Let, 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 200 sectors and 1 sector can store 1 MB of data. Also consider disk has a seek time of 2 milliseconds and rotates at the speed of 6000 rpm.
Sunday, 11 January 2005
Now, Calculation is following:
Given:
Disk = 8 platters * 2 Surfaces =16 platters
Each platter = 4084 tracks
Each track = 200 sectors
Each sector = 1 MB
Calculation:
One track = 200 sectors
= 200 x 1 MB
= 200 MB
One platter = 4084 tracks
= 4084 x 200 MB
= 4 x1024 x 200 MB
= 4 x 200 x 1024 MB
= 4 x 200 GB
= 800 GB
16 platters = 16 x 800 GB
= 12800 GB
= 12 x 1024 GB
= 12 TB
The storage capacity of this disk in Bytes = 12 TB
Given:
Disk Rotation Speed = 6000 rpm
Average seek time = 10 millisecond
Thus, the time for one rotation = 1 minute/6000 rotations
=10ms per rotation
Average seek time = 2 millisecond (given)
Average Rotational delay = ½ × Rotation in milliseconds
Average Rotational delay = ½ × (60/6000) ×1000 ms
= ½ × 10 ms
= 5 millisecond
Rotational latency = Average seek time + Average Rotational delay
Rotational latency = 2 millisecond + 5 millisecond
= 2 + 5 milliseconds
= 7 milliseconds.
