Answer:
The average time to access a Sector on the disk is 6.005 ms
Explanation:
Rotational rate = 15000 RPM = (60/15000) = 4 × 10⁻³ s
Rotation time = 1/2 × Rotational rate = 1/2 × 4 × 10⁻³ = 2 × 10⁻³ s
Rotation time = 2 × 10⁻³ × 1000 ms = 2 ms
Transfer time = Rotational rate x (1/Average number of sectors/track)
= 4 × 10⁻³ x 1/800 = 5 × 10⁻⁶
Transfer time = 5 × 10⁻⁶ × 1000 ms = 0.005 ms
Average time to access a Sector on the disk = Average seek time + Rotation time + transfer time
Average time to access a Sector on the disk = 4ms + 2ms + 0.005 ms = 6.005 ms
The average time to access a Sector on the disk is 6.005 ms
In this exercise we have to use computer knowledge to describe the average number of rotations of a disk, so we have to:
The average time on the disk is 6.005 ms
Organizing the information given in the statement we have that:
Calculating the rotation time we find that:
[tex]Rotation time = 1/2* Rotational rate = 1/2 * 4 * 10^{-3} = 2 * 10^{-3}\\Rotation time = 2 * 10^{-3} * 1000 ms\\ = 2 ms[/tex]
The transferred time can be calculated as:
[tex]Transfer time = Rotational rate * (1/Average number of sectors/track)\\ = 4* 10^{-3}* 1/800 = 5 * 10^{-6}\\= 5 * 10^{-6} * 1000 ms = 0.005 ms[/tex]
so knowing that Average time to access a Sector on the disk = Average seek time + Rotation time + transfer time, we can put he numbers and:
[tex]4ms + 2ms + 0.005 ms = 6.005 ms[/tex]
See more about disk time at brainly.com/question/16614000
