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Lecture 5. This lecture is about: Introduction to Queuing Theory Queuing Theory Notation Bertsekas/Gallager: Section 3.3 Kleinrock (Book I) Basics of Markov Chains Bertsekas/Gallager: Appendix A Kleinrock (Book I) Markov Chains by J. R. Norris. Queuing Theory.

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Lecture 5
Lecture 5

  • This lecture is about:

  • Introduction to Queuing Theory

    • Queuing Theory Notation

    • Bertsekas/Gallager: Section 3.3

    • Kleinrock (Book I)

  • Basics of Markov Chains

    • Bertsekas/Gallager: Appendix A

    • Kleinrock (Book I)

    • Markov Chains by J. R. Norris

Queuing theory
Queuing Theory

  • Queuing Theory deals with systems of the following type:

  • Typically we are interested in how much queuing occurs or in the delays at the servers.

Server Process(es)




Queuing theory notation
Queuing Theory Notation

  • A standard notation is used in queuing theory to denote the type of system we are dealing with.

  • Typical examples are:

    • M/M/1 Poisson Input/Poisson Server/1 Server

    • M/G/1 Poisson Input/General Server/1 Server

    • D/G/n Deterministic Input/General Server/n Servers

    • E/G/ Erlangian Input/General Server/Inf. Servers

  • The first letter indicates the input process, the second letter is the server process and the number is the number of servers.

  • (M = Memoryless = Poisson)

The m m 1 queue
The M/M/1 Queue

  • The simplest queue is the M/M/1 queue.

  • Recall that a Poisson process has the following characteristics:

  • Where A(t) is the number of events (arrivals) up to time t.

  • Let us assume that the arrival process is a Poisson with mean  and the service process is a Poisson with a mean 

Poisson processes a refresher
Poisson Processes (a refresher)

  • Interarrival times are i.i.d. and exponentially distributed with parameter .

  • tn is the time of packet n and n= tn+1 - tn then:

  • For every t  0 and   0:

Poisson processes a refresher1
Poisson Processes (a refresher)

  • If two or more Poisson processes (A1,A2...Ak) with different means(1, 2... k) are merged then the resultant process has a mean  given by:

  • If a Poisson process is split into two (or more) by independently assigning arrivals to streams then the resultant processes are both Poisson.

  • Because of the memoryless property of the Poisson process, an ideal tool for investigating this type of system is the Markov chain.

On the buses a paradoxical property of poisson processes
On the Buses (a paradoxical property of Poisson Processes)

  • You are waiting for a bus. The timetable says that buses are every 30 minutes. (But who believes bus timetables?)

  • As a mathematician, you have observed that, in fact, the buses are a Poisson process with a mean arrival rate such that the expectation time between buses is 30 minutes.

  • You arrived at a random time at the bus stop. What is your expected wait for a bus? What is the expected time since the last bus?

    • 15 minutes. After all, they are, on average, 30 minutes apart.

    • 30 minutes. As we have said, a Poisson Process is memoryless so logically, the expected waiting time must be the same whether we arrive just after a previous bus or a full hour since the previous bus.

Introduction to markov chains
Introduction to Markov Chains

  • Some process (or time series) {Xn| n= 0,1,2,...} takes values in nonnegative integers.

  • The process is a Markov chain if, whenever it is in state i, the probability of being in state j next is pij

  • This is, of course, another way of saying that a Markov Chain is memoryless.

  • pij are the transition probabilities.

Visualising markov chains the confused hippy hitcher example










Visualising Markov Chains (the confused hippy hitcher example)

A hitchhiking hippy begins at A

town. For some reason he has

poor short-term memory and

travels at random according

to the probabilities shown. What

is the chance he is back at A after 2

days? What about after 3 days? Where is he likely to end up?

The hippy hitcher continued










The Hippy Hitcher (continued)

  • After 1 day he will be in B town with probability 3/4 or C town with probability 1/4

  • The probability of returning to A via B after 1 day is 3/12 and via C is 2/12 total 5/12

  • We can perform similar

    calculations for 3 or 4 days

    but it will quickly

    become tricky and

    finding which city he

    is most likely to end up

    in is impossible.

Transition matrix










Transition Matrix

  • Instead we can represent the transitions as a matrix

Prob of going to B from A

Prob of going to A from C

Markov chain transition basics
Markov Chain Transition Basics

  • pijare the transition probabilities of a chain. They have the following properties:

  • The corresponding probability matrix is:

Transition matrix1
Transition Matrix

  • Define n as a distribution vector representing the probabilities of each state at time step n.

  • We can now define 1 step in our chain as:

  • And clearly, by iterating this, after m steps we have:

The return of the hippy hitcher
The Return of the Hippy Hitcher

  • What does this imply for our hippy?

  • We know the initial state vector:

  • So we can calculate n with a little drudge work.

  • (If you get bored raising P to the power n then you can use a computer)

  • But which city is the hippy likely to end up in?

  • We want to know

Invariant or equilibrium probabilities
Invariant (or equilibrium) probabilities)

  • Assuming the limit exists, the distribution vector  is known as the invariant or equilibrium probabilities.

  • We might think of them as being the proportion of the time that the system spends in each state or alternatively, as the probability of finding the system in a given state at a particular time.

  • They can be found by finding a distribution which solves the equation:

  • We will formalise these ideas in a subsequent lecture.

Some notation for markov chains
Some Notation for Markov Chains

  • Formally, a process Xn is Markov chain with initial distribution  and transition matrix P if:

    • P{X0=i} = i (where i is the ith element of )

    • P{Xn+1=j| Xn=i, Xn-1=xn-1,...X0=x0}= P{Xn+1=j| Xn=i }=pij

  • For short we say Xn is Markov (,P)

  • We now introduce the notation for an n step transition:

  • And note in passing that:

This is the Chapman-Kolmogorov equation