1. 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.

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 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 systems and 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 queue as a single “black box”

11. Basic Quantities Tthe length of the observation period A the number of arrivals during the observation period Bthe total amount of busy times during the observation period C the number of completions during the observation period

12. ACB T

13. Derived quantities = A/Tthe arrival rate X = C/Tthe output rate U = B/Tthe utilization S = B/Cthe mean service time

14. Example Checkout lane: T = 2 hours A = 31 customers B = 108 minutes C = 30 customers We have = 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 = S

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 Number of tasks in the system Time T W N = W/T 0 The green area and the area delimited by the dashed lines have equal sizes

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) Arrivals Departures Open network

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

24. Operational quantities Over the observation period, we measure C = the number of job completions C k = the number of tasks completed by device k We define X 0 = C/T = the system throughput X k = C k /T = the output rate at server k V k = C k /C = the visit count at server k

25. Relations C k = V k C 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 X k = V k X 0 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 X i is the same as the total input rates to device i A i = C i True for all observation periods long enough such that |A i - C i | << C i Output rates X i are then throughputs

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

28. System response time (II) Applying Little’s law, we have R = Nbar/X 0 and nbar i = R i X i which we can rewrite nbar i = R i V i X 0 Hence R = Σ i V i R i

29. Application: Bottleneck analysis A system has one CPU and one disk drive It processes transactions such that V CPU = 12 and S CPU = 5ms V Disk = 11 and S DISK = 8ms What is the maximum system throughput?

30. Bottleneck analysis (cont’d) Let us compute maximum device throughputs Maximum X CPU = 1/0.005 = 200 requests/s Maximum X Disk = 1/0.008 = 125 requests/s Since X i = V i X 0 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 V i S i 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. Systems with terminals M 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 ) X 0 then R = M/X 0 – 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 of 5 s?

36. Answer Applying R = M/X 0 – Z, we compute a lower bound for the response time R min = M/X 0,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/X 0 – Z, we have X 0 = (R + Z)/M Hence X 0 = (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 V disk = 20 visits to the single disk per user interaction U disk = 0.30 S disk = 25 ms ( http://www.owlnet.rice.edu/~elec428/handouts/Op.Analysis.pdf ) http://www.owlnet.rice.edu/~elec428/handouts/Op.Analysis.pdf

40. Answer Let us compute first the throughput X 0 Since X k = U k /S k X disk = 0.30/0.025 = 12 disk requests/s Since X k = V k X 0, we have X 0 = X k /V k X 0 = 12/20 = 0.6 interactions/s The response time is then R = M/X 0 – Z = 25/0.6 – 18 = 23.7s

41. Demand Rather than expressing CPU workloads by the product V CPU S CPU, we can use the total demand of the task D CPU = V CPU S CPU Since X k = U k /S k and X k = V k X 0 U CPU /D CPU = X CPU /V CPU = X 0

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 X 0 = U CPU /D CPU = 0.5/0.4 = 1.25 tasks/s Since U disk = X disk S disk and X disk = V disk X 0 X disk = 1.25 x 60 = 75 disk requests/s U disk = 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/handou ts/Op.Analysis.pdf http://www.owlnet.rice.edu/~elec428/handou ts/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.The Operational Analysis of Queueing Network Models Remains the fundamental text on operational analysis