24191 os virtualmemory final

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MEMORY MANAGEMENT:

VIRTUAL MEMORYPrepared by: EJ Antonio

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Introduction:

In a very simple operating system, each process is given itsown segment of RAM for its own use. Though this systemis straightforward and easy to implement, it brings upsome issues.

How does the operating system decide how much memory toallocate to each process?What if the total memory required by active processesexceeds the amount of RAM available?How can the system prevent a process from reading oraltering the memory supposedly reserved for other processes?

To address these issues, and to address several others thatwe'll discuss later, modern operating systems use virtualmemory ..

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Basics

The idea of virtual memory is to create a virtual addressspace that doesn't correspond to actual addresses inRAM. The system stores the official copy of memory ondisk and caches only the most frequently used data in

RAM.VM is divided into equal pages . (Usually 4K in size)MM is divided into equal Page FramesThe system also maintains a page table , stored in RAM,which is an array of entries, one for each page, storinginformation about the page.

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TERMSInput Queue

Collection of jobs waiting to be brought into the Memory Unit forexecution

Logical AddressAddress generated by the CPU

Physical AddressAddress generated by the MU

Dynamic LoadingLoading of job/process is stopped until called

Dynamic LinkingLinking is postponed until executed

SwappingTransferring of data from MU into temporary storage for the use of differentprocess/job

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Virtual Memory Schemes

Types of Virtual Memory Schemes:PagingDemand PagingSegmentation

Segmentation / Demand Paging Allocation

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Paged Memory AllocationBefore a job is loaded into memory, it is divided into parts called pages that will be loaded into memory locations called pageframes.Paged memory allocation is based on the concept of dividing eachincoming job into pages of equal size.

SectorsThe sections of a Virtual Memory (disk)

Page FramesSections of main memory are called page frames.

The scheme works quite efficiently when the pages, sectors, andpage frames are all the same size. The exact size (the number ofbytes that can be stored in each of them) is usually determined bythe disks sector size.

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Paging Requirements

1.Job Table (JT) - 2 entries for each active job. Job number and job sizeDynamic grows/shrinks as jobs loaded/completed.

2.Page Map Table (PMT) - 1 entry per page.Page number & corresponding page frame memory address.Page numbers are sequential (Page 0, Page 1 )

3.Memory Map Table (MMT) - 1 entry for each page frame.Location & free/busy status.

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Example:

1. Job 1 = 350 bytes longPage size = 100 bytes

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Displacement / Offset

The displacement , or offset , of a byte is the factor used tolocate that byte within its page frame. It is a relative factor.In the simplified example, bytes 0, 100, 200, and 300 are thefirst bytes for pages 0, 1, 2, and 3, respectively, so each has adisplacement of zero.Likewise, if the operating system needs to access byte 214, itcan first go to page 2 and then go to byte 14 (the fifteenthline).The first byte of each page has a displacement of zero, and thelast byte, has a displacement of 99. So once the operatingsystem finds the right page, it can access the correct bytesusing its relative position within its page

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Address CalculationThe operating system uses an algorithm to calculate the page anddisplacement; it is a simple arithmetic calculation.To find the address of a given program instruction, the bytenumber is divided by the page size, keeping the remainder as aninteger.The resulting quotient is the page number, and the remainder isthe displacement within that page. When it is set up as a longdivision problem, it looks like this:

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Example:

1. For example, if we use 100 bytes as the page size, find thepage number and the displacement of byte 214.

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Example:

2. Lets try another example with a more common pagesize of 256 bytes. Say we are seeking the location of byte384.

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Locating a Line in the Memory Address Resolution / Address TranslationProcedure:

1. Determine the page no. and the displacement by thealgorithm discussed.

2. Refer to jobs PMT to find the corresponding pageframe.

3. Get the address of the beginning of the page frame bymultiplying the page frame no. by the page frame size.ADDR_PAGE_FRAME = PAGE_FRAME_NUM * PAGE_SIZE

4. Finally add the displacement to the starting address of

the page frame.INSTR_ADDR_IN_MEM = ADDR_PAGE_FRAME + DISPL

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Example:

1. Page size = 512 bytesPage Frame size = 512 bytesFind the exact location of line byte 518.

Page Map Table:

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Pros and Cons of Paged Memory Allocation

Allows jobs to be allocated in non-contiguous memorylocations.Memory used more efficiently; more jobs can fit.

Size of page is crucial (not too small, not too large).

Increased overhead occurs.Reduces, but does not eliminate, internal fragmentation.

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DEMAND PAGINGBring a page into memory only when it is needed, so lessI/O & memory needed.

Faster response.

Takes advantage that programs are written sequentially sonot all pages are necessary at once. For example:

User-written error handling modules.Mutually exclusive modules.Certain program options are either mutually exclusive or not

always accessible.Many tables assigned fixed amount of address space eventhough only a fraction of table is actually used.

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DEMAND PAGING

Demand paging made virtual memory widely available.Can give appearance of an almost-infinite or nonfinite amountof physical memory.

Requires use of a high-speed direct access storage devicethat can work directly with CPU.

How and when the pages are passed (or swapped)depends on predefined policies that determine when tomake room for needed pages and how to do so.

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Table Requirements in Demand Paging Job Table.Page Map Table (with 3 new fields).1. Determines if requested page is already in memory.2. Determines if page contents have been modified.3. Determines if the page has been referenced recently.

Used to determine which pages should remain in main memory andwhich should be swapped out.

Memory Map Table.

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Page StatusBit

ReferencedBit

ModifiedBit

PageFrame

0 1 1 1 5

1 1 0 0 92 1 0 0 7

3 1 1 0 12


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Page Statusbit

Referencedbit

Modifiedbit

Page frame

0 1 1 1 5

1 1 0 0 9

2 1 0 0 7

3 1 1 0 12


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Example:

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Trashing | Page FaultTrashing an excessive amount of page swapping back andforth between main memory and secondary storage.

Operation becomes inefficient.Caused when a page is removed from memory but is calledback shortly thereafter.Can occur across jobs, when a large number of jobs arevying for a relatively few number of free pages.Can happen within a job (e.g., in loops that cross pageboundaries).

Page fault a failure to find a page in memory.

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Page Replacement Policies

Policy that selects page to be removed iscrucial to system efficiency.Selection of algorithm is critical.

Types:First-in first-out (FIFO) policyLeast-recently-used (LRU) policyMost recently used (MRU) policy

Least frequently used (LFU) policyOptimal PolicyHardware Matrix LRU

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First-In First-Out (FIFO)The best page to be replaced when the main memory isfull is the first page that was allocated in the memory.

Example: Reference String: ABC AED CAB ECBThree (3) Page Frame Allocation

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A B C A E D C A B E C B


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Least Recently Used (LRU)In this policy, the least recently used page is the page tobe removed or displaced from the memory, when thememory is full.


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Most Recently Used (MRU)In this policy, the most recently used page is the page tobe removed or displaced from the memory, when thememory is full.


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Least Frequently Used (LFU)This technique uses counter, the least frequently usedpage is the page to be replaced when the main memory isfull


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Optimal PolicyThe Optimal policy selects for replacement the page that will not beused for longest period of time.Impossible to implement (need to know the future) but serves as astandard to compare with the other algorithms.


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Hardware Matrix Implementation (LRU)Example: Pages are referenced in the order 0, 1, 2, 3, 2, 1, 0, 3, 2, 3Four (4) Page Frame Allocation

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Segmented Memory AllocationBased on common practice by programmers of structuringtheir programs in modules ( logical groupings of code).

A segment is a logical unit such as: main program,subroutine, procedure, function, local variables, globalvariables, common block, stack, symbol table, or array.

Main memory is not divided into page frames because sizeof each segment is different.

Memory is allocated dynamically.

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Segment Map Table (SMT)When a program is compiled, segments are set upaccording to programs structural modules.

Each segment is numbered and a

Segment Map Table (SMT)is generated for each job.

Contains segment numbers,their lengths, access rights, status, and(when each is loaded into memory)its location in memory.

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Tables Used in Segmentation

Memory Manager needs to track segments in memory:1. Job Table (JT)

Lists every job in process (one for whole system).

2. Segment Map Table (SMT)

Lists details about each segment (one for each job).

3. Memory Map Table (MMT)Monitors allocation of main memory (one for whole system).

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Example1:Segment Map Table track for each segment for Job1.

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Example2:Use segmentation to allocate each segment in the MM.

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Pros and Cons of SegmentationNeed for Compaction.Memory is allocated dynamically.External fragmentation.

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Segmented/Demand Paged Memory Allocation

It is a combination of segmentation and demand paging,and it offers the logical benefits of segmentation, as wellas the physical benefits of paging.This allocation scheme doesnt keep each segment as a

single contiguous unit but subdivides it into pages of equalsize, smaller than most segments, and more easilymanipulated than whole segments. Therefore, many of theproblems of segmentation (compaction, external

fragmentation, and secondary storage handling) areremoved because the pages are of fixed length.

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Tables used in S/DP Memory Allocation1. Job Table lists every job in process (one for whole

system).2. Segment Map Table lists details about each segment (one

for each job).

E.g., protection data, access data.3. Page Map Table lists details about every page (one for

each segment).E.g., status, modified, and referenced bits .

4. Memory Map Table monitors allocation of page framesin main memory (one for whole system).

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Memory LocationTo access a location in memory, the system must locatethe address, which is composed of three entries: segmentnumber, page number within that segment, anddisplacement within that page. It is a three-dimensionaladdressing scheme:SEGMENT_NO. & PAGE_NO. & DISPLACEMENT

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Example;

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Pros & Cons of S/DP Memory AllocationOverhead required for the extra tablesTime required to reference segment table and page table.Logical benefits of segmentation.Physical benefits of paging

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Virtual Memory (VM)Even though only a portion of each program is stored inmemory, virtual memory gives appearance that programsare being completely loaded in main memory during theirentire processing time.

Shared programs and subroutines are loaded ondemand, reducing storage requirements of mainmemory.

VM is implemented through demand paging andsegmentation schemes.

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Virtual Paging versus Virtual SegmentationVIRTUAL MEMORY WITH

PAGINGVIRTUAL MEMORY WITH

SEGMENTATION

Allows internal fragmentationwithin page frames

Doesnt allow internalfragmentation

Doesnt allow externalfragmentation Allows external fragmentation

Programs are divided intoequal-sized pages

Programs are divided into unequal-sized segments

Absolute address calculatedusing page number anddisplacement

Absolute address calculated usingsegment number and displacement

Requires PMT Requires SMT

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The Virtual Page Number through


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The Virtual Page Number throughExamples

For some examples of this, provided are some virtualaddresses with the page size of the system. Then, tell whatthe page is, and what the page offset is.

1. Virtual address 0x000A502F, with page size of 4K bytes.

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The Virtual Page Number throughExamples2. Virtual address 0x000A502F, with page size of 2K bytes.

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The Virtual Page Number throughExamples3. Virtual address 0x009B18A0, with page size of 4K bytes.

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The Virtual Page Number throughExamples4. Virtual address 0x009B18A0, with page size of 2K bytes.

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Advantage of VMWorks well in a multiprogramming environment becausemost programs spend a lot of time waiting.

Jobs size is no longer restricted to the size of main memory(or the free space within main memory).Memory is used more efficiently.

Allows an unlimited amount of multiprogramming.Eliminates external fragmentation when used with paging andeliminates internal fragmentation when used withsegmentation.

Allows a program to be loaded multiple times occupying adifferent memory location each time.Allows the sharing of code and data.Facilitates dynamic linking of program segments.

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Disadvantage of VMIncreased processor hardware costs.

Increased software complexity to prevent thrashing.

24191 os virtualmemory final

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