Presentation on theme: "Virtual Memory Background Demand Paging Performance of Demand Paging"— Presentation transcript:
1Virtual Memory Background Demand Paging Performance of Demand Paging Page ReplacementPage-Replacement AlgorithmsAllocation of FramesThrashingOther ConsiderationsDemand Segmentation
2BackgroundVirtual 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.Need to allow pages to be swapped in and out.Virtual memory can be implemented via:Demand pagingDemand segmentation
3Demand Paging Bring a page into memory only when it is needed. Less I/O neededLess memory neededFaster responseMore usersPage is needed reference to itInvalid reference abortNot-in-memory bring to memory
4Valid-Invalid BitWith each page table entry, a valid–invalid bit is associated. (1 in-memory, 0 not-in-memory)Initially valid–invalid bit is set to 0 on all entries.
5Valid-Invalid Bit (cont) Example of a page table snapshot.During address translation, if valid–invalid bit in page table entry is 0 page fault.1...Frame #valid-invalid bitpage table1
6Page FaultIf there is ever a reference to a page, first reference will trap to OS page faultOS looks at another table to decide:Invalid reference abort.Just not in memory.Get empty frame.Swap page into frame.Reset table, validation bit = 1.
7Page Fault (cont)Restart instruction – the process now accesses the page as though it had always been in memory.Paging works because of the locality of reference principle – programs stay in certain localities in a program and then move on to the next locality.
8No Free Frame?Page replacement – find some page in memory, but not really in use, swap it out.AlgorithmPerformance – want an algorithm which will result in minimum number of page faults.Same page may be brought into memory several times.
9Performance of Demand Paging Page Fault Rate – 0 p 1.0If p = 0 no page faultsIf p = 1, every reference is a faultEffective Access Time (EAT)EAT = (1 – p) memory access+ p (page fault overhead+ [swap page out ]+ swap page in+ restart overhead)
10Demand Paging Example Memory access time = 100 nanoseconds. Page fault service time is 25 milliseconds.EAT = (1 – p) p (25 milliseconds)= (1 – p) p 25,000,000= ,999,900 pThe page fault rate has to be kept as small as possible.
11Page ReplacementPrevent over-allocation of memory (too much multiprogramming) 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.
12Page Fault Service Routine Find the location of the desired page on the disk.Find a free frame:If there is a free frame, use it.Otherwise, use a page replacement algorithm to select a victim frame.Write the victim page to the disk; change the page and frame tables accordingly.Read the desired page into the (newly) free frame; change the page and frame tables.Restart the user process.
13Page-Replacement Algorithms Want lowest page-fault rate.Evaluate algorithm by running it on a particular string of page references (reference string) and compute the number of page faults on that string.In all our examples, the reference string is1, 2, 3, 4, 1, 2, 5, 1, 2, 3, 4, 5
14First-In-First-Out Algorithm Reference string: 1, 2, 3, 4, 1, 2, 5, 1, 2, 3, 4, 53 frames (3 pages can be in memory at a time per process)123459 page faults
15FIFO Algorithm (cont) 4 frames FIFO Replacement – Belady’s anomaly. More frames less page faults1235410 page faults
16Optimal AlgorithmReplace page that will not be used for longest period of time.4 frames example1, 2, 3, 4, 1, 2, 5, 1, 2, 3, 4, 512346 page faults5
17Optimal Algorithm (cont) In practice, how do you determine the page to be replaced?This algorithm is used only for measuring how well other algorithms perform.
18Least Recently Used (LRU) Algorithm Reference string: 1, 2, 3, 4, 1, 2, 5, 1, 2, 3, 4, 5Counter implementationEvery page entry has a counter; every time the page is referenced through this entry, the clock is copied into the counter.The page to be replaced is the one that has the smallest value.123455
19LRU Algorithm (Cont.)Stack implementation – keep a stack of page numbers in a doubly linked form:Page referenced:Move it to the topRequires six pointers to be changedNo search for replacement because it is at the bottom of the stack.
20LRU Approximation Algorithms Reference bitWith each page associate a bit, initially = 0When a page is referenced, the bit set to 1.Replace the page whose value is 0 (if one exists). We do not know the order, however.
21LRU Approximation (cont) Second chanceNeed reference bit.Clock replacement.If page to be replaced (in clock order) has reference bit = 0, replace it. Else:Set reference bit 0.Leave the page in memory.Replace the next page (in clock order), subject to same rules.
22Belady’s AnomalyNeither the optimal algorithm nor LRU replacement suffers from Belady’s anomaly.
23Counting AlgorithmsKeep a counter of the number of references that have been made to each page.LFU (Least Frequently Used) Algorithm: replace the page with the smallest count.MFU (Most Frequently Used) Algorithm: based on the argument that the page with the smallest count was probably just brought in and has yet to be used.
24Allocation of Frames Each process needs minimum number of pages. Example: IBM 370 – 6 pages to handle SS MOVE instruction:Instruction is 6 bytes, might span 2 pages.2 pages to handle from.2 pages to handle to.Two major allocation schemes.Fixed allocationPriority allocation
25Fixed AllocationEqual allocation – e.g., if 100 frames and 5 processes, give each 20 pages.Proportional allocation – Allocate according to the size of process.
26Priority AllocationUse a proportional allocation scheme using priorities rather than size.If process Pi generates a page fault:Select for replacement one of its frames.Select for replacement a frame from a process with lower priority number.
27Global vs. Local Allocation Global replacement – process selects a replacement frame from the set of all frames; one process can take a frame from another.Local replacement – each process selects from only its own set of allocated frames.
28ThrashingIf a process does not have “enough” pages, the page-fault rate is very high. This leads to:Low CPU utilization.Operating system thinks that it needs to increase the degree of multiprogramming.Another process added to the system.Thrashing a process is busy swapping pages in and out.
30Trashing (cont) Why does paging work? Locality model Process migrates from one locality to another.Localities may overlap.Why does thrashing occur? size of locality > total memory size
31Working-Set Model working-set window a fixed number of page references Example: 10,000 instructionWSSi (working set of Process Pi) = total number of pages referenced in the most recent (varies in time)If too small will not encompass entire locality.If too large will encompass several localities.If = will encompass entire program.
32Working-Set Model (cont) D = WSSi total demand framesIf D > m thrashingPolicy if D > m, then suspend one of the processes.
33Keeping Track of the Working Set Approximate with interval timer + a reference bitExample: = 10,000Timer interrupts after every 5000 time units.Keep in memory 2 bits for each page.Whenever a timer interrupts copy and sets the values of all reference bits to 0.If one of the bits in memory = 1 page in working set.
34Keeping Track of the Working Set (cont) Why is this not completely accurate?Improvement = 10 bits and interrupt every 1000 time units.
35Page-Fault Frequency Scheme Establish “acceptable” page-fault rate.If actual rate too low, process loses frame.If actual rate too high, process gains frame.
36Other ConsiderationsPrepaging – bring into memory at one time all the pages that will be needed.Page size selection:FragmentationTable sizeI/O overheadLocality
37Other Considerations (cont) Program structureArray A[1024, 1024] of integerEach row is stored in one pageOne frameProgram 1 for j := 1 to 1024 do for i := 1 to 1024 do A[i,j] := 0; 1024 x 1024 page faults
38Other Considerations (cont) Program 2 for i := 1 to 1024 do for j := 1 to 1024 do A[i,j] := 0; 1024 page faultsI/O interlock and addressing – allow pages to be locked in memory when they are used for I/O.
39Demand SegmentationDemand paging is generally considered the most efficient virtual memory system.Demand segmentation can be used when there is insufficient hardware to implement demand paging.OS/2 allocates memory in segments, which it keeps track of through segment descriptors.
40Demand Segmentation (cont) Segment descriptor contains a valid bit to indicate whether the segment is currently in memory.If segment is in main memory, access continues.If segment is not in memory, a segment fault occurs.