CS457 - System Performance Evaluation - Winter 2008

Questions and Comments

  1. Tutorial: MC4045, Monday March 10, 18.00 to 19.00

Lecture 27

Analytic Queueing Theory

Text: Chapters 30--36.

Concepts We Know


With some change in notation.

  1. \lambda - arrival rate

    1/\lambda - mean interrival time

  2. 1/\mu - mean service time

    \mu - service rate

Carefully distinguish between

Stable system

Throughput, X = \lambda, arrival rate


U = \lambda / \mu

Stability requires \lambda <= \mu

Little's Law

XE(r) = E(n)

E(r) - mean response time

E(n) - mean number of jobs in the system

Note the change in notation

New Concepts

Stochastic process

A sequence of random variables indexed by time: S0, S1, S2, ...

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 one birth ) = 1 - \mu \Delta t + o(\Delta t)
  5. P(more than one birth and/or deatth) = o(\Delta t)

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

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