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 this 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.

Solution

1. Storage Capacity of the Disk

• Calculation:

Storage Capacity = Number of Platters × Surfaces per Platter × Tracks per Surface × Sectors per Track × Data per Sector (in Bytes)
Storage Capacity = 8 × 2 × 4084 × 400 × 1 MB × (10242 Bytes)

Result: The storage capacity of the disk is 27,407,260,057,600 Bytes or approximately 27.41 TB.

2. Access Time of the Disk

• Seek Time: 2 milliseconds (given).
• Rotational Latency:

Rotational Latency = (60 / RPM) × (1000 / 2)
Rotational Latency = (60 / 6000) × (1000 / 2) = 5 ms

• Access Time:

Access Time = Seek Time + Rotational Latency
Access Time = 2 ms + 5 ms = 7 ms

Result: The access time for the disk is 7 milliseconds.