Question

What is the average access time (in milliseconds, to the nearest hundredth of a millisecond) for a hard disk which rotates at 5400rpm and has an edge-to-edge seek time of 8ms? (Assume that seek time is proportional to the number of tracks to seek.)

Answer 1

The number of rotations are given as 5400 rpm and the given hard disk has an edge to edge seek time which is 8 ms. The seek time will be proportional to the number of tracks to seek.

The average access time can be computed as follows:

Average access time = average latency + average seek time + transfer time          … (1)

Since number of rotations are 5400 rmp. Therefore, the average rotaional latency can be computed as follows:

Average rotational latency = 0.5 rotations /(5400 rotations/minute)

                                              = 5.55 ms

The average seek time can be computed by taking 1/3^rd or 1/4^th of the given edge to edge seek time. The required average seek time can be calculated as follows:

Average seek time = 1/3 * 8 => 2.67 ms

Since the transfer time is not given in the problem. So, it can be assuemed 0 in this case. Now, the average access time for the given hard disk can be calculated by putting all the necessary values in the equation (1) as follows:

Average access time = average latency + average seek time + transfer time

                                   = (5.55 + 2.67 + 0) ms

                                   = 8.22 ms

Hence, the required average access time for a hard disk which rotates at 5400 rpm and has an edge to edge seek time of 8ms is 8.22 ms.