A hard disk system has the following parameters : Number of tracks = 500…

Solution: compute the average access time as the sum of average seek time, average rotational latency, and transfer time.

• Rotation speed = 600 rpm = 10 revolutions/sec → rotation period = 1/10 s = 0.1 s = 100 ms. Average rotational latency = 100 ms / 2 = 50 ms.

• Bytes per track = sectors/track × bytes/sector = 100 × 500 = 50,000 bytes. Transfer time for 250 bytes = (250 / 50,000) × 100 ms = 0.5 ms.

• Average seek time: time to move one track = 1 ms. For uniformly chosen start and target tracks, average number of tracks moved = (number of tracks − 1) / 2 = (500 − 1) / 2 = 249.5. So average seek = 249.5 ms.

• Total average time = average seek + average rotational latency + transfer time = 249.5 ms + 50 ms + 0.5 ms = 300.0 ms.

Therefore, the average time taken for transferring 250 bytes from the disk is 300.0 ms.