2: Using relationship between Xk[m] and Xk-1[n], we obtain X1[n]. 4 X n1[ N  n  1]  3  X n1[ N  n  2]  4 X n [ N  n]

X 1[n]   n X 1[0]  n n1 X 0 [0]  4    3   4 

Step 3: Recursively, we can obtain X k [n  1]    X k [n]    X k 1[n  2],

  

1  k  N  2, 0  n  N  k  2 k

X k [n]    X k [0]  n   n

n

 (ni!(n2i i)!1)!  ( )  X i

k  i [0],

i 1


0  k  N  1, 0  n  N  k  1.

Equilibrium Probabilities under FBFS Step 4: By symmetry, we get the inverse transforms for the lower region k

Yk [n]    X N  k [0]  n   n

n

 i 1

(n  2i  1)!  ( ) i  X N  k  i [0], i!(n  i)!

0  k  N  1, 1  n  N  k  1.

Step 5: Using remaining balance equations, we express all Xk[n] in terms of X0[0].(Toeplitz matrix structure) 1   B [1]   B[ 2]  .  .   .   B[ N  3]  B[ N  2]   C[1] 

A[ N  3] A[ N  4] A[ N  5] . . . 1 B[1]

A[ N  2] A[ N  3] A[ N  4] . . . A[1] 1

... C[ N  2]

C[ N  1]

A[1] 1 B[1] . . . B[ N  4] B[ N  3]

A[ 2] A[1] 1

... ... ...

B[ N  5] B[ N  4]

... ...

C[ 2]

C[3]

A[ N  1]   A[ N  2] A[ N  3]  .  .   .  A[ 2]  A[1]     1   3 

 X 1[0]    B[1]       X 2 [0]    B[ 2]   X 3 [0]    B[3]      . .       . .  X 0 [0]     . .      X N  2 [0]  B[ N  2] X [0]    B[ N  1]   N 1     X N [0]    C[0] 

( 2n  2)! n!( n  1)! ( 2n  2)! B[ n]     n   n 1  n!( n  1)!

where, A[ n]     n   n 1 

N n2

C[ n] 

 i 1

i  i 

( 2 N  2n  i  3)!  (   ) N  n  i 1   N  n 1. ( N  n  i  1)!( N  n  1)!


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 A[1] A[2] ...  1  1 A[1] ...  B[1]  B[2] B[1] 1 ...  . .  . .   . .   B[ N  3] B[ N  4] B[ N  5] ...  B[ N  2] B[ N  3] B[ N  4] ...   C[1] C[2] C[3] ...  (2n  2)! where, A[n]     n   n 1  n!(n  1)! (2n  2)! B[n]     n   n 1  n!(n  1)! N n2

C[n] 

 i 1

i  i 

A[ N  3] A[ N  2] A[ N  1]   A[ N  4] A[ N  3] A[ N  2] A[ N  5] A[ N  4] A[ N  3]  . . .  . . .   . . .  1 A[1] A[2]  B[1] 1 A[1]    C[ N  2] C[ N  1]  1  3 

 X 1[0]    B[1]       X 2 [0]    B[2]   X 3[0]    B[3]      . .       . .  X 0 [0]     . .      X N  2 [0]  B[ N  2] X [0]    B[ N  1]   N 1     X N [0]    C[0] 

(2 N  2n  i  3)!  (   ) N  n  i 1   N  n 1. ( N  n  i  1)!( N  n  1)!


Concluding Remarks  LBFS : Indeed obtained a closed form solution  FBFS : 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 works  Attempting to obtain a closed-form expression for the inverse of the

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