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Assignment No. 04
Semester: Spring 2012

CS609: System Programming

 

Total Marks: 20

 

Due Date: 18/06/2012

Instructions

Please read the following instructions carefully before submitting assignment:

It should be clear that your assignment will not get any credit if:

  • The assignment is submitted after due date.
  • The submitted assignment does not open or file is corrupt.
  • Solution is copied from any other source.

 

Objective

The objective of this assignment is to enhance your knowledge about;

  • Data storage on a hard disk
  • Schemes for addressing storage units.

 

Assignment                                                                  

 

Part a)  

05 Marks

You have studied in this course that Tracks are the circular division of the disk and the Sectors are the longitudinal division of the disk. A typical track is shown in yellow; a typical sector is shown in blue. A sector contains a fixed number of bytes -- for example, 256 or 512.

 

As per division like above, there is a clear difference in size of sectors on inner and outer tracks which causes under utilization of the larger, outer tracks of the disk. What mechanism is used to improve disk surface area utilization?

 

Part b)

15 Marks

Suppose, we have a Hard disk having 16 (Read/Write) Heads per Cylinder, and Maximum number of Sectors per Track is 63. It has total 1024 cylinders and Sector size is 512 Bytes. Answer the following questions.

Q#1. Find the LBA address of the CHS address (4, 6, 12)?

Q#2. Find the CHS address of the LBA address 4256?

Q#3. Find the Total Data Storage Capacity of this Hard Disk?

 

Submission

You are required to submit a single MS word file through LMS containing:

  • Justify your answer with solid reasons while answering the question given in Part (a).
  • Give answers to the questions given in Part (b) with brief explanation.

 

 

 

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Replies to This Discussion

plz anyone help for cs 609 solutionnnnnnnnnnnnnnnnnnnn

plz tell me any 1 is that correct or not,>>>>>>>>>>>>>

LBA address of the CHS address (4, 6, 12)

LBA Address=(c*h'+h)*S'+s-1

c=4, h=6, S=12, H'=16, S' = 63

=(4*16+6)*63+12-1

=(64+6)63+12-1

=4421

.............................

LBA to CHS>>>>>>>>>>

cylender =LBA/(heads per cylender*sectors per track)

LBA=4421, heads per cylender=16, sector per track is 63

So we have

4421/(16*63)

=4.386

.......please behno bhaio next solve e kr den agr kisi ko ata hay n ye b bta den k agr is me koi galti shlti hay to wo b sahi kr den.....:( :( :(

LBA theek hai bt yahan CHS k liye jo LBA value given hai wo use krni hai aur wo hai 4256

Cylinder = 4256/16*63 = 4 

formula sb handouts pg#135 sy daikho values me btati hn

temp = 4256/16*63 = 224

heads = 224/63 = 3

sectors = 224/63+1

sectors = 36 total capacity jo bios function k according di v hai buk me  wohi hai 63*16*1024*512 = 504MB

nazma plz check it total data storage capacity.

 cylinders x heads x sectors x sector size

 1024 * 16 * 63 * 512 = 528482304 = 516 MB approx

yes, i agree with this statement that in book answer is 504MB, but if we calculate mannualy in excel and divided by 1024000, which is value of MB  than answer is 516MB.

 

Vivek who told you that I MB = 102400 Bytes?

I MB = 1024 * 1024 Bytes

Thanks Nazma 

Please koi Part a k solution me b help kr day...........

li8 to aati nhe banda help kya karay

For Part a:

many factors affects the HDD storage capacity.
some of these are disks (platters), increase in surface area, density of megnetic media etc etc

ya thanks Osma it's right, sorry it's my mistake

 

 

guys according to handouts correct formulas ye hain:
temp = LBA % (heads per cylender*sectors per track)
sector = temp % sectors per track+1

where % means MOD.
change ur PC's calculator into scientific view, u can easily find MOD button in ur calculator.
thanx
so,
temp=4256%16*63
= 224
sec=224%63+1
= 36

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