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Question No 1

 

Marks: 5

 

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.

Question No 2

 

Marks: 5 

 

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?

 

Question No 3

 

Marks: 5 

 

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

 


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Please Discuss here about this assignment.Thaks

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check ur 3rd q solution is wrong 2048*2048*2048 is equal to 6GB

nahi sir g, wrong nahi hai, 2*2*2= 8

if we multiply 2048 four time than 8192 mean 8GB 

am i right 

100 % accurate solution ..

see pg # 312, 324, 325 respectively for Q 1, 2 and 3. u can solve all three questions very easily

Thanks saadali....

thnx to saadali once again 

Idea solution same of other you can find..from Hand out 

Example 3

A hard disk with 5 platters has 1024 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=0.2MB/track
0.2MB x 1024 tracks=0.2GB/platter
Therefore the hard disk has the total capacity of 5 x 0.2=1GB

You are right Amjad

But in the lecture, the total capacity of hard disk is stated by the Lecturer is

5 x 2 x 0.2 = 2GB

As platters has 2 sides. so 0.2 GB/platter is of 1 side only.

Questions 1
Answer

48 / 2 = 24K

*10 = 245760 instruction per second.

 

Question 2

Answer

512 bytes * 512 sector = 0.2 MB / track

0.2 MB * 512 tracks = 0.1 GB /platter

10 * 0.1 = 1 GB

 

Question 3

Answer

2048 * 2048 * 2048 = 8 GB

80 / 8 = 1

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