A 2.5 inch diameter disk has 8 platters with each platter having two data recording surfaces, each platter on disk has 4084 tracks, each track has 400 sectors and one sector can store 1 MB of data. Calculate the storage capacity of this disk in Bytes. If thia disk has a seek time of 2 milli-seconds and rotates at the speed of 6000 rpm, find the access time for the disk. Make suitable assumptions, if any.
Answers
Given, disk = 8 platters * 2 Surfaces =16 platters
Given, Platter = 4084 tracks
Given, Track = 400 sectors
Given, Sector = 1 MB
One track = 400 sectors
= 400 MB
One platter = 4084 tracks
= 4084 * 400 MB
= 1600 GB
16 platters = 16 * 1600 GB
= 25600 GB
= 25 TB
The storage capacity = 25TB.
Given, Disk rotation speed = 6000rpm.
Average Seek time = 2 m/s.
We know that Average rotational delay = 1/2 * time of rotation in m/s
= 1/2 * (60/6000) * 1000
= 5 ms.
Rotational latency = Average seek time + Average rotational latency
= 2 + 5
= 7 ms.
The access time for the disk = 7 milliseconds.