Advance Computer Architecture - CS501 assignment no 5 idea solution

[rabeel]

[FONT=&quot]Question No 1[/FONT]
[FONT=&quot]Marks: 5[/FONT]


Consider a floppy drive having a transfer rate of 48 KB per second is attached to a 32 bit, 10MIPS CPU using an interrupt driven interface. The drive has 16-bit data bus.

Suppose that the interrupt overhead is 10 instructions. Calculate the fraction of CPU time required to service this drive when it is active. [FONT=&quot]Question No 2[/FONT]
[FONT=&quot]Marks: 5 [/FONT]



A hard disk with 10 platters has 512 tracks per platter, 512 sectors per track and 512 bytes/sector. What is the total capacity of the disk? [FONT=&quot]Question No 3[/FONT]
[FONT=&quot]Marks: 5 [/FONT]


How many platters are required for an 80GB disk if there are 2048 bytes/sector, 2048 sectors per track and 2048 tracks per platter?

[Malaika khan]

Q.1 SOLUTION:

The floppy drive has a 16-bit data bus; it can transfer two bytes at one time.
Its transfer rate is 48/2 = 24K per second.
This corresponds to an overhead of 10 instructions
or 24K x 10 = 24 x 210 x 10 = 245760 instructions per second.

Q.2 SOLUTION:

512 bytes x 512
Sectors=0.2MB/track
0.2MB x 512 tracks
=0.1GB/platter
so the hard disk has the total capacity of 10 x 0.1=1GB


Q.3 SOLUTION:

The capacity of one platter
= 2048 x 2048 x 2048
= 8GB
For an 80GB hard disk, we need 80/8
= 10 such platters.

[Vuhelper]

Thanks for sharing this solution \

[sadd]

your 3rd question is wrong

[Vuhelper]

correct the 3rd question