CS457 - System Performance Evaluation - Winter 2008


Questions and Comments

  1. Tutorial: MC4061, Monday March 17, 18.00 to 19.00

Lecture 28

Analytic Queueing Theory

Text: Chapters 30--36.

Markov process

A stochastic process in which S(n+1) is independent of S(0), ..., S(n-1), but may depend on S(n).

In performance evaluation, when we talk of a Markov process we usually also mean that the next transition occurs at a time distributed by an exponential distribution.

Another useful property of the exponential distribution

Birth-death process

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.

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 ) = (\lambda \Delta t) * exp(-\lambda \Delta t) = \lambda \Delta t + o(\Delta t)
  2. P(exactly one death ) = \mu \Delta t + o(\Delta t)
  3. P(exactly zero births ) = exp(-\lambda \Delta t) = 1 - \lambda \Delta t + o(\Delta t)
  4. P(exactly zero deaths ) = 1 - \mu \Delta t + o(\Delta t)
  5. P(more than one birth and/or death) = o(\Delta t)

o(t) is any function that has the property lim(t->0) o(t)/t = 0.

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

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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