Consider a disk I/O system in which an I/O request arrives at the rate of 80 IOPS. The disk service time is 6ms.

a. Compute the following:

Utilization of I/O controller

Total response time

Average queue size

Total time spent by a request in a queue

b. Compute the preceding parameter if the service time is halved.

a. Arrival Rate= a=80 IOPS

Average service Time=Rs= 6ms

Hence, the utilization of I/O controller (U) is given by:

U = a * Rs

= 80 * 6 * 10^-3

= 0.48

Total Response Time (R) is given by:

$R = \frac{Rs} {1 – U}$

$ = \frac{6}{1-0.48}$

= 11.538 …