A Hard drive disk spins at 3000 RPM, has 20 sectors per track, and has 30 tracks

Question

A Hard drive disk spins at 3000 RPM, has 20 sectors per track, and has 30 tracks.

a) What is the average rotational delay for randomly placed data?

b) What is a single sector transfer time in milliseconds?

c) What will be the average number of tracks crossed per seek for randomly placed data?

d) Let the average track to track seek time be 10 milliseconds, and let each sector contain 5 records with the disk at full capcity:

i) If tracks are ordered randomly and the 1st record on each track is located randomly, how much time should it take to access all records in order if each track has the records in sorted order?

Explanation / Answer

a) Average rotational delay = 0.5 / rotational speed
Given 3000RPM hard disk,
Average rotational delay = 0.5 / 3000 rotations/minute
       = 0.5 x 60 /3000 rotations/seconds
= 0.01 seconds = 10 x 10^-3
= 10 ms

b)single sector transfer time:
data size:
Transfer time= (1/number of sectors)/ rotational speed
Given 20 sectors per track and 3000RPM hard disk
   = (1/20) / (60/3000) rotations/seconds
   = 0.001 seconds
= 1 ms

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