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Memory Management and Virtual Memory

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Memory Management and Virtual Memory Background Overlays versus Swapping Contiguous Allocation Operating System Concepts – PowerPoint PPT presentation

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Title: Memory Management and Virtual Memory


1
Memory Management and Virtual Memory
  • Background
  • Overlays versus Swapping
  • Contiguous Allocation

2
Memory-Management Unit (MMU)
  • The run-time mapping from virtual to physical
    addresses is done by the memory-management unit
    (MMU), which is a hardware device.
  • The user program deals with logical addresses it
    never sees the real physical addresses.
  • Two different types of addresses
  • Logical range 0 to max.
  • Physical range R0 to Rmax where R is a base
    value.
  • The user generates only logical addresses and
    thinks that the process runs in locations 0 to
    max.

3
Memory-Management Unit (MMU) (Cont.)
  • In MMU scheme, the value in the relocation
    register is added to every address generated by a
    user process at the time it is sent to memory.
  • Example Dynamic relocation using a relocation
    register.

4
Overlays
  • The entire program and data of a process must be
    in the physical memory for the process to
    execute.
  • The size of a process is limited to the size of
    physical memory.
  • If a process is larger than the amount of memory,
    a technique called overlays can be used.
  • Overlays is to keep in memory only those
    instructions and data that are needed at any
    given time.
  • When other instructions are needed, they are
    loaded into space that was occupied previously by
    instructions that are no longer needed.
  • Overlays are implemented by user, no special
    support needed from operating system, programming
    design of overlay structure is complex.

5
Overlays Example
  • Example Consider a two-pass assembler.
  • Pass1 constructs a symbol table.
  • Pass2 generates machine-language code.
  • Assume the following
  • To load everything at once, we need 200k of
    memory.

Size (k 1024 bytes)
Pass1 70k
Pass2 80k
Symbol table 20k
Common routines 30k
Total size 200k
6
Overlays Example (Cont.)
  • If only 150K is available, we cannot run our
    process.
  • Notice that Pass1 and Pass2 do not need to be in
    memory at same time.
  • So, we define two overlays
  • Overlay A symbol table, common routines, and
    Pass1.
  • Overlay B symbol table, common routines, and
    Pass2.
  • We add overlay driver 10k and start with overlay
    A in memory.
  • When finish Pass1, we jump to overlay driver,
    which reads overlay B into memory overwriting
    overlay A and transfer control to Pass2.

7
Overlays Example (Cont.)
  • Overlay A needs 130k and Overlay B needs 140k.

Symbol table
Common routines
Overlay driver

8
Swapping
  • A process can be swapped temporarily out of
    memory to a backing store, and then brought back
    into memory for continued execution.
  • Backing store fast disk large enough to
    accommodate copies of all memory images for all
    users must provide direct access to these memory
    images.
  • Roll out, roll in swapping variant used for
    priority-based scheduling algorithms
    lower-priority process is swapped out so
    higher-priority process can be loaded and
    executed.
  • Normally, a process that is swapped out will be
    swapped back into the same memory space that it
    occupied previously.
  • Example In a multiprogramming environment with
    Round Robin CPU scheduling, when time slice
    expires, the memory manager swap out the process
    that just finished, and swap in another process
    to the memory space that has been freed.

9
Schematic View of Swapping
  • Major part of swap time is transfer time total
    transfer time is directly proportional to the
    amount of memory swapped.
  • The context-switch time in swapping system is
    high.
  • Modified versions of swapping are found on many
    systems, i.e., UNIX and Microsoft Windows.

10
Example 1 Swapping
  • Let Process P1 size is 100KB and the transfer
    rate of a hard disk is 1MB/second.
  • To transfer P1 (100KB) to or from memory takes
    100K/1000K per second, which is 1/10 second 100
    milliseconds.
  • Assume the average latency is 8 milliseconds,
    then to swap in or out takes 108 milliseconds.
    The total swap (in and out) is 216 milliseconds
    for 100KB process.
  • For efficient CPU utilization, we want our
    execution time for each process to be long
    relative to the swap time.
  • A Round Robin scheduling algorithm, the time
    slice should be larger than 216 milliseconds
    (from the above example).

11
Contiguous Allocation
  • Main memory usually is divided into two
    partitions
  • For the resident operating system
  • For the user processes.

O.S.
User
12
Example 2 Swapping
  • A computer system has 1MB of main memory. The
    O.S. takes 100KB.
  • Maximum size of user process is 900KB.
  • User process may be smaller than 900KB.
  • From previous example if process P1 is 100K the
    swap time is 108 milliseconds. But, if P1 is 900K
    then the swap time is 908 milliseconds.
  • As the size of a process increases the swap time
    increases too.

O.S. 100K
User 900K
Main Memory
13
Single Partition Allocation
  • Single-partition allocation
  • Needs Protection.
  • Protect O.S. code and data from changes by the
    user processes.
  • Protect user processes from one another.
  • We can provide protection by using a
    relocation-register with a limit register.
  • Relocation register contains value of smallest
    physical address.
  • Limit register contains range of logical
    addresses.
  • Each logical address must be less than the limit
    register.

14
Hardware Support for Relocation Limit Registers
15
Multiple - Partition Allocation
  • Several user processes residing in memory at the
    same time.
  • Memory can be divided into a number of
    fixed-sized partitions, where each partition may
    contain exactly one process. Therefore, the
    degree of multiprogramming is bound by the number
    of partitions.
  • Or all memory is available for user processes as
    one large block (hole).
  • When a process arrives and needs memory, we
    search for a hole large enough for this process.
  • If we find a space, we allocate only as much
    memory as is needed, keeping the rest available
    to satisfy future requests.

16
Multiple - Partition Allocation (Cont.)
  • Hole block of available memory holes of
    various size are scattered throughout memory.
  • Operating system maintains information abouta)
    allocated partitions b) free partitions (hole)

17
Multiple - Partition Allocation (Cont.)
  • When no available block of memory (hole) is large
    enough to hold process, the O.S. waits until a
    large block is available.
  • In general, there is at any time a set of holes,
    of various sizes, scattered throughout memory.
  • When a process arrives and needs memory, we
    search this set for a hole that is large enough
    for this process.
  • If the hole is too large, it is split into two
  • One part is allocated to the arriving process.
  • The other is returned to the set of holes.
  • When a process terminates, it releases its block
    of memory, which is then placed back in the set
    of holes.
  • If the new hole is adjacent to other holes, we
    merge these adjacent holes to form one larger
    hole.

18
Dynamic Storage-Allocation Problem
  • First-fit, best-fit, and worst fit are the most
    common strategies used to select a free hole from
    the set of available holes.
  • First-fit Allocate the first hole that is big
    enough. Searching starts at the beginning of the
    set of holes. We can stop searching as soon as we
    find a free hole that is large enough.
  • Best-fit Allocate the smallest hole that is big
    enough must search entire list, unless ordered
    by size. Produces the smallest leftover hole.
  • Worst-fit Allocate the largest hole must also
    search entire list, unless ordered by size.
    Produces the largest leftover hole, which may be
    more useful than the smaller leftover hole from
    best-fit approach.
  • First-fit and best-fit better than worst-fit in
    terms of speed and storage utilization.

19
Fragmentation
  • External fragmentation total memory space
    exists to satisfy a request, but it is not
    contiguous storage is fragmented into a large
    number of small holes.
  • Example We have a total external fragmentation
    of (300260)560KB.
  • If P5 is 500KB, then this space would be large
    enough to run P5. But the space is not
    contiguous.
  • The selection of first-fit versus best-fit can
    affect the amount of fragmentation.

0 400K O.S.
1000K P1
1700K P4
2000K Free
2300K P3
2560K Free
20
Fragmentation (Cont.)
  • Internal fragmentation Memory that is internal
    to a partition, but is not being used, because it
    is too small.
  • Example Assume next request is for 18462 bytes.
  • If we allocate exactly the requested block, we
    are left with a hole of 2 bytes.
  • The overhead to keep track of this hole will be
    larger than the hole itself. So, we ignore this
    small hole (internal fragmentation).

O.S.
P7
Hole of 18464 bytes Free
P4
21
Virtual Memory
  • Background
  • Demand Paging
  • Performance of Demand Paging
  • Page Replacement
  • Page-Replacement Algorithms

22
Background (Cont.)
  • Virtual memory is the separation of user logical
    memory from physical memory. This separation
    allows an extremely large virtual memory to be
    provided for programmers when only a smaller
    physical memory is available.
  • 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

23
Diagram virtual memory larger than physical
memory






Physical Memory
Page 0
Page 1
Page 2

Page n
Virtual Memory




Memory Map
24
Transfer of a Pages Memory to Contiguous Disk
Space








Main Memory
25
Valid-Invalid bit
  • We need hardware support to distinguish between
    those pages that are in memory and those pages
    are on the disk.
  • The valid-invalid bit scheme can be used
  • Valid indicates that the associated pages is
    both legal and in memory.
  • Invalid indicates that the page either is not
    valid (not in logical address space) or is valid
    but is currently on the disk.
  • What happens if the process tries to use a page
    that was not brought into memory?
  • Access to a page marked invalid causes a
    page-fault trap.

26
Valid-Invalid Bit (Cont.)
  • With each page table entry a validinvalid bit is
    associated(1 ? in-memory, 0 ? not-in-memory)
  • Initially validinvalid but is set to 0 on all
    entries.
  • Example of a page table snapshot.
  • During address translation, if validinvalid bit
    in page table entry is 0 ? page fault.

27
Page Table when some pages are not in main memory
0 A
1 B
2 C
3 D
4 E
5 F
6 G
7 H
Logical Memory Logical Memory
0
1
2
3
4 A
5
6 C
7
8
9 F

Physical Memory Physical Memory
0 4 v
1 i
2 6 v
3 i
4 i
5 9 v
6 i
7 i
28
Steps in Handling a Page Fault (Cont.)
  1. We check an internal table for this process, to
    determine whether the reference was a valid or
    invalid memory access.
  2. If the reference was invalid, we terminate
    process. If it was valid, but we have not yet
    brought in that page, we now page in the latter.
  3. We find a free frame.
  4. We schedule a disk operation to read the desired
    page into the newly allocated frame.
  5. When the disk read is complete, we modify the
    internal table kept with the process and the page
    table to indicate that the page is now in memory.
  6. We restart the instruction that was interrupted
    by the illegal address trap. The process can now
    access the page as though it had always been in
    memory.

29
What happens if there is 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.

30
Performance of Demand Paging
  • Effective Access Time (EAT) for a demand-paged
    memory.
  • Memory Access Time (ma) for most computers now
    ranges from 10 to 200 nanoseconds.
  • If there is no page fault, then EAT ma.
  • If there is page fault, then
  • EAT (1 p) x (ma) p x (page-fault time).
  • p the probability of a page fault (0 ? p ? 1),
  • we expect p to be close to zero ( a few page
    faults).
  • If p0 then no page faults, but if p1 then
    every reference
  • is a fault
  • If a page fault occurs, we must first read the
    relevant page from disk, and then access the
    desired word.

31
Performance of Demand Paging (Cont.)
  • We are faced with three major components of the
    page-fault service time
  • Service the page-fault interrupt.
  • Read in the page.
  • Restart the process.
  • A typical hard disk has
  • An average latency of 8 milliseconds.
  • A seek of 15 milliseconds.
  • A transfer time of 1 milliseconds.
  • Total paging time (8151) 24 milliseconds,
    including hardware and software time, but no
    queuing (wait) time.

32
Demand Paging Example 1
  • Assume an average page-fault service time of 25
    milliseconds (10-3), and a Memory Access Time of
    100 nanoseconds (10-9). Find the Effective Access
    Time?
  • Solution Effective Access Time (EAT)
  • (1 p) x (ma) p x (page fault time)
  • (1 p) x 100 p x 25,000,000
  • 100 100 x p 25,000,000 x p
  • 100 24,999,900 x p.
  • Note The Effective Access Time is directly
    proportional to the page-fault rate.

33
Page Replacement
  • Example Assume each process contains 10 pages
    and uses only 5 pages.
  • If we had 40 frames in physical memory then we
    can run 8 processes instead of 4 processes.
    Increasing the degree of multiprogramming.
  • If we run 6 processes (each of which is 10 pages
    in size), but uses only 5 pages. We have higher
    CPU utilization and throughput, also 10 frames to
    spare (i.e., 6x530 frames needed out of 40
    frames).
  • It is possible each process tries to use all 10
    of its pages, resulting in a need for 60 frames
    when only 40 are available.

34
Need for Page Replacement
0 H
1 Load M
2 J
3 M
Logical Memory for user 1 Logical Memory for user 1
0 monitor
1
2 D
3 H
4 Load M
5 J
6 A
7 E
Physical Memory Physical Memory
0 3 v
1 4 v
2 5 v
3 i
Page table for user 1 Page table for user 1
0 A
1 B
2 D
3 E
Logical Memory for user 2 Logical Memory for user 2
0 6 v
1 i
2 2 v
3 7 v
Page table for user 2 Page table for user 2
35
The Operating System has Several Options
  • Terminate user process.
  • Swap out a process, freeing all its frames, and
    reducing the level of multiprogramming.
  • Page replacement takes the following approach
  • If no frames is free, find one that is not
    currently being used and free it.
  • We can free a frame by writing its contents to
    swap space, and changing the page table to
    indicate that the page is no longer in memory.

36
Page Replacement
Swap out victim page

victim

Physical Memory

O i
F v

Page table Page table
37
Page Replacement (Cont.)
  • The page-fault service time is now modified to
    include page replacement
  • 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.

38
Page Replacement (Cont.)
  • Note If no frames are free, two page transfers
    (one out and one in) are required. This doubles
    the page-fault service time and will increase the
    effective access time accordingly.
  • This overhead can be reduced by the use of a
    modify (dirty) bit.
  • Each page or frame may have a modify bit
    associated with it in the hardware.
  • To implement demand paging, we must develop
  • Frame-allocation algorithm, to decide how many
    frames to allocate to each process.
  • Page-replacement algorithm, to select the frames
    that are to be replaced.

39
Page-Replacement Algorithms
  • We want a page replacement algorithm with the
    lowest page-fault rate.
  • We evaluate an algorithm by running it on a
    particular string of memory references (reference
    string) and computing the number of page faults
    on that string.
  • The string of memory references is called a
    reference string.
  • We can generate reference strings by tracing a
    given system and recording the address of each
    memory reference.
  • In our examples, the reference string is
  • 1, 2, 3, 4, 1, 2, 5, 1, 2, 3, 4, 5.

40
Page-Replacement Algorithms (Cont.)
  • Example If we trace a particular process, we
    might record the following address sequence
  • 0100, 0432, 0101, 0102, 0609, 0601, 0612
  • This is reduced to the following reference
    string 1, 4, 1, 6
  • As the number of frames available increases, the
    number of page faults will decrease.
  • From the above example If we had 3 or more
    frames, we would have only 3 faults, one fault
    for the first reference to each page. If we had
    only one frame, we would have a replacement with
    every reference resulting in 4 faults.

41
First-In-First-Out (FIFO) Algorithm
  • A FIFO algorithm associates with each page the
    time when that page was brought into memory.
  • When a page must be replaced, the oldest page is
    chosen.
  • Or we can use a FIFO queue to hold all pages in
    memory.
  • We replace the page at the head of the queue.
  • When a page is brought into memory, we insert it
    at the tail of the queue.


42
Example FIFO Algorithm
  • Reference string 1, 2, 3, 4, 1, 2, 5, 1, 2, 3,
    4, 5
  • Let 3 frames are initially empty (3 pages can be
    in memory at a time per process).
  • The first 3 references (1, 2, 3) cause page
    faults, and are brought into these empty frames.
  • 4 frames

43
Optimal (OPT) Algorithm
  • An optimal algorithm has the lowest page-fault
    rate of all algorithms.
  • An optimal algorithm will never suffer from
    Beladys anomaly.
  • Replace the page that will not be used for the
    longest period of time.
  • This algorithms guarantees the lowest possible
    page-fault rate for a fixed number of frames.
  • The optimal algorithm is difficult to implement,
    because it requires future knowledge of the
    reference string.
  • Similar situation with Shortest-Job-First in CPU
    scheduling.

44
Example OPT Algorithm
  • Initially 4 frames empty.
  • Reference String 1, 2, 3, 4, 1, 2, 5, 1, 2, 3,
    4, 5
  • How do you know this?
  • Used for measuring how well your algorithm
    performs.

45
Least Recently Used (LRU) Algorithm
  • The key distinction between FIFO and OPT
    algorithms is that FIFO uses the time when a page
    was brought into memory the OPT uses the time
    when a page is to be used (future).
  • LRU algorithm uses the time when a page has not
    been used for the longest period of time (Past).
  • LRU replacement associates with each page the
    time of that pages last use.

46
Example LRU Algorithm
  • Looking backward in time.
  • Reference string 1, 2, 3, 4, 1, 2, 5, 1, 2, 3,
    4, 5
  • Note Number of page faults (on same reference
    string) using
  • LRU is 8.
  • FIFO is 10.
  • OPT is 6.
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