CS457 - System Performance Evaluation - Winter 2010


Public Service Announcement

  1. pdf available
  2. Another pdf available.
  3. Assignment 4

Lecture 32 - Queueing Networks

General Markov Processes

Solution method

  1. Draw state transition diagram
  2. Obtain balance condition equations: rate in equals rate out
  3. Solve balance equations with normalizing equation
  4. Determine performance metrics from pn

Simple in principle. May be complex in practice.


Analytic Queueing Theory - Queueing Networks

Collection of interacting service providers.

One new concept:

Operational Analysis

Very similar to what we did near the beginning of the course

Open Networks in Steady State

Jobs enter the network and depart from it

  1. Number of requests in a server is a random variable, N_i
  2. External arrivals can appear anywhere.
  3. Extra destination for departures:

Definitions at server i

  1. external arrivals \gamma_i
  2. internal arrivals
  3. total arrivals \lambda_i = \gamma_i + internal arrivals
  4. throughput X_i = \lambda_i
  5. total arrivals \lambda_i = \gamma_i + \sum_j X_j q_ji
  6. conservation of requests

Solution procedure

  1. Write balance equations
  2. Solve balance equations
  3. Derive performance metrics

Utilization

U_i = \lambda_i s_i

Throughput

X - rate of jobs departing from the network

  1. X = \sum \gamma_i because the network is stable
  2. X = \sum \lambda_i q_i(M+1)
  3. Not too hard to prove that these are the same

Closed Networks in Steady State

  1. No external arrivals.
  2. No departures
  3. Number of jobs in the system, N, is constant

A. Single server per service centre

1. Total arrival rate

Example: pdf

2. Utilization

3. Throughput

Arbitrary decision

B. Interactive System in Steady State

1. Call the set of terminals server 0

2. Throughput

Cut at entrance to terminals: X = \lambda_0

Define visit ratio: number of visits to server i per request

3. Utilization

U_i = \lambda_i s_i = X V_i s_i = X D_i

4. Upper bound on throughput

D = \sum D_i

X(n): throughput with N users

5. Bottleneck analysis

  1. b - server with highest demand
  2. D_b = max{D_i} = D_max
  3. U_b = X D_b <= 1
  4. X <= 1/D_max

Therefore, X(N) <= min( 1/D_max, N / (D+Z) )

6. Lower bound on mean response time

  1. Treat entire system as one server. Then R(N) = N/X - Z
  2. R(N) >= D = sum_i D_i because X(N) <= N / (D+Z)
  3. X(N) <= 1 / D_max, so that R(N) <= ND_max - Z

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