Queue Networks

8/10/2019 Queue Networks

1/25

Queuing Networks: BurkesTheorem, KleinrocksApproximation, and Jacksons

Theorem

Wade Trappe


2/25

Lectu re Overview

Network of Queues Introduction

Queues in Tandem Product Form Solutions

Burkes Theorem

What is reversibility?

Kleinrocks Approximation Quick Jackson Theorem


3/25

In many networking scenarios, a customer or packet must receiveservice from many servers before its final task is completed.

Hence, departures from a queue might become arrivals at anotherqueue.

All that discussion we did in M/G/1 queues becomes veryimportant for networks of queues!

Consider two queues in tandem:

Departures from first queue become arrivals at second queue

First queues arrival process is Poisson with rate andservice time is exponential with rate

Service time at the second queue is exponential with rateand independentof first servers service time

How do we model these two queues together?

Whats the state and state diagram?

Network o f Queues: Setup

2

1


4/25

Two Queues in Tandem

Queue2

We need to keep track of N1(t) and N2(t) to describe the

state of the system!

Queue1 1 2


5/25

Two Queues in Tandem : State Diagram

Our state diagram must keep track of both N1(t) and N2(t),and the man transitions that are ossible

n1=0 1 2 3

n2=0

1

2

3

2

2

22

2

2 2

2

2 2

2

2

1 1 1

1

111

1 1


6/25

Define (N1(t), N2(t)) to be the state vector for the two queues in

tandem

Notice, now, that we have a Markov Process in terms of the statevector

Recall the global balance equations for M/M/1 queues:

What this means is: At steady state, the amount entering a state is

equal to the amount leaving the state.

We similarly may find the global balance equations for this twoqueue system

Global Balance Equations fo r 2 Queues, pg 1

,...2,1j,ppp

pp

1j1jj

10


7/25

Global Balance Equations fo r 2 Queues, pg 2

Global Balance Equations: Case 1

Case 2: n>0, m=0

Case 3: m>0, n=0

Case 4: n>0, m>0

1N,0NP0N,0NP 21221

0N,1nNP1N,nNP0N,nNP 21212211

1mN,1NP1mN,0NPmN,0NP 211212212

mN,1nNP

1mN,1nNP

1mN,nNPmN,nNP

21

211

2122121


8/25

Steady State PMFs fo r Two M/M/1 Queues

We may show that the following joint probability mass function

satisfies the global balance equations

where ri=/i

So, how do we get P(N1=n)?

Easy, its just an M/M/1!

So P(N1=n) = (1-r1)r1n for n=0,1,2

How do we get P(N2=m)?

Answer: Its a marginal. Integrate out the joint pmf!

Sum over all n to get:

m222 1mNP rr

m22n

1121 11mN,nNP rrrr

Check

This!


9/25

Steady State PMFs fo r Two Queues, pg 2

This looks interesting

means

Thus, the number of customers at queue 1 and the number atqueue 2 at a particular time are independentrandom variables!

The steady state at queue 2 is the same as for an M/M/1 queue

with Poisson arrival rate and exponential service time 2.

Definition:A network of queues is said to have a product-form

solutionwhen the joint pmf of the number of customers at eachqueue is the product of the marginal pmfs of the number of

customers at each queue.

m22n1121 11mN,nNP rrrr

mNPnNPmN,nNP 2121


10/25

Burkes Theorem

Burkes Theorem is the fundamental result describing product formsolutions

Burkes Theorem:Consider an M/M/1, M/M/m, M/M/infinity queuingsystem at steady state with arrival rate , then

The departure process is Poisson with rate ;

At each time t, the number of customers in the system N(t) is independentof the sequence of departure times prior to t.

What Burkes Theorem implies:

Two queue problem follows from Burkes Thm (arrivals to queue 2 arePoisson with rate ).

Arrivals to queue 2 prior to time t are departures from queue 1 prior totime t, thus Burkes theorem says queue-1s departures (queue-2s

arrivals) are independent of N1(t).

N2(t) is determined by the sequence of arrivals from queue-1 prior to timet and independent of service times, then N1(t) and N2(t) are independentas random variables.

Note: N1(t) and N2(t) are not independent as processes!


11/25

Consider the network of queues:

Here Queue 1 is driven by a Poisson process with rate 1,, andthe departures are randomly routed to queues 2 and 3.

Queue 3 has an additional, independent Poisson arrival processwith rate 2.

Example Application of Burkes Thm

1

2

3

1/2

1/2

1

2


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Example Application of Burkes Thm, pg 2

Burkes Theorem says: N

1

(t) and N2

(t) are independent

N1(t) and N3(t) are independent

Recall that the random split of a Poisson process yieldsindependent Poisson processes Hence inputs to Queue 2 and Queue 3 are independent

Input to Queue 2 is Poisson with rate 1/2

Input to Queue 3 is Poisson with rate 1/2 + 2

Thus

where r1=1/1, r2=1/22 , r3=(1/2 2/3. All queues areassumed to be stable.

n

33

m

22

k

11321 111n)t(N,m)t(N,k)t(NP rrrrrr


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Reversible Markov Processes

In order to prove Burkes theorem, we need the concept of thereversibility of a Markov process.

A stationary Markov process X(t), with a countable state space (i.e. aMarkov chain will do), is reversibleif X(t) and Y(t)=X(-t) have thesame joint distribution at arbitrarily chosen instants {t1, t2, , tN}.

A necessary and sufficient condition for reversibility is

where {pi} and {pij} are the stationary probabilities and transitionprobabilities of X(t)

This condition can be easily shown for M/M/1 queuesbut we willshow it in more general form

In fact, it holds for any birth-death process, and N(-t) is statisticallyidentical to N(t)

i,jjj,ii pppp


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Reversible Markov Processes, pg 2

Time-Reversal Theorem:Let {X(t): t>=0} be a stationary

Markov process with (infinitesimal) generator P=[pij], andwith initial distribution equal to stationary distribution. Thenfor all T>0, the time-reversed process

is equivalent to a stationary Markov process with(infinitesimal) generator given by:

for all state pairs (i,j)

Tt0:)tT(X

i

j ij

ij p

ppp~

ijP~


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Proof:Let Q(t)=[qij(t)] denote the transition probabilities of

X(t),

We need to show X(T-t) is a Markov process with transitionprobabilities

Then we obtain the necessary and sufficient condition bydifferentiating this and setting t=0. (Its an infinitesimal

generator)

i)0(X|j)t(XP)t(qij

i

j ij

ijp

)t(qp)t(q~)t(Q

~


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Consider the interval (0,t+s] and divide it into (0,t] and

(t,t+s], i.e. set T=t+s.

The joint probability of the three random variables

is

Similarly we have

)0(X)st(X~

)s(X)t(X~

)st(X)0(X~

)t(q)s(qp

i)st(X,i)s(X,i)0(XP

i)st(X~,i)t(X~,i)0(X~P

01122 i,ii,ii

012

210

)t(qp

i)ts(X,i)s(XPi)t(X~,i)0(X

~P

011 i,ii

0110


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The conditional probability

Hence, we have shown that is a Markov process with

generator (after differentiation).P~

j,i

12

i

i,ii

012

q~i)t(X~|i)st(X~P

p

)t(qp

i)0(X~,i)t(X

~|i)st(X

~P

1

122

tX~


18/25

B-D Processes are Reversib le

It is now easy to show the following

Time Reversibility of B-D Processes:The stationary B-D process N(t)with generator Pand steady state probabilities pis a time-reversibleMarkov process. Thus, the time-reversal of the death process is a birth

process.

Burkes Theorem follows from this:

Interdeparture times of the forward-time system are the interarrivaltimes of the time-reversed system... Hence we have Poisson withrate coming out of the system.

Fix a time t, then the departures before time t from the forwardsystem are arrivals after time t in the reverse system.

Arrivals in reverse system are Poisson, and thus arrivals in reversesystem after time t do not depend on N(t)

Consequently, departures after time t in forward system do notdepend on N(t)


19/25


20/25

Network o f Queues, M/D/1 f irst queue

The interarrival times of the second queue must be at least 1/

Why?

Now, each packet arriving at either queue takes 1/time toprocess.

The first packet being finished by first queue is immediately

sent to second server

It takes at leastanother 1/amount of time for first queue to

get and finish the next packet/customer.

So, first packet will be finished by second server at or before

the next packet arrives to second server.

Result: No queue (waiting) at second system!


21/25

Two Queues, co rrelated service t imes

Earlier we considered the service times independent of each other

and independent of the arrival times. Reality:A big packet at the first system is probably still a big

packet at the second system!

Interarrival times at the second queue are strongly correlatedwith packet lengths!

Long packets at first system will typically find the queue at thesecond server more empty

Shorter packets from the first system will typically find the queueat the second server more busy because the second server is

processing some prior big packet

It is tough to find an analytical solution for joint pmf underdependence assumptions!


22/25

The Kleinroc k Independence Approximat ion

We have argued that in practice there is dependence upon the

interarrival times and service times. Independence is lost after the first system!

Reality hurt us, but reality provides us one more gift

Reality:Real networks typically involve more than one stream of

packets merging at a node The combination of multiplestreams helps restoreindependence in many cases!

This observation is due to Kleinrock.

Kleinrocks Approximation:It is often appropriate to useM/M/1 queues for each communication link when the arrivals atentry points are Poisson, packet lengths are roughly exponentially

distributed, network is dense and traffic is heavy.


23/25

Quick Look at Jacksons Theorem

Many queuing networks, a packet/customer may visit a queue

more than once. Burkes theorem does not apply!

Typical example: Queue with feedback

If the arrival rate is much less than departure rate, then netarrival process has a few, isolated external arrivals followed bya burst of feedback arrivals (dependent on packet length).

a

p

1-p


24/25

Jacksons Theorem, (Open) pg 1.

Consider a queuing network consisting of M separate service stations,each with its own queue.

Define the vector process:

In an open queuing network, customers may arrive from an externalsource and eventually may leave the network (as depicted in previousslide).

A closed network has no arrivals or departures from the system (totalcustomers is fixed just they may move around)

Assumptions: The rate of the source (birth process) is , and a customer goes to

station i with probability qsi.

Service time at station k is exponential with rate k.

Customers are routed according to a Markov chain: Probabilitythat a customer departing station i goes to station j is qij.

)t(N,),t(N),t(NtN M21


25/25

Jacksons Theorem, (Open) pg 2.

Jacksons Decomposition Theorem:For an open queue as

described, the joint distribution of the queue vector n(t) isgiven by:

where P(Nk=nk) is the steady state pmf of an M/M/1 systemwith arrival rate kand service rate k, i.e.

Where rkdescribes the utilization factor of system k.

KK2211 nNPnNPnNPnP

kn

kkk 1nNP k rr

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Queue Networks