chapter 6 product-form queuing network models

Chapter 6

Product-Form Queuing

Network Models

Prof. Ali Movaghar

2

Product-Form Queuing Networks

Consider a modular system consisting of multiple subsystems

The more independent subsystems, the easier overall system analysis

For example, each subsystem can be modeled as a queue and the overall system analysis can be obtained from the product of terms corresponding to each queue!

We call such queuing networks product-form networks.

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Product-Form Queuing Networks (Con.)

There are several properties that lead to

product-form (PF) queuing networks

Local balance

Reversibility

Quasi-reversibility

Station balance

MM

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Local Balance

The effective rate at which the system leaves

state S due to the service completion (of a

chain r customer) at station i equals the

effective rate at which the system enters

state S due to an arrival (of chain r

customers) to station i

n n+1

Effective arrival rate

Effective service completion

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Reversible Markov Process

Consider a film we make from the behavior of an isolated service station and then run the film backward

The departing customers in the real system will be seen as arrival in backward run.

Let X(t) denote the random process describing the number of customers at the station

Let Xb(t) denote the corresponding process in the backward run of the film

If X(t) and Xb(t) are statistically identical, we say X(t) is reversible.

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Reversible Markov Process (Con.)

A random process X(t) is reversible,

If for any finite sequence of time instants t1, …, tk

and a parameter , the joint distribution of X(t1),

…, X(tk) is the same as that of X(-t1),…, X(-tk)

M/M/c queue is a reversible system

M/M/1 queue with batch arrival is not reversible

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A reversible process must be stationary

X(+t1),…, X(+tk) has the same distribution as

X(-t1),…, X(-tk) which has the same distribution as

X(t1),…, X(tk).

Thus all joint distribution are independent of time

shift as required for stationary

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Reversible Markov processes can be

characterized by an important property

known as detailed balance.

Lemma : Let X(t) be a stationary, discrete

parameter Markov chain with one-step

transition matrix Q[q(i,j)]. If X(t) is reversible

than it satisfy detailed balance equation :

P(i)q(i,j) = P(j) q(j,i)

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Similarly, a continuous Markov chain is

reversible if i,j. P(i)qi,j = P(j) qi,j where qi,j is

the transition rates.

Conversely if we can find probabilities P(i)’s

with P(i)=1 satisfying detailed balance

property, then X(t) is reversible and P(i)’s

form its stationary distribution.

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Lemma (Kolmogorove criterion)

A stationary Markov chain is reversible if and only

if for every finite sequence of states i0, …, ik the

transition probability (or rates in the continuous

parameter case) satisfy the following equation:

q(i0, i1) q(i1, i2)…q(ik-1, ik) q(ik, i0)=

q(ik, ik-1) q(ik-1, ik-2)…q(i1, i0) q(i0, ik)

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Lemma: Let X(t) denote an ergodic Markov

chain with transition matrix Q. Then X(-t) has

the same stationary distribution as X(t). Let

Q*=[q*(i,j)] denote the transition matrix of X(-t),

then :

q*(i,j)=P(j)q(j,i)/P(i)

In other words, for a reversible process Q=Q*

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Example : Consider M/M/c queue

Global balance results

P(n)(λ+cμ)=P(n-1)λ+P(n+1)cμ

Local balance results P(n)(λ)=P(n+1)cμ

Local balance is contained in global balance

Thus Piqi,j = Pj qj,i and is reversible

Number of leaves has Poisson distribution with rate λ

1

λ

0

μ

n

λ

c

cμ

n+1

λ

cμ

…

…

λ

2μ

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Reversibility of M/M/c results in following

lemma.

The departure process of M/M/c system is

Poisson and the number in the queue at any time

t is independent of the departure process prior to

t.

Thus we can easily analyze feed-forward network

of M/M systems

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Example : Find P(n1, n2)

P(n1) = (1-1)n1

P(n2) = (1-2)n2

The arrival to second queue (departure from first

queue) is independent of number of customers at

first queue (Previous Lemma).

P(n1, n2) = (1-1) (1-2) n1 n2

M/M/c λ μ1

M/M/c μ2

Number of customers = n1 Number of customers = n2

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Example

P(n1, n2,n3) = (1-1) (1-2) (1-3) n1 n2 n3

Avg. Queue Length =

1/(1- 1)+ 2/(1- 2)+ 3/(1- 3)

W = Avg. Queue Length /λ

What if there is a feed-back from third queue to first queue:

Quasi-reversibility preserves product-form property

M/M/c λ μ1

M/M/c μ2

n1 n2

M/M/c μ3

n3

Burke’s Theorem

Theorem 16.5 (Burke) Consider an M/M/1

system with arrival λ. Suppose the state

starts in a steady state. Then the following

are true:

1. The departure process is Poisson (λ).

2. At each time t, the number of jobs in the

system at time t is independent of the

sequence of departure times prior to time t.

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General Acyclic Networks with

Probabilistic Routing

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Quasi-Reversible Queuing Systems

Let X(t) denote the queue length process at a

queuing system. Then X(t) is quasi-reversible

if for any time instant t0, X(t0) is independent

of

Arrival time of customers after t0

Departure times of customer prior to t0

Jackson Network Definition

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Theorem 17.1

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Example: Web Server

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Example: Web Server (cont.)

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Classed Jackson Networks

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Product Form Theorems

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The Distribution of Jobs

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CPU-Bound and I/O-Bound Jobs

Example

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Solution:

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Solution (cont.):

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Solution (cont.):

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Solution (cont.):

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Closed Networks of Queues

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The Corresponding CTMC

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Solving Closed Batch Jackson

Networks

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Example

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Example (cont.)

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Example (cont.)

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Networks with Process Sharing Servers

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Example 1 – Single Server

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Example 2 – Sever Farm

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Example 2 (cont.)

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Network of PS Severs with Probabilistic

Routing

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Open Single-Chain Product-Form

Networks

qsi : Probability that an external arrival is directed to station i (similarly qid is defined)

(n) : Generation of customers at source with Poisson distribution .

i(n): External arrival to station i (qsi(n)= i(n))

μi : basic service rate of station i

Ci(n) : capacity of station i at load n

i

j

src sink

Network qs1

qs2

qsM

q1d

q2d

qMd

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Networks (Con.)

vi : the visit ratio of station i relative to the source node

λi(n) : the arrival rate of station i when there n customers in

the entire network (λi(n) = vi (n) )

Visit ratio to source and so destination should be 1

By flow balance we have :

M M

si i id

i 1 i 1

M M

id ij i si j ji

j 1 j 1

q 1 and v q 1

q 1 q and v q v q 1

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Networks (Con.)

n=(n1,…,nM) : state of network where ni is the

number of customers at station i.

ei=(0,…,0,1,0,…,0) : 1 at position i

The rate at which the system leaves state n due to

an arrival or completion at station i is qsi(n)+

μi(ni)qid+ μi(ni)qij or qsi(n)+μi(ni)

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Networks (Con.)

The global balance equation results

n

n-ei+ej

n-ei n+ei

μj(nj+1)qij μi(ni)qij

(n-1)qsi

μi(ni)qid

(n)qsi

μi(ni+1)qid

M M M

si i i j j i j ji

i 1 i 1 j 1

M

i si i i i id

i 1

P(n) [ (n)q (n )] (n 1)P(n e e )q

(n 1)P(n e )q (n 1)P(n e )q

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Networks (Con.)

Since all stations are individually qausi-

reversible, the entire network should be

qausi-reversible :

μi(ni)P(n)= λi(n-1)P(n-ei) , i=1,..,M

We can find P(n) for station i and then the

desired product-form solution for network:

inn 1 Mi

k 0 i 1 i i

uP(n) (k) P(0)

V (n )

chapter 6 product-form queuing network models

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