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AN INTRODUCTION TO THE OPERATIONAL ANALYSIS OF QUEUING NETWORK MODELS

AN INTRODUCTION TO THE OPERATIONAL ANALYSIS OF QUEUING NETWORK MODELS. Peter J. Denning, Jeffrey P. Buzen, The Operational Analysis of Queueing Network Models. ACM Computing Surveys 10(3): 225-261, 1978. Queuing Models. Computer systems contain many queues Ready queue I/O device queues

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AN INTRODUCTION TO THE OPERATIONAL ANALYSIS OF QUEUING NETWORK MODELS

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  1. AN INTRODUCTION TO THE OPERATIONAL ANALYSISOF QUEUING NETWORK MODELS Peter J. Denning, Jeffrey P. Buzen, The Operational Analysis of Queueing Network Models.ACM Computing Surveys 10(3): 225-261, 1978.

  2. Queuing Models • Computer systems contain many queues • Ready queue • I/O device queues • Message queues • … • Bottlenecks and queuing delays have a major impact on computer system performance • Queuing models capture this impact

  3. Operational Analysis • Operational Analysis • Provides good bounds of performance indices of many computer systems • Only makes verifiable assumptions of behavior of these systems • Is easier to teach and to understand than stochastic analysis

  4. Applications • “Back of the envelope” estimations of system performance • Fast and easy • Validation of results of a simulation study • Checks for reasonableness of results • Communication with non-specialists

  5. Other queuing models • Stochastic models • Also known as Markov models • More powerful • Often make unrealistic assumptions about investigated systems • Disk service times are not exponentially distributed! • Provide surprisingly good estimates of computer system performance

  6. Note • Stochastic models can also be used to estimate • Availability of computer systems:fraction of time system will be operational • Reliability of computer systems:probability that the system will never fail over a time interval of duration t • Essential for real-time systemsand storage systems

  7. BASIC CONCEPTS

  8. Hypothesis testability • All hypotheses made by OA can be tested by • Observing the behavior of actual systems • Over finite time periods

  9. Operational Quantities • Can be • Basic Quantities:Directly measured over a finite observation period • Number of arrivals, … • Derived Quantities:Computed from basic quantities • Server utilization, mean service time, …

  10. A single server • Server has its associated queue • We will treat the server and its queueas a single “black box”

  11. Basic Quantities • T the length of the observation period • A the number of arrivals during the observation period • B the total amount of busy times during the observation period • C the number of completions during the observation period

  12. T A B C

  13. Derived quantities • l = A/T the arrival rate • X = C/T the output rate • U = B/T the utilization • S = B/C the mean service time

  14. Example • Checkout lane: • T = 2 hours • A = 31 customers • B = 108 minutes • C = 30 customers • We have • l = 15.5 customers/hour • …

  15. Utilization law • U = B/T = (C/T )(B/C) = XS • Can compute utilization knowing both • Completion rate • Mean service time When you think about it, it is fairly intuitive

  16. Job flow balance • For most systems, arrivals and completions balance each other over any long observation period • Systems have finite buffers • We will assume that A = C • Can check validity of assumption over any specific observation period • If A = C then U = lS

  17. Steady state • If the system has a steady state, we can define • N = average number of tasks in system • R = average residency of tasks in system • R is also known as the response time

  18. Little’s law • If W is the total time spent by all tasks inside the system over the observation period • Measured in requests × time units • Then • N = W/T • R = W/C • Since W/T = (C/T)(W/C) = XR, N = XR This is an important result

  19. Explanation The green area and the area delimited by thedashed lines have equalsizes Number of tasks in the system N = W/T W Time 0 T

  20. An example • An FBI agent has observed people entering and leaving a secret meeting • Over a period of two hours he has seen 10 people staying an average of 30 minutes each • Having 10 people staying 30 minutes each corresponds to 300 people x minutes • This is the same as having 300/120 = 2.5 people staying over the whole duration of the meeting

  21. NETWORKS OF SERVERS

  22. Network of servers (I) Open network Arrivals Departures

  23. Network of servers (II) Closed network Arrivals Departures

  24. Operational quantities • Over the observation period, we measure • C = the number of job completions • Ck = the number of tasks completed by device k • We define • X0= C/T = the system throughput • Xk = Ck/T=the output rate at server k • Vk = Ck/C=the visit count at server k

  25. Relations • Ck = VkC • The total number of visits of server k is equal to the number of visits of server k by job completion times the number of job completions • Xk = Vk X0 • The output rate of server k is equal to the number of visits of server k by job completion times the job throughput

  26. Principle of job flow balance • For each device i, the output rate Xiis the same as the total input rates to device i Ai= Ci • True for all observation periods long enough such that |Ai- Ci | << Ci • Output rates Xiare then throughputs

  27. System response time (I) • We define • Nbar = average number of jobs in the system • nbari = average number of jobs at device i • Nbar = Σi nbari

  28. System response time (II) • Applying Little’s law, we have R = Nbar/X0 and nbari = RiXi which we can rewrite nbari = RiViX0 • Hence R = Σi ViRi

  29. Application:Bottleneck analysis • A system has one CPU and one disk drive • It processes transactions such that • VCPU = 12 and SCPU= 5ms • VDisk = 11 and SDISK= 8ms • What is the maximum system throughput?

  30. Bottleneck analysis (cont’d) • Let us compute maximum device throughputs • Maximum XCPU= 1/0.005 = 200 requests/s • Maximum XDisk = 1/0.008 = 125 requests/s • Since Xi = ViX0 • Maximum throughput compatible with CPU workload is 200/12 = 16.7 transactions/s • Maximum throughput compatible with disk workload is 125/11 = 11.4 transactions/s

  31. Bottleneck analysis (cont’d) • The disk is this the bottleneck • It has highest ViSi product • Identifying feature of any bottleneck device • Increasing the system throughput might require • Sharing disk requests with a second disk • Increasing the efficiency of the system I/O buffer Important

  32. Bottleneck analysis (cont’d) • In addition, the maximum throughput of 11.4 transactions/s can only be achieved with a 100% disk utilization • It would result in very large queues at the disk • In practice, we want to limit the disk utilization to 60 to 80%

  33. M Terminals Systems with terminals Whole system

  34. Interactive response time formula • We have • M terminals • Think time Z between the completion of a job and the submission of the next job • Applying Little’s law to the whole system M = (R + Z ) X0 then R = M/X0 – Z Very Important

  35. Problem 1 • A system • Can process up to 5 transactions/s • Has 60 client workstations • Client think time is 5s • Can the system achieve a response time of5 s?

  36. Answer • Applying R = M/X0 – Z, we compute a lower bound for the response time Rmin = M/X0,max – Z = 60/5 – 5 =7 • Our answer is no

  37. Problem 2 • We have • M = 50 terminals • Z = 20 s • R = 4s • What is the system throughput?

  38. Answer • From R = M/X0 – Z, we have X0 = (R + Z)/M Hence X0 = (20 + 4)/50 = 0.48 tasks/s

  39. Problem 3 • Compute the response time of a system knowing the following parameters • M = 25 terminals (users) • Z = 18 s • Vdisk = 20 visits to the single disk per user interaction • Udisk = 0.30 • Sdisk = 25 ms (http://www.owlnet.rice.edu/~elec428/handouts/Op.Analysis.pdf)

  40. Answer • Let us compute first the throughput X0 • Since Xk= Uk/Sk Xdisk= 0.30/0.025 = 12 disk requests/s • Since Xk = Vk X0 , we have X0 = Xk/Vk X0 = 12/20 = 0.6 interactions/s • The response time is then R = M/X0 – Z = 25/0.6 – 18 = 23.7s

  41. Demand • Rather than expressing CPU workloads by the product VCPU SCPU , we can use the total demand of the task DCPU = VCPU SCPU • Since Xk = Uk/Sk and Xk = Vk X0 UCPU /DCPU = XCPU /VCPU= X0

  42. Problem 4 • A computer system has a single disk • It processes tasks with an average CPU demand of 400 ms. • Each task requires 60 disk accesses • Each disk access takes 10 ms • If the CPU utilization is 50%, what are • The system throughput? • The disk utilization?

  43. Answer • X0 = UCPU /DCPU = 0.5/0.4 = 1.25 tasks/s • Since Udisk = Xdisk Sdisk and Xdisk = Vdisk X0 Xdisk= 1.25 x 60 = 75 disk requests/s Udisk= 75 x 0.010 = .75 or 75% (a rather high disk utilization)

  44. Load–dependent behavior • Device service times may depend on number of jobs at device • Disk drives optimize disk accesses • Number of visits to a device may depend on workload • Swap device We did not cover that topic

  45. Resources • http://www.owlnet.rice.edu/~elec428/handouts/Op.Analysis.pdf • Offers a concise summary of the topics that were discussed in class • Peter J. Denning, Jeffrey P. Buzen, The Operational Analysis of Queueing Network Models. ACM Computing Surveys 10(3): 225-261, 1978. • Remains the fundamental text on operational analysis

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