Queuing-theoretic framework.

Consider a system of objects where count of objects at a time t is given by . The colony of objects grows and shrink through “birth” and “death” processes, respectively.

Birth and death model.

A birth or a death event may occur during a time interval . If is the birthrate and is the death rate then

■ probability that there will be a single birth

during is

■ probability that there will be no birth during

is

■ probability of multiple births during is zero.

■ probability of a single death during is .

■ probability of no death during is

■ probability of multiple death in is zero.

Let be the probability that the system is in state at time .

Then,

That is,

For , we get

… (1)

This is Birth-death equation of a Markov system. Why is it a Markov system?

Some specific cases.

1. Pure birth process. No death in the system. .

Then eqn (1) becomes

Check that the solution of this differential equation is

This is Poisson probability distribution. For example,

● number of packets received at a router in a 5 min period

● number of flaws on an IC panel

● number of typos per page typed

will most likely be Poisson distributed.

Notice average level or value of

and variance in ,

Example. The number of intrusion events as detected by a SNORT type system roughly amounts to 2.1 per day (i.e. the arrival rate per day). What is the probability that it will detect 4 intrusion events on a given day?

P(4 intrusions) =

Consider the equilibrium solution of eqn(1) i.e. . This implies:

M/M/1 queue

■ Poisson arrival (Indicated by M)

■ Exponential service time (time between two

departures from server)

■ Single server

■ Unlimited queue-size

■ Infinite population

What is G/G/m/k/N? What is D/M/m/k/?

Symbolic representation of a queuing system.

State diagram (transition diagram)

At state 0: the balance equation is

Flow-in = Flow-out

… (2a)

At state 1: the balance equation is

Flow-in = Flow-out

(

… (2b)

We see that in general it must be of the form

Define . Then .

Since we get if

Therefore,

or

… (3)

From here, we get,

…(4)

The average number of customers in the system, the resident-density is

=

But,

Discouraged arrivals:

More customers in the system lower arrival rate.

State Transition diagram:

Balance equations:

Continuing this way, we get

Since, we have

i.e. and

Probability that the system is busy = system utilization =

Expected number of customers in the system, L,

=

To use Little theorem once again,

average arrival rate in the system.

= utilization factor in terms of average

arrival rate

Thus,

Using this, and the Little theorem, average time spent in the system (Residence time)

Kendal’s notation for queues:

A/B/C/D/E

Where:

A: Distribution of interarrival times of customers

B: Distribution of service times

C: Number of servers

D: Max number of customers which can be in the system

E: Calling population size

A and B can take any of the following distribution.

: Exponential distribution (Markovian)

: Degenerate or Deterministic distribution

: Erlang distribution (k = shape parameter)

: General distribution (arbitrary distribution)

If G is used for A, it is often appears as GI (General and Independent).

e.g. M/M/m/k/N

Exponential interarrival time (Poison arrival), exponential service time, m servers, buffer size k, population size N

M/M/c queue. Multiserver queue. A queue with c servers.

Balance equations.

At state 0,

At state 1,

At state 2,

At state

At state

Rewriting, we get

… …

This gives, for and

for

From these two, we get (from the requirement )

By Little’s formula, and

Where

= …