CS457 - System Performance Evaluation - Winter 2010


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Lecture 26 - Statistical Mechanics

Two State System

A Birth/Death Process


Analytic Queueing Theory

New Concepts

Stochastic process

The example above is a stochastic process. Why?

Markov process

The example above is a Markov process. Why?


Birth-death processes

The example above is a Birth-death process. Why?

A special process where only transitions to neighbouring states are possible That is, if we are in S(j) then the next state can be

  1. S(j-1), a death occurs.

    Death rate \mu_j.

  2. S(j+1), a birth occurs

    Birth rate \lambda_j.

  3. S(j), neither a birth nor a death occurs, or both a birth and death occur.

We now want to examine what happens, in a birth-death process, in the short time between t and t + \Delta t. Use the Poisson distribution.

  1. P(exactly one birth | n(t) = j ) = (\lambda_j \Delta t) * exp(-\lambda_j \Delta t) = \lambda_j \Delta t + terms that are quadratic or higher order in \Delta t
  2. P(exactly one death | n(t) = j ) = \mu_j \Delta t + o( (\Delta t)^2 )
  3. P(exactly zero births | n(t) = j ) = exp(-\lambda_j \Delta t) = 1 - \lambda \Delta t + o( (\Delta t)^2 )
  4. P(exactly zero deaths | n(t) = j ) = 1 - \mu_j \Delta t + o( (\Delta t)^2 )
  5. P(more than one birth and/or death) = o( (\Delta t)^2 )

Exercise for the reader. Explain how it is that this is equivalent to only Arrival and Departure events affecting the system state.

What happens between t and \Delta t ? Four possibilities

  1. Exactly one birth
  2. Exactly one death
  3. Neither a birth nor a death
  4. Something else

The book's notation: P(n(t) = j) = pj(t)

Then

This equation looks as though you could solve it!

Steady State Solutions

Reminder about transient versus steady state.

Definition of steady state: dp_j(t) / dt = 0.

Then

Solve this iteratively

  1. l0 p0 = m1 p1 => p1 = (l0/m1) p0
  2. (l1 + m1) p1 = l0 p0 + m2 p2 => (l1 + m1) (l0/m1) p0 = l0 p0 + m2 p2 => p2 = p0 (l1*l0 / m1*m0)
  3. ...
  4. pn = p0 (l_n-1*...*l1*l0 / m_n-1 *...*m1*m0)
  5. Set p0 by the condition sum_n pn = 1.

This looks like you could solve it further, but you can't. How could you possibly solve the differential equation above, in that case?

What do we do? Try simplified examples.

The Simplest Example M/M/1

First M: Markovian (Exponential) birth (interarrival) times

Second M: Markovian (Exponential) life/death (service) times

1: one server

Assumptions:

  1. \lambda_j = \lambda
  2. \mu_j = \mu
  3. Define r = \lambda / \mu

Then

  1. pn = r^n p0
  2. p0 = 1 / (1 + r + r^2 + ...) = 1 - r.
  3. pn = (1 - r) * r^n.

The mean number of jobs in the system is E(r) = sum_n n * (1 - r ) * r^n = r / (1 - r)

Little's law


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