CS501 4 Advanced Computer Architecture assignment solution fall january 2012

Question No.1

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.

Solution:

Since the floppy drive has a 16-bit data bus, it can transfer two bytes at one time.

Thus its transfer rate is 48/2 = 24 K half-words (16-bits each) per second.

This corresponds to an overhead of 10 instructions or 24K x 10 = 24 x 2³ x 10 = 240000 instructions per second.

Question No.2

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?

Solution:

512 bytes x 512

Sectors=0.25MB/track

0.25MB x 512 tracks=0.125GB/platter

Therefore the hard disk has the total capacity of 10 x 0.125=1.25GB

Question No.3

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

Solution:

The capacity of one platter

= 2048 x 2048 x 2048

= 8GB

For an 80GB hard disk, we need 80/8

= 10 such platters.