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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

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### Lecture 5

• 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 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)

Input

Process

Output

### 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/1Poisson Input/Poisson Server/1 Server

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

• D/G/nDeterministic 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 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)

• 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 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)

• 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

• 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.

A

1/3

1/4

1/2

3/4

2/3

B

C

1/2

### 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?

A

1/3

1/4

1/2

3/4

2/3

B

C

1/2

### 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.

A

1/3

1/4

1/2

3/4

2/3

B

C

1/2

### 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

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

• The corresponding probability matrix is:

### 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

• 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)

• 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

• 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