Answer:
Disk Capacity = 45056000 bytes
Optimal skew = 26.455 ≈ 27
Maximum Data transfer rate = 262.5 GB per second
Explanation:
given data
heads = 16
cylinders = 400
cylinder zones = 100
each sector contains = 512 bytes
average seek time = 1 msec
disk rotates = 7200 RPM
to find out
disk capacity, optimal track skew, and maximum data transfer rate
solution
first we get total number of sectors that is
total number of sector = number of zones × (number of sectors in different zones
total number of sector = 100 × (160+200+240+280)
total number of sector = 88000
so
Disk Capacity = total number of sectors × size of each sector
Disk Capacity = 88000 × 512
Disk Capacity = 45056000 bytes
and
Rotation time =
Rotation time = 8.33 milli seconds
so Optimal number of sectors in a track = average of ( 160,200,240,280 )
Optimal number of sectors in a track = 220
now New sector is read every i.e = 0.0378 ms
here Optimal skew = seek time ÷ new sector read time
Optimal skew =
Optimal skew = 26.455 ≈ 27
and
here we know that for maximum transfer rate we will select cylinder with maximum number of sectors i.e here 280 sectors
so
capacity of one track with maximum = 280 × 512
capacity of one track with maximum = = 143360 bytes
and Number of rotations in 1 second is =
Number of rotations in 1 second is = 120
so Data transfer rate = Number of heads × Capacity of one track × Number of rotations in one second
Data transfer rate = 16 × 143360 × 120
Data transfer rate = 275251200 bytes per second
Data transfer rate = 262.5 GB per second