1 Multiple class queueing networks Mean Value Analysis - Open queueing networks - Closed queueing networks

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

3 Incoming request class Different kind of requests should be in a system (queueing network) that need different services by the servers, i.e: -a database server is subject to two type of transactions: -simple query (that needs only read activities on the disks) -updating transactions (that needs read and write activities on the disks) -a web server is subject to two type of requests: -Read of a little file -Uploading of a big file

4 Definitions K: number of queues i: queue identification r: class identification (from 1 to R)  r : arrival rate for class r request = ( 1, 2,..., R ) V i,r : average number of visits a class r request makes to server i from its generation to its service time (request goes out from the system if open network)

Definitions S i,r : average class r request service time at the server i W i,r : average class r request waiting time in the queue i R i,r : average class r request answer time in the queue i R i,r = S i,r + W i,r

6 Definitions R ’ i,r : average class r request residence time in the queue i from its creation to its service completion time (request goes out from the system in case of open network) R ’ i,r = V i,r R i,r D i,r : request class r service demand to a server in a queue i from its creation to its service completion time (request goes out from the system in case of open network) D i,r = V i,r S i,r

Input parameters D i,r, r Equations. U i,r ( ) = r V i,r S i,r = r D i,r. U i ( ) =  R r=1 U i,r ( ) total utilization factor. R ’ i,r ( ) = D i,r delaying resource R ’ i,r ( ) = D i,r / (1-U i ( )) queuing resource Formulas for multiple class open QNs

. R 0,r ( ) =  K i=1 R ’ i,r ( ). n i,r ( ) = U i,r ( ) / (1-U i ( )) NOTE: total utilization in the denominator. n i, ( ) =  R r=1 n i,r ( ) Formulas for multiple class open QNs

9 DB Server (example ) Class 1 trx: query  = 5 requests per second (tps) D CPU = 0,1 sec Service demand at CPU D DISK1 = 0.08 Service demand at disk 1 D DISK2 = 0.07 Service demand at disk 2 Class 1 trx: updating trx  = 2 requests per second (tps) D CPU = 0,15 sec Service demand at CPU D DISK1 = 0.20 Service demand at disk 1 D DISK2 = 0.10 Service demand at disk 2 CPU DISK1 DISK2

10 DB Server (example ) CPU DISK1 DISK2 Service demand x QueryUpdates CPU0,10,15 DISK10,080,20 DISK20,070,10

11 Utilizations (%) CPU5030 Disk140 Disk Residence times (sec) CPU0, Disk10,401,00 Disk 20,0160,22 Response times (sec)1,061,97

12 DISK TAPE CPU M clients Multiclass closed queue networks (finite number of users)

Notations N r : fixed number of requests in the system for each class (r) N: (N 1, N 2,..., N R ) 1 r : vector where all components are zero except for the r-th component, which is equal to 1

Formulas -> Residence Time Equation for class r R ’ i,r (N)= D i,r [1+n i (N – 1 r )] -> Throughput equation for class r X 0,r = N r /  K r=1 R ’ i,r (N) -> Queue lenght equation for class r n i,r (N) = X 0,r (N) R ’ i,r -> Queue equation n i (N)=  R r=1 n i,r (N)

Example with 2 classes Residence Time Equation for class r R ’ i,r (N)= D i,r [1+n i (N – 1 r )] for example, to evaluate the formulas, when the state is N=(3,4), i.e. 3 customers of class 1 and 4 customers of class 2, we need to know: the average number of users in queue i when there are 2 customers of class 1 and 4 customer of class 2 the average number of users in queue i when there are 3 customers of class 1 and 4 customer of class 3 R ’ i,1 (3,4)= D i,r [1+n i (2,4)] R ’ i,2 (3,4)= D i,r [1+n i (3,3)]