Virtual Memory Names, Virtual Addresses & Physical Addresses - PowerPoint PPT Presentation

Virtual Memory

Names, Virtual Addresses & Physical Addresses Source Absolute Program Module Name Space Name Space Virtual Virtual Address Space Address Space Compile/Link tools

Names, Virtual Addresses & Physical Addresses Source Absolute Executable Program Module Image Name Space Name Space Virtual Virtual Address Space Address Space Physical Physical Address Space Address Space ? t : Virtual Address Space ? Physical Address Space

Virtual Memory • Uses dynamic address relocation/binding – Generalization of base-limit registers – Physical address corresponding to a compile- time address is not known until run time • Idea is that only part of the address space is loaded as process executes • This works because of program and data locality

Virtual Memory (cont) • Use a dynamic virtual address map, ? t Virtual Physical ? t Address Address Space Space

Address Formation • Translation system produces an address space, but address are virtual instead of physical • A virtual address, x: – Is mapped to y = ? t (x) if x is loaded at physical address y – Is mapped to ? if x is not loaded • The map changes as the program executes ? t : Virtual Address ? Physical Address ? { ? } •

Size of Blocks of Memory • Fixed size: Pages are moved back and forth between primary and secondary memory • Variable size: Programmer-defined segments are the unit of movement • Paging is the commercially dominant form of virtual memory today

Paging • A page is a fixed sized block of virtual addresses • A page frame is a fixed size block of physical memory, the same size as a page • When a virtual address 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

Addresses • Suppose there are G=2 g+h virtual addresses and H=2 j+h physical addresses – Each page/page frame is 2 h addresses – There are 2 g pages in the virtual address space – 2 j page frames are allocated to the process – Rather than map individual addresses, ? t maps the 2 g pages to the 2 j page frames

Address Translation • Let N = {d 0 , d 1 , … d n-1 } be the pages • Let M = {b 0 , b 1 , …, b m-1 } be page frames • Virtual address, i, satisfies 0 ? i<G= 2 g+h • Physical address, k = U2 h +V (0 ? V<G= 2 h ) – U is page frame number – V is the line number within the page – ? t :[0:G-1] ? <U, V> ? { ? } – Since every page is size c=2 h • page number = U = ? i/c ? • line number = V = i mod c

Address Translation (cont) g bits h bits Page # Line # Virtual Address ? t Missing Page j bits h bits Physical Address Frame # Line # CPU Memory MAR

Demand Paging • 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

Modeling Page Behavior • Let ? = r 1 , r 2 , r 3 , …, r i , … be a page reference stream – r i is the i th 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, S t (m), is set of pages loaded – S t (m) = S t-1 (m) ? X t - Y t • X t is the set of fetched pages at time t • Y t is the set of replaced pages at time t

More on Demand Paging • If r t was loaded at time t-1, S t (m) = S t-1 (m) • If r t was not loaded at time t-1 and there were empty page frames – S t (m) = S t-1 (m) ? {r t } • If r t was not loaded at time t-1 and there were no empty page frames – S t (m) = S t-1 (m) ? {r t } - {y} • The alternative is prefetch paging

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

Random Replacement • Replaced page, y, is chosen from the m loaded page frames with probability 1/m Let page reference stream, ? = 2031203120316457 Frame 2 0 3 1 2 0 3 1 2 0 3 1 6 4 5 7 0 1 2

Random Replacement • Replaced page, y, is chosen from the m loaded page frames with probability 1/m Let page reference stream, ? = 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

Random Replacement • Replaced page, y, is chosen from the m loaded page frames with probability 1/m Let page reference stream, ? = 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

Random Replacement • Replaced page, y, is chosen from the m loaded page frames with probability 1/m Let page reference stream, ? = 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

Random Replacement • Replaced page, y, is chosen from the m loaded page frames with probability 1/m Let page reference stream, ? = 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 3 1 0 0 1 1 1 1 2 3 3 3 0 0

Random Replacement • Replaced page, y, is chosen from the m loaded page frames with probability 1/m Let page reference stream, ? = 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 3 3 3 1 0 0 1 1 1 1 1 1 2 3 3 3 0 0 0 2

Random Replacement • Replaced page, y, is chosen from the m loaded page frames with probability 1/m Let page reference stream, ? = 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 3 3 3 0 1 0 0 1 1 1 1 1 1 1 2 3 3 3 0 0 0 2 2

Random Replacement • Replaced page, y, is chosen from the m loaded page frames with probability 1/m Let page reference stream, ? = 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 3 3 3 0 0 1 0 0 1 1 1 1 1 1 1 3 2 3 3 3 0 0 0 2 2 2

Random Replacement • Replaced page, y, is chosen from the m loaded page frames with probability 1/m Let page reference stream, ? = 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 3 3 3 0 0 0 0 1 0 0 1 1 1 1 1 1 1 3 3 6 2 3 3 3 0 0 0 2 2 2 1 1

Random Replacement • Replaced page, y, is chosen from the m loaded page frames with probability 1/m Let page reference stream, ? = 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 3 3 3 0 0 0 0 4 1 0 0 1 1 1 1 1 1 1 3 3 6 6 2 3 3 3 0 0 0 2 2 2 1 1 1

Random Replacement • Replaced page, y, is chosen from the m loaded page frames with probability 1/m Let page reference stream, ? = 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 3 3 3 0 0 0 0 4 4 1 0 0 1 1 1 1 1 1 1 3 3 6 6 5 2 3 3 3 0 0 0 2 2 2 1 1 1 1

Random Replacement • Replaced page, y, is chosen from the m loaded page frames with probability 1/m Let page reference stream, ? = 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 3 3 3 0 0 0 0 4 4 7 1 0 0 1 1 1 1 1 1 1 3 3 6 6 5 5 2 3 3 3 0 0 0 2 2 2 1 1 1 1 1

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

Belady’s Optimal Algorithm • Replace page with maximal forward distance: y t = max xeS t-1(m) FWD t (x) Let page reference stream, ? = 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

Belady’s Optimal Algorithm • Replace page with maximal forward distance: y t = max xeS t-1(m) FWD t (x) Let page reference stream, ? = 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 FWD 4 (2) = 1 FWD 4 (0) = 2 FWD 4 (3) = 3

Belady’s Optimal Algorithm • Replace page with maximal forward distance: y t = max xeS t-1(m) FWD t (x) Let page reference stream, ? = 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 3 1 FWD 4 (2) = 1 FWD 4 (0) = 2 FWD 4 (3) = 3

Belady’s Optimal Algorithm • Replace page with maximal forward distance: y t = max xeS t-1(m) FWD t (x) Let page reference stream, ? = 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 3 1 1 1

Belady’s Optimal Algorithm • Replace page with maximal forward distance: y t = max xeS t-1(m) FWD t (x) Let page reference stream, ? = 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 3 1 1 1 1 FWD 7 (2) = 2 FWD 7 (0) = 3 FWD 7 (1) = 1

Belady’s Optimal Algorithm • Replace page with maximal forward distance: y t = max xeS t-1(m) FWD t (x) Let page reference stream, ? = 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 3 1 1 1 1 1 1 1 FWD 10 (2) = ? FWD 10 (3) = 2 FWD 10 (1) = 3

Belady’s Optimal Algorithm • Replace page with maximal forward distance: y t = max xeS t-1(m) FWD t (x) Let page reference stream, ? = 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 3 1 1 1 1 1 1 1 1 1 FWD 13 (0) = ? FWD 13 (3) = ? FWD 13 (1) = ?