Consider a single-platter disk with the following parameters: rotation speed: 7200 rpm; number of tracks on one side ofplatter: 30,000; number of sectors per track: 600; seek time: one ms for every hundred tracks traversed. Let the diskreceive a request to access a random sector on a random track and assume the disk head starts at track 0.a. What is the average seek time?b. What is the average rotational latency?c. What is the transfer time for a sector?d. What is the total average time to satisfy a request?

Given Data:

Rotation Speed = 7200 rpm

No. of tracks on one side of platter = 30000

No. of sectors per track = 600

Seek time for every 100 track traversed = 1 ms

To find:

Average Seek Time.

Average Rotational Latency.

Transfer time for a sector.

Total Average time to satisfy a request.

Explanation:

a) As given, the disk head starts at track 0. At this point the seek time is 0.

Seek time is time to traverse from 0 to 29999 tracks (it makes 30000)

Average Seek Time is the time taken by the head to move from one track to another/2

29999 / 2 = 14999.5 ms

As the seek time is one ms for every hundred tracks traversed.  So the seek time for 29,999 tracks traversed is

14999.5 / 100 = 149.995 ms

b) The rotations per minute are 7200

1 min = 60 sec

7200 / 60 = 120 rotations / sec

Rotational delay is the inverses of this. So

1 / 120 = 0.00833 sec

          = 0.00833 * 100

          = 0.833 ms

So there is  1 rotation is at every 0.833 ms

Average Rotational latency is one half the amount of time taken by disk to make one revolution or complete 1 rotation.

So average rotational latency is: 1 / 2r

8.333 / 2 = 4.165 ms

c) No. of sectors per track = 600

Time for one disk rotation = 0.833 ms

So transfer time for a sector is: one disk revolution time / number of sectors

8.333 / 600 = 0.01388 ms = 13.88 μs

d)  Total average time to satisfy a request is calculated as :

Average seek time + Average rotational latency + Transfer time for a sector

= 149.99 ms + 4.165 ms + 0.01388 ms

= 154.168 ms

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