CS457 - System Performance Evaluation - Winter 2010

Public Service Announcement

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

Lecture 33 - Queueing Networks: Operational Analysis

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

• e.g., Little's law
• The important new idea is that where a request goes after receiving service is no longer deterministic

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.
   • Average rate of external arrivals at server i, \gamma_i, is deterministic
   • Actual arrivals are random
3. Extra destination for departures:
   • q_i(M+1) = P( service is complete for a request immediately after service at i )

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
   • because of stability
5. total arrivals \lambda_i = \gamma_i + \sum_j X_j q_ji
   • The second term is the internal arrivals
   • Remember: job can come back to where it just finished.
   • Therefore, X_i = \gamma_i + \sum_j X_j q_ji
6. conservation of requests
   • \sum_i^M \lambda_i = \sum_i^M \gamma_i + \sum_i^M \sum_j^M \lambda_j q_ji
   • \sum_i^M \lambda_i = external input + \sum_j^M \lambda_j (1 - q_j(M+1))
   • \sum_i^M \lambda_i = external input + \sum_j^M \lambda_j - sum_j^M \lambda_j q_j(M+1))
   • \sum_i^M \lambda_i = external input + \sum_j^M \lambda_j - external output
   • \sum_i^M \gamma_i = sum_j^M \lambda_j q_j(M+1))

Solution procedure

1. Write balance equations
   • These are different balance equations.
2. Solve balance equations
3. Derive performance metrics

Utilization

U_i = \lambda_i s_i

• s_i, the mean service time, is a parameter assumed to be known
• s_i = 1 / \mu_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
   • In fact, we did so just above

Closed Networks in Steady State

1. No external arrivals.
   • \gamma_i = 0
2. No departures
3. Number of jobs in the system, N, is constant
   • stability is assured

A. Single server per service centre

1. Total arrival rate

• \lambda_i = sum_j \lambda_j q_ji
• \sum_i q_ji = 1
  • \sum_i \lambda_i = \aum_i sum_j \lambda_j q_ji = \sum_j \lambda_j
• Balance equations give M equations, which are not linearly independent
• Can only solve for ratios of \lambda_i
  • Because the system is closed the number of requests is conserved

Example: pdf

2. Utilization

• Can only solve for ratios of U_i
• U_i / U_j = (\lanbda_i / \lambda_j) * s_i/s_j: the factor in parenthesis 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

• If we cut the network into two disjoint pieces the average flow across the cut is zero.

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

• V_i = \lambda_i / X = \lambda_i / \lambda_0
• This is what we solve for.

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 request at server i
  • demand measured in seconds
  • units of D_i is seconds per request
  • units of X is requests per second
  • U is unitless
• 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

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

Queueing Networks: Markov Analysis

Assumptions

1. Service times (S_i = 1 / M_i) are exponentially distributed, with mean s_i = 1 / m_i
2. A product form solution exists: p(n_1, ..., n_M) = (1/G) {(r_1)^(n_1) (r_2)^(n_2) (r_3)^(n_3)... (r_M)^(n_M) }
3. (Open networks) Interarrival times (\Gamma_i) are exponentially distributed with mean 1 / \gamma_i
4. (Closed networks) Think times (\Gamma) are exponentially distributed with mean 1 / \gamma

Solution Method

1. Define states
   • Usually (n_1, n_2, ..., n_M)
2. Draw the state diagram
   • Flow into p(n_1, ..., n_i, ..., n_M)
     1. external request to server i: p(n_1, ..., n_i - 1, ..., n_M) \gamma_i
     2. Departure from system from server i: p(n_1, ..., n_i + 1, ..., n_M) \mu_i * q_{i(M+1)}
     3. Departure from server j, arrival at server i: p(n_1, ..., n_i - 1, ..., n_j + 1, ..., n_M) * q_ji \mu_j
     4. Departure from server i, arrival at server i: p(n_1, ..., n_i, ..., n_M) * q_ii \mu_i
   • Flow out of p(n_1, ..., n_i, ..., n_M)
     1. external request to server i: p(n_1, ..., n_i + 1, ..., n_M) \gamma_i
     2. Departure from system from server i: p(n_1, ..., n_i - 1, ..., n_M) \mu_i * q_{i(M+1)} \mu_i
     3. Departure from server i, arrival at server j: p(n_1, ..., n_i - 1, ..., n_j + 1, ..., n_M) * q_ij /mu_j
     4. Departure from server i, arrival at server i: p(n_1, ..., n_i, ..., n_M) * q_ii \mu_i
3. Solve the balance equations
   • Flow into p(n_1, ..., n_i, ..., n_M)
     1. \sum_i \gamma_i * p(p(n_1, ..., n_i - 1, ..., n_M)
     2. \sum_i \mu_i * q_{i(M+1) * p(n_1, ..., n_i + 1, ..., n_M)
     3. \sum_{i!=j} \mu_j * q_ji * p(n_1, ..., n_i - 1, ..., n_j + 1, ..., n_M)
     4. \mu_i * q_ii * p(n_1, ..., n_i, ..., n_M)
   • Flow out of p(n_1, ..., n_i, ..., n_M)
     1. \sum_i { \gamma_i + \mu_i q_{i(M+1)} + \sum_{j!=i} /mu_j * q_ij + \mu_i * q_ii } p(n_1, ..., n_i, ..., n_M)
        = \sum_i { \gamma_i + \mu_i } p(n_1, ..., n_i, ..., n_M
4. Verify product form solution
   • p(n_1, ..., n_i + 1, ..., n_M) / p(n_1, ..., n_i, ..., n_M) = r_i
   • Therefore divide balance equations by p(n_1, ..., n_i, ..., n_M)
   • Flow in is
     1. \sum_i { \gamma_i /r_i + \mu_i * q_{i(M+1) r_i + \sum_{j!=i} \mu_j * q_ji * {r_j / r_i} } + \mu_i q_ii
   • Flow out is
     1. \sum_i { \gamma_i + \mu_i }
   • Network is stable
     • \sum_i \gamma_i = \sum_i \mu_i * q_i{M+1} r_i
5. Evaluation of G
   • \sum_{all n_i} (1/G) =
   • Therefore, G = 1 / { (1-r_1)(1-r_2)...(1-r_M)

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