# CS457 - System Performance Evaluation - Winter 2008

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

# Analytic Queueing Theory

Text: Chapters 30--36.

## Concepts We Know

#### Parameters

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

• 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

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.