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Components performance modelling

Components performance modelling. Outline of queue networks - Mean Value Analisys (MVA) for open and close queue networks. outgoing requests. incoming requests. CPU. DISK. DISK. TAPE. TAPE. M clients. CPU. Open queue network. Closed queue network (number finite of users).

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Components performance modelling

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  1. Components performance modelling • Outline of queue networks- Mean Value Analisys (MVA) for open and close • queue networks

  2. outgoing requests incoming requests CPU DISK DISK TAPE TAPE M clients CPU Open queue network Closed queue network (number finite of users)

  3. Kind of resources in a queue network S(n) R(n) Load independent n S(n) R(n) Load dependent n S(n) R(n) Delay n

  4. Definitions K: number of queues X0: network average throughput. If open network in a stationary condition X0 =  Vi: average number of visits a generic request makes to i server from its generation to its service time (request goes out from the system if open network) Si: average request service time at the server i Wi: average request waiting time in the queue i Ri: average request answer time in the queue i Ri = Si + Wi

  5. Definitions Xi: throughput for the i-th queue Xi =X0 Vi R’i: average request residence time in the queue i from its creation to its service time (request goes out from the system if open network) R’i =Vi Ri Di: request service demand to a server in a queue i from its creation to its service time (request goes out from the system if open network) Di =Vi Si

  6. Qi: total time a request spends waiting in the queue i from its creation to its service time (request goes out from the system if open network) Qi =Vi Wi ------------------------------- R’i =Vi Ri =Vi (Wi +Si) =Wi Vi + Si Vi = Qi +Di ------------------------------- R0: average request answer time from the whole system R0 = ki=1 R’i ni: average number of requests waiting or in service at the queue i N: average number of requests in the system N = ki=1 ni

  7. Open queue network outgoing requests incoming requests CPU DISK TAPE 7

  8. Open networks (Single Class) Equations: Arrival theorem (for open networks): the average number of requests in a queue i that an incoming request find in the same queue (nai), is equal to the average number of requests in the queue i (ni). Ri(n) = Si + Wi(n) = Si + ni  Si Using Little’s Law (ni = Xi Ri) and Ui = XiSi: Ri = Si _ (1-Ui) Ri = Si (1 + ni) = Si + Si Xi Ri Ri (1- Ui) = Si

  9. Open networks (Single Class) Equations: Then: . R’i = Vi Ri = Di _ (1-Ui) Besides: . ni = Ui _ (1-Ui) Because Ui = Xi Si

  10. Open networks(Single Class) Calculation of the greatest  : In an open network the average frequency of users incoming into the network is fixed. For  too much big the network will become unstable, we are then interested in the greatest value of  that we can apply to the network. Because: Ui = Xi Si =  Vi Si then:  = Ui / Di because Di = Vi Si Ui = 1 for the greatest use of the queue i, then we can calculate the greatest  that doesn’t make unstable the system: 1 _ maxki=1 Di

  11. DISK2 DISK1 CPU DB Server(example 1) 10.800 request per hour = X0 DCPU = 0,2 sec Service demand at CPU VDISK1 = 5 VDISK2 = 3 SDISK1 = SDISK2 = 15 msec DDISK1 = VDISK1 * SDISK1 = 5 * 15 msec = 75 msec Service demand at disk 1 DDISK2 = VDISK2 * SDISK2 = 3 * 15 msec = 45 msec Service demand at disk 2

  12. DISK2 DISK1 CPU DB Server(example 1) . Service Demand Law UCPU = DCPU * X0 = 0,2 sec/req * 3 req/sec = 0,6 CPU utilization UD1 = DDISK1 * X0 = = 0,225 Disk1 utilization UD2 = = 0,135 Disk2 utilization Residence time R’CPU = DCPU / (1- UCPU ) = 0,5 sec R’D1 = DDISK1 / (1- UDISK1 ) = 0,097 sec R’D2 = DDISK2 / (1- UDISK2 ) = 0,052 sec

  13. Total response time R0 = R’CPU + R’D1 + R’D2 = 0,649 sec Average number of requests at each queue nCPU = UCPU / (1- UCPU ) = 0,6 / (1-0,6) = 1,5 nDISK1 = = 0,29 nDISK2 = = 0,16 Total number of requests at the server N = nCPU + nDISK2 +nDISK2 = 1,95 requests RMaximum arrival rate  = 1 _ = 1 _ = 5 req /sec maxki=1 Di max (0,2; 0,075; 0,045)

  14. DISK TAPE M clients CPU Closed queue network (number finite of users) 14

  15. Closed networks(Mean Value Analysis) • Allows calculating the performance indexes (average response time, throughput, average queue lenght, etc…) for a closed network • Iterative method based on the consideration that a queue network results can be calculated from the same network results with a population reduced by one unit. • Useful also for hybrid queue networks Definitions .X0: average queue network throughput. . Vi: average number of visits for a request at a queue i. . Si: average service time for a request on the server i. . Ri: average stay time for a request at the queue i.

  16. Closed networks(Mean Value Analysis) Definitions . R’i: total average stay time for a request at the queue i considering all its visits at the queue. Equal to Vi Ri . Di: total average service time for a request at the queue i considering all its visits at the queue. Equal to Vi Si . R0: average response time of the queue network. Equal to the sum of theR’i . nia: average number of the requests found by a request incoming in the queue. Forced Flow Law Then we have: . Xi = X0 Vi

  17. Mean Value Analysis (Single class) Equations: Ri(n) = Si + Wi(n) = Si + nia(n) Si = Si (1+ nia(n) ) Arrival Theorem: the average number of requests (nia) in a queue i thatan incoming request find in the same queue, is equal to the average number of requests in the queue i if n-1 requests are in the queue network (ni(n-1) that is n minus what wants the service on the i-th queue) in other words: nia(n) = ni(n-1) then: Ri = Si(1+ni(n-1)) and multiplying both members for Vi →R’i = Di(1+ni(n-1))

  18. Mean Value Analysis (Single class) Equations: Applying Little’s Law to the whole “queue network” system (n=X0R0), we have: →X0 = n / R0(n) = n / Kr=1 R’i(n) Applying Little’s Law and Forced Flow Law: → ni(n) = Xi(n) Ri(n) = X0(n) Vi Ri(n) = X0(n) R’i(n)

  19. Mean Value Analysis (Single class) Three equations: →Residence Time equation R’i = Di[1+ni(n-1)] →Throughput equation X0 = n / Kr=1 R’i(n) →Queue lenght equation ni(n) = X0(n) R’i(n) 19

  20. Mean Value Analysis (Single class) Iterative procedure: • We know that ni(n) = 0 for n=0: if no message is in the queue network, then no message will be in every single queue. • Given ni(0) it’s possible to evaluate all R’i(1) • Given all R’i(1) it’s possible to evaluate all ni(1) and X0(1) • Given all ni(1) it’s possible to evaluate all R’i(2) • The procedure continues until all ni(n), R’i(n) and X0(n) are found, where n is the number of requests inside the network.

  21. DB Server(example 2) • Requests from 50 clients • Every request needs 5 record read from (visit to) a disk • Average read time for a record (visit) = 9 msec • Every request to DB needs 15 msec CPU DCPU = SCPU = 15 msecCPU service demand DDISK = SDISK * VDISK = 9 * 5 = 45 msecDisk service demand

  22. DB Server(example 2) UsingMVA Equations n = 0; Number of concurrent requests R’CPU = 0; Residence time for CPU R’DISK = 0; Residence time for disk R0 = 0; Average response time X0 = 0; Throughput nCPU = 0; Queue lenght at CPU nDISK = 0 Queue lenght at disk n = 1; R’CPU = DCPU (1+ nCPU(0)) = DCPU = 15 msec; R’DISK = DDISK (1+ nDISK(0)) = DDISK = 45 msec; R0 = R’CPU + R’DISK = 60 msec; X0 = n/ R0 = 0,0167 tx/msec nCPU = X0 * R’CPU = 0,250 nDISK = 0,750

  23. DB Server(example 2) n = 1; R’CPU = DCPU (1+ nCPU(0)) = DCPU = 15 msec; R’DISK = DDISK (1+ nDISK(0)) = DDISK = 45 msec; R0 = R’CPU + R’DISK = 60 msec; X0 = 1 / R0 = 0,0167 tx/msec nCPU = X0 * R’CPU = 0,250 nDISK = 0,750 n = 2; R’CPU = DCPU (1 + nCPU(1)) = 15 * 1,25 = 18,75 msec; R’DISK = DDISK (1 + nDISK(1)) = 45 * 1,750 = 78,75 msec; R0 = R’CPU + R’DISK = 97,5 msec; X0 = 2 / R0 = 0,0333 tx/msec nCPU = X0 * R’CPU = 0,625 nDISK = X0 * R’DISK = 2,625 23

  24. Closed networks (Single Class) - Bounds Bottleneck identification (1/3) Usually the queue network throughput will reach saturation if requests increase inside the system; we are then interested in finding the component in the system that causes saturation. → in open networks: 1 _ maxki=1 Di and replacing  with X0 (n): X0 (n) 1 _ maxki=1 Di

  25. Closed networks (Single Class) - Bounds Bottleneck identification (2/3) • from throughput equation of MVA, remembering that R’i (n)= Di [1 + ni(n)] → R’i  Di for every queue i, then we have: X0 (n) = nn _ Kr=1R’iKr=1Di

  26. X0 n Closed networks (Single Class) - Bounds Bottleneck identification (3/3) • Combining the preceding two equations we obtain: → X0 (n)  min n _ , 1 _ Kr=1Di maxki=1 Di For little n the throughput will increase at the most in a linear way with n, then becomes flat around the value 1/maxki=1 Di

  27. Closed networks (Single Class) - Bounds Average response time (1/2) When throughput reaches its greatest value (that is for n big) the average response time is equivalent to: R0 (n)  n _ max throughput Then for n big the response time increases in a linear way with n: → R0 (n)  n maxki=1 Di On the contrary, for small values of n (n near to 1) the average response time will be: → R0 (n) = Kr=1Di considering that all waiting times are null.

  28. Closed networks (Single Class) - Bounds Average response time (2/2) We can establish a lower bound on average response time equal to: → R0(n)  max Ki=1Di , n ∙maxki=1 Di

  29. DB Server(Example 3) New scenarios with regard to preceding example: • index variation in DB (# of disk access equal to 2,5 (before was 5)) • 60% faster Disk (average service time = 5,63 msec) • faster CPU (service demand = 7,5 msec)

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