1. Background
Program must be brought into memory and placed within a process for it to be run.
Input queue – collection of processes on the disk that are waiting to be brought into memory to run the program.
User programs go through several steps before being run.

2. Binding of Instructions and Data to Memory
Compile time: If memory location known a priori, absolute code can be generated; must recompile code if starting location changes.
Load time: Must generate relocatable code if memory location is not known at compile time.

3. Execution time: Binding delayed until run time if the process can be moved during its execution from one memory segment to another. Need hardware support for address maps (e.g., base and limit registers).

4. Logical vs. Physical Address Space
The concept of a logical address space that is bound to a separate physical address space is central to proper memory management.
Logical address – generated by the CPU; also referred to as virtual address.
Physical address – address seen by the memory unit.
Logical and physical addresses are the same in compile-time and load-time address-binding schemes; logical (virtual) and physical addresses differ in execution-time address-binding scheme.

5. Memory-Management Unit (MMU)
Hardware device that maps virtual to physical address.
In MMU scheme, the value in the relocation register is added to every address generated by a user process at the time it is sent to memory.
The user program deals with logical addresses; it never sees the real physical addresses.

6. Dynamic relocation using a relocation register

7. Dynamic loading to reduce memory requirements
Routine is not loaded until it is called
Better memory-space utilization; unused routine is never loaded.
Useful when large amounts of code are needed to handle infrequently occurring cases.

8. Dynamic Linking Linking postponed until execution time.
Small piece of code, stub, used to locate the appropriate memory-resident library routine.
Stub replaces itself with the address of the routine, and executes the routine.
Operating system needed to check if routine is in processes’ memory address.
Dynamic linking is particularly useful for libraries.

9. Overlays
Keep in memory only those instructions and data that are needed at any given time.
Needed when process is larger than amount of memory allocated to it.
Implemented by user, no special support needed from operating system, programming design of overlay structure is complex

10. Swapping
A process can be swapped temporarily out of memory to a backing store, and then brought back into memory for continued execution.
Backing store – fast disk large enough to accommodate copies of all memory images for all users; must provide direct access to these memory images.

11. Schematic View of Swapping

12. Contiguous Allocation
Main memory usually into two partitions:
Resident operating system, usually held in low memory with interrupt vector.
User processes then held in high memory.
Single-partition allocation
Relocation-register scheme used to protect user processes from each other, and from changing operating-system code and data.
Relocation register contains value of smallest physical address; limit register contains range of logical addresses – each logical address must be less than the limit register.

13. Hardware Support for Relocation and Limit Registers

14. Contiguous Allocation (Cont.)
Multiple-partition allocation
Hole – block of available memory; holes of various size are scattered throughout memory.
When a process arrives, it is allocated memory from a hole large enough to accommodate it.
Operating system maintains information about: a) allocated partitions b) free partitions (hole)

15. OS
process 5
process 8
process 2

16. OS OS
process 5
process 2
process 5
process 8
process 2

17. OS OS
process 5
process 2 OS
process 5
process 2
process 9
process 5
process 8
process 2

18. OS OS
process 5
process 2 OS
process 5
process 2
process 9 OS
process 5
process 9
process 2
process 10
process 5
process 8
process 2

19. Fragmentation
External Fragmentation – total memory space exists to satisfy a request, but it is not contiguous.
Internal Fragmentation – allocated memory may be slightly larger than requested memory; this size difference is memory internal to a partition, but not being used.

20. Fragmentation Reduce external fragmentation by compaction
Shuffle memory contents to place all free memory together in one large block.
Compaction is possible only if relocation is dynamic, and is done at execution time.

21. Paging
Logical address space of a process can be noncontiguous; process is allocated physical memory whenever the latter is available.
Divide physical memory into fixed-sized blocks called frames (size is power of 2, between 512 bytes and 16MB bytes).
Divide logical memory (i.e., memory addresses generated by CPU) into blocks of same size called pages.

22. Paging OS keeps track of all free frames.
To run a program of size n pages, need to find n free frames and load program.
Set up a page table to translate logical to physical addresses.
Fragmentation?

23. Paging OS keeps track of all free frames.
To run a program of size n pages, need to find n free frames and load program.
Set up a page table to translate logical to physical addresses.
Fragmentation? Possible to have internal fragmentation, not external.

24. Address Translation Scheme
Address generated by CPU is divided into:
Page number (p) – used as an index into a page table which contains base address of each page in physical memory.
Page offset (d) – concatenated with base address to determine the physical memory address that is sent to the memory unit.

25. Paging System Example Memory consists of 32 Page Frames.
Pages are 1024 bytes.
How many bits in physical address?

26. Paging System Example Memory consists of 32 Page Frames.
Pages are 1024 bytes.
How many bits in physical address?
1) How many bits needed to access all pages?

27. Paging System Example Memory consists of 32 Page Frames.
Pages are 1024 bytes.
How many bits in physical address?
1) How many bits needed to access all pages? 5: 2^5 = 32.

28. Paging System Example Memory consists of 32 Page Frames.
Pages are 1024 bytes.
How many bits in physical address?
1) How many bits needed to access all pages? 5: 2^5 = 32.
2) How many bits needed to access all memory locations within a page?

29. Paging System Example Memory consists of 32 Page Frames.
Pages are 1024 bytes.
How many bits in physical address?
1) How many bits needed to access all pages?
5: 2^5 = 32.
2) How many bits needed to access all memory locations within a page?
10: 2^10 = 1024.
This machine using 15 bit addresses.

30. Paging System Example
Compiler generates relative (or logical) addresses.
Relative to the beginning of the program.
Same number of bits in physical and logical addresses.
Assume process P0 consists of 5096 bytes.
Consider variable located at relative address

31. Physical Memory
Process P0: 5096 bytes. How many pages in logical address space? 1 2 29 30 31

32. Physical Memory
Process P0: 5096 bytes. How many pages in logical address space? 5 1 2 29 30 31

33. Physical Memory
Process P0: 5096 bytes. Logical Address space. 1 2 1
Contiguous 2 29 3 30 31 4

34. 1 Contiguous 2 3 4 Process P0: 5096 bytes. Logical Address space.
What is page number and offset within page for logical address 1029? 1
Contiguous 2 3 4

35. Process P0: 5096 bytes. Logical Address space.
What is page number and offset within page for logical address 1029?
Pages are 1024 bytes, so 1029 falls in logical page 1 with offset of 5. 1 2
Contiguous 3 4

36. Physical Memory
Process P0: 5096 bytes. Logical Address space. 1 2
P0.0 1 1 2 29 3 30 31 4

37. Physical Memory
Process P0s logical address space. 1 2
P0.0 1
P0.1 1 3 2 29 3 30 31 4

38. Physical Memory
Process P0: 5096 bytes. Logical Address space. 1 2
P0.2
P0.0 1 3
P0.1 1 3 2 29 3 30 31 4

39. Physical Memory
Process P0: 5096 bytes. Logical Address space. 1 2
P0.2
P0.0 1 3
P0.1 1 3 2 29 3 30 30 31
P0.3 4

40. Physical Memory
Process P0: 5096 bytes. Logical Address space. 1 2 29 30 31
P0.0
P0.1 3
P0.2
P0.3
P0.4 1 1 3 2 3 30 4 2

41. Physical Memory
This is the Page Table for process P0. Maps logical pages to physical pages. 1 2 29 30 31
P0.0
P0.1 3
P0.2
P0.3
P0.4 1 2 3 4 30

42. Physical Memory 1 2 29 30 31
P0.0
P0.1 3
P0.2
P0.3
P0.4
Frame Physical Ad.
2048 – 3071
3072 – 4095
Logical address 1029 stored in physical location 3077.

43. Page Table for Process P0:1 2 3 4 30
CPU generates Logical Address 1029.
Address broken into page number:offset within page.

44. Page Table for Process P01 2 3 4 30
Page# Offset within page.

45. Page Table for Process P0 1 2 3 4 30
Page# Offset within page.
How it works:
Page number used as index into page table.

46. Index into Page Table
00001 1 2 3 4 30
Page Offset

47. Index into Page Table
00001
= 3077 1 2 3 4 30
Physical page number concatenated with offset is physical address.

48. Paging Examples
Assume a page size of 1K and a 15-bit logical address space.
How many pages are in the system?

49. Paging Examples
Assume a page size of 1K and a 15-bit logical address space.
How many pages are in the system?
How many bits are required to address each byte within a 1024-byte page?

50. Paging Examples
Assume a page size of 1K and a 15-bit logical address space.
How many pages are in the system?
How many bits are required to address each byte within a 1024-byte page? 10 (2^10 = 1024).
This leaves 5 bits for page number.
So, How many pages are in the system?

51. More Paging Examples
Assume a page size of 1K and a 15-bit logical address space.
How many pages are in the system?
How many bits are required to address each byte within a byte page? 10 (2^10 = 1024).
This leaves 5 bits for page number.
So, How many pages are in the system?
32 (2^5 = 32)

52. Now consider a 15-bit address space with 8 logical pages
Now consider a 15-bit address space with 8 logical pages. How large are the pages?

53. Assuming a 15-bit address space with 8 logical pages
Assuming a 15-bit address space with 8 logical pages. How large are the pages?
Answer: 2^12 = 4K.
It takes 3 bits to reference 8 logical pages (2^3 = 8).
This leaves 12 bits for the page size and thus pages are 2^12.

54. Assume a 15-bit address space with a page size of 1K
Assume a 15-bit address space with a page size of 1K. (5 bits for page, 10 for offset).
What is the physical address to which logical address 2049 will be mapped?
Page Table for P0
Logical Pages 1 2 3 4 8 3
X (not used)

55. Assume a 15-bit address space with a page size of 1K.
Consider logical address 2049 and the following page table for some process P0.
Assume a 15-bit address space with a page size of 1K.
What is the physical address to which logical address 2049 will be mapped?
Logical Pages
Step 1. Convert logical address to binary:
Logical address: 1 2 3 4 8 3

56. Step2. Determine the logical page number:
Logical Pages
Step2. Determine the logical page number:
Since there are 5-bits allocated to the logical page, the address is broken up as follows:
Logical page number offset within page 1 2 3 4 8 3

57. 8 3
00010
00011
Step 3. Use logical page number as an index into the page table to get physical page number.
Logical Address:

58. 8 3
00010
Step 4. Concatenate offset with physical page frame number
Logical Address

59. 1 2 3 4 8 3
P0.1
1024
2048
3072
P0.2
4096
= 3073

60. Multiple processes have their pages also in memory.1 2 3 4 8 3
P0.1
1024
P1.2
P1.0
2048
P0.2
4096
P2.0
= 3073
Multiple processes have their pages also in memory.

61. Address Translation Architecture

62. Yet Another Example

63. Implementation of Page Table
Problem: Each instruction requires two memory accesses.
One to access page table
One to access physical page
VERY slow

64. Implementation of Page Table
Solved by the use of a special fast-lookup hardware cache called
associative memory or translation look-aside buffers (TLBs)

65. Associative Memory Associative memory – parallel search
Address translation (Page Num, Page Frame)
Page #
Frame #

66. Associative Memory
Page #
Frame #
If Page Num is in associative register, get frame # out
TLB Hit
Otherwise get frame # from page table in memory
TLB Miss

67. Paging Hardware With TLB

68. Memory Protection
Memory protection implemented by associating protection bit with each frame.
Valid-invalid bit attached to each entry in the page table:
“valid” indicates that the associated page is in the process’ logical address space, and is thus a legal page.
“invalid” indicates that the page is not in the process’ logical address space.

69. Valid (v) or Invalid (i) Bit In A Page Table

70. Implementation of Page Table
Page table is kept in main memory.
Page-table base register (PTBR) points to the page table.
Page-table length register (PRLR) indicates size of the page table.
In this scheme every data/instruction access requires two memory accesses. One for the page table and one for the data/instruction.

71. Page Table Implementation Issues
Consider 32-bit address space with 4K pages.
2^32/2^12 pages = 2^20 pages. Each entry 4 bytes.
Need 4 MB for each processes page table (in worst case).
Consider logical address space of 64-bit machines with 4K pages (2^12).
2^64/2^12 pages (2^52 pages). Four byte entry gives 2^54 MB for each processes page table (in the worst case).

72. Page Table Implementation Issues
Consider 32-bit address space with 4K pages.
2^32/2^12 pages = 2^20 pages. Each entry 4 bytes.
Need 4 MB for each processes page table (in worst case).
Consider logical address space of 64-bit machines with 4K pages (2^12).
2^64/2^12 pages (2^52 pages). Four byte entry gives 2^54 MB for each processes page table (in the worst case).
If anyone asks, 2^54 is a big number.

73. Other Page Table Structures
Hierarchical Paging
Hashed Page Tables
Inverted Page Tables

74. Hierarchical Page Tables
One way to deal with large page tables is to divide the page table into smaller pieces. The page table itself is paged.
That is, have to use a paging mechanism to find the individual pages of the page table.
A simple technique is a two-level page table.

75. Two-Level Paging Example
A logical address (on 32-bit machine with 4K page size) is divided into:
a page number consisting of 20 bits.
a page offset consisting of 12 bits.
In a two-level paging system, the 20-bit page number is divided between an outer page table (termed a page directory) and an inner page table that has the number of the actual physical page frame.

76. Two-Level Paging Example
Pi is an index into the page directory.
This contains the base address of one page of the processes page table.
P2 is a displacement into this page of the processes page table.
page number
page offset pi p2 d 10 12

77. Address-Translation Scheme
Address-translation scheme for a two-level 32-bit paging architecture. Each page of the page table maps 1024 virtual pages to physical page frames.

78. Outer Page Table (Page Directory).
Physical Memory 2 1 15 5
Frame 0
Frame 1
Frame 2 1 2 3 4
1022
1023
Base address of Page 0 of page tab.
Base address of Page 1 of page tab.
Base address of Page 2 of page tab.
Base address of Page 3 of page tab.
Page 1 of PT 19 8 3 9 91
Outer Page Table (Page Directory).
Page 0 of PT
Frame N

79. Outer Page Table (Page Directory).
Physical Memory
Assume Logical Address:
P P D
Frame 0
Frame 1
Frame 2
Outer Page Table (Page Directory). 1 2 3 4
1022
1023
Base address of Page 0 of page tab.
Base address of Page 1 of page tab.
Base address of Page 2 of page tab.
Base address of Page 3 of page tab. 15 5 19 8 9 91
Page 1 of PT
Page 0 of PT
Frame N

80. Outer Page Table (Page Directory).
Physical Memory
Assume Logical Address:
P D
Frame 0
Frame 1
Frame 2
Outer Page Table (Page Directory). 1 2 3 4
1022
1023
Base address of Page 0 of page tab.
Base address of Page 1 of page tab.
Base address of Page 2 of page tab.
Base address of Page 3 of page tab. 15 5 19 8 9 91
Page 1 of PT
Page 0 of PT
Frame N

81. Outer Page Table (Page Directory).
Physical Memory
Assume Logical Address:
P D
Frame 0
Frame 1
Frame 2 2 1 15 5 1 2 3 4
1022
1023
Base address of Page 0 of page tab.
Base address of Page 1 of page tab.
Base address of Page 2 of page tab.
Base address of Page 3 of page tab.
Page 1 of PT
Outer Page Table (Page Directory).
Frame N

82. Outer Page Table (Page Directory).
Physical Memory
Assume Logical Address: D
Frame 0
Frame 1
Frame 2 2 1 15 5 1 2 3 4
1022
1023
Base address of Page 0 of page tab.
Base address of Page 1 of page tab.
Base address of Page 2 of page tab.
Base address of Page 3 of page tab.
Page 1 of PT
Outer Page Table (Page Directory).
Frame N

83. Outer Page Table (Page Directory).
Physical Memory
Assume Logical Address: D
Frame 0
Frame 1
Frame 2 2 1 15 5 1 2 3 4
1022
1023
Base address of Page 0 of page tab.
Base address of Page 1 of page tab.
Base address of Page 2 of page tab.
Base address of Page 3 of page tab.
Page 1 of PT
Outer Page Table (Page Directory).
Frame N

84. } d Outer Page Table (Page Directory). Page 1 of PT
Physical Memory
Assume Logical Address:
Frame 0
Frame 1
Frame 2 2 1 15 5
} d 1 2 3 4
1022
1023
Base address of Page 0 of page tab.
Base address of Page 1 of page tab.
Base address of Page 2 of page tab.
Base address of Page 3 of page tab.
Page 1 of PT
Outer Page Table (Page Directory).
Frame N

85. Inverted Page Table One entry for each real page of memory.
Virtual address is now <pid, virtual page#, offset>
PT Entry consists of the virtual address of the page stored in that real memory location, process id.
What is the maximum size of the inverted page table?
Decreases memory needed to store each page table, but increases time needed to search the table when a page reference occurs.
Use hash table to limit the search to one — or at most a few — page-table entries.

86. Inverted Page Table Architecture

87. Shared Pages Shared code
One copy of read-only (reentrant) code shared among processes (i.e., text editors, compilers, window systems).
Reentrant code: Non-Self-Modifying code.
Each users page table maps to the same physical location.
Private code and data
Each process keeps a separate copy of its own code and data.
These pages can be mapped to any available page frame.

88. Shared Pages Example

89. Shared Pages
Example of system that uses shared pages for multiple users?

90. Shared Pages
Example of system that uses shared pages for multiple users?
Windows: The DLLs are shared among all processes.

91. Shared Pages
Example of system that uses shared pages for multiple users?
Windows: The DLLs are shared among all processes.
How can the paging mechanism protect the shared pages (i.e, no process can write to the page)?

92. Shared Pages
Example of system that uses shared pages for multiple users?
Windows: The DLLs are shared among all processes.
How can the paging mechanism protect the shared pages (i.e, no process can write to the page)?
Provide additional protection bits to the page table entry (i.e., permissions for the use of the shared page.

93. Shared Pages
Example of system that uses shared pages for multiple users?
Windows: The DLLs are shared among all processes.
How can the paging mechanism protect the shared pages (i.e, no process can write to the page)?
Provide additional protection bits to the page table entry (i.e., permissions for the use of the shared page.
How can shared memory between processes be implemented?

94. Shared Pages
Example of system that uses shared pages for multiple users?
Windows: The DLLs are shared among all processes.
How can the paging mechanism protect the shared pages (i.e, no process can write to the page)?
Provide additional protection bits to the page table entry (i.e., permissions for the use of the shared page.
How can shared memory between processes be implemented?
Have two processes’ page table pointing to the same physical page frame.
Allow read/write access to the shared pages.

95. Segmentation
Memory-management scheme that supports user view of memory.
A program is a collection of segments. A segment is a logical unit such as:
main program,
procedure,
function,
method,
object,
local variables, global variables,
common block,
stack,
symbol table, arrays

97. Logical View of Segmentation1 4 2 3 1 3 2 4
physical memory space
user space

98. Segmentation Architecture
Logical address consists of a two tuple:
<segment-number, offset>
Segment table – maps two-dimensional physical addresses; each table entry has:
base – contains the starting physical address where the segments reside in memory.
limit – specifies the length of the segment.
Segment-table base register (STBR) points to the segment table’s location in memory.
Segment-table length register (STLR) indicates number of segments used by a program;
segment number s is legal if s < STLR.

99. Segmentation Hardware