our project for OPERATING SYSTEM..

1. Paging
And Segmentation

2.  Involve swapping block of data from second
memory
 Memory needs to be allocated to ensure a reasonable
supply of ready processes to consume available
processor time
 Memory management requirements:
 Relocation
 Protection
 Sharing
 Logical organisation
 Physical organisation
Memory management

3.  Paging is a partition memory into small equal fixed-
size chunks and divide each process into the same
size chunks
 Pages-The chunks of a process
 Frame or pages frame-The chunks of a memory
 Page table:
 Contains the frame location for each page in the
process
 Memory address consist of a page number and offset
within the page
PAGING

4. 
Process Pages to
free frames
 Some of the frames in memory are use and some are free.
 A list of frames is maintained by the OS.
 Process A stored on disk and it consist of four pages.
 OS will find the four free frames and loads the
four pages of process A into four frames.

5. 
…
 Process B consisting of three pages and Process C consisting of four
pages are subsequently loaded.
 Process B is suspended and swapped out to main memory.
 All the process in main memory will blocked
 OS need to bring a new process, which is process D that contain five
pages.

6. 
 The five pages of process D are loaded into frames
5,6,11, and 12.

7. 
 A pages tables contain one entry for each pages of the process.
 The table easily indexed by the pages number (starting at 0)
 Each pages table entry contains the number of the frame in
main memory that hold the corresponding page.
 A free frame list, available for pages, is maintained
 Thus we will se the simple paging is similar to fixed partioning.

8. 
 manages the main memory by dividing the memory into
regions/fixed/variable size is called partitioning
 Two types of partitions
1. Static Partition
Whole memory is divided into fixed sized frame
2. Dynamic Partition
Memory is divided into variable sized frame according to
page size
Partitioning

9. 
Partitioning example :
P1
P2
P4
P3
P3
P1
P4
P3
P2
Static
Partition
Dynamic
Partition
0
10
32
25
15
0
42
49
34
20
10
51
66
70 70
28
Proces
s
Space
P1 10
P2 10
P3 14
P4 15
Total space for
those of partition

10. 
pAGING
CPU
p d f d
Physical
memory
f
p
Logical
address
Physical
address
Page table
 Visual address of a process divided into
1. page number (p) – index to page table
2. page offset (d) – offset into page/frame

11. 
PAGING EXample
0 0 1 0 1 1
10
01
11
00
000
001
010
111
110
101
100
011
000
010
111
110
101
100
011
001
00
01
10
11
Offset
page frame
Physical
memory
Logical
memory
 Page size 4 bytes
 Memory size 32 bytes (8 pages)

12. 
Paging hardware
 Address space 2m
 Page offset 2n
 Page number 2m-n
p d
Page
number
Page
offset
m-n n
Note :
Not losing any
space !
Step : address translation
1. Extract the page number as the n nits of the logical address
2. Use the page number as an index into process page table to
find the frame number, k
3. The starting physical address of the frame is k * 2m , the
physical address of the referenced bytes is that number plus
offset.
Physical
memory
2m bytes

13. 
Segmentation
Step : Address translation from virtual to physical
address
1. Extract the segment number as the n bits of the
logical address
2. Use the segment number as an index into the
process segment table to find the starting physical
address of the segment
3. Compare the offset, expressed in the m bits, to the
length of the segment. If offset is “>=“ to the
length, the address is invalid.
4. The desired physical address is the sum of the
starting physical address of the segment plus
offset.

14. 
Segmentation
CPU s d
< +
limit base
s
Physical memory
Segment
table
yes
Trap, addressing error
Physica
l
address
no
Logical
address
main
stack

16. Paging
logical address
page number = 1 [000001]
offset = 478 [0111011110]
Physical address
Residing frame = 6 [000110]
offset = 478 [0111011110]
Physical address
= 0001100111011110

18. Segmentation
logical address
page number = 1 [0001]
offset = 752 [001011110000]
Physical address
Residing frame = 4 [0010000000100000]
offset = 752 [001011110000]
Physical address
= 0010000000100000 + 001011110000
= 0010001100010000

19. 
The End..
pROjECT: oPERATing sYSTem
From:
FATIN NABILAH BT ABDUL RAMAN
CHONG LEE MAN
HAFIZOH BINTI MD ISA
SITI FATIHAH BT MOHD SOHAIMI
Lecturer :
DR NURUL AZMA ZAKARIA