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Given that

Rotation speed = 8000 rpm

Number of tracks on one side of platter=10,000

Number of sectors per track=500

Sector size=1KB

Seel tine =1ms/100 tracks

Time for one rotation =(1000 ms/sec × 60 sec/ minute) / 8000RPM=7.5ms

(a)Given that the track starts at track 0

If the request track is track 0, then the seek time is0.

If the request track is track 9,999( start track 0 to9,999 equal to 10,000) then the seek time is the time to traverse 9,999 tracks.

Average the number of tracks traversed = 9,999/2=4,999.5 tracks

Average Seek time=4999.5/100=49.995.

(b)(average seek time) + (average rotational latency) + (transfer time [2 full rotations])

The rotations per minute are 8000

1 min = 60 sec

8000 / 60 = 133.3 rotations / sec

Rotational delay is the inverses of this. So 1 / 133.3 = 0.0075 sec = 0.0075*100 = 0.75 msec.

So there is 1 rotation is at every 0.75 ms

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

average rotational latency is: 1 / 2r

7.5/2=3.75 ms.

The number of sectors per track=500

time for one complete revolution of 7.5 ms

For 2 complete rotations 2*7.5=15

=(average seek time) + (average rotational latency) + (transfer time [2 full rotations])

=49.995+3.75+15

=68.745.

(c)(average seek time) + (average rotational latency) + (transfer time [1/1000 of a full rotation])

=49.995+3.75+0.007

=53.745.

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