Queuing Networks. - Outline of queuing networks. - Mean Value Analisys (MVA) for open and closed queuing networks

Transcription

1 Queuing Networks - Outline of queuing networks - Mean Value Analisys (MVA) for open and closed queuing networks 1

2 incoming requests Open queuing networks DISK CPU CD outgoing requests Closed queuing networks (finite number of users) DISK M clients CPU CD 2

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

4 Definitions K: number of queues X 0 : network average throughput. If open network in a stationary condition X 0 = V i : 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) S i : average request service time at the server i W i : average request waiting time in the queue i R i : average request response time in the queue i R i = S i + W i 4

5 Definitions X i : throughput for the i-th queue X i = X 0 V i R i: average request residence time in the queue i from its creation to its service completion time (request goes out from the system if open network) R i = V i R i D i : request service demand to a server in a queue i from its creation to its service completion time (request goes out from the system if open network) D i = V i S i 5

6 Q i : 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) Q i = V i W i R i = V i R i =V i (W i + S i ) = W i V i + S i V i = Q i + D i R 0 : total average request response time ((from the whole system) R 0 = k i=1 R i n i : average number of requests waiting or in service at the queue i N: average number of requests in the system N = k i=1 n i 6

7 Open queuing networks incoming requests CPU DISK TAPE outgoing requests 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 (n a i), is equal to the average number of requests in the queue i (n i ). R i (n) = S i + W i (n) = S i + n i S i Using Little s Law (n i = X i R i ) and U i = X i S i : R i = S i _ (1-U i ) given that R i = S i (1 + n i ) = S i + S i X i R i = S i + U i R i R i (1- U i ) = S i 8

9 Open networks (Single Class) Equations: Then: R i = V i R i = D i _ (1-U i ) besides: n i = U i _ (1-U i ) because n i = X i R i R i = S i / (1- U i ) U i = X i S i 9

10 Calculation of the greatest : Given: Open networks (Single Class) 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. U i = X i S i = V i S i then: = U i / D i because D i = V i S i U i = 1 is the greatest utilization factor of a queue (i.e.= i), then we can calculate the greatest that doesn t make unstable the system as: 1 _ max k i=1 D i 10

11 DB Server (example 9.1) CPU DISK1 DISK2 = requests per hour = 3 requests per sec = X 0 D CPU = 0,2 sec V DISK1 = 5 V DISK2 = 3 S DISK1 = S DISK2 = 15 msec Service demand at CPU D DISK1 = V DISK1 * S DISK1 = 5 * 15 msec = 75 msec Service demand at disk 1 D DISK2 = V DISK2 * S DISK2 = 3 * 15 msec = 45 msec Service demand at disk 2 11

12 DB Server (example 1). Service Demand Law CPU DISK1 DISK2 U CPU = D CPU * X 0 = 0,2 sec/req * 3 req/sec = 0,6 CPU utilization U D1 = D DISK1 * X 0 = = 0,225 Disk1 utilization U D2 = = 0,135 Disk2 utilization Residence time R CPU = D CPU / (1- U CPU ) = 0,5 sec R D1 = D DISK1 / (1- U DISK1 ) = 0,097 sec R D2 = D DISK2 / (1- U DISK2 ) = 0,052 sec 12

13 Total response time R 0 = R CPU + R D1 + R D2 = 0,649 sec Average number of requests at each queue n CPU = U CPU / (1- U CPU ) = 0,6 / (1-0,6) = 1,5 n DISK1 = = 0,29 n DISK2 = = 0,16 Total number of requests at the server N = n CPU + n DISK2 + n DISK2 = 1,95 requests RMaximum arrival rate = 1 _ = 1 _ = 5 req /sec max k i=1 D i max (0,2; 0,075; 0,045) 13

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

15 Closed networks (Mean Value Analysis) Allows calculating the performance indeces (average response time, throughput, average queue lenght, etc ) for a closed network Iterative method based on the consideration that a queuing network results can be calculated from the same network results with a population reduced by one unit. Useful also for hybrid queuing networks Definitions. X 0 : average queuing network throughput.. V i : average number of visits for a request at a queue i.. S i : average service time for a request on the server i.. R i : average response time for a request at the queue i (service+waiting time) 15

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 V i R i. D i : total average service time for a request at the queue i considering all its visits at the queue. Equal to V i S i. R 0 : average response time of the queuing network. Equal to the sum of the R i. n ia : average number of the requests found by a request incoming in the queue. Forced Flow Law Then we have: X i = X 0 V i 16

17 Mean Value Analysis (Single class) Equations: R i (n) = S i + W i (n) = S i + n ia (n) S i = S i (1+ n ia (n) ) Arrival Theorem: the average number of requests (n ia ) in a queue i that an incoming request finds in the same queue is equal to the average number of requests in the queue i when n-1 requests are in the queuing network (n i (n-1) that is n minus the incoming request that wants the service on the i-th queue) in other words: n ia (n) = n i (n-1) (i.e n i is function of n-1) then: R i = S i (1+n i (n-1)) and multiplying both members for V i R i = D i (1+n i (n-1)) 17

18 Mean Value Analysis (Single class) Equations: Applying Little s Law to the whole queuing network system (n=x 0 R 0 ), we have: X 0 = n / R 0 (n) = n / K r=1 R i(n) Applying Little s Law and Forced Flow Law: n i (n) = X i (n) R i (n) = X 0 (n) V i R i (n) = X 0 (n) R i(n) 18

19 Mean Value Analysis (Single class) Three equations: Residence Time equation R i (n)= D i [1+n i (n-1)] Throughput equation X 0 (n)= n / K r=1 R i(n) Queue lenght equation n i (n) = X 0 (n) R i(n) 19

20 Iterative procedure: Mean Value Analysis (Single class) 1. We know that n i (n) = 0 for n=0: if no users is in the queuing network, then no users (requests) will be in every single queue. 2. Given n i (0) it s possible to evaluate all R i(1) 3. Given all R i(1) it s possible to evaluate all n i (1) and X 0 (1) 4. Given all n i (1) it s possible to evaluate all R i(2) 5. The procedure continues until all n i (n), R i(n) and X 0 (n) are found, where n is the total number of users (requests) inside the network. 20

21 DB Server (example 9.3) 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 D CPU = S CPU = 15 msec D DISK = S DISK * V DISK = 9 * 5 = 45 msec CPU service demand Disk service demand 21

22 Using MVA Equations DB Server (example 2) n = 0; Number of concurrent requests R CPU = 0; Residence time for CPU R DISK = 0; Residence time for disk R 0 = 0; Average response time X 0 = 0; Throughput n CPU = 0; Queue lenght at CPU n DISK = 0 Queue lenght at disk n = 1; R CPU = D CPU (1+ n CPU (0)) = D CPU = 15 msec; R DISK = D DISK (1+ n DISK (0)) = D DISK = 45 msec; R 0 = R CPU + R DISK = 60 msec; X 0 = n/ R 0 = 0,0167 tx/msec n CPU = X 0 * R CPU = 0,250 n DISK = X 0 * R DISK = 0,750 22

23 DB Server (example 2) n = 1; R CPU = D CPU (1+ n CPU (0)) = D CPU = 15 msec; R DISK = D DISK (1+ n DISK (0)) = D DISK = 45 msec; R 0 = R CPU + R DISK = 60 msec; X 0 = 1 / R 0 = 0,0167 tx/msec n CPU = X 0 * R CPU = 0,250 n DISK = 0,750 n = 2; R CPU = D CPU (1 + n CPU (1)) = 15 * 1,25 = 18,75 msec; R DISK = D DISK (1 + n DISK (1)) = 45 * 1,750 = 78,75 msec; R 0 = R CPU + R DISK = 97,5 msec; X 0 = 2 / R 0 = 0,0205 tx/msec n CPU = X 0 * R CPU = 0,38 n DISK = X 0 * R DISK = 1,62 23

24 The related Markov process 50/Z 49/Z 1/Z X 0 (1) X 0 (2) X 0 (50) 24

25 Closed networks (Single Class) - Bounds Bottleneck identification (1/3) Usually the queuing 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 _ max k i=1 D i and replacing with X 0 (n): X 0 (n) 1 _ max k i=1 D i 25

26 Closed networks (Single Class) - Bounds Bottleneck identification (2/3) from throughput equation of MVA, remembering that R i (n) = D i [1 + n i (n-1)] R i D i for every queue i, then we have (from Little s formula): X 0 (n) = n n _ K r=1 R i K r=1 D i 26

27 Closed networks (Single Class) - Bounds Bottleneck identification (3/3) Combining the preceding two equations we obtain: X 0 (n) min n _, 1 _ K r=1 D i max k i=1 D i For little n the throughput will increase at the most in a linear way with n, then becomes flat around the value 1/ max k i=1 D i X 0 n 27

28 Average response time (1/2) Closed networks (Single Class) - Bounds When throughput reaches its greatest value (that is for n big) the average response time is equivalent to: R 0 (n) n _ max throughput Then for n big the response time increases in a linear way with n: R 0 (n) n max k i=1 D i On the contrary, for small values of n (n near to 1) the average response time will be: R 0 (n) = K r=1 D i considering that all waiting times are null. 28

29 Closed networks (Single Class) - Bounds Average response time (2/2) We can establish a lower bound on average response time equal to: R 0 (n) max K i=1 D i, n max k i=1 D i 29

30 DB Server (Example 9.4) New scenarios with regard to previous example: a. index variation in DB (# of disk access equal to 2,5 (before was 5)) b. 60% faster Disk (average service time = 5,63 msec) c. faster CPU (service demand = 7,5 msec) Scenario Service demand Service demand D i 1/ maxd i Bottleneck D CPU D DISK a 15 2,5 * 9 = 22,5 37,5 0,044 disk b 15 5*5,63 = 28,15 43,15 0,036 disk c 15/2 = 7, ,5 0,022 disk a+b 15 2,5*5,63 = 14,08 29,08 0,067 CPU a+c 15/2 = 7,5 2,5 * 9 = 22,5 30,0 0,044 disk 30