Chapter 9: Virtual Memory PDF

Chapter 9: Virtual Memory Operating System Concepts – 9th Edition Silberschatz, Galvin and Gagne ©2013  Background Code needs to be in memory to execute, but entire program rarely used  Error code, unusu...

Chapter 9: Virtual Memory Operating System Concepts – 9th Edition Silberschatz, Galvin and Gagne ©2013  Background Code needs to be in memory to execute, but entire program rarely used  Error code, unusual routines, large data structures Entire program code not needed at same time Consider ability to execute partially-loaded program  Program no longer constrained by limits of physical memory  Each program takes less memory while running -> more programs run at the same time  Increased CPU utilization and throughput with no increase in response time or turnaround time  Less I/O needed to load or swap programs into memory -> each user program runs faster Operating System Concepts – 9th Edition 9.2 Silberschatz, Galvin and Gagne ©2013  Background (Cont.) 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  More programs running concurrently  Less I/O needed to load or swap processes Operating System Concepts – 9th Edition 9.3 Silberschatz, Galvin and Gagne ©2013  Background (Cont.) Virtual address space – logical view of how process is stored in memory  Usually start at address 0, contiguous addresses until end of space  Meanwhile, physical memory organized in page frames  MMU must map logical to physical 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  Basic Concepts With swapping, pager guesses which pages will be used before swapping out again Instead, pager brings in only those pages into memory How to determine that set of pages?  Need new MMU functionality to implement demand paging If pages needed are already memory resident  No difference from non demand-paging If page needed and not memory resident  Need to detect and load the page into memory from storage  Without changing program behavior  Without programmer needing to change code Operating System Concepts – 9th Edition 9.7 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 9.8 Silberschatz, Galvin and Gagne ©2013  Page Table When Some Pages Are Not in Main Memory Operating System Concepts – 9th Edition 9.9 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 9.10 Silberschatz, Galvin and Gagne ©2013  Steps in Handling a Page Fault Operating System Concepts – 9th Edition 9.11 Silberschatz, Galvin and Gagne ©2013  Aspects of Demand Paging Extreme case – start process with no pages in memory  OS sets instruction pointer to first instruction of process, non- memory-resident -> page fault  And for every other process pages on first access  Pure demand paging Actually, a given instruction could access multiple pages -> multiple page faults  Consider fetch and decode of instruction which adds 2 numbers from memory and stores result back to memory  Pain decreased because of locality of reference Hardware support needed for demand paging  Page table with valid / invalid bit  Secondary memory (swap device with swap space)  Instruction restart Operating System Concepts – 9th Edition 9.12 Silberschatz, Galvin and Gagne ©2013  Performance of Demand Paging Stages in Demand Paging (worse case) 1. Trap to the operating system 2. Save the user registers and process state 3. Determine that the interrupt was a page fault 4. Check that the page reference was legal and determine the location of the page on the disk 5. Issue a read from the disk to a free frame: 1. Wait in a queue for this device until the read request is serviced 2. Wait for the device seek and/or latency time 3. Begin the transfer of the page to a free frame 6. While waiting, allocate the CPU to some other user 7. Receive an interrupt from the disk I/O subsystem (I/O completed) 8. Save the registers and process state for the other user 9. Determine that the interrupt was from the disk 10. Correct the page table and other tables to show page is now in memory 11. Wait for the CPU to be allocated to this process again 12. Restore the user registers, process state, and new page table, and then resume the interrupted instruction Operating System Concepts – 9th Edition 9.13 Silberschatz, Galvin and Gagne ©2013  Performance of Demand Paging (Cont.) Three major activities  Service the interrupt – careful coding means just several hundred instructions needed  Read the page – lots of time  Restart the process – again just a small amount of time 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 9.14 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 9.15 Silberschatz, Galvin and Gagne ©2013  What Happens if There is no Free Frame? Used up by process pages Also in demand from the kernel, I/O buffers, etc How much to allocate to each? 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 9.16 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 9.17 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 9.18 Silberschatz, Galvin and Gagne ©2013  Page Replacement Operating System Concepts – 9th Edition 9.19 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 9.20 Silberschatz, Galvin and Gagne ©2013  Graph of Page Faults Versus The Number of Frames Operating System Concepts – 9th Edition 9.21 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 9.22 Silberschatz, Galvin and Gagne ©2013  FIFO Illustrating Belady’s Anomaly Operating System Concepts – 9th Edition 9.23 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 9.24 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 9.25 Silberschatz, Galvin and Gagne ©2013  LRU Algorithm (Cont.) Counter implementation  Every page entry has a counter; every time page is referenced through this entry, copy the clock into the counter  When a page needs to be changed, look at the counters to find smallest value  Search through table needed Stack implementation  Keep a stack of page numbers in a double link form:  Page referenced:  move it to the top  requires 6 pointers to be changed  But each update more expensive  No search for replacement LRU and OPT are cases of stack algorithms that don’t have Belady’s Anomaly Operating System Concepts – 9th Edition 9.26 Silberschatz, Galvin and Gagne ©2013  Use Of A Stack to Record Most Recent Page References Operating System Concepts – 9th Edition 9.27 Silberschatz, Galvin and Gagne ©2013  End of Chapter 9 Operating System Concepts – 9th Edition Silberschatz, Galvin and Gagne ©2013