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Virtual Memory

Virtual Memory

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Virtual Memory

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  1. Virtual Memory

  2. Executable Image Physical Address Space Yt: Virtual Address Space  Physical Address Space Names, Virtual Addresses & Physical Addresses Dynamically Source Program Absolute Module Name Space Pi’s Virtual Address Space

  3. <1% 15% 35% 20% <1% 30% Execution time Locality Address Space for pi • Address space is logically partitioned • Text, data, stack • Initialization, main, error handle • Different parts have different reference patterns: Initialization code (used once) Code for 1 Code for 2 Code for 3 Code for error 1 Code for error 2 Code for error 3 Data & stack

  4. Physical Address Space 0 n-1 • Each address space is fragmented Primary Memory • Fragments of the virtual address space are dynamically loaded into primary memory at any given time Virtual Memory Secondary Memory Virtual Address Space for pi Virtual Address Space for pj Virtual Address Space for pk • Complete virtual address space is stored in secondary memory

  5. Virtual Memory • Every process has code and data locality • Code tends to execute in a few fragments at one time • Tend to reference same set of data structures • Dynamically load/unload currently-used address space fragments as the process executes • Uses dynamic address relocation/binding • Generalization of base-limit registers • Physical address corresponding to a compile-time address is not bound until run time

  6. Virtual Memory (cont) • Since binding changes with time, use a dynamic virtual address map, Yt Virtual Address Space Yt

  7. Address Formation • Translation system creates an address space, but its address are virtual instead of physical • A virtual address, x: • Is mapped to physical address y = Yt(x) if x is loaded at physical address y • Is mapped to W if x is not loaded • The map, Yt, changes as the process executes -- it is “time varying” • Yt: Virtual Address  Physical Address  {W}

  8. Size of Blocks of Memory • Virtual memory system transfers “blocks” of the address space to/from primary memory • Fixed size blocks: System-defined pages are moved back and forth between primary and secondary memory • Variable size blocks: Programmer-defined segments – corresponding to logical fragments – are the unit of movement • Paging is the commercially dominant form of virtual memory today

  9. Paging • A page is a fixed size, 2h, block of virtual addresses • A page frame is a fixed size, 2h, block of physical memory (the same size as a page) • When a virtual address, x, in page i is referenced by the CPU • If page i is loaded at page frame j, the virtual address is relocated to page frame j • If page is not loaded, the OS interrupts the process and loads the page into a page frame

  10. Addresses • Suppose there are G= 2g2h=2g+h virtual addresses and H=2j+h physical addresses assigned to a process • Each page/page frame is 2h addresses • There are 2g pages in the virtual address space • 2j page frames are allocated to the process • Rather than map individual addresses • Yt maps the 2g pages to the 2j page frames • That is, page_framej = Yt(pagei) • Address k in pagei corresponds to address k in page_framej

  11. Page-Based Address Translation • Let N = {d0, d1, … dn-1} be the pages • Let M = {b0, b1, …, bm-1} be page frames • Virtual address, i, satisfies 0i<G= 2g+h • Physical address, k = U2h+V (0V<G= 2h ) • U is page frame number • V is the line number within the page • Yt:[0:G-1]  <U, V>  {W} • Since every page is size c=2h • page number = U = i/c • line number = V = i mod c

  12. Address Translation (cont) g bits h bits Virtual Address Page # Line # “page table” Missing Page Yt j bits h bits Physical Address Frame # Line # CPU Memory MAR

  13. Demand Paging Algorithm • Page fault occurs • Process with missing page is interrupted • Memory manager locates the missing page • Page frame is unloaded (replacement policy) • Page is loaded in the vacated page frame • Page table is updated • Process is restarted

  14. Modeling Page Behavior • Let w = r1, r2, r3, …, ri, … be a page reference stream • ri is the ith page # referenced by the process • The subscript is the virtual time for the process • Given a page frame allocation of m, the memory state at time t, St(m), is set of pages loaded • St(m) = St-1(m)  Xt - Yt • Xt is the set of fetched pages at time t • Yt is the set of replaced pages at time t

  15. More on Demand Paging • If rt was loaded at time t-1, St(m) = St-1(m) • If rt was not loaded at time t-1 and there were empty page frames • St(m) = St-1(m)  {rt} • If rt was not loaded at time t-1 and there were no empty page frames • St(m) = St-1(m)  {rt} - {y} • The alternative is prefetch paging

  16. Static Allocation, Demand Paging • Number of page frames is static over the life of the process • Fetch policy is demand • Since St(m) = St-1(m)  {rt} - {y}, the replacement policy must choose y -- which uniquely identifies the paging policy

  17. 13 page faults • No knowledge of v not perform well • Easy to implement Random Replacement • Replaced page, y, is chosen from the m loaded page frames with probability 1/m Let page reference stream, v = 2031203120316457 Frame 2 0 3 1 2 0 3 1 2 0 3 1 6 4 5 7 0 1 2 2 2 0 2 0 3 2 1 3 2 1 3 2 1 0 3 1 0 3 1 0 3 1 2 0 1 2 0 3 2 0 3 2 0 6 2 4 6 2 4 5 2 7 5 2

  18. Belady’s Optimal Algorithm • Replace page with maximal forward distance: yt = max xeS t-1(m)FWDt(x) Let page reference stream, v = 2031203120316457 Frame 2 0 3 1 2 0 3 1 2 0 3 1 6 4 5 7 0 2 2 2 1 0 0 2 3

  19. Belady’s Optimal Algorithm • Replace page with maximal forward distance: yt = max xeS t-1(m)FWDt(x) Let page reference stream, v = 2031203120316457 Frame 2 0 3 1 2 0 3 1 2 0 3 1 6 4 5 7 0 2 2 2 1 0 0 2 3 FWD4(2) = 1 FWD4(0) = 2 FWD4(3) = 3

  20. Belady’s Optimal Algorithm • Replace page with maximal forward distance: yt = max xeS t-1(m)FWDt(x) Let page reference stream, v = 2031203120316457 Frame 2 0 3 1 2 0 3 1 2 0 3 1 6 4 5 7 0 2 2 2 2 1 0 0 0 2 31 FWD4(2) = 1 FWD4(0) = 2 FWD4(3) = 3

  21. Belady’s Optimal Algorithm • Replace page with maximal forward distance: yt = max xeS t-1(m)FWDt(x) Let page reference stream, v = 2031203120316457 Frame 2 0 3 1 2 0 3 1 2 0 3 1 6 4 5 7 0 2 2 2 2 2 2 1 0 0 0 0 0 2 31 1 1

  22. Belady’s Optimal Algorithm • Replace page with maximal forward distance: yt = max xeS t-1(m)FWDt(x) Let page reference stream, v = 2031203120316457 Frame 2 0 3 1 2 0 3 1 2 0 3 1 6 4 5 7 0 2 2 2 2 2 2 2 1 0 0 0 0 0 3 2 31 1 1 1 FWD7(2) = 2 FWD7(0) = 3 FWD7(1) = 1

  23. Belady’s Optimal Algorithm • Replace page with maximal forward distance: yt = max xeS t-1(m)FWDt(x) Let page reference stream, v = 2031203120316457 Frame 2 0 3 1 2 0 3 1 2 0 3 1 6 4 5 7 0 2 2 2 2 2 2 2 2 2 0 1 0 0 0 0 0 3 3 3 3 2 31 1 1 1 1 1 1 FWD10(2) =  FWD10(3) = 2 FWD10(1) = 3

  24. Belady’s Optimal Algorithm • Replace page with maximal forward distance: yt = max xeS t-1(m)FWDt(x) Let page reference stream, v = 2031203120316457 Frame 2 0 3 1 2 0 3 1 2 0 3 1 6 4 5 7 0 2 2 2 2 2 2 2 2 2 0 0 0 1 0 0 0 0 0 3 3 3 3 3 3 2 31 1 1 1 1 1 1 1 1 FWD13(0) =  FWD13(3) =  FWD13(1) = 

  25. Belady’s Optimal Algorithm • Replace page with maximal forward distance: yt = max xeS t-1(m)FWDt(x) Let page reference stream, v = 2031203120316457 Frame 2 0 3 1 2 0 3 1 2 0 3 1 6 4 5 7 0 2 2 2 2 2 2 2 2 2 0 0 0 0 4 4 4 1 0 0 0 0 0 3 3 3 3 3 3 6 6 6 7 2 31 1 1 1 1 1 1 1 1 1 1 5 5 10 page faults • Perfect knowledge of v perfect performance • Impossible to implement

  26. Least Recently Used (LRU) • Replace page with maximal forward distance: yt = max xeS t-1(m)BKWDt(x) Let page reference stream, v = 2031203120316457 Frame 2 0 3 1 2 0 3 1 2 0 3 1 6 4 5 7 0 2 2 2 1 0 0 2 3 BKWD4(2) = 3 BKWD4(0) = 2 BKWD4(3) = 1

  27. Least Recently Used (LRU) • Replace page with maximal forward distance: yt = max xeS t-1(m)BKWDt(x) Let page reference stream, v = 2031203120316457 Frame 2 0 3 1 2 0 3 1 2 0 3 1 6 4 5 7 0 2 2 2 1 1 0 0 0 2 3 3 BKWD4(2) = 3 BKWD4(0) = 2 BKWD4(3) = 1

  28. Least Recently Used (LRU) • Replace page with maximal forward distance: yt = max xeS t-1(m)BKWDt(x) Let page reference stream, v = 2031203120316457 Frame 2 0 3 1 2 0 3 1 2 0 3 1 6 4 5 7 0 2 2 2 1 1 1 0 0 0 2 2 3 3 3 BKWD5(1) = 1 BKWD5(0) = 3 BKWD5(3) = 2

  29. Least Recently Used (LRU) • Replace page with maximal forward distance: yt = max xeS t-1(m)BKWDt(x) Let page reference stream, v = 2031203120316457 Frame 2 0 3 1 2 0 3 1 2 0 3 1 6 4 5 7 0 2 2 2 1 1 1 1 0 0 0 2 2 2 3 3 3 0 BKWD6(1) = 2 BKWD6(2) = 1 BKWD6(3) = 3

  30. Least Recently Used (LRU) • Replace page with maximal forward distance: yt = max xeS t-1(m)BKWDt(x) Let page reference stream, v = 2031203120316457 Frame 2 0 3 1 2 0 3 1 2 0 3 1 6 4 5 7 0 2 2 2 1 1 1 3 3 3 0 0 0 6 6 6 7 1 0 0 0 2 2 2 1 1 1 3 3 3 4 4 4 2 3 3 3 0 0 0 2 2 2 1 1 1 5 5

  31. Least Recently Used (LRU) • Replace page with maximal forward distance: yt = max xeS t-1(m)BKWDt(x) Let page reference stream, v = 2031203120316457 Frame 2 0 3 1 2 0 3 1 2 0 3 1 6 4 5 7 0 2 2 2 2 2 2 2 3 2 2 2 2 6 6 6 6 1 0 0 0 0 0 0 0 0 0 0 0 0 4 4 4 2 3 3 3 3 3 3 3 3 3 3 3 3 5 5 3 1 1 1 1 1 1 1 1 1 1 1 1 7 • Backward distance is a good predictor of forward distance -- locality

  32. Least Frequently Used (LFU) • Replace page with minimum use: yt = min xeS t-1(m)FREQ(x) Let page reference stream, v = 2031203120316457 Frame 2 0 3 1 2 0 3 1 2 0 3 1 6 4 5 7 0 2 2 2 1 0 0 2 3 FREQ4(2) = 1 FREQ4(0) = 1 FREQ4(3) = 1

  33. Least Frequently Used (LFU) • Replace page with minimum use: yt = min xeS t-1(m)FREQ(x) Let page reference stream, v = 2031203120316457 Frame 2 0 3 1 2 0 3 1 2 0 3 1 6 4 5 7 0 2 2 2 2 1 0 0 1 2 3 3 FREQ4(2) = 1 FREQ4(0) = 1 FREQ4(3) = 1

  34. Least Frequently Used (LFU) • Replace page with minimum use: yt = min xeS t-1(m)FREQ(x) Let page reference stream, v = 2031203120316457 Frame 2 0 3 1 2 0 3 1 2 0 3 1 6 4 5 7 0 2 2 2 2 2 2 1 0 0 1 1 1 2 3 3 3 0 FREQ6(2) = 2 FREQ6(1) = 1 FREQ6(3) = 1

  35. Least Frequently Used (LFU) • Replace page with minimum use: yt = min xeS t-1(m)FREQ(x) Let page reference stream, v = 2031203120316457 Frame 2 0 3 1 2 0 3 1 2 0 3 1 6 4 5 7 0 2 2 2 2 2 2 1 0 0 1 1 1 2 3 3 3 0 FREQ7(2) = ? FREQ7(1) = ? FREQ7(0) = ?

  36. First In First Out (FIFO) • Replace page that has been in memory the longest: yt = max xeS t-1(m)AGE(x) Let page reference stream, v = 2031203120316457 Frame 2 0 3 1 2 0 3 1 2 0 3 1 6 4 5 7 0 2 2 2 1 1 0 0 0 2 3 3

  37. First In First Out (FIFO) • Replace page that has been in memory the longest: yt = max xeS t-1(m)AGE(x) Let page reference stream, v = 2031203120316457 Frame 2 0 3 1 2 0 3 1 2 0 3 1 6 4 5 7 0 2 2 2 1 0 0 2 3 AGE4(2) = 3 AGE4(0) = 2 AGE4(3) = 1

  38. First In First Out (FIFO) • Replace page that has been in memory the longest: yt = max xeS t-1(m)AGE(x) Let page reference stream, v = 2031203120316457 Frame 2 0 3 1 2 0 3 1 2 0 3 1 6 4 5 7 0 2 2 2 1 1 0 0 0 2 3 3 AGE4(2) = 3 AGE4(0) = 2 AGE4(3) = 1

  39. First In First Out (FIFO) • Replace page that has been in memory the longest: yt = max xeS t-1(m)AGE(x) Let page reference stream, v = 2031203120316457 Frame 2 0 3 1 2 0 3 1 2 0 3 1 6 4 5 7 0 2 2 2 1 1 0 0 0 2 3 3 AGE5(1) = ? AGE5(0) = ? AGE5(3) = ?

  40. Belady’s Anomaly Let page reference stream, v = 012301401234 Frame 0 1 2 3 0 1 4 0 1 2 3 4 0 0 0 0 3 3 3 4 4 4 4 4 4 1 1 1 1 0 0 0 0 0 2 2 2 2 2 2 2 1 1 1 1 1 3 3 Frame 0 1 2 3 0 1 4 0 1 2 3 4 0 0 0 0 0 0 0 4 4 4 4 3 3 1 1 1 1 1 1 1 0 0 0 0 4 2 2 2 2 2 2 2 1 1 1 1 3 3 3 3 3 3 3 2 2 2 • FIFO with m = 3 has 9 faults • FIFO with m = 4 has 10 faults

  41. Stack Algorithms • Some algorithms are well-behaved • Inclusion Property: Pages loaded at time t with m is also loaded at time t with m+1 Frame 0 1 2 3 0 1 4 0 1 2 3 4 0 0 0 0 3 1 1 1 1 2 2 2 LRU Frame 0 1 2 3 0 1 4 0 1 2 3 4 0 0 0 0 0 1 1 1 1 2 2 2 3 3

  42. Stack Algorithms • Some algorithms are well-behaved • Inclusion Property: Pages loaded at time t with m is also loaded at time t with m+1 Frame 0 1 2 3 0 1 4 0 1 2 3 4 0 0 0 0 3 3 1 1 1 1 1 2 2 2 0 LRU Frame 0 1 2 3 0 1 4 0 1 2 3 4 0 0 0 0 0 0 1 1 1 1 1 2 2 2 2 3 3 3

  43. Stack Algorithms • Some algorithms are well-behaved • Inclusion Property: Pages loaded at time t with m is also loaded at time t with m+1 Frame 0 1 2 3 0 1 4 0 1 2 3 4 0 0 0 0 3 3 3 1 1 1 1 0 0 2 2 2 2 1 LRU Frame 0 1 2 3 0 1 4 0 1 2 3 4 0 0 0 0 0 0 0 1 1 1 1 1 1 2 2 2 2 2 3 3 3 3

  44. Stack Algorithms • Some algorithms are well-behaved • Inclusion Property: Pages loaded at time t with m is also loaded at time t with m+1 Frame 0 1 2 3 0 1 4 0 1 2 3 4 0 0 0 0 3 3 3 4 1 1 1 1 0 0 0 2 2 2 2 1 1 LRU Frame 0 1 2 3 0 1 4 0 1 2 3 4 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 2 2 2 2 2 4 3 3 3 3 3

  45. Stack Algorithms • Some algorithms are well-behaved • Inclusion Property: Pages loaded at time t with m is also loaded at time t with m+1 Frame 0 1 2 3 0 1 4 0 1 2 3 4 0 0 0 0 3 3 3 4 4 4 2 2 2 1 1 1 1 0 0 0 0 0 0 3 3 2 2 2 2 1 1 1 1 1 1 4 LRU Frame 0 1 2 3 0 1 4 0 1 2 3 4 0 0 0 0 0 0 0 0 0 0 0 0 4 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 4 4 4 4 3 3 3 3 3 3 3 3 3 2 2 2

  46. Stack Algorithms • Some algorithms are well-behaved • Inclusion Property: Pages loaded at time t with m is also loaded at time t with m+1 Frame 0 1 2 3 0 1 4 0 1 2 3 4 0 0 0 0 3 3 3 4 4 4 4 4 4 1 1 1 1 0 0 0 0 0 2 2 2 2 2 2 2 1 1 1 1 1 3 3 FIFO Frame 0 1 2 3 0 1 4 0 1 2 3 4 0 0 0 0 0 0 0 4 4 4 4 3 3 1 1 1 1 1 1 1 0 0 0 0 4 2 2 2 2 2 2 2 1 1 1 1 3 3 3 3 3 3 3 2 2 2

  47. Implementation • LRU has become preferred algorithm • Difficult to implement • Must record when each page was referenced • Difficult to do in hardware • Approximate LRU with a reference bit • Periodically reset • Set for a page when it is referenced • Dirty bit

  48. Dynamic Paging Algorithms • The amount of physical memory -- the number of page frames -- varies as the process executes • How much memory should be allocated? • Fault rate must be “tolerable” • Will change according to the phase of process • Need to define a placement & replacement policy • Contemporary models based on working set

  49. Working Set • Intuitively, the working set is the set of pages in the process’s locality • Somewhat imprecise • Time varying • Given k processes in memory, let mi(t) be # of pages frames allocated to pi at time t • mi(0) = 0 • i=1k mi(t)  |primary memory| • Also have St(mi(t)) = St(mi(t-1))  Xt - Yt • Or, more simply S(mi(t)) = S(mi(t-1))  Xt - Yt

  50. Placed/Replaced Pages • S(mi(t)) = S(mi(t-1))  Xt - Yt • For the missing page • Allocate a new page frame • Xt = {rt} in the new page frame • How should Yt be defined? • Consider a parameter, , called the window size • Determine BKWDt(y) for every yS(mi(t-1)) • if BKWDt(y) , unload y and deallocate frame • if BKWDt(y) < do not disturb y