**BACHELOR OF COMPUTER APPLICATIONS**

Course Code : BCS-011

Course Title : Computer Basics and PC Software

Assignment Number : BCA(I)/011/Assignment/ 2015

Maximum Marks : 100

Weightage : 25%

**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.**

*Given:*

Disk = 8 platters * 2 Surfaces =16 platters

each platter = 4084 tracks

each track = 400 sectors

each sector = 1 MB

*Calculate:*

one track = 400 sectors

= 400 * 1 MB

= 400 MB

one platter = 4084 tracks

= 4084 * 400 MB

= 4 *1024 * 400 MB

= 4 * 400 * 1024 MB

= 4 * 400 GB

= 1600 GB

16 platters = 16 * 1600 GB

= 25600 GB

= 25 * 1024 GB

= 25 TB

**The storage capacity of this disk in Bytes = 25 TB**

*Given:*

*Disk Rotation Speed = 6000 rpm*

*Average seek time = 10 millisecond*

Sample Convertion is Shown Below:

*Rotational latency = Average seek time + Average Rotational delay + Controller Overheads*

Average seek time = 2 millisecond (given)

Calculate

*Average Rotational delay = ½ Time of Rotation in milliseconds*

* *

Average Rotational delay = ½ * (60/6000) *1000 millisecond

= ½ * 10 millisecond

= 5 millisecond

*Rotational latency = Average seek time + Average Rotational delay *

**Rotational latency = **2 millisecond + 5 millisecond

= 2 + 5 millisecond

= 7 millisecond.