1. Memory Management
Chapter 7

2. Memory Management Subdividing memory to accommodate multiple processes
Memory needs to allocated efficiently to pack as many processes into memory as possible

3. Memory Management Requirements (1)
Relocation
Protection
Sharing
Logical organization
Physical organization

4. Memory Management Requirements (2)
Relocation
programmer does not know where the program will be placed in memory when it is executed
while the program is executing, it may be swapped to disk and returned to main memory at a different location
the processor hardware and operating system software must be able to translate the memory references found in the code of the program into actual physical memory address,reflecting the current location of the program in main memory

5. Memory Management Requirements (3)
Protection
processes should not be able to reference memory locations in another process without permission
because the location of a program in main memory is unknown, it is impossible to check absolute addresses in programs since the program could be relocated
all memory references generated by a process must be checked during execution to ensure that they refer only to the memory space allocated to that process.

6. Memory Management Requirements (4)
Sharing
any protection mechanism must have the flexibility to allow several processes to access the same portion of memory
If a number of process are executing the same program, it is advantageous to allow each process (person) access to the same copy of the program rather than have their own separate copy

7. Memory Management Requirements (5)
Logical Organization
memory is organized as a linear address space. While this organization closely mirrors the actual machine hardware, it does not correspond t the way in which programs are typically constructed.
most programs are written in modules
different degrees of protection given to modules (read-only, execute-only)
Segmentation is a tool to satisfies these requirement.

8. Memory Management Requirements (6)
Physical Organization
memory available for a program plus its data may be insufficient
overlaying allows various modules to be assigned the same region of memory
secondary memory cheaper, larger capacity, and permanent

9. Memory Partitioning Fixed Partitioning Dynamic Partitioning
Simple Paging
Simple Segmentation
Virtual-Memory Paging
Virtual-Memory Segmentation

10. Fixed Partitioning
Partition available memory into regions with fixed boundaries
Equal-size partitions
any process whose size is less than or equal to the partition size can be loaded into an available partition
if all partitions are full, the operating system can swap a process out of a partition
a program may not fit in a partition. The programmer must design the program with overlays

11. Fixed Partitioning
Main memory use is inefficient. Any program, no matter how small, occupies an entire partition. This is called internal fragmentation.
Operating System
8 M
8 M
8 M
8 M
8 M

12. Fixed Partitioning Unequal-size partitions
lessens the problem with equal-size partitions
Operating System
8 M
2 M
4 M
6 M
8 M
8 M
12 M

13. Placement Algorithm with Partitions
Equal-size partitions
because all partitions are of equal size, it does not matter which partition is used
Unequal-size partitions
can assign each process to the smallest partition within which it will fit
queue for each partition
processes are assigned in such a way as to minimize wasted memory within a partition

14. One Process Queue per Partition
Operating
System
New
Processes

15. One Process Queue per Partition
When its time to load a process into main memory the smallest available partition that will hold the process is selected
Operating
System
New
Processes

16. Dynamic Partitioning Partitions are of variable length and number
Process is allocated exactly as much memory as required
Eventually get holes in the memory. This is called external fragmentation
Must use compaction to shift processes so they are contiguous and all free memory is in one block

17. Example Dynamic Partitioning
Operating
System
Operating
System
Process 1
320 K
576 K
Operating
System
Process 1
320 K
Process 2
128 K
224 K
896 K
352 K

18. Example Dynamic Partitioning
Operating
System
Operating
System
Operating
System
Process 1
320 K
Process 3
288 K
64 K
Process 4
128 K
96 K
320 K
320 K
Process 1
Process 1
224 K
224 K
Process 2
Process 3
288 K
Process 3
288 K
64 K
64 K

19. Example Dynamic Partitioning
Operating
System
Operating
System
320 K
Process 2
224 k
96 K
128 K
128 K
Process 4
Process 4
96 K
96 K
Process 3
288 K
Process 3
288 K
64 K
64 K

20. Dynamic Partitioning Placement Algorithm
Operating system must decide which free block to allocate to a process
Best-fit algorithm
chooses the block that is closest in size to the request
worst performer overall
since smallest block is found for process, the smallest amount of fragmentation is left memory compaction must be done more often

21. Dynamic Partitioning Placement Algorithm
First-fit algorithm
starts scanning memory from the beginning and chooses the first available block that is large enough.
fastest
may have many process loaded in the front end of memory that must be searched over when trying to find a free block

22. Dynamic Partitioning Placement Algorithm
Next-fit
starts scanning memory from the location of the last placement and chooses the next available block that is large enough
more often allocate a block of memory at the end of memory where the largest block is found
the largest block of memory is broken up into smaller blocks
compaction is required to obtain a large block at the end of memory

23. Dynamic Partitioning Placement Algorithm8K 8K
12K
First Fit
12K
22K 6K
Best Fit
Last
allocated
block (14K)
18K 2K 8K 8K 6K 6K
Allocated block
14K
Free block
14K
Next Fit
36K
20K
Before
After

24. Why Buddy System? - Buddy Systems (1)
A fixed partitioning scheme limits the number of active processes and may use space inefficiently if there is a poor match between available partition size and process size
A dynamic partitioning scheme is more complex to maintain and includes the overhead of compaction.

25. Memory Blocks in a Buddy System - Buddy Systems (2)
In a buddy system, memory blocks are available of size 2k, L <= K <= U,
where:
2L = smallest block that is allocated.
2U = largest size block that is allocated.
Generally, 2U is the size of the entire memory available for allocation.

26. Allocation Method - Buddy Systems (3)
To begin, the entire space available for allocation is treated as a single block of size 2U. If a request of size that 2 U-1 < s <= 2U is made, then the entire block is allocated.
Otherwise, the block is split into two equal buddies of size 2 U-1. If 2 U-2 < s<= 2U-1, then the request is allocated to one of the two buddies.
Otherwise, one of the buddies is split in half again.
This process continues until the smallest block greater than or equal to s is generated and allocated to the request.

27. Remove Unallocated Blocks - Buddy Systems (4)
At any time, the buddy system maintains a list of holes (unallocated blocks) of each size 2 i.
A hole may be removed from the (i + 1) list by splitting it in half to create two buddies of size 2i in the i list.
Whenever a pair of buddies on the i list both become unallocated, they are removed from that list and coalesced into a single block on the
(i + 1) list.

28. Example of Buddy System - Buddy Systems (5)
1 Megabyte Block M
Request 100K (A) A=128K 128K K K
Request 240K (B) A=128K K B=256K K
Request 256K (D) A=128K K B=256K D=256K K
Release B A=128K K K D=256K K
Release A K D=256K K
Request 75K (E) E=128K 128K K D=256K K
Release E K D=256K K
Release D M

29. Tree Representation of Buddy System - Buddy Systems (6)
512K
256K
128K
A=128K 128K B=256K D=256K K

30. Fibonacci Buddy System (1)
The Fibonacci sequence is defined as follow:
F0 = 0, F1 = 1, F n+2 = F n+1 + Fn (n>=0)
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, ….

31. Fibonacci Buddy System (2)
Show the results of the following sequence in a figure using Fibonacci Buddy system:
Request 60 (A);
Request 135 (B);
Request 70 (C);
Return A;
Request 20 (D);
Return B;
Return C;
Return D.

32. Fibonacci Buddy System (3)
377 Byte Block ( )
Request (A) A=
Request (B) A= B=144
Request (C) A= C= B=144
Return A C= B=144
Request (D) D= C= B=144
Return B D= C=
Return D C=
Return C

33. Fibonacci Buddy System (4)
377
(55+89)+ (89+144)
128K
64K
D= C=

34. Relocation
When program loaded into memory the actual (absolute) memory locations are determined
A process may occupy different partitions which means different absolute memory locations during execution (from swapping)
Compaction will also cause a program to occupy a different partition which means different absolute memory locations

35. Addresses Logical Relative Physical
reference to a memory location independent of the current assignment of data to memory
translation must be made to the physical address
Relative
address expressed as a location relative to some known point, usually the beginning of the program
Physical
the absolute address or actual location

36. Hardware Support for Relocation
Relative address
Process Control
Block
Base Register
Adder
Program
Absolute
address
Bounds Register
Comparator
Data
Interrupt to
operating system
Stack
Process image in
main memory

37. Registers Used during Execution
Base register
starting address for the process
Bounds register
ending location of the process
These values are set when the process is loaded and when the process is swapped in

38. Registers Used during Execution
The value of the base register is added to a relative address to produce an absolute address
The resulting address is compared with the value in the bounds register
If the address is not within bounds, an interrupt is generated to the operating system

39. Paging
Partition memory into small equal-size chunks and divide each process into the same size chunks
The chunks of a process are called pages and chunks of memory are called frames
Operating system maintains a page table for each process
contains the frame location for each page in the process
memory address consist of a page number and offset within the page

40. Paging (a) (b) (c) Frame Number 1 2 3 4 5 6 7 8 9 10 11 12 13 14 1 2 31 2 3 4 5 6 7 8 9 10 11 12 13 14 1 2 3 4 5 6 7 8 9 10 11 12 13 14
A.0
A.1
A.2
A.3 1 2 3 4 5 6 7 8 9 10 11 12 13 14
A.0
A.1
A.2
A.3
B.0
B.1
B.2
(a)
(b)
(c)

41. Paging 1 2 3 4 5 6 7 8 9 10 11 12 13 14
A.0
A.1
A.2
A.3
B.0
B.1
B.2
C.0
C.1
C.2
C.3 1 2 3 4 5 6 7 8 9 10 11 12 13 14
A.0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
A.0
A.1
A.1
A.2
A.2
A.3
A.3
D.0
D.1
D.2
C.0
C.0
C.1
C.1
C.2
C.2
C.3
C.3
D.3
D.4
(d)
(e)
(f)

42. Page Tables for Example
--- 7 1 2 3 4 4 13 1 1 1
--- 1 8 5 14 2 2 2
--- 2 9 6
Free Frame List 3 3 3 10 11
Process B 12
Process C
Process A
Process D
Data Structures for the Example at Time Epoch (f)

43. Logical Address of Paging
Relative Address = Logical Address =
User Process (2700 Bytes) Page# =1, Offset = 478
Page
Page
Page 3
Internal Fragmentation
(a) Partitioning (b) Paging (Page Size = 1K)

44. Segmentation
All segments of all programs do not have to be of the same length
There is a maximum segment length
Addressing consist of two parts - a segment number and an offset
Since segments are not equal, segmentation is similar to dynamic partitioning

45. Logical Address of Segmentation
Relative Address = Logical Address =
User Process (2700 Bytes) Segment# =1, Offset = 752
Segment 0
750 Bytes
752
Segment 1
1950 Bytes
(a) Partitioning (b) Segmentation

46. Question 1 - Exercise/Home Work (1)
A 1 Megabyte block of memory is allocated using buddy system.
A. Show the results of the following sequence in a figure:
Request 70K (A); Request 35K (B);
Request 80K (C); Return A;
Request 60K (D); Return B;
Return C; Return D.
B. Show the binary tree representation following Return B.

47. Question 2 - Exercise/Home Work (1)
The Fibonacci sequence is defined as follow:
F0 = 0, F1 = 1, F n+2 = F n+1 + Fn (n>=0)
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, ….
A. Could this sequence be used to establish a buddy system?
B. What would be the advantage of this system over the binary buddy system?

48. Question 2 - Exercise/Home Work (2)
C. Show the results of the following sequence in a figure using Fibonacci buddy system:
Request 70 (A); Request 35 (B);
Request 80 (C); Return A;
Request 60 (D); Return B;
Return C; Return D.
D. Show the binary tree representation following Return B.