Queueing Network Beamer

7/29/2019 Queueing Network Beamer

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Introduction to Queueing Networks

Prof. John C.S. LuiDepartment of Computer Science & Engineering

The Chinese University of Hong Kong

John C.S. Lui () Computer System Performance Evaluation 1 / 79


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Outline

Outline

1 Introduction

Illustration

2 Jackson Network

Example

Theory on Jackson Networks

Examples3 Closed Queueing Network

Example

Theory of Closed Queueing Network

Computation Methods

Convolution Algorithm

Multiclass Queueing Networks

BCMP Networks

Mean Value Analysis (MVA)



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Introduction Illustration

Example: Network of Queues

0

1!

0,0 1,0 2,0

0,1 1,1

0,2

,0k0

k0-1,1

! ! ! !

! !!

!

1 0

1

0

1

1

0

1 0

1

1

!

!

.........

.........

.........

.........

.........



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for k0 > 0, k1 > 0:

(0 +1 +)p(k0, k1) = 0p(k0 +1, k1 1)+1p(k0, k1 +1)+p(k0 1, k1)

for k0 > 0, k1 = 0:

(0

+ )p(k0

, 0) = 1

p(k0

, 1) + p(k0

1, 0)

for k0 = 0, k1 > 0:

(1 + )p(0, k1) = 0p(1, k1 1) + 1p(0, k1 + 1)

for k0 = 0, k1 = 0: p(0, 0) = 1p(0, 1)



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Normalization :

k0=0

k1=0

p(k0, k1) = 1

p(k0, k1) = (1 0)k00 (1 1)

k11 for k0, k1 = 0, 1,

where 0 =0

; 1 =1

;

P[N0 = k0] = (1 0)k0

0

; P[N1 = k1] = (1 1)k1

1


J k N k E l


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Jackson Network Example

Open Queueing Network [Jackson 57]

It allows "feedback" and "product-form" can still be maintained.

0

1

!

P0

P1

!0

!0

!1

CPU I/O

p(k0, k1) = (1 0)k00 (1 1)

k11

0 = + 1

1 = 0p1


J k N t k E l


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Therefore,

0 = 1 p1

1 =p1

1 p1

0 = 0

0=

(1 p1)0=

p00

1 =11

=p1

p01

What is the average response time of a job?




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By Littles formula, T = N

where N is the average no. of jobs in the

"system".

N = N0 + N1 =

k0=0

k1=0

(k0 + k1)p(k0, k1)

=

k0=0

k1=0

k0p(k0, k1) +

k0=0

k1=0

k1p(k0, k1)

=

k0=0

k0(1 0)k00

k1=0

(1 1)k11

+

k0=0

(1 0

)k00

k1=0

k1

(1 1

)k11

N =0

1 0+

11 1

; E[T] =N

=

0

1 0+

11 1

1




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Types of service centersFCFS and service time is exponentially distributed.Processor sharing (PS)Last come first serve pre-emptive resume (LCFS-PR)Infinite server (IS) or delay nodes

We also allow a state dependent service rate (i(n) = service rateat the ith node where there is ncustomer).

Single server fixed rate (SSFR) where i(n) = uiInfinite server (IS) , i(n) = ni

Queue length dependent (QLD) with service rate i(n)


Jackson Network Theory on Jackson Networks


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Jackson Network

A queueing network with M nodes (labeled i = 1, 2, , M) s.t.

Node i is QLD with rate i(n) when it has n customers.

A customer completing service at a node makes a probabilisticchoice of either leaving the network or entering another node,

independent of past history.

The network is open and any external arrivals to node i is from a

Poisson stream.




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Jackson Network : Continue

State space S = {(n1, n2, , nM) | ni 0}

Routing probability matrix Q = (qij | i,j = 1, 2, M)

qi0 = 1

Mj=1

qij

Let be the "TOTAL" external arrival rate to the open queueingnetwork, the rate to node i is i = q0i for i = 1, 2, M. So

=M

i=1

i (

q0i = 1)




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All nodes in the Jackson network are QLD with exponential

service time. Pictorially we can have:

!1

"1

"2

"3

!2

!3

!4

M=4

Let i = mean arrival rate to node i,(i = 1, 2, , M)

i = i +M

j=1

jqji there is unique solution to{i}



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y

Jacksons Theorem: For a Jackson Network in steady state with

arrival rate i to node i,

The no. of customer at any node is independent of the number of

customers at every other node.Node i behaves "stochastically" as if it were subjected to Poisson

arrival rates iLet (n) = Prob[(n1, n2, nM)] where ni 0, for SSFR:

(n) =M

i=1

1

ii

ii

ni

For QLD, M/M/c queues, define ui(r) = i min(r, ci) for r 0,

i = 1,.., M, and i = i/i for i = 1, ..M.

(n) =M

i=1

Ci

nii

nir=1 i(r)

; Ci=

ci1r=0

rir!

+

ciici!

1

1 i/ci

1



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Comment

1

The arrival to each node, in general (unless its only feed forward),is NOT a Poisson process.

2 How can we compute the T and each node utilization?

3 Optimal allocation : assume that the open network of SSFR nodes

with arrival rate i

and i

for each node (i = 1, 2, M)

Min N =

Mi=1

ii

1 ii

s.t.

Mi=1

i = C = constant

4 Application : Network, distributed system


Jackson Network Examples


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Example 1

Consider a switching facility that transmits messages to a required

destination. A NACK is sent by the destination when a packet has not

been properly received. If so, the packet in error is retransmitted as

soon as the NACK is received.

Assume the time to send a message and the time to receive a NACK

are both exponentially distributed with parameter . Assume thatpackets arrive at the switch according to a Poisson process with rate

0. Let p, 0 < p 1, be the probability that a message is received

correctly. Derive mean response time T.



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Example 2

Similar to last example but now the switching facility is composed of Knodes in series, each model as M/M/1 queue with switching rate .What is the response time T?

We have 0i = 0 for i = 2,..., K (no external arrival to nodes

2,..., K), i = for i = 1, 2,..., K, pi,i+1 = 1 for i = 1,..., K 1, andpK,0 = p and pK,1 = 1 p.

1 = 0 + (1 p)K, i = i1 for i = 2,..., K. So

i = 0/p i = 1,..., K.




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By the Jacksons theorem, we have

(n) =

p 0

p

K

0

p

n1++nKn= (n1, n2, . . . , nK) INK

provided that 0 < p. Let E[Ni] be the average number of packets in

queue i:

E[Ni] =0

p 0i = 1, . . . , K.

Let E[T] be the average response time:

E[T] =K

i=1

E[Ni] = K

1p 0

.




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Example 3: open central server network

A computer system with one CPU and two I/O devices. New jobs

enter the system, wait for CPU resource, then possibly an I/O

requests. When a job finishes its I/O, it may return for more CPUresource. Eventually a job completes and leave the system.

It means that K = 3 (three nodes). 0i = 0 for i = 2, 3,p2,1 = p3,1 = 1, while p1,0 > 0.

The traffic equations are: 1 = 01 + 2 + 3, 2 = 1p1,2,

3 = 1p1,3. Solving, we have: 1 = 01/p1,0, i =

01p1,i/p1,0 for

i = 2, 3. Thus,

(n)=

1

011p1,0

011p1,0

n1 3

i=2

1

01p1,i

ip1,001p1,i

ip1,0

ni

n IN

3

E[T] =1

1p1,0 01

+3

i=2p1,i

ip1,0 01p1,i


Closed Queueing Network Example


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Example of Queueing Network

We fixed the total number of jobs be n in the system, where n is

also called the degree of multiprogramming.

0 1

P0

P1!0

!1

CPU I/O

new program path

State representation (k0, k1). Now with the constant thatk0 + k1 = n




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Unlike the open queueing network, the state space is finite. The

flow balance equations are:

(1 + 0p1)p(k0, k1) = 0p1p(k0 + 1, k1 1) + 1p(k0 1, k1 + 1)

1p(0, n) = 0p1p(1, n 1)

0p1p(n, 0) = 1p(n 1, 1)



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The normalization factor C(n) is chosen s.t

k0+k1=n& k0,k10

p(k0, k1) = 1

The choice of a(where 0 =a0

, 1 =ap11

) is very arbitrary in that

the value of p(k0, k1) will not change with a.

If we define 0 = a and 1 = ap1, then (0, 1) as the relative

throughputof the corresponding nodes.

We can choose, for example, a= 1 or a= 0. Assume we choosea= 0 , then 0 = 1, 1 =

0p1

. Since

p(k0, k1) =1

C(n) k

00 k

11 =1

C(n) k

11

1 =1

C(n)

nk1=0

k11 =1

C(n)

1 n+11

1 1




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C(n) = 1n+1

111

where 1 = 1

n+ 1 where 1 = 1(LHospital Rule)



If e choose a 1 then 1 p1


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If we choose a= 1,then 0 =10

, 1 =p11

p(k0, k1) =1

C(n)

k00 k11 =

1

C(n) 1

0

k0

p11

k1

, summing all k0, k1

1 =n

k0=0

nk0k1=nk0

1

C(n)

1

0

k0 p11

k1

=

n

k0=0

1

C(n) 1

0k0 p1

1nk0

C(n) =

nk0=0

1

0

k0 p11

nk0= (

p1

1)n

nk0=0

1

0

k0 p11

k0

=

p11

n nk0=0

1p10

k0

C(n) =

p11

n1 ( 1p10

)n+1

1 1p10




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p(k0, k1) =1

C(n) 1

0k0

p11

k1

=1

C(n)

1

0

k0 p11

nk0=

1

C(n)

p10

n 1p10

k0

p(k0, k1) = 1p

1

n

1 1p10

1 ( 1p10

)n+1p1

1

n

1

p1

0

k0

= (p1)k0

1 p11

1 p(n+1)1

= (p1)(nk1)

pn

1

1 p1

1 pn+11

= (p1)k1

1 p1

1 pn+11



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Closed Queueing Network Theory of Closed Queueing Network

G d N ll t k (1967)


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Gordon - Newell network (1967)

A Gordon-Newell network has M nodes (i = 1, 2, M) s.t.

Node i is QLD with rate i(n) when there is n customers.a customer completing service at a node chooses a node to enternext probabilistically, independent of past historyThe network is CLOSED and has a fixed population K

State space:S = {(n1, n2, nM) | ni ,M

i=1 ni = K}

|S| =

K + M 1

M 1

VERY LARGE NUMBER

If M = 5, K = 10, |S| = 1, 001.If M = 10, K = 35, |S| = 52, 451, 256.


Closed Queueing Network Theory of Closed Queueing Network

Si th i t l i l


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Since there is no external arrival, = .Routing probabilities qij satisfy:

Mj=1

qij = 1

traffic equations are:

i =

Mj=1

jqji i = 1, 2, , M

The number of solutions {i} that satisfy the traffic equations isinfinite.

All solutions differ by a multiplicative factor C

Let (e1, e2, eM) be any non-zero solution, that is ei = Ci (visitrate). Define: xi =

eii

(e1, e2, eM) is chosen by fixing one component to a convenientvalue, such as e1 = 1



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Closed Queueing Network Computation Methods

Computation of G (assume SSFR)


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Computation of G (assume SSFR)

Define S(m, n) = {(n1, nm) | ni 0,mi=1 ni = n}

G(m, n) =

nS(m,n)

mi=1

xi(ni) where xi =eii

G(m, n) =

nS(m,n);

nm=0

mi=1

xnii +

nS(m,n);

nm>

mi=1

xnii

= nS(m1,n)

m1

i=1

xi(ni) + xm nS(m,n);

ki=ni(i=m);

km=nm1

m

i=1

xi(ki)

G(m, n) = G(m 1, n) + xmG(m, n 1) m, n>




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In summary, for SSFR queueing network, we have

G(m, n) = G(m 1, n) +em

mG(m, n 1) m, n> 0

G(m, 0) = 1 m > 0G(0, n) = 0 n 0




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mn 0 1 2 3 i M-1 M

0 1 1 1 1 1

0e1

1

e11

+e22

e11

( )2

..........

.

.

.

.

.

.

.

.

.

.

0

0

1

2

K-1

K

0

0

e11( )

K-1

e11

( )K

.

.

.

.

.

.

.

.

.

.

ei

ix

+

+

..........



Performance Measures


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Performance Measures

Idle Probability for node i:

P(NM = 0) =1

G(M, K)

nS(M,K);

nM=0

M1i=1

eii

ni=

G(M 1, K)

G(M, K)

Another way to express it:

P(Ni = 0) =1

G(M, K) nS(,k);ni=0

i1

j=1

eij

nj Mk=i+1

ekk

nk

=G(M\i, K)

G(M, K)



Utilization of node i or (Ui ):


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Utilization of node i or (Ui):

Ui = 1 P[Ni = 0] = 1 G(M\i, K)

G(M, K)

How about P[Ni k]:

P[Ni k] =1

G(M, K)

nS(M,K);

nik

Mj=1

ej

j

nj

=1

G(M, K)

eii

k mj=nj(j=i);

mi=nik;

nS(M,K);

nik

Mj=1

ej

j

mj=

eii

k G(M, K k)G(M, K)

For P[Ni 1]:

P[Ni 1] = i =

eii

G(M, K 1)

G(M, K)

(more useful)John C.S. Lui () Computer System Performance Evaluation 45 / 79


The throughout of node i is


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g

i = iUi = ieii

G(M, K 1)

G(M, K)= ei

G(M, K 1)

G(M, K)

Prob[ni = k] = i(k) = P[Ni k] P[Ni k + 1]

i(k) =1

G(M, K) nS(M,K);

ni=k

M

j=1

ej

j

nj

=( eii

)k

G(M, K)

nS(M,Kk);

ni=0

Mj=1

ej

j

nj=

eii

kG(M\i, K k)

G(M, K)

i(k) = P[Ni k] P[Ni k + 1]

=

eii

kG(M, K k) eii G(M, K k 1)G(M, K)



Expected no. of customer in node i = Li(K)


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( )

Li(K) =K

j=1

jP[Ni = j] =K

j=1

P[Ni = j]K

k=1

I(kj)

=K

k=1

Kj=k

P[Ni = j] =K

k=1

P[Ni k]

Li(K) =1

G(M, K)

Kk=1

ei

ik

G(M, K k) i = 1, 2, , M

Derivation of above expression:

LI(K) = P[Ni = 1] +

P[Ni = 2] + P[Ni = 2] +

P[Ni = 3] + P[Ni = 3] + P[Ni = 3] + +

P[Ni = K] + P[Ni = K] + P[NI = K] + + P[Ni = K]


Closed Queueing Network Convolution Algorithm

Garden Newell Convolution Algorithm


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Garden Newell Convolution Algorithm

Assume M QLD nodes and K customers

Define

fi(Z) =

k=0

Xi(k)Zk = 1 + Xi(1)Z + Xi(2)Z

2 + Xi(3)Z3 +

= 1 +

ei

i(1)

Z +

e2i

i(1)i(2)

Z2 +

e3i

i(1)i(2)i(3)

Z3 +

f(Z) = f1(Z)f2(Z) . . . fM(Z)

The coefficient of Zk in f(Z) the sum of products of the formX1(n1)X2(n2) . . . XM(nM) such that

i ni = k




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Thus

f(Z) = 1+G(M, 1)Z+G(M, 2)Z2+G(M, 3)Z3+ +G(M, k)Zk+

*IDEA: build up f(Z) from partial products gi(Z) so that G(i, k) is

the coefficient of Zk in gi(Z)

g1(Z) = f1(Z)

gi(Z) = gi1(Z)fi(Z) i = 2, 3, , M



Therefore,


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G(1, k) = coefficient of the Zk term in g1(Z)

G(1, k) = X1(k) =

ek1ki=1 1(i)

G(i, k) =k

j=0

G(i 1,j)xi(k j) (convolution!)

Example :

G(2, k) =k

j=0G(1,j)x2(k j)

= G(1, 0)x2(k) + G(1, 1)x2(k 1) + + G(1, k)x2()

=ek2

kj=1 x2(j)

+e1

1(1)

ek12

k1j=1 2(j)

+




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k\ 1 2 0 1 1

1 e11(1)

2e2

11(1)1(2)

3 e3

11(1)1(2)1(3)

......

Kek

1

1(1)...1(k)



Performance Measure


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Prob[node i has k customers] = i(k):

i(k) =1

G(M, K)

nS(,k);

ni=k

X1(n1)X2(n2) . . . XM(nM)

i(k) = xi(k)

G(M, K)G(M\i, K k)

Expected number of customers in node i

E[Li] =

Kk=0

ki(k) = very involve

Prob[node i 1 customer] = ?




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Utilization of a node, in general, is very involved.

Ui(K) = Prob[ni 1] =K

j=1

Prob[ni = j]

For a node that is SSFR, we have

Ui(K) =

eii

G(M, K 1)

G(M, K)

This holds even the other nodes are QLD servers!!



Example :


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.

.

.

.

.

.

front end communication

transaction processing

query processing

0P

1P

2P

3

P

4P

terminal

!"

!average think time = 1/

eT = eFp0; eF = eT + eCp2; eC = eFp1 + eD + eP;eD = eCp3; eP = eCp4

eT = 1; eF =

1

p0 ; eC =

1p0

p0p2 ; eD =

(1p0)p3

p0p2 ; eP =

(1p4)p4

p0p2

T =eT

= 1

; F =1

p0F; C =

(1p0)p0p1c

; D =(1p0)p3

p0p20P =

(1p0)p4p0p2p

(nT, nF, nC, nD, np) =1

G

(T)nT

nT!(F)

nF(C)nC(D)

nD(P)nP



UF = ?


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Since UF is a SSFR, we have

UF = F G(5, K 1)G(5, K)

= 1p0F

G(5, K 1)G(5, K)

Average throughput or rate of request completion is:

= Fp0UF =G(5, k 1)

G(5, k)

but Littles Result (N = T)

T = the expected response time = K

= KG(5,k)G(5,K1)

T = average think + average processing time =KG(5,k)

G(5,K1)

T = 1 + average processing time =KG(5,k)

G(5,K1)

Therefore, average processing time =KG(5,k)

G(5,K1) 1


Closed Queueing Network Multiclass Queueing Networks

Multiclass Open/Closed/Mixed Jackson Networks


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Setting

We have K customers and M nodes with i exponential servicerate for i = 1,.., M.

R, an arbitrary but finite number of classes of customers.

Let pi,r;j,s be the probability that a customer of class r completes

service at node i will become class s in node j.

The pairs (i, r) and (j, s) belong to the same subchain if thesame customer can visit node i in class r and node j in class s.

Let m be the number of subchains, let E1, . . . , Em be the set ofstates in each subchains.



Setting: continue


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Let nir be the number of customers of class r at node i. A closed

system is characterized by(i,r)Ej

= constant. j = 1, . . . , m.

For an open system, a Poisson process with 0ir is the external

arrival rate of class r to node i. Customer may leave the systemwith pi,r;0 so that

j,s pi,r;j,s + pi,r;0 = 1.

Define Q(t) = (Q1(t), . . . ,QM(t)) whereQi(t) = (Qi1(t), . . . , QiR(t)) with Qir(t) being the number of class rcustomers at node i. Note that Q(t) is NOT a CTMC because theclass of a customer leaving a node is not known.

Instead, we define Xi(t) = (Ii1(t), . . . , Ii,|Qi(t)|(t)) whereIij(t) {1, 2,..., R} is the class of the customer in position j innode i at time t. Then (X1(t),...,XM(t)), t 0) is a CTMC.



Multiclass Open/Closed/Mixed Jackson Networks: For k {1,..., m} suchthat Ek is an open subchain, let (ir )(i r)E be the "unique" strictly positive


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that Ek is an open subchain, let (ir)(i,r)Ek be the unique strictly positivesolution of the traffic equations

ir =

0

ir +

(j,s)Ekjspj,s;i,r (i, r) Ek.

For k {1,..., m} such that Ek is a closed subchain, let (ir)(i,r)Ek be anynon-zero solution of

ir =

(j,s)Ekjspj,s;i,r (i, r) Ek.

If

r:(i,r)belongs to an open subchain ir < i for all i = 1, 2, ..., M, then

(n) = 1G

Mi=1

ni!

Rr=1

1nir!

iri

nir

for all n= ((n1),...,(nM)) in the state space, where (ni) = (ni1,..., niR) INR

and ni = Rr=1 nir. Here, G is the normalization constant.John C.S. Lui () Computer System Performance Evaluation 60 / 79



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Example

There are M = 2 nodes and R = 2 classes of customers. There isno external arrival to node 2. External customers enter node 1 in

class 1 with rate . Upon completion at node 1, a customer ofclass 1 is routed to node 2 with the probability 1. Upon completion

at node 2, a customer of class a leaves with probability 1.There are always K customers of class 2 in the system. Upon

service completion at node 1 (resp. node 2), customer of class 2

is routed back to node 2 (resp. node 1) in class 2 with probability

1. Let i be the service rate at node i = 1, 2.




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Example: continue

The state space S is

S = {(n11, n12, n21, n22) IN4 : n11 0, n21 0, n12 + n22 = K}.

There are two subchains: E1 (open), and E2 (closed), with

E1 = {(1, 1), (2, 1)} and E2 = {(1, 2), (2, 2)}.

We find 11 = 2,1 = and 12 = 22. Take 12 = 22 = 1 for instance.We have

(n) =1

G

n1

n11

n2

n22

1

n11

2

n21

1

1

n12

1

2

n22

with < i, i = 1, 2 (stability condition).



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Extension to M/M/c

Let ci 1 be the number of servers at node i and definei(j) = min(ci,j) for i = 1, ..., M. Hence ii(j) is the service rate atnode i when there are j customers. We have

(n) =1

G

Mi=1

ni

j=1

1

i(j)

ni!

R

r=1

1

nir!

iri

nir .




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Extension to M/M/c with state-depending routing

Let the total number of customer be M(n) = Mi=1 ni. Let the externalarrival rate of class r customer at node i be 0ir(M(n)), where is anarbitrary function from IN into [0, ). We have

(n) =d(n)

G

M

i=1

ni

j=1

1

i(j)ni!

R

r=1

1

nir!ir

i

nir

.where

d(n) =

M(n)1

j=0

(j).

and d(n) = 1 if the network is closed.


Closed Queueing Network BCMP Networks


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A classic piece by F. Baskett, K.M. Chandy, R.R. Muntz and F.G.

Palacios on Open, Closed, and Mixed Networks of Queues withDifferent Classes of Customers, JACM, 22(2), 1975.

Terminology

FCFS: First come first serve M/M/c queue.

PS: Processor sharing queue.

LCFS: Last come first serve single server queue.

IF: Infinite server queue




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Characterization

If node i is of type FCFS, then ir = ir/i for r = 1, . . . , R, whereR is the number of classes of customer, and i is the exponentialservice times in node i.

If node i is of type PS, LCFS, or IS, then ir = ir/ir forr = 1, . . . , R, and

iris the mean service time for customer of type

r in node i.

For nodes of types PS, LCFS, or IS, the service time distribution is

arbitrary.

ir is the solution of the traffic equations.



Theorem


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For a BCMP network with M nodes and R classes of customer, which

is open, closed, or mixed in which each node is of type FCFS, PS,

LCFS, or IS, the steady state probabilities are:

(n) =d(n)

G

M

i=1fi(ni).

wheren= (n1, , nM) in the state space S withni = (ni1, ni2, , niR) where nir is the number of jobs of class r atnode i. Moreover, |ni| =

Rr=1 nir for i = 1, 2, . . . , M.

G < is the normalization constant such that

nS

(n) = 1,d(n) =

M(n)1j=0 (j) if the arrivals depend on the total number of

customers M(n) =M

i=1 |ni|, and d(n) = 1 if the network is closed.



fi(ni) for different types of nodes


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If node i is of type FCFS:

fi(ni) = |ni|!|ni|

j=1

1

i(j)

Rr=1

nirirnir!

If node i is of type PS or LCFS:

fi(ni) = |ni|!R

r=1

nirirnir!

If node i is of type IS:

fi(ni) =R

r=1

nirirnir!




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Comment

Solve the traffic equations ir for i = 1, ..., M and r = 1,..., R.Use queueing network package to solve for G.

When using queueing network package, all we need to enter arethe topology of the network:

K, the number and the types of nodes,R classes, androuting matrix [pi,r;j,s].external arrival ratesservice rates, e.g., ii(j), for j = 0, 1,... for a node that is FCFS.service rates ir for customers of class r = 1,...R for a node that is

PS, LCFS, or IS.


Closed Queueing Network Mean Value Analysis (MVA)

Mean value analysis [S. Lavenberg & M. Reiser]


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Derive expected performance measures while avoiding derivation

of steady state probability

0

b

a!

!

1

!2CLOSED

NETWORKArrival

Departure

Replace an arc by 1 node 2




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Whenever a customer arrives at node (via along 1), it departsfrom the network and is replaced by a

"stochastically" identical customer who has the same routing

probability along arc 2

This open network behaves exactly as the closed one (except

node )



Mean Value analysis for state-dependent service rates


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The mean queue length is:

Li(K) =K

j=1

ji(j|K)

By definition,i(0|0) = 1 and

i(j|K) =

viT(K)i(j)

i(j 1|K 1) j = 1, 2, . . . , k

1

Kk=1 i(k|K) j =



For the we chosen, let us define the network throughput T asthe average rate customers pass along arc in steady state.


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t e a e age ate custo e s pass a o g a c steady state

Now we can view T as the external arrival rate () in the open

network.Suppose that we have M nodes in the network. Define

vi = average number of visit to node i by a customer= (visitation rate)

i = average arrival rate of customer to node i

i = Tvi , v0 = vaqab

But since customers visit node EXACTLY ONCE, v0 = 1,therefore:

va =1

qab; vj =

i=1

viqij

Once one vi is found,we can find other vis.



Looking at node i, let Li= average no. of customers, we have:

L w L Tv w


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Li = iwi , Li = Tviwi

But sinceM

i=1 Li = K = TM

i=1 viwi, therefore the systemthroughput

T =K

Mi=1 viwi

If node i is infinite server,then

wi =1

i

If node i is single server fixed rate (SSFR),

wi = 1i

[Yi + 1]

where Yi is the mean number of customers seen by an arrival to

node i.



Example:

T (K )


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i(j|K) = viT(K)i(j)

i(j 1|K 1) j = 1, 2, . . . , k

1 K

k=1 i(k|K) j =

Assume it is node 1 (that is , i=1)

1(1|1) =

v1T(1)

1(1) 1(|) =

v1T(1)

1(1)

1(|1) = 1 1(1|1)

1(1|2) =v1T(2)

1(1)1(|1)

1(2|2) = v1T(2)2(2)

1(1|1)

1(|2) = 1 (1|2) (2|2)


Queueing Network Beamer

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