Virtual Memory Background Demand Paging Performance of Demand Paging

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Presentation transcript:

Virtual Memory Background Demand Paging Performance of Demand Paging Page Replacement Page-Replacement Algorithms Allocation of Frames Thrashing Other Considerations Demand Segmentation

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. Need to allow pages to be swapped in and out. Virtual memory can be implemented via: Demand paging Demand segmentation

Demand Paging Bring a page into memory only when it is needed. Less I/O needed Less memory needed Faster response More users Page is needed  reference to it Invalid reference  abort Not-in-memory  bring to memory

Valid-Invalid Bit With 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.

Valid-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 bit page table 1

Page Fault If there is ever a reference to a page, first reference will trap to OS  page fault OS 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.

Page 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.

No Free Frame? Page replacement – find some page in memory, but not really in use, swap it out. Algorithm Performance – want an algorithm which will result in minimum number of page faults. Same page may be brought into memory several times.

Performance of Demand Paging Page Fault Rate – 0  p  1.0 If p = 0 no page faults If p = 1, every reference is a fault Effective Access Time (EAT) EAT = (1 – p)  memory access + p (page fault overhead + [swap page out ] + swap page in + restart overhead)

Demand Paging Example Memory access time = 100 nanoseconds. Page fault service time is 25 milliseconds. EAT = (1 – p)  100 + p (25 milliseconds) = (1 – p)  100 + p  25,000,000 = 100 + 24,999,900  p The page fault rate has to be kept as small as possible.

Page Replacement Prevent 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.

Page 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.

Page-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 is 1, 2, 3, 4, 1, 2, 5, 1, 2, 3, 4, 5

First-In-First-Out Algorithm Reference string: 1, 2, 3, 4, 1, 2, 5, 1, 2, 3, 4, 5 3 frames (3 pages can be in memory at a time per process) 1 2 3 4 5 9 page faults

FIFO Algorithm (cont) 4 frames FIFO Replacement – Belady’s anomaly. More frames  less page faults 1 2 3 5 4 10 page faults

Optimal Algorithm Replace page that will not be used for longest period of time. 4 frames example 1, 2, 3, 4, 1, 2, 5, 1, 2, 3, 4, 5 1 2 3 4 6 page faults 5

Optimal 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.

Least Recently Used (LRU) Algorithm Reference string: 1, 2, 3, 4, 1, 2, 5, 1, 2, 3, 4, 5 Counter implementation Every 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. 1 2 3 4 5 5

LRU Algorithm (Cont.) Stack implementation – keep a stack of page numbers in a doubly linked form: Page referenced: Move it to the top Requires six pointers to be changed No search for replacement because it is at the bottom of the stack.

LRU Approximation Algorithms Reference bit With each page associate a bit, initially = 0 When 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.

LRU Approximation (cont) Second chance Need 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.

Belady’s Anomaly Neither the optimal algorithm nor LRU replacement suffers from Belady’s anomaly.

Counting Algorithms Keep 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.

Allocation 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 allocation Priority allocation

Fixed Allocation Equal allocation – e.g., if 100 frames and 5 processes, give each 20 pages. Proportional allocation – Allocate according to the size of process.

Priority Allocation Use 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.

Global 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.

Thrashing If 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.

Thrashing Diagram

Trashing (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

Working-Set Model   working-set window  a fixed number of page references Example: 10,000 instruction WSSi (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.

Working-Set Model (cont) D =  WSSi  total demand frames If D > m  thrashing Policy if D > m, then suspend one of the processes.

Keeping Track of the Working Set Approximate with interval timer + a reference bit Example:  = 10,000 Timer 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.

Keeping Track of the Working Set (cont) Why is this not completely accurate? Improvement = 10 bits and interrupt every 1000 time units.

Page-Fault Frequency Scheme Establish “acceptable” page-fault rate. If actual rate too low, process loses frame. If actual rate too high, process gains frame.

Other Considerations Prepaging – bring into memory at one time all the pages that will be needed. Page size selection: Fragmentation Table size I/O overhead Locality

Other Considerations (cont) Program structure Array A[1024, 1024] of integer Each row is stored in one page One frame Program 1 for j := 1 to 1024 do for i := 1 to 1024 do A[i,j] := 0; 1024 x 1024 page faults

Other Considerations (cont) Program 2 for i := 1 to 1024 do for j := 1 to 1024 do A[i,j] := 0; 1024 page faults I/O interlock and addressing – allow pages to be locked in memory when they are used for I/O.

Demand Segmentation Demand 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.

Demand 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.