CS457 - System Performance Evaluation - Winter 2008

Questions and Comments

Tutorial: MC4045, Monday March 10, 18.00 to 19.00

Lecture 27

Analytic Queueing Theory

Text: Chapters 30--36.

Concepts We Know

Parameters

With some change in notation.

\lambda - arrival rate
1/\lambda - mean interrival time
1/\mu - mean service time
\mu - service rate

Carefully distinguish between

Throughput - X - metric - jobs done per second
Service rate - \mu - parameter - jobs that can be done per second when the server is processing

Stable system

Throughput, X = \lambda, arrival rate

for the steady state phase of processing

Utilization

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, ...

E.g. state of a system at time t. Why?

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.

In a Markov process with exponentially distributed transitions the mean transition rate is the only parameter that can vary with state.
That is, if in state S(j) then the transition rate is \lambda_j

Another useful property of the exponential distribution

Y1 - arrival time - exponential - \lambda
Y2 - service time - exponential - \mu
Ym - when the first of an arrival or end of service occurs - Ym = min(Y1, Y2)
P(Ym < y) = P(Y1 < y AND Y2 < y) = F1(y) * F2(y) = exp(-\lambda y) * exp( -\mu y) = exp( -(\lambda + \mu) * y )
Ym is exponentially distributed with mean 1 / (\lambda + \mu)

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

S(j-1), a death occurs.
Death rate \mu_j.
S(j+1), a birth occurs
Birth rate \lambda_j.
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

P(exactly one birth ) = (\lambda \Delta t) * exp(-\lambda \Delta t) = \lambda \Delta t + o(\Delta t)
P(exactly one death ) = \mu \Delta t + o(\Delta t)
P(exactly zero births ) = exp(-\lambda \Delta t) = 1 - \lambda \Delta t + o(\Delta t)
P(exactly one birth ) = 1 - \mu \Delta t + o(\Delta t)
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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