The document summarizes key concepts about virtual memory, including: 1) Virtual memory allows processes to execute even if not entirely in physical memory by automatically allocating storage upon request, creating the illusion of large contiguous memory spaces. 2) Common virtual memory implementations include paging, segmentation, and paging with segmentation. Paging divides memory into fixed-size pages while segmentation uses multiple logical segments. 3) Issues in virtual memory design include address mapping, placement, replacement, load control, and sharing. Translation lookaside buffers help speed up address translation.

1. Operating Systems Principles Memory Management Lecture 8: Virtual Memory 主講人：虞台文

3. Operating Systems Principles Memory Management Lecture 8: Virtual Memory Principles of Virtual Memory

5. Principles of Virtual Memory 00000000 00000000 FFFFFFFF FFFFFFFF 00000000 3FFFFFF 4 G 4 G Virtual Memory Virtual Memory Physical Memory 64 M . . . . . . . . .

6. Principles of Virtual Memory 00000000 00000000 FFFFFFFF FFFFFFFF 00000000 3FFFFFF 4 G 4 G Virtual Memory Virtual Memory Physical Memory 64 M Address Map Address Map . . . . . . . . .

10. Operating Systems Principles Memory Management Lecture 8: Virtual Memory Implementations of Virtual Memory

12. Paged Virtual Memory Page No. 0 1 2 p P  1 0 w 2 n  1 1 2 Page size 2 n Virtual Memory Offset Virtual Address ( p , w )

13. Physical Memory Frame No. 0 1 2 f F  1 0 w 2 n  1 1 2 Frame size 2 n Physical Memory Offset Physical Address ( f , w ) The size of physical memory is usually much smaller than the size of virtual memory .

14. Virtual & Physical Addresses The size of physical memory is usually much smaller than the size of virtual memory . va pa Memory Size Address Page number p Offset w Frame number f Offset w | p | bits | w | bits | f | bits | w | bits

15. Address Mapping Address Map Given ( p , w ) , how to determine f from p ? Page No. 0 1 2 p P  1 Virtual Memory Frame No. 0 1 2 f F  1 Physical Memory ( p , w ) ( f , w )

16. Address Mapping Address Map Each process (pid = id ) has its own virtual memory. Given ( id , p , w ) , how to determine f from ( id , p ) ? Page No. 0 1 2 p P  1 Virtual Memory Frame No. 0 1 2 f F  1 Physical Memory ( id , p , w ) ( f , w )

17. Frame Tables frame f  1 frame f frame f +1 Physical Memory ID p f pid page Each process (pid = id ) has its own virtual memory. Given ( id , p , w ) , how to determine f from ( id , p ) ? ID p w ID p Frame Table FT 0 1 f F  1 w

18. Address Translation via Frame Table address_map(id,p,w){ pa = UNDEFINED; for( f =0; f <F; f++) if( FT [ f ].pid==id && FT [ f ].page==p) pa = f +w; return pa ; } Each process (pid = id ) has its own virtual memory. Given ( id , p , w ) , how to determine f from ( id , p ) ?

21. Page Tables f Page Tables frame f  1 frame f frame f +1 PTR Page Table Register Physical Memory p w p w

26. Contiguous Allocation Per Segment Segment Tables Segment x Segment s Segment y Physical Memory STR Segment Table Register s w s w

29. Paging with segmentation

31. Paging of System Tables

34. Translation Look-Aside Buffers (TLB)

35. Translation Look-Aside Buffers (TLB) When the search of ( s , p ) fails, the complete address translation is needed. Replacement strategy: LRU .

36. Translation Look-Aside Buffers (TLB) The buffer is searched associatively (in parallel )

37. Operating Systems Principles Memory Management Lecture 8: Virtual Memory Memory Allocation in Paged Systems

44. Example: Optimal Replacement Algorithm (MIN) RS = c a d b e b a b c d Replace page that will not be referenced for the longest time in the future . c a d b e b a b c d Problem: Reference String not known in advance. 2 page faults Time t RS Frame 0 Frame 1 Frame 2 Frame 3 IN t OUT t 0 1 2 3 4 5 6 7 8 9 10 a b c d a b c d a b c d a b c d a b c d a b c e d b c e e d a b c e a b c e a b c e a b c e d a

48. Example: FIFO Replacement Algorithm RS = c a d b e b a b c d c a d b e b a b c d Replace oldest page.            5 page faults Time t RS Frame 0 Frame 1 Frame 2 Frame 3 IN t OUT t 0 1 2 3 4 5 6 7 8 9 10 a b c d a b c d a b c d a b c d a b c d e b c d d a b c e a e b c d e a c d e a b d e a b c d e a b b c c d

49. Example: Belady’s Anormaly RS = dcbadcedcbaecdcbae 17 page faults 14 page faults Time t RS Frame 0 Frame 1 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 0 8 d c b a d c e d c b a e c d c b a e d d c b c b a d a d c e c e d c d c b a b e b c b c d c d b d b a e a Time t RS Frame 0 Frame 1 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 0 8 Frame 2 d c b a d c e d c b a e c d c b a e d d c d c a c a d b b b a d c e d c e d c e d c e b c e b a e b a e c a e d a c d a c b a c b a c e a

50. Example: Belady’s Anormaly 14 page faults 15 page faults Time t RS Frame 0 Frame 1 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 0 8 Frame 2 d c b a d c e d c b a e c d c b a e d d c d c a c a d b b b a d c e d c e d c e d c e b c e b a e b a e c a e d a c d a c b a c b a c e a d c b a d c e d c b a e c d c b a e d d c d c d c b b a d c b a d c b a e c b a e d b a e d c a e d c b a d c b a e c b a e c b a e d b a e d c b e d c b a d c b a e c Time t RS Frame 0 Frame 1 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 0 8 Frame 2 Frame 3

52. Example: Least Recently Used Replacement (LRU) RS = c a d b e b a b c d Replace Least Recently Used page. 3 page faults c a d b e b a b c d Time t RS Frame 0 Frame 1 Frame 2 Frame 3 IN t OUT t 0 1 2 3 4 5 6 7 8 9 10 Queue end Queue head a b c d a b c d a b c d a b c d a b c d a b e d a b d c e c a b e d a b e d a b e d a b e c d e c d d c b a c d b a a c d b d a c b b d a c e b d a b e d a a b e d b a e d c b a e d c b a

61. Example: Optimal Page Replacement Algorithm (VMIN) RS = c c d b c e c e a d 5 page faults  = 3 Time t RS Page a IN t OUT t 0 1 2 3 4 5 6 7 8 9 10 c c d b c e c e a d Page b Page c Page d Page e                                                        c b d b c e e a a d d

62. Example: Optimal Page Replacement Algorithm (VMIN) RS = c c d b c e c e a d 5 page faults  = 3 Time t RS Page a IN t OUT t 0 1 2 3 4 5 6 7 8 9 10 c c d b c e c e a d Page b Page c Page d Page e                                                        c b d b c e e a a d d

65. Example: Working Set Model RS = e d a c c d b c e c e a d 5 page faults  = 3 Time t RS Page a IN t OUT t 0 1 2 3 4 5 6 7 8 9 10 c c d b c e c e a d Page b Page c Page d Page e                                                        c b a b e a d a e d

66. Example: Working Set Model RS = e d a c c d b c e c e a d 5 page faults  = 3 Time t RS Page a IN t OUT t 0 1 2 3 4 5 6 7 8 9 10 c c d b c e c e a d Page b Page c Page d Page e                                                        c b a b e a d a e d

69. Example: Page Fault Frequency (PFF) Replacement RS = c c d b c e c e a d 5 page faults  = 2 Time t RS Page a IN t OUT t 0 1 2 3 4 5 6 7 8 9 10 c c d b c e c e a d Page b Page c Page d Page e                                                        c b a , e e a d a e b , d

70. Example: Page Fault Frequency (PFF) Replacement RS = c c d b c e c e a d 5 page faults  = 2 Time t RS Page a IN t OUT t 0 1 2 3 4 5 6 7 8 9 10 c c d b c e c e a d Page b Page c Page d Page e                                                        c b a , e e a d a e b , d

79. Evaluation of Paging

80. Evaluation of Paging A process requires a certain percentages of its pages within the short time period after activation . Prepaging is important.

81. Evaluation of Paging Smaller page size is beneficial. Another advantage with a small page size: Reduce memory waste due to internal fragmentation . However, small pages require lager page tables .

82. Evaluation of Paging W : Minimum amount of memory to avoid thrashing . Load control is important.