# CS457 - System Performance Evaluation - Winter 2008

### Questions and Comments

# Lecture 32

# Analytic Queueing Theory - Queueing Networks

Text: Chapters 30--36

Collection of interacting service providers.

One new concept:

- p_ij = P( job just finished at i will get immediate further processing
at j ) 1 <= i,j <= M

## 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

- Number of jobs is a random variable
- External arrivals can appear anywhere
- Extra destination for departures: p_i(M+1) = P( job departs from
network immediately after service at i )

Definitions at server i

- external arrivals \gamma_i
- internal arrivals
- total arrivals \lambda_i = \gamma_i + internal arrivals
- throughput X_i = \lambda_i
- total arrivals \lambda_i = \gamma_i + \sum_j X_j p_ji
- Remember: job can come back to where it just finished.

#### Solution procedure

- Write balance equations
- Solve balance equations
- Derive performance metrics

Example: pdf

#### Utilization

U_i = \lambda_i s_i

- s_i, the mean service time, is a parameter assumed to be known

#### Throughput

X - rate of jobs departing from the network

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

### Closed Networks in Steady State

- No external arrivals.
- No departures
- Number of jobs in the system, N, is constant

#### A. Single server per service centre

1. Total arrival rate

- \lambda_i = sum_j \lambda_j p_ji
- \sum_i p_ji = 1
- Balance equations give M equations, which are not linearly
independent
- Can only solve for ratios of \lambda_i

Example: pdf

2. Utilization

- Can only solve for rations of U_i
- U_i / U_j = (\lanbda_i / \lambda_j) * s_i/s_j: the factor in
paraenthesis is known

3. Throughput

Arbitrary decision

- Cut the system somewhere, e.g. at the arrival of server i
- Then X is the rate of jobs passing across the cut

#### 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 service i per job

- V_i = \lambda_i / X = \lambda_i / \lambda_0

3. Utilization

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

- D_i = X s_i is the demand per job at server i
U_i / U_j = D_i / D_j

4. Upper bound on throughput

D = \sum D_i

X(n): throughput with N users

- X(1) = 1 / (D + Z)
- X(N) <= N/(Z + D) : equality when there is no contention

5. Bottleneck analysis

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

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

6. Lower bound on mean response time

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

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