Question

7. [1+1+1-3 points] Compute the (a) transfer time, (b) average rotational latency, and(c) disk access time for the following input data for a disk? Rotational speed- 7200 rpm, sector size- 0.5 KB, average seek time 5.7 ms, transfer rate 24.5 MB per sec, controller overhead -1 ms

Answer 1

7) (a) Transfer speed = 24.5 MB = 24.5 * 1024 KB [1 MB = 1024 KB]

Transfer time = sector size * transfer speed = (0.5 * (24.5/1024)) seconds = 0.01196 seconds = 11.96 milliseconds

7) (b) Rotational speed 7200 rpm.

Number of rotations in one second = 7200/60 = 120 [1 minute = 60 seconds]

So, one single rotation takes 1/120 seconds = 8.33 ms

Average rotational latency = 8.33 Milliseconds

7) (c) Disk access time = seek time + rotational latency + transfer time = 5.7 + 8.33 + 11.96 milliseconds = 25.99 milliseconds

Hope this helps.