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CS604 –Operating System Dear Students, Assignment No.4 has been uploaded on LMS...SOLUTION Discus Here Due Date:- 10-02-2016

Assignment No. 04
Semester: Fall 2015

CS604 –Operating System

Total Marks: 20

 

Due Date:- 10-02-2016

 

Assignment Objective:

  • Implementation of logical addressing
  • To know how to change logical to physical address
  • Page table entries for mapping to frames
  • Finding the physical address from logical address
  • Uploading instructions:

 

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Assignment

 

 

Consider logical memory address space 32 bytes in which page size is of 8 bytes and physical memory is of 0.125 K bytes

Find   

  1. Number of pages
  2. Number of bits required for pages
  3. Number of memory frames
  4. Number of bits required for memory frames
  5. Show the physical memory mapping in the form of table from the following logical memory and page table.
  6. Find physical address according to the following page number and offset (p,d)
  • (2,3)  
  • (3,6) 

 

Logical Memory       Page Table               Physical Memory

 

 

0

3

1

8

2

13

3

10

 

0

0

1

1

2

2

3

3

4

4

5

5

6

6

7

7

8

8

9

9

10

10

11

11

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12

13

13

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20

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31

 

 

 

Deadline: Your assignment must be uploaded/submitted on or before 10-02-2016

 

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Replies to This Discussion

Could anyone please upload complete solution of this assignment.

Any one plz Upload assignment solution bhi kis na kar lia ha upload kar do

there are 4 pages in logical address space so 2^2=4

so number of bits required for pages are 2 bits?

m i right?

  1. Number of pages

The page size is defined by the CPU hardware. If the size of logical address space is 2^m and a page size is 2^n addressing units (bytes or words) , then the high-order m-n bits of a logical address designate the page number and the n low order bits designate offset within
the page.

32 bytes = 2^5, 8 bytes = 2^3

p= 5-3= 2 bits 

Kiran number of bits req for p = ceiling[log2 4] = 2 bits asa kea hoa hy book main

Please koe 5 and 6 ka smjh de 

6 sol

  1. 1.       (P , d)  = (2 , 3)  = 1011 it is logical address that is given.now physical address becomes (13, 3) =110111.
  2. .       (P , d)  = (3, 6)  = 11110  it is logical address that is given.now physical address becomes (10, 6) =1010110.

Number of bits required for pages= ceiling [log24]

Number of bits required for pages= ceiling [2]

Number of bits required for pages= 2bits

Number of memory frames = physical memory size/ frame size

put the values in this formula

plz upload the solution its getting late

plz upload the solution

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