Virtual Memory 2.0

8/10/2019 Virtual Memory 2.0

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Course 10: Virtual Memory


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9.2 Silberschatz, Galvin and Gagne 2008Operating System Concepts 8th Edition

Course 9: Review

Memory management concept: a logicaladdress space is bound to a separatephysical address space

Logical address generated by theCPU; also referred to as virtualaddress

Physical address address seen bythe memory unit

Logical and physical addresses are thesame in compile-time and load-timeaddress-binding schemes; logical (virtual)

and physical addresses differ in execution-time address-binding scheme


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Course 9:

Review

-

MMU/ classic model

MMU - Hardware devicethat maps virtual tophysical address

In classic MMU scheme,the value in the relocationregister is added to everyaddress generated by auser process at the time itis sent to memory

The user program dealswith logicaladdresses; it

never sees the realphysical addresses


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Course 9:

Review

-

Paging

Page table is kept in mainmemory

In this scheme everydata/instruction accessrequires two memoryaccesses. One for thepage table and one forthe data/instruction.

The two memory accessproblem can be solved bythe use of a special fast-

lookup hardware cachecalled associativememory or translationlook-aside buffers(TLBs)


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Course 9:

Review

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Paging


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Course 9:

Review

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Segmentation

Logical address consistsof a two tuple: ,

Segment table mapstwo-dimensional physicaladdresses; each tableentry has:

base contains thestarting physicaladdress where thesegments reside in

memory limit specifies the

length of the segment


7/529.7 Silberschatz, Galvin and Gagne 2008Operating System Concepts 8th Edition

Course 9: Review - Segmentation

A reference to byte 53 ofsegment 2 is mapped ontolocation 4300 + 53 = 4353

A reference to segment 3rdbyte 852, is mapped to 3200(the base of segment 3) +852 = 4052

A reference to byte 1222 ofsegment 0 would result in atrap to the operating system,as this segment is only 1,000bytes long.



Chapter 10: Virtual Memory

Background

Demand Paging

Copy-on-Write

Page Replacement

Allocation of Frames



Objectives

To describe the benefits of a virtual memory system

To explain the concepts of demand paging, page-replacementalgorithms, and allocation of page frames



Background



Background

Virtual memory separation of user logical memory from physicalmemory.

Only part of the program needs to be in memory for execution

Logical address space can therefore be much larger thanphysical 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



Virtual Memory That is Larger Than Physical Memory



Shared Library Using Virtual Memory



Demand paging



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

Lazy swapper never swaps a page into memory unless page willbe needed

Swapper that deals with pages is a pager



Transfer of a Paged Memory to Contiguous Disk Space



Valid-Invalid Bit

With each page table entry a validinvalid bit is associated

(v in-memory, i not-in-memory) Initially validinvalid bit is set to i on all entries

Example of a page table snapshot:

During address translation, if validinvalid bit in page table entry

is I page fault

v

vv

v

i

i

i

.

Frame # valid-invalid bit

page table



Page Table When Some Pages Are Not in Main Memory



Page Fault

If there is a reference to a page, first reference to thatpage will trap to operating system:

page fault

1. Operating system looks at another table to decide:

Invalid reference abort

Just not in memory2. Get empty frame

3. Swap page into frame

4. Reset tables

5. Set validation bit = v

6. Restart the instruction that caused the page fault



Steps in Handling a Page Fault



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 = (1p) x memory access+ p (page fault overhead

+ swap page out

+ swap page in

+ restart overhead

)



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!!


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Process creation copy on write


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Copy-on-Write

Copy-on-Write (COW) allows both parent and child processes toinitially share the same pages in memory

If either process modifies a shared page, only then is the pagecopied

COW allows more efficient process creation as only modifiedpages are copied

Free pages are allocated from a pool of zeroed-out pages


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Before Process 1 Modifies Page C


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After Process 1 Modifies Page C


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Page Replacement


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What happens if there is no free frame?

Page replacement find some page in memory, but notreally in use, swap it out

algorithm

performance want an algorithm which will result inminimum number of page faults

Same page may be brought into memory several times


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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 onlymodified pages are written to disk

Page replacement completes separation between logical memoryand physical memory large virtual memory can be provided on asmaller physical memory


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Need For Page Replacement


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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

3. Bring the desired page into the (newly) free frame;update the page and frame tables

4. Restart the process


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Page Replacement


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Page Replacement Algorithms

Want lowest page-fault rate

Evaluate algorithm by running it on a particular string of memoryreferences (reference string) and computing the number of pagefaults on that string


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Graph of Page Faults Versus The Number of Frames


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First-In-First-Out (FIFO) 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)

4 frames

Beladys Anomaly: more frames more page faults

1

2

3

1

2

3

4

1

2

5

3

4

9 page faults

1

2

3

1

2

3

5

1

2

4

5 10 page faults

44 3


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FIFO Page Replacement


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FIFO Illustrating Beladys Anomaly


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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

How do you know this? Used for measuring how well your algorithm performs

1

2

3

4

6 page faults

4 5


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Optimal Page Replacement


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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 page is referencedthrough this entry, copy the clock into the counter

When a page needs to be changed, look at the counters todetermine which are to change

5

2

4

3

1

2

3

4

1

2

5

4

1

2

5

3

1

2

4

3


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LRU Algorithm (Cont.)

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

No search for replacement


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Use Of A Stack to Record The Most Recent Page References


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LRU Approximation Algorithms

Reference bit

With each page associate a bit, initially = 0

When page is referenced bit set to 1

Replace the one which is 0 (if one exists)

We do not know the order, however

Second chance

Need reference bit

Clock replacement

If page to be replaced (in clock order) has reference bit = 1then:

set reference bit 0

leave page in memory

replace next page (in clock order), subject to same rules


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Second-Chance (clock) Page-Replacement Algorithm


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Counting Algorithms

Keep a counter of the number of references that have been

made to each page

LFU Algorithm: replaces page with smallest count

MFU Algorithm: based on the argument that the page with

the smallest count was probably just brought in and has yetto be used


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Allocation of frames


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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

i i


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Fixed Allocation

Equal allocation For example, if there are 100 frames and 5

processes, give each process 20 frames.

Proportional allocation Allocate according to the size of process

mS

spa

m

sS

ps

iii

i

ii

forallocation

framesofnumbertotal

processofsize

5964137

127

564137

10

127

10

64

2

1

2

a

a

s

s

m

i

P i i All i


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Priority Allocation

Use a proportional allocation scheme using priorities ratherthan size

If process Pigenerates a page fault,

select for replacement one of its frames

select for replacement a frame from a process withlower priority number

Gl b l L l All i


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Global vs. Local Allocation

Global replacement process selects a replacementframe from the set of all frames; one process can take aframe from another

Local replacement each process selects from only itsown set of allocated frames


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End of Course 10

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Virtual Memory 2.0