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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:

 

  • Your assignment must be in .doc format (Any other formats like scan images, PDF, bmp, etc will not be accepted).
  • Save your assignment with your ID (e.g. bc020200786.doc).
  • No assignment will be accepted through email.

 

Rules for Marking:

 

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 corrupted.
  • Your assignment is copied from internet, handouts or from any other student

      (Strict disciplinary action will be taken in this case).

 

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

12

12

13

13

14

14

15

15

16

16

17

17

18

18

19

19

20

20

21

21

22

22

23

23

24

24

25

25

26

26

27

27

28

28

29

29

30

30

31

31

 

 

 

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

 

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Replies to This Discussion

solution dy do .................. jinho ny kr li plzzzzzzzzzzzzzzz

CS604 assignment solution 2016

Attachments:

Sample

 

                                   

Q1) In a  paging system, what is the effective access time (EAT) assuming that a memory access takes 200 ns, the TLB search takes 30 ns, and the hit ratio is 90% ?

 

EAT     = hit ratio * (time to search TLB + time for memory access)

+ miss ratio * (time to search TLB + 2*time for memory access)

 

            = 0.9*(30+200) + 0.1*(30+2*200)

            = 0.9*230 + 0.1*430

            = 207+43

            = 250 ns

 

Q2) Consider that the   Physical memory is 128K, logical memory (per process): 32K and page size: 2K. Calculate the following values:

 

a)    Pages per process?

Logical memory size / page size = 32K/2K = 16 pages

 

b)      Frames in physical memory?

Physical memory size / frame size = 128K/2K = 64 frames

 

c)      Bits indicating page number?

Since there are 16=24 pages, hence 4 bits are needed for the page number.

 

d) Bits indicating offset?

Since the page size = 2K = 2*210= 211, hence 11 bits are required for the offset.

Q3) Consider a logical address space of eight pages mapped onto a physical memory of 32 frames. Given that the frame size is 1024 bytes, calculate the following:

 

a)      How many bits for page number?

Since there are 8=23 pages, hence 3 bits are needed for the page number.

 

b)      How many bits offset?

Since the page size = 1024 = 210, hence 10 bits are required for the offset.

 

c)      How many bits are there in the logical address?

Number of bits for page number + number of bits for offset = 3+10 = 13 bits

 

d)      The logical address space?

Number of pages in logical address space * page size = 8*1024 = 8K

 

e)      The physical address space?

Number of frames in physical memory * frame size = 32*1024 = 32K

 

Q4) Consider a simple segmentation system that has the following segment table:

 

Process X

Segment #

Base

Limit

0

3000

450

1

2000

350

2

2650

350

3

1050

350

For each of the following logical addresses, determine the physical address or indicate if a segment fault occurs:

a) (0, 240)                    b) (2, 350)                   c) (1, 000)

a) Check if (offset < segment limit) (240<450) OK

Physical address = 3000 + 240 = 3240

 

b) Check if (offset < segment limit) (350<350) ERROR

 illegal reference, trap to operating system

 

c) Check if (offset < segment limit) (000<350) OK

Physical address = 2000 + 000 = 2000

Sample

 

                                               

Q1) In a  paging system, what is the effective access time (EAT) assuming that a memory access takes 200 ns, the TLB search takes 30 ns, and the hit ratio is 90% ?

 

EAT        = hit ratio * (time to search TLB + time for memory access)

+ miss ratio * (time to search TLB + 2*time for memory access)

 

                = 0.9*(30+200) + 0.1*(30+2*200)

                = 0.9*230 + 0.1*430

                = 207+43

                = 250 ns

 

Q2) Consider that the    Physical memory is 128K, logical memory (per process): 32K and page size: 2K. Calculate the following values:

 

a)    Pages per process?

Logical memory size / page size = 32K/2K = 16 pages

 

b)      Frames in physical memory?

Physical memory size / frame size = 128K/2K = 64 frames

 

c)      Bits indicating page number?

Since there are 16=24 pages, hence 4 bits are needed for the page number.

 

d) Bits indicating offset?

Since the page size = 2K = 2*210= 211, hence 11 bits are required for the offset.

Q3) Consider a logical address space of eight pages mapped onto a physical memory of 32 frames. Given that the frame size is 1024 bytes, calculate the following:

 

a)      How many bits for page number?

Since there are 8=23 pages, hence 3 bits are needed for the page number.

 

b)      How many bits offset?

Since the page size = 1024 = 210, hence 10 bits are required for the offset.

 

c)      How many bits are there in the logical address?

Number of bits for page number + number of bits for offset = 3+10 = 13 bits

 

d)      The logical address space?

Number of pages in logical address space * page size = 8*1024 = 8K

 

e)      The physical address space?

Number of frames in physical memory * frame size = 32*1024 = 32K

 

Q4) Consider a simple segmentation system that has the following segment table:

 

Process X

Segment #

Base

Limit

0

3000

450

1

2000

350

2

2650

350

3

1050

350

For each of the following logical addresses, determine the physical address or indicate if a segment fault occurs:

a) (0, 240)                            b) (2, 350)                            c) (1, 000)

a) Check if (offset < segment limit) (240<450) OK

Physical address = 3000 + 240 = 3240

 

b) Check if (offset < segment limit) (350<350) ERROR

 illegal reference, trap to operating system

 

c) Check if (offset < segment limit) (000<350) OK

Physical address = 2000 + 000 = 2000

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Code:

#include<graphics.h>
#include<conio.h>
#include<stdio.h>
void main()
{
int gd=DETECT,gm;
int i,j,k,t,q;
float x,y;
initgraph(&gd,&gm,"c:\\tc\\bgi");
setcolor(2);
int left=0, int right=0, int right=INT_MAX, int bottom=INT_MAX;
&nbsp;setcolor(2);
&nbsp;i=0;
for(t=0;t<getmaxx();t+=120)
{
line(t,250,t+60,170);
line(t+60,170,t+120,250);
}
line(0,400,getmaxx(),350);
setfillstyle(11,CYAN);
floodfill(2,420,2);
setfillstyle(4,LIGHTGREEN);
floodfill(1,300,2);
i=0;
while(i!=150)
{
setcolor(BLACK);
setfillstyle(SOLID_FILL,BLACK);
fillellipse(k,j,30,30);
setfillstyle(SOLID_FILL,LIGHTRED);
fillellipse(170+i,235-i,30,30);
j=235-i;
k=170+i;
i++;
setcolor(2);
for(t=0;t<getmaxx();t+=120)
{
line(t,250,t+60,170);
line(t+60,170,t+120,250);
}
setfillstyle(1,GREEN);
floodfill(202,200,GREEN);
delay(25);
}
for(i=36;i<80;i++)
for(j=0;j<=360;j+=20)
{
x=319+i*cos(((float)j*3.14)/180);
y=86+i*sin(((float)j*3.14)/180);
putpixel(x,y,LIGHTRED);
delay(1);
}
getch();
}

WA

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