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Consider a disk I/O system in which an I/O request arrives at the rate of 80 IOPS.

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.

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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 ms

Average queue size$= \frac{U * U}{(1 - U)}$

$=\frac{ 0.48 * 0.48}{(1 - 0.48)}$

= 0.44

Thus, the total time spent by a request in a queue

= U * R

=0.48 * 11.538

=5.538 ms

b.
Now if the controller power is doubled, the service time is halved. Consequently,

Rs= 3ms in this scenario.

Therefore, a=80 IOPS

Utilization (U) = a * Rs

= 80 * 3 * 10^-3

= 0.24

Total Response Time (R) $=\frac{ Rs}{1-U}$

$=\frac{3}{1-0.24}$

= 4 ms

Average Queue Size $=\frac{ U * U}{(1 - U)}$

$=\frac{0.24 * 0.24}{1- 0.24}$

= 0.0757

Hence, Time spent by request in a queue = U * R= 0.24 * 4= 0.96ms.