# CS457 - System Performance Evaluation - Winter 2010

## Public Service Announcements

• class on assignment 3
• possible tutorial
2. Issues out of which on which exam questions could be based.
• anything in the notes with a question mark

# Analytic Queueing Theory

## 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

XE(r) = E(n)

## 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).

• We could say something like, "The state of the system incorporates the past completely."
• Discrete event simulation implicitly assumes that the system is Markovian.

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. Why? (You should already know this.)

• 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)
• Why does this matter?

# Birth-death processes

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
• Write `terms that are quadratic or higher order in \Delta t' as o( (\Delta t)^2 )
• lim_x->0 o( x^2 ) / x = 0
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
• j-1 -> j
• P(n(t+\Delta t) = j AND n(t) = j-1) = P(n(t+\Delta t) = j | n(t) = j-1)P(n(t) = j-1) = (\lambda_(j-1) \Delta t + o(\Delta t))P(n(t) = j-1)
• Abbreviate the notation P(n(t) = j) = pj(t) (This is the notation used in the text.
• P(n(t+\Delta t) = j AND n(t) = j-1) = (\lambda_(j-1) \Delta t) p_j-1(t)
2. Exactly one death
• j+1 -> j
• P(n(t+\Delta t) = j AND n(t) = j+1) = (\mu_(j+1) \Delta t) p_(j+1)(t)
3. Neither a birth nor a death
• j -> j
• P(n(t+\Delta t) = j AND n(t) = j) = 1 - (\lambda_j + \mu_j) p_j(t)
4. Something else
• Nothing. Why? Add P(1) + P(2) + P(3)

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

Then

• p_j(t+dt) = p_j(t) + ( \lambda_j-1 p_j-1(t) + \mu_j+1 p_j+1(t) - (\lambda_j + \mu_j) p_j(t) )dt + o(dt)
• (p_j(t+dt) - p_j(t) ) / dt = \lambda_j-1 p_j-1(t) + \mu_j+1 p_j+1(t) - (\lambda_j + \mu_j) p_j(t) + o(dt) / dt
• In the limit dt -> 0 dp_j(t) / dt = \lambda_j-1 p_j-1(t) + \mu_j+1 p_j+1(t) - (\lambda_j + \mu_j) p_j(t)

This equation looks as though you could solve it!

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

Then

• \lambda_j-1 p_j-1(t) + \mu_j+1 p_j+1(t) - (\lambda_j + \mu_j) p_j(t) = 0

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)

• Goes to infinity as r -> 1. Why?

Little's law

• Mean response time: E(r) = (1/\lambda) * E(n) = 1 / (\mu - \lambda)
• Goes to infinity as \lambda -> \mu from below. Why?
• What happens when \lambda > \mu?