For the sake of simplicity, consider that the database server services only one type of transaction: query. The DB server has one CPU and two disks. During the peak period, the DB server receives requests from the application server at a rate of 57,600 queries per hour. Based on measurements collected during the operation of the call center application, the analyst obtains the following data: on average, each query needs 50 msec of CPU and performs four I/Os on disk 1 and two I/Os on disk 2. Each I/O takes an average of 8 msec.
The throughput, X0, is equal to the average arrival rate l, given by: 57,600/3,600 = 16 queries/sec. The service demand at the CPU, DCPU, is 0.050 sec. The service demands at disks 1 and 2 are:
From the Service Demand Law (see Chapter 3), Ui = Di x X0. Therefore, the utilization of the CPU and the disks are given by UCPU = DCPU x X0 = 0.05 x 16 = 80%, Udisk1 = Ddisk1 x X0 = 0.032 x 16 = 51.2%, and Udisk2 = Ddisk2 x X0 = 0.016 x 16 = 25.6%, respectively. Using the residence time equation for open queuing networks of Chapter 13, the residence times at the CPU and disks are
see Example 4.4.
For the sake of simplicity, consider that the database server services only one type of transaction: query. The DB server has one CPU and two disks. During the peak period, the DB server receives requests from the application server at a rate of 57,600 queries per hour. Based on measurements collected during the operation of the call center application, the analyst obtains the following data: on average, each query needs 50 msec of CPU and performs four I/Os on disk 1 and two I/Os on disk 2. Each I/O takes an average of 8 msec.
The throughput, X0, is equal to the average arrival rate l, given by: 57,600/3,600 = 16 queries/sec. The service demand at the CPU, DCPU, is 0.050 sec. The service demands at disks 1 and 2 are:
From the Service Demand Law (see Chapter 3), Ui = Di x X0. Therefore, the utilization of the CPU and the disks are given by UCPU = DCPU x X0 = 0.05 x 16 = 80%, Udisk1 = Ddisk1 x X0 = 0.032 x 16 = 51.2%, and Udisk2 = Ddisk2 x X0 = 0.016 x 16 = 25.6%, respectively. Using the residence time equation for open queuing networks of Chapter 13, the residence times at the CPU and disks are
Actual Result :