Question

Assume that a disk has an average seek time of 9 msec and a rotation speed of 7000 RPM. Assume also that there are 40 sectors in a track. What is the average time expected to retrieve a single sector?

Answer 1

Solution:

Given,

=>Average seek time = 9 msec

=>Rotation speed = 7000 RPM

=>Number of sectors = 40

Explanation:

Calculating average rotational delay:

=>Rotation speed = 7000 RPM

=>Rotation speed = 7000/60 RPS as 1 minute = 60 sec

=>Rotation delay in one rotation = 1/(7000/60) sec

=>Rotation delay in one rotation = 60*1000/7000 msec as 1 sec = 1000 msec

=>Rotation delay in one rotation = 8.57 msec

=>Average rotational delay = rotational delay/2

=>Average rotational delay = 8.57/2 msec

=>Average rotational delay = 4.29 msec

Calculating sector read time:

=>Sector read time = rotational delay/number of sectors

=>Sector read time = 8.57 msec/40

=>Sector read time = 0.0357 msec

Calculating total delay to read a sector:

=>Total delay to read one sector = seek time + average rotational delay + sector read time

=>Total delay to read one sector = 9 msec + 4.29 msec + 0.0357 msec

=>Total delay to read one sector = 13.3 mesc

I have explained each and every part with the help of statements attached to the answer above.