Multiple class queueing networks Mean Value Analysis Open

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

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

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 3

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) Vi, 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) 4

Definitions Si, r: average class r request service time at the server i Wi, r: average class r request waiting time in the queue i Ri, r: average class r request answer time in the queue i Ri, r = Si, r + Wi, r

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 = Vi, r Ri, r Di, 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) Di, r = Vi, r Si, r 6

Formulas for multiple class open QNs Input parameters Di, r , r Equations . Ui, r ( ) = r Vi, r Si, r = r Di, r. Ui ( ) = Rr=1 Ui, r ( ) . R’i, r ( ) = Di, r / (1 -Ui ( )) total utilization factor delaying resource queuing resource

Formulas for multiple class open QNs. R 0, r ( ) = Ki=1 R’i, r ( ). ni, r ( ) = Ui, r ( ) / (1 -Ui ( )). ni, ( ) = Rr=1 ni, r ( ) NOTE: total utilization in the denominator

DB Server (example ) DISK 1 CPU Class 1 trx: query DISK 2 l 1 = 5 requests per second (tps) DCPU = 0, 1 sec DDISK 1 = 0. 08 DDISK 2 = 0. 07 Service demand at CPU Service demand at disk 1 Service demand at disk 2 Class 1 trx: updating trx l 1 = 2 requests per second (tps) DCPU = 0, 15 sec DDISK 1 = 0. 20 DDISK 2 = 0. 10 Service demand at CPU Service demand at disk 1 Service demand at disk 2 9

DB Server (example ) DISK 1 CPU Service demand x Query DISK 2 Updates • CPU 0, 15 • DISK 1 0, 08 0, 20 • DISK 2 0, 07 0, 10 10

Utilizations (%) CPU 50 30 Disk 1 40 40 Disk 2 35 20 CPU 0, 50 0. 75 Disk 1 0, 40 1, 00 Disk 2 0, 016 0, 22 Response times (sec) 1, 06 1, 97 Residence times (sec) 11

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

Notations Nr : fixed number of requests in the system for each class (r) N: (N 1 , N 2 , . . . , NR) 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)= Di, r[1+ni(N – 1 r)] -> Throughput equation for class r X 0, r = Nr / Kr=1 R’i, r(N) -> Queue lenght equation for class r ni, r(N) = X 0, r(N) R’i, r -> Queue equation ni(N)= Rr=1 ni, r(N)

Example with 2 classes Residence Time Equation for class r R’i, r(N)= Di, r[1+ni(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)= Di, r[1+ni(2, 4)] R’i, 2(3, 4)= Di, r[1+ni(3, 3)]