Chapter 9: Virtual Memory PDF

Summary

This document is Chapter 9 from the Operating System Concepts textbook, specifically covering Virtual Memory. It explains the benefits of virtual memory, the concepts of demand paging and page replacement algorithms, as well as allocation of page frames, and gives specific examples.

Full Transcript

Chapter 9: Virtual Memory Operating System Concepts – 9th Edition Silberschatz, Galvin and Gagne ©2013 Chapter 9: Virtual Memory Background Demand Paging Page Replacement Operating System Concepts – 9th Edition 9.2...

Chapter 9: Virtual Memory Operating System Concepts – 9th Edition Silberschatz, Galvin and Gagne ©2013 Chapter 9: Virtual Memory Background Demand Paging Page Replacement Operating System Concepts – 9th Edition 9.2 Silberschatz, Galvin and Gagne ©2013 Objectives To describe the benefits of a virtual memory system To explain the concepts of demand paging, page-replacement algorithms, and allocation of page frames Operating System Concepts – 9th Edition 9.3 Silberschatz, Galvin and Gagne ©2013 Background Virtual memory – separation of user logical memory from physical memory Only part of the program needs to be in memory for execution Logical address space can therefore be much larger than physical address space Allows address spaces to be shared by several processes Allows for more efficient process creation Virtual memory can be implemented via: Demand paging Demand segmentation Operating System Concepts – 9th Edition 9.4 Silberschatz, Galvin and Gagne ©2013 Virtual Memory That is Larger Than Physical Memory Operating System Concepts – 9th Edition 9.5 Silberschatz, Galvin and Gagne ©2013 Demand Paging Could bring entire process into memory at load time Or bring a page into memory only when it is needed Less I/O needed, no unnecessary I/O Less memory needed Faster response More users Similar to paging system with swapping (diagram on right) Page is needed  reference to it invalid reference  abort not-in-memory  bring to memory Lazy swapper – never swaps a page into memory unless page will be needed Swapper that deals with pages is a pager Operating System Concepts – 9th Edition 9.6 Silberschatz, Galvin and Gagne ©2013 Valid-Invalid Bit With each page table entry a valid–invalid bit is associated (v  in-memory – memory resident, i  not-in-memory) Initially valid–invalid bit is set to i on all entries Example of a page table snapshot: During MMU address translation, if valid–invalid bit in page table entry is i  page fault Operating System Concepts – 9th Edition 9.7 Silberschatz, Galvin and Gagne ©2013 Page Table When Some Pages Are Not in Main Memory Operating System Concepts – 9th Edition 9.8 Silberschatz, Galvin and Gagne ©2013 Page Fault If there is a reference to a page, first reference to that page will trap to operating system: page fault 1. Operating system looks at another table to decide: Invalid reference  abort Just not in memory 2. Find free frame 3. Swap page into frame via scheduled disk operation 4. Reset tables to indicate page now in memory Set validation bit = v 5. Restart the instruction that caused the page fault Operating System Concepts – 9th Edition 9.9 Silberschatz, Galvin and Gagne ©2013 Steps in Handling a Page Fault Operating System Concepts – 9th Edition 9.10 Silberschatz, Galvin and Gagne ©2013 Performance of Demand Paging Page Fault Rate 0  p  1 if p = 0 no page faults if p = 1, every reference is a fault Effective Access Time (EAT) EAT = (1 – p) x memory access + p (page fault overhead + swap page out + swap page in ) Operating System Concepts – 9th Edition 9.11 Silberschatz, Galvin and Gagne ©2013 Demand Paging Example Memory access time = 200 nanoseconds Average page-fault service time = 8 milliseconds EAT = (1 – p) x 200 + p (8 milliseconds) = (1 – p x 200 + p x 8,000,000 = 200 + p x 7,999,800 If one access out of 1,000 causes a page fault, then EAT = 8.2 microseconds. This is a slowdown by a factor of 40!! If want performance degradation < 10 percent 220 > 200 + 7,999,800 x p 20 > 7,999,800 x p p <.0000025 < one page fault in every 400,000 memory accesses Operating System Concepts – 9th Edition 9.12 Silberschatz, Galvin and Gagne ©2013 What Happens if There is no Free Frame? Page replacement – find some page in memory, but not really in use, page it out Algorithm – terminate? swap out? replace the page? Performance – want an algorithm which will result in minimum number of page faults Same page may be brought into memory several times Operating System Concepts – 9th Edition 9.13 Silberschatz, Galvin and Gagne ©2013 Page Replacement Prevent over-allocation of memory by modifying page- fault service routine to include page replacement Use modify (dirty) bit to reduce overhead of page transfers – only modified pages are written to disk Page replacement completes separation between logical memory and physical memory – large virtual memory can be provided on a smaller physical memory Operating System Concepts – 9th Edition 9.14 Silberschatz, Galvin and Gagne ©2013 Need For Page Replacement Operating System Concepts – 9th Edition 9.15 Silberschatz, Galvin and Gagne ©2013 Basic Page Replacement 1. Find the location of the desired page on disk 2. Find a free frame: - If there is a free frame, use it - If there is no free frame, use a page replacement algorithm to select a victim frame - Write victim frame to disk if dirty 3. Bring the desired page into the (newly) free frame; update the page and frame tables 4. Continue the process by restarting the instruction that caused the trap Note now potentially 2 page transfers for page fault – increasing EAT Operating System Concepts – 9th Edition 9.16 Silberschatz, Galvin and Gagne ©2013 Page Replacement Operating System Concepts – 9th Edition 9.17 Silberschatz, Galvin and Gagne ©2013 Page and Frame Replacement Algorithms Frame-allocation algorithm determines How many frames to give each process Which frames to replace Page-replacement algorithm Want lowest page-fault rate on both first access and re-access Evaluate algorithm by running it on a particular string of memory references (reference string) and computing the number of page faults on that string String is just page numbers, not full addresses Repeated access to the same page does not cause a page fault Results depend on number of frames available In all our examples, the reference string of referenced page numbers is 7,0,1,2,0,3,0,4,2,3,0,3,0,3,2,1,2,0,1,7,0,1 Operating System Concepts – 9th Edition 9.18 Silberschatz, Galvin and Gagne ©2013 Graph of Page Faults Versus The Number of Frames Operating System Concepts – 9th Edition 9.19 Silberschatz, Galvin and Gagne ©2013 First-In-First-Out (FIFO) Algorithm Reference string: 7,0,1,2,0,3,0,4,2,3,0,3,0,3,2,1,2,0,1,7,0,1 3 frames (3 pages can be in memory at a time per process) 15 page faults Can vary by reference string: consider 1,2,3,4,1,2,5,1,2,3,4,5 Adding more frames can cause more page faults!  Belady’s Anomaly How to track ages of pages? Just use a FIFO queue Operating System Concepts – 9th Edition 9.20 Silberschatz, Galvin and Gagne ©2013 FIFO Illustrating Belady’s Anomaly Operating System Concepts – 9th Edition 9.21 Silberschatz, Galvin and Gagne ©2013 Optimal Algorithm Replace page that will not be used for longest period of time 9 is optimal for the example How do you know this? Can’t read the future Used for measuring how well your algorithm performs Operating System Concepts – 9th Edition 9.22 Silberschatz, Galvin and Gagne ©2013 Least Recently Used (LRU) Algorithm Use past knowledge rather than future Replace page that has not been used in the most amount of time Associate time of last use with each page 12 faults – better than FIFO but worse than OPT Generally good algorithm and frequently used But how to implement? Operating System Concepts – 9th Edition 9.23 Silberschatz, Galvin and Gagne ©2013 Page Replacement Exercise Operating System Concepts – 9th Edition 9.24 Silberschatz, Galvin and Gagne ©2013 End of Chapter 9 Operating System Concepts – 9th Edition Silberschatz, Galvin and Gagne ©2013

Use Quizgecko on...
Browser
Browser