1 TCOM 501: Networking Theory & Fundamentals Lecture 8 March 19, 2003 Prof. Yannis A. Korilis.
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8-2 Topics
Closed Jackson Networks Convolution Algorithm Calculating the Throughput in a Closed Network Arrival Theorem for Closed Networks Mean-Value Analysis Norton’s Equivalent
8-3 Closed Jackson Networks
Closed Network: K nodes with exponential servers No external arrivals (γi=0) , no departures (ri0=0) Fixed number M of circulating customers
Appropriate model for systems with “limited” resources, e.g., flow control mechanisms
Steady-state distribution will be shown to be of “product-form” type
1
2
3
4
5
1
2
43
551r
53r
M
8-4 Closed Jackson Network
Aggregate arrival rates
Relative arrival rates – visit ratios Can only be determined up to a constant Use an additional equation to obtain unique solution to the above system, e.g.
Set λj=1, for some node j
Set λj=μj, for some node j
Set λ1+ λ2+…+ λK=1
ni: number of customers at node i
Possible states for the closed network n=(n1, n2,…,nK), with
Let F(M) denote the set of all such states
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2
3
4
5
1 2
43
551r
53r
M
1, 1,...,
K
i j jijr i K
10 and | |
K
i iin n n M
8-5 Closed Jackson Network
Let xi be the number of customers at station i, at steady state
Random variables x1, x2,…, xK are not independent – their sum must be equal to M
The state x=(x1, x2,…, xK) of the closed network can take values n=(n1, n2,…,nK), with
Let F(M) denote the set of all such states Define ρi ≡ λi/μi – this is not the actual utilization factor of station i Jackson’s theorem for closed networks gives the stationary distribution
10 and | |
K
i iin n n M
1 1( ) { } { , , }K Kp n P x n P x n x n
8-6
Theorem 1: The stationary distribution of a closed Jackson network is
where the normalization constant G(M) is a function of M G(M) guarantees that {p(n)} is a valid probability distribution
This stationary distribution is said to have a product-form However: at steady-state the queues are not independent
{pi(ni)}: marginal stationary distribution of queue i
Jackson’s Theorem for Closed Networks
1
1( ) , for all ( ) { : 0, | | }
( )i
Kn
i ii
p n n M n n n MG M
F
( ) ( ) 1
( ) 1 ( ) i
Kn
in M n M i
p n G M
F F
1 1( ) ( ) ( )K Kp n p n p n
8-7 Jackson’s Theorem for Closed Networks (proof)
Theorem 2: The reversed chain of a stationary closed Jackson network is also a stationary closed Jackson network with the same service rates and routing probabilities:
Proof of Theorems 1 & 2: Show that for the corresponding forward and reversed chains
Need to prove only for m=Tijn
Verify, exactly as in the open-network case, that:
* /ij j ji ir r
*( ) ( , ) ( ) ( , ), , ( ),p m q m n p n q n m n m M n m F
* * *( , ) ( , ) { 1 0} ( / )ij i j j ji i j i ij jq T n n q n e e n r n r 1*
1
( ) ( , ) ( ) ( , ) ( ) ( / ) ( ) 1{ 0}
( ) ( )1{ 0}ij ij ij ij j i ij j i ij i
ij j i i
p T n q T n n p n q n T n p T n r p n r n
p T n p n n
( , ) { 0}ij i ij iq n T n r n 1
*( , ) ( , ) 1{ 0}, ( )i im n m n i
q n m q n m n n M
F
8-8 Closed Jackson Network
Example: Closed network model for CPU (rate μ1) and I/O (rate μ2) system. Upon service completion in 1, customer routed to 2 with probability p2=1-p1, or back to 1 with probability p1. M =fixed number of customers
Stationary distribution: n customers in 2 and M-n in 1
Normalization constant
Utilization factor and throughput of node 1:
1
2
1p
2p
2
1
1 1 1 2 2 2 1 1 1 2 2 1, . Choose solution: and p p p 2 1
1 22
1, p
1 2 2
1 1( , ) , 0,1,...,
( ) ( )M n n np M n n n M
G M G M
12
202
1( )
1
MM nn
G M
2 ( 1)( ) 1 (0, ) 1
( ) ( )
M G MU M p M
G M G M
1 1 1
( 1)( ) ( )
( )
G MM U M
G M
8-9 Closed Networks: Normalization Constant
Normalization constant as a function of M and K:
All performance measures of interest – throughput, average delay – can be obtained in terms of G(M,K)
Computational complexity is exponential in M and K:
Recursive methods can be used to reduce complexity Iterative algorithm [due to Buzen]
Normalization constant will be treated as a function of both M and K and denoted G(M,K) only in the context of the iterative algorithm
1 2
1
1 2( ) 1
0
( , ) i K
K
i
Kn n n n
i Kn M i n n M
n
G M K
F
1 terms in the summation
M K
M
8-10 Iterative Computation of G(M)
For any m and k (m=0,…, M; k=1,…, K) define:
For a closed network of single-server queues G(M,K) can be computed iteratively using the following recursive relation:
with boundary conditions:
1 2
1
1 2( ) 1
( , ) i k
k
kn nn n
i kn m i n n m
G m k
F
( , ) ( , 1) ( 1, )kG m k G m k G m k
1( ,1) , 0,1,...,
(0, ) 1, 1,2,...,
mG m m M
G k k K
8-11 Iterative Algorithm (proof)For 0 and 1m k we split the sum into two sums over disjoint sets of states, corresponding to 0,kn and 0kn .
1 2
1
1 2 1 2
1 1
11 2 1 2
1 1 1
1 2
1 2 1 2
0 0
1 2 1 1 2
0
( , ) k
k
k k
k k
k k
k k
k k
k
nn nk
n n M
n nn n n nk k
n n m n n mn n
n nn n n nk k
n n m n n mn
G m k
Note that the first sum is ( , 1).G m k For the second sum, observing that 0,kn we define
1,k kn n where 0.kn Then:
1 2 1 2
1 1
1 2
1
11 2 1 2
10 0
1 21
0
( 1, )
k k
k k
k k
k
k
k
n nn n n nk k
n n m n n mn n
nn nk k k
n n mn
G m k
Therefore: ( , ) ( , 1) ( 1, )kG m k G m k G m k
8-12 Iterative Algorithm – Example
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551
1
2r
53
1
2r
M
2 22
1
Visit ratios λi determined up to a multiplicative constant
Letting λ1= 2μ, we have: Calculation of G(M,5) based on the iterative algorithm using these values
1 2 3 4
51 53 1 3
5 1 3 1
,
2
r r
1 2 1 3 4 3 1
5 5 1 1
/ 2 , / 2
/ 2 / 4 / , with /
1 2 3 4 51, 2, 4 /
8-13 Iterative Algorithm – Example
Boundary conditions:
Iteration:
1 2 3 4 5
0 1 1 1 1 1 1 1 2 4 6 6+4/ρ 2 1 3 11 23 23+(6+4/ρ)(4/ρ) 3 1 4 26 72 72+[23+(6+4/ρ)(4/ρ)](4/ρ)
km
1 2 3 4 51, 2, 4 /
1( ,1) , 0,1,...,
(0, ) 1, 1,2,...,
mG m m M
G k k K
( , ) ( , 1) ( 1, )kG m k G m k G m k
Example:
2
3
2
(1,2) (1,1) (0,2) 1 1 2
(1,3) (1,2) (0,3) 2 2 4
(2,2) (2,1) (1,2) 1 2 3
G G G
G G G
G G G
8-14 Marginal Distribution
Proposition 1: In a closed Jackson network with M customers, the probability that at steady-state, the number of customers in station j greater than or equal to m is:
Proof 1:
( ){ } , 0
( )m
j j
G M mP x m m M
G M
1
1
1
1
1
1
1
( )
1
1
0
{ } ( )( )
( )
( )
( )( )
j K
j K
j j
j K
j K
j
j K
j K
j
nn nj K
jn M n n n M
n m n m
n mn nj K
n n m n Mn m m
mnj n nj K
n n n M mn
mj
P x m p nG M
G M
G M
G M mG M
F
8-15 Marginal Distribution
Proposition 2: In a closed Jackson network with M customers, the probability that in steady state there are m customers at station j is:
Proof 2:
Proposition 3: In a closed Jackson network with M customers, the average number of customers at queue j is:
Proof 3:
( ) ( 1){ } , 0
( )m i
j j
G M m G M mP x m m M
G M
{ } { } { 1}j j jP x m P x m P x m
1 1
( )[ ] { }
( )
M Mm
j j j jm m
G M mN E x P x m
G M
1
( )( )
( )
Mm
j jm
G M mN M
G M
8-16
Proposition 4: In a closed Jackson network with M customers, the average throughput of queue j is:
Proof 4: Average throughput is the average rate at which customers are serviced in the queue. For a single-server queue the service rate is μj when there are one or more customers in the queue, and 0 when the queue is empty. Thus:
Average Throughput
( 1)( )
( )j j
G MM
G M
( 1) ( 1)( ) { 1}
( ) ( )j j j j j j
G M G MM P x
G M G M
8-17 Example: ./M/1 Queues in Tandem
1 2 K
Choose 1 2 1, 1,...,K i i K
1 21 2
( ) ( )
1 ( 1)!( ) 1
!( 1)!Kn n n
Kn M n M
M K M KG M
M M K
F F
1
1 1 !( 1)!( ) , for all ( )
1( ) ( 1)!i
Kn
ii
M Kp n n M
M KG M M K
M
F
M
K21
( )M
8-18 Example: ./M/1 Queues in Tandem (cont.)
Average throughput:
For queue j=1,…,K:
Average time-delay:
M
K21
( )M
( 1) ( 2)! !( 1)!( ) ( )
( ) ( 1)!( 1)! ( 1)!
1
j j
G M M K M KM M
G M M K M K
M
M K
( )
( ) 1 1 1 1( )
( )
j
jj
j
MN M
KN M M M K M K
T MM K M K
1( ) ( )jj
M KT M T M
8-19 Arrival Theorem for Closed Networks
Theorem: In a closed Jackson network with M customers, the occupancy distribution seen by a customer upon arrival at queue j is the same as the occupancy distribution in a closed network with the arriving customer removed.
Corollary: In a closed network with M customers, the expected number of customers found upon arrival by a customer at queue j is equal to the average number of customers at queue j, when the total number of customers in the closed network is M-1.
Intuition: an arriving customer sees the system at a state that does not include itself.
8-20 Arrival Theorem (proof) 1( ) ( ( ),..., ( ))Kx t x t x t state of the network at time t
( )ijT t probability that a customer moves from station i to j, at time t+
For any state ( )n MF with 0in , find the conditional probability that a customer
moving from node i to node j finds the network at state n
1
1
( )0
1
1( ) ( )0 0
{ ( ) , ( )} { ( ) } { ( ) | ( ) }( ) { ( ) | ( )}
{ ( )} { ( ) } { ( ) | ( ) }
( )
( )
i
i K
i K
i i
ij ijij ij
ij ijm M
m
nn ni ij i K
mm mi ij i K
m M m Mm m
P x t n T t P x t n P T t x t nn P x t n T t
P T t P x t m P T t x t m
p n r
p m r
F
F F
Changing index 1, 0i i im m m in the sum in the denominator:
1
1
1
1 1
1
1
11
111 0
1 11 1
11
0
( )
( 1)
i K
i K
i K
i
i iK K
i K
i K
i
nn ni K
ij mm mi K
m m m Mm
n nn n n ni K i K
mm mi K
m m m Mm
n
G M
This is the probability of state 1( , , 1, )i Kn n n in an identical network with M-1
customers
8-21 Mean-Value Analysis Closed network with M customers; performance measures
Nj(M): average number of customers in queue j
Tj(M): average a customer spends (per visit) in queue j
γj(M): average throughput of queue j
Mean-Value Analysis: Calculates Nj(M) and Tj(M) directly, without first computing G(M) or deriving the stationary distribution of the network
Iterative calculation:
with initial condition:
Average throughput
1
1 ( 1)( ) , 1,..., ; 1,...,
( )( ) , 1,..., ; 1,...,
( )
jj
j
j jj K
i ii
N mT m j K m M
T mN m m j K m M
T m
(0) 0, 1,...,jN j K
( )( ) , 1,..., ; 1,...,
( )j
jj
N mm j K m M
T m
8-22
Arrival Theorem → expected number of customers that an arrival finds at queue j is Nj(m-1). Service rate for all customer at the queue μj.
λ1,…,λK: visit ratios – a solution to flow conservation equations Actual throughput of queue j: Using Little’s Theorem:
Summing for all j and noting that ∑j Nj(m) = m:
Then:
Mean Value Analysis (proof)
1 ( 1)( ) , 1,..., ; 1,...,j
jj
N mT m j K m M
( ) ( 1) / ( ) ( )j j jm G m G m m
( ) ( ) ( ) ( ) ( )j j j j jN m m T m m T m
1 11
( ) ( ) ( ) ( )( )
K K
i i i Ki i i ii
mm N m m T m m
T m
1
( )( ) , 1,..., ; 1,...,
( )j j
j K
i ii
T mN m m j K m M
T m
8-23 Example: ./M/1 Queues in Tandem
M
K21
( )M
1 2 1, 1,...,K i i K
1
1 (0) (1) (1)1 1(1) , (1) 1 , (1)
(1)(1)
j j jj j jK
ji
i
N T NT N K
K TT
1 (1) (2)1 1/ 1 1 2 2
(2) , (2) , (2)(2) 1
j jj j j
j
N NK KT N
K K T K
1 (2) (3)1 2 / 2 1 3 3
(3) , (3) , (3)(3) 2
j jj j j
j
N NK KT N
K K T K
1 (3) (4)1 3/ 3 1 4 4
(4) , (4) , (4)(4) 3
j jj j j
j
N NK KT N
K K T K
1 1
( ) , ( ) , ( )1j j j
M K M MT M N M M
K K M K
8-24 State-Dependent Service Rates
Theorem: The stationary distribution of a closed Jackson network where the nodes have state-dependent service rates is
where the normalization constant G(M) is a function of M, the fixed number of customers in the network
Normalization constant:
Proof similar to the one for open networks
1
1( ) , for all ( ) { : 0, | | }
( ) (1) ( )
inKi
ii i i i
p n n K n n n KG M n
F
( ) 1 ( )
( ) ( ) 1 (1) ( )
inKi
n K i n Ki i i
G M p nn
F F
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