MeanValue Analysis MVA and Related Techniques

1
Chapter 34

• Mean-Value Analysis (MVA) and Related Techniques

2
Analysis of Open Queuing Network

• Open queuing network are used to represent
  transaction processing system.
• The transaction arrival rate is not dependent on
  the load on the system.
• Transaction arrivals are modeled as Poisson
  process.

3
Devices in Computer System

• All devices in the computer system can be molded
  as either
• fixed-capacity service centers
• Single server with exponentially distributed
  service time
• Delay server
• Infinite server with exponentially distributed
  service time

4
Fixed-capacity Service Center

• Fixed-capacity service centers in an open queuing
  network, the response time is
• RiSi(1Qi)
• On arrival of the ith device, the job see Qi
  jobs ahead and expect to wait QiSi seconds.
  Including the service to itself, the job should
  expect a total response time of Si(1Qi)
• No operational law, service is memoryless.

5
Fixed-capacity Service Center (ii)

• Assume job flow balance, the throughput of the
  system is equal to the arrival rate
• X?
• The throughput of the ith device using the forced
  flow law is
• Xi X Vi
• The utilization of the ith device using the
  utilization law is
• Ui Xi SiX Vi Si?Di

6
Fixed-capacity Service Center (iii)

• The queue length of the ith device using Littles
  law is
• Qi XiRi Xi Si(1Qi)Ui(1Qi)
• or
• Qi Ui/(1-Ui)
• Substitute to RiSi(1Qi)
• RiSi/(1-Ui)

7
Delay Service Center

• In delay center, there are infinite server,
  response time is equal to service time regardless
  the queue length.
• RiSi
• The queue length of denotes the number of job
  receiving service since there is no waiting
  required.
• Qi RiXi Ri X ViX Di Ui
• The utilization of the delay center represent the
  mean number of job receiving service, and does
  not need to be less than 1.

8
Example 34.1

• The figure represents a queuing model of a file
  consisting a CPU and two disks, A and B.
  Measurements on a distributed system with six
  clients systems making requests to the file
  server produced the data in next page.

9
Example 34.1 (ii)

• Observation Period 3600 s
• Number of client requests 10, 800
• CPU busy time 1728 s
• Disk A busy time 1512 s
• Disk B busy time 2592 s
• No. of visits (I/O requests) to disk A75, 600
• No. of visits (I/O requests) to disk B86,400

10
Example 34.1 (iii)

• Service Demand and Visit Ratio are
• X 10800/36003 request/s
• VA75600/108007 visits/request
• VB86400/108008 visits/request
• VCPU17816
• DCPU 1728/108000.16
• DA 1512/108000.14
• DB 2592/108000.24
• SCPU 0.16/160.01
• SA 0.14/70.02
• SB 0.24/80.03

11
Example 34.1 (iv)

• Utilizations are
• UiX Di, X 3, DCPU 0.16, DA 0.14, DB 0.24
• UCPU 3 x 0.16 0.48
• UA 3 x 0.14 0.42
• UB 3 x 0.24 0.72
• RiSi/(1-Ui), SCPU 0.01,SA 0.02, SB 0.03
• RCPU 0.01/(1-0.48)0.0192
• RA 0.02/(1-0.42)0.0345
• RB 0.03/(1-0.72)0.107
• R?ViRi
• R16x0.01927x0.03458x0.1071.406

12
Example 34.1 (v)

• We study the impact of following changes
• The impact to increase number of client to 8.
• The impact to use cache for disk B with hit rate
  50, increase CPU overhead by 30, increase disk
  B service time by 10.
• The impact to have a lower cost server with only
  one disk A and direct all I/O request to it.

13
Example 34.1 (vi)

• The impact to increase number of client to 8
• Request rate10800x8/614400
• X 14400/36004 request/s
• UCPU 4 x 0.16 0.64
• UA 4 x 0.14 0.56
• UB 4 x 0.24 0.96
• RCPU 0.01/(1-0.64)0.0278
• RA 0.02/(1-0.56)0.0455
• RB 0.03/(1-0.96)0.75
• R16x0.02787x0.04458x0.756.76
• The serve response time will degrade by a factor
  of 6.76/1.4064.8

14
Example 34.1 (vii)

• The impact to use cache for disk B with hit rate
  50, increase CPU overhead by 30, increase disk
  B service time by 10
• VB 8 x 0.5 4
• SCPU 1.3 x 0.01 0.013, DCPU 1.3 x
  0.160.208 s
• SB 1.1 x 0.03 0.033, DB 0.24x 1.1x0.5
  0.132 s
• Therefore
• UCPU 3 x 0.208 0.624
• UA 3 x 0.14 0.42
• UB 3 x 0.132 0.396
• RCPU 0.013/(1-0.624)0.0346
• RA 0.02/(1-0.42)0.0345
• RB 0.033/(1-0.396)0.0546
• R16x0.03467x0.03454x0.05461.013
• Server response time will improve by
  (1.406-1.013)/1.406 28.

15
Example 34.1 (viii)

• The impact to have a lower cost server with only
  one disk A and direct all I/O request to it
• VB 0
• VA 78 15
• DCPU 0.16
• DA 15 x 0.020.3
• UCPU 3 x 0.16 0.48
• UA 3 x 0.3 0.9
• RCPU 0.01/(1-0.48)0.0192
• RA 0.02/(1-0.90)0.2
• R16x0.019215x0.23.31
• The server response time will degrade by a factor
  3.31/1.4062.35

16
MVA

• MVA solves closed queuing network.
• It gives mean performance.
• The variance computation is not possible using
  this method.

17
Closed Queuing Network

• Given a closed queuing network with N jobs,
  Reiser and Lavenberg (1980) showed that the
  response time of the ith device is given by
• Ri (N) Si 1Qi(N-1)
• The mean queue length is
• Qi(N-1)
• As before, on arrival of the ith device, the job
  see Qi(N-1) jobs ahead and expect to wait
  Qi(N-1)Si seconds. Including the service to
  itself, the job should expect a total response
  time of Si 1Qi(N-1)

18
Closed Queuing Network (ii)

• Given the performance for N-1 users, we can
  compute the performance for N users.
• Since the performance with no users (N0) can be
  easily computed, performance for any number of
  users can be compute iteratively.

19
Closed Queuing Network (iii)

• The system response time using the general time
  law is
• R(N)ViRi(N)
• The system throughput using the interactive
  response time law is
• X(N)N/(R(N)Z)
• The device throughputs measured in terms of jobs
  per second are
• Xi(N)X(N)Vi

20
Closed Queuing Network (iv)

• The device queue length with N jobs in the
  network using Littles Law are
• Qi(N)Xi(N)Ri(N)X(N)Vi Ri(N)
• The equation is developed based on assumption
  that all are fixed-capacity queuing center.
• If a device is delay center, the queue length
  denotes the number of job receiving service.
• Reponses time for delay server is
• Ri(N)Si

21
MVA Algorithm

• For i 1 to M DO Qi0
• For n 1 to N DO
• BEGIN
• FOR i1 to M do RiSi(1Qi) fixed capacity
• or Si Delay centers
• R?RiVi
• XN/(ZR)
• FOR i1 to M DO QiXViRi
• END
• Device throughputs Xi XVi
• Device utilizations UiXSiVi

22
Example 34.2

• Consider the queuing network model. Each user
  make ten I/O request to disk A and five I/O
  request to disk B. The service times per visit
  to disk A and disk B are 300 and 200 ms,
  respectively. Each request takes 2 seconds of
  CPU time and the user think time is 4 second.

23
Example 34.2 (ii)

• Consider the queuing network model. Each user
  make ten I/O request to disk A and five I/O
  request to disk B. The service times per visit
  to disk A and disk B are 300 and 200 ms,
  respectively. Each request takes 2 seconds of
  CPU time and the user think time is 4 second.
• SA0.3, VA10 gt DA3
• SB0.2, VB5 gt DB1
• DCPU2, VCPU105116 gtSCPU0.125
• Z4, N20

24
Example 34.2 (iii)

• SA0.3, VA10, DA3
• SB0.2, VB5 , DB1
• DCPU2, VCPU16, SCPU0.125
• Z4, N20
• Initialization
• Number of users N0
• Device queue length QCPU0, QA0, QB0

25
Example 34.2 (iv)

• SA0.3, SB0.2, SCPU0.125
• VA10, VB5 , VCPU16, Z4, N20
• Iteration 1
• Number of users N1
• RCPUSCPU(1QCPU)0.125(10)0.125
• RA0.3(10)0.3 RB0.2(10)0.2
• R0.125x160.3x100.2x56
• XN/(RZ)1/(64)0.1
• QCPUX RCPU VCPU0.1 x 0.125 x 160.2
• QA0.1x 0.3x100.3 QB0.1x0.2x50.1

26
Example 34.2 (v)

• SA0.3, SB0.2, SCPU0.125
• VA10, VB5 , VCPU16, Z4, N20
• Iteration 2
• Number of users N2
• RCPUSCPU(1QCPU)0.125(10.2)0.15
• RA0.3(10.3)0.39 RB0.2(10.1)0.22
• R0.15x160.39x100.22x57.4
• XN/(RZ)2/(7.44)0.175
• QCPUX RCPU VCPU0.175 x 0.15 x 160.421
• QA0.175x 0.39x100.684
• QB0.175x0.22x50.193

27
Examples 34.2 (vi)
28
Approximate MVA

• MVA is a recursive algorithm.
• Computation of performance of N jobs in the
  network require knowledge of performance with N-1
  jobs
• N0, performance is trivially known.
• Start with N0, we computer N1,2,
• For small N, it is not computationally expensive
• For large N, approximate analysis technique would
  be preferred.

29
Approximate MVA

• Based on assumption that as the no. of job
  increase, the queue length at each device
  increase proportionally.
• Qi(N)/Nai
• Particular,
• Qi(N-1)/(N-1)Qi(N)/N
• Qi(N-1) (N-1)/N Qi(N)

30
Approximate MVA (ii)

• The MVA equation can therefore be written as
  follows
• The initial value of queue length will not affect
  final result.
• The initial value affect number of iteration.
• One alternative is start with all queue length
  being equal.

31
Approximate Algorithm

• Initialization
• X0
• FOR i1 TO M DO QiN/M
• Iteration
• WHILE maxiQi-XRiVigte DO
• BEGIN
• FOR i1 to M DO RiSi(1(N-1)/N Qi or Si
• R?RiVi
• XN/(ZR)
• FOR i1 to M DO QiX Vi Ri
• END
• Device throughputs Xi X Vi
• Device utilizations UiX Si Vi

32
Example 34.3

• Consider the time sharing system of Example 34.2.
  Lets analyze it with 20 users and the maximum
  error is less than 0.01.
• SA0.3, VA10, DA3
• SB0.2, VB5 , DB1
• DCPU2, VCPU16, SCPU0.125
• Z4, N20
• Initialization
• Device queue length QCPUQAQB20/36.67

33
Example 34.3 (ii)

• SA0.3, SB0.2, SCPU0.125
• VA10, VB5 , VCPU16, Z4, N20
• Iteration 1
• RCPUSCPU(1QCPU)0.125(16.77)0.92
• RA0.3(16.77)2.2 RB0.2(16.67)1.47
• R0.92x162.1x101.47x544
• XN/(RZ)20/(444)0.42
• QCPUX RCPU VCPU0.42 x 0.92 x 166.11
• QA0.42x2.20x109.17 QB0.42x1.47x53.06
• ?Q max6.67-6.11,6.67-9.17,6.67-3.06
• max0.56,2.5, 3.613.61

34
Example 34.3 (iii)
35
Comparison of MVA

• Exact MVA
• No. of computation depends on value of N
• If N is large, no. of iterations can be large
• Memory required will be substantial
• It is used for small no. job classes.
• Approximate MVA
• No. of iteration is independent on N
• If accuracy requirement is not high, Approximate
  MVA should be choice for quick and economical
  solution.

36
Throughput Comparison
37
Response Time Comparison
38
CPU queue length Comparison
39
Balance System

• A system without a bottleneck device is called a
  balance system.
• Total service time demand on all device are
  equal.
• An unbalanced system performance can be improved
  by replacing the bottleneck device with a faster
  device.