ECE668 .1

Csaba Andras Moritz

UNIVERSITY OF MASSACHUSETTSDept. of Electrical & Computer

Engineering

Computer Architecture ECE 668

Review of MMU and Virtual Memory

ECE668 .2

Background

Program must be brought into memory and placed within a process for it to be run.Assume (for the time being) that the program must fit in memory or the program itself needs to deal with when not. This is different from virtual memory (will be covering later) when this problem (i.e. fit or not in memory) will be managed by the OS!

User programs go through several steps before being run. Let us review these steps….

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Logical vs. Physical Address Space

The concept of a logical address space that is bound to a separate physical address space is central to proper memory management. Logical address – generated by the CPU; also

referred to as virtual address. Physical address – address seen by the memory

unit.

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Memory-Management Unit (MMU)

Hardware device that maps virtual address to physical address.

In a simple 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.

There are many different methods for accomplishing such a mapping, as we will see later.

The user program deals with logical addresses; it never sees the real physical addresses.

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Dynamic relocation using a relocation register

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Paging Logical address space of a process can

be non-contiguous; process is allocated physical memory whenever the latter is available. Divide physical memory into fixed-sized blocks

called frames (size is power of 2, between 512 bytes and 8192 bytes).

Divide logical memory into blocks of same size called pages.

Keep track of all free frames. To run a program of size n pages, need to find n

free frames and load program. Set up a page table to translate logical to

physical addresses. Internal fragmentation possible.

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Address Translation Scheme

Address generated by CPU is divided into: Page number (p) – used as an index into a page

table which contains base address of each page in physical memory.

Page offset (d) – combined with base address to define the physical memory address that is sent to the memory unit.

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Address Translation Architecture

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Paging Example

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Paging Example

Address in physical memory of 32 bytesFrame size of 4 bytes

Address in logical memory Page size of 4 bytes

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Free Frames

Before allocation After allocation

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Implementation of Page Table

Page table is kept in main memory. Page-table base register (PTBR) points to the

page table. Page-table length register (PRLR) indicates

size of the page table. In this scheme every data/instruction access

requires two memory accesses. One for the page table and one for the data/instruction!

The two memory access problem can be solved by the use of a special fast-lookup hardware cache called translation look-aside buffers (TLBs)

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Paging Hardware With TLB

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Memory Protection

Memory protection implemented by associating protection bit (called valid bit) with each frame.

Valid-invalid bit attached to each entry in the page table: “valid” indicates that the associated page is in

the process’ logical address space, and is thus a legal page.

“invalid” indicates that the page is not in the process’ logical address space.

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Valid (v) or Invalid (i) Bit In A Page Table

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Page Table Structure

Hierarchical Paging

Hashed Page Tables

Inverted Page Tables

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Two-Level Paging Example A logical address (on 32-bit machine with 4K page size) is

divided into: a page number consisting of 20 bits. a page offset consisting of 12 bits.

Since the page table is paged, the page number is further divided into: a 10-bit page number. a 10-bit page offset.

Thus, a logical address is as follows:

where pi is an index into the outer page table, and p2 is the displacement within the page of the outer page table.

page number page offset

pi p2 d

10 10 12

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Two-Level Page-Table Scheme

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Address-Translation Scheme

Address-translation scheme for a two-level 32-bit paging architecture

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Hashed Page Tables

Common in address spaces > 32 bits.

The virtual page number is hashed into a page table. This page table contains a chain of elements hashing to the same location.

Virtual page numbers are compared in this chain searching for a match. If a match is found, the corresponding physical frame is extracted.

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Hashed Page Table

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Inverted Page Table

One entry for each real page of memory. Entry consists of the virtual address of

the page stored in that real memory location, with information about the process that owns that page.

Decreases memory needed to store each page table, but increases time needed to search the table when a page reference occurs.

Use hash table to limit the search to one — or at most a few — page-table entries.

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Inverted Page Table Architecture

ECE668 .24

Segmentation

Memory-management scheme that supports user view of memory.

A program is a collection of segments. A segment is a logical unit such as:

main program,

procedure,

function,

method,

object,

local variables, global variables,

common block,

stack,

symbol table, arrays

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User’s View of a Program

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Logical View of Segmentation

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user space physical memory space

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Segmentation Architecture

Logical address consists of a two tuple:

<segment-number, offset>, Segment table – maps two-dimensional

physical addresses; each table entry has: base – contains the starting physical address where the

segments reside in memory. limit – specifies the length of the segment.

Segment-table base register (STBR) points to the segment table’s location in memory.

Segment-table length register (STLR) indicates number of segments used by a program;

segment number s is legal if s < STLR.

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Segmentation Architecture (Cont.)

Relocation. dynamic by segment table

Sharing. shared segments same segment number

Allocation. first fit/best fit external fragmentation

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Segmentation Architecture (Cont.)

Protection. With each entry in segment table associate: validation bit = 0 illegal segment read/write/execute privileges

Protection bits associated with segments; code sharing occurs at segment level.

Since segments vary in length, memory allocation is a dynamic storage-allocation problem.

A segmentation example is shown in the following diagram

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Segmentation Hardware

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Example of Segmentation

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Sharing of Segments

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Segmentation with Paging – MULTICS

The MULTICS system solved problems of external fragmentation and lengthy search times by paging the segments.

Solution differs from pure segmentation in that the segment-table entry contains not the base address of the segment, but rather the base address of a page table for this segment.

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MULTICS Address Translation Scheme

ECE668 .35

Segmentation with Paging – Intel 386

As shown in the following diagram, the Intel 386 uses segmentation with paging for memory management with a two-level paging scheme.

ECE668 .36

Intel 30386 Address Translation

ECE668 .37

Virtual Memory - Background

Virtual memory – separation of user logical memory from physical memory (even size does not need to match). Only part of the program needs to be in memory for

execution. Logical address space can therefore be much larger

than physical 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

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Virtual Memory That is Larger Than Physical Memory

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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

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Transfer of a Paged Memory to Contiguous Disk Space

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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. Example of a page table snapshot.

During address translation, if valid–invalid bit in page table entry is 0 page fault.

111

1

0

00

Frame # valid-invalid bit

page table

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Page Table When Some Pages Are Not in Main Memory

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Page Fault If there is ever a reference to a page, first reference

will trap to OS page fault (BDW can we get a page fault in a non-demand based paging environment?)

OS looks at another table to decide if what happened is an: Invalid reference abort. Or page just not in memory.

What happens during a fault? Get empty frame. Swap page into frame. Reset tables, validation bit = 1.

Restart instruction: Least Recently Used But what happens during a block move instruction –(e.g., in

PDP11 was such an instruction)? Page fault in the middle of transfer… things get complicated as some memory locations are already changed.

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Steps in Handling a Page Fault

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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.

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Architectural Support for VM

From Jacob et al Micro 2005 paper

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MIPS R10000

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Alpha 21164

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Power PC604

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PA-RISC

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Ultra Sparc I

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Pentium II

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Backup Slides

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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) x memory access

+ p (page fault overhead

+ swap page out

+ swap page in

+ restart overhead)

PageFault

Service time

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Demand Paging Example

Memory access time = 100 nsec = 0.1usec Page fault service time = 25 msec =

25,000usec

EAT (usec) = (1 – p) x 0.1 + p (25,000)

Assuming we have p=0.1 (10% of the case page fault)

EAT(usec) = 0.9 x 0.1 + 0.1* 25,000 ~2500usec=2.5msec

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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 – 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.

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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. Read 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 memory references (reference string) and computing 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.

ECE668 .61

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

FIFO Replacement – Belady’s Anomaly more frames less page faults?

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FIFO Page Replacement

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FIFO Illustrating Belady’s 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.

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6 page faults

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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

referenced through this entry, copy the clock into the counter.

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

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LRU Page Replacement

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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 = 1. then:» 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

Idea: Keep a counter of the number of references that

have been made to each page.

LFU (Least Frequently Used) Algorithm: replaces page with smallest count. Argument is that an actively used bae should have a

large reference count

MFU (Most Frequently Used) Algorithm: replaces page with the largest count based on the argument that the page with the

smallest count was probably just brought in and has yet to be used.

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Allocation of Frames How are we allocating the frames to multiple processes? (For example, simplest is to partition memory equally between processes, but are there any constraints that prevent us doing this….?)

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 in general. fixed allocation (equal or proportional) priority allocation (use priorities)

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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.

mSs

pa

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ps

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for allocation

frames of number total

process of size

5964137127

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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.

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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.

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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.

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Thrashing

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

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Locality In A Memory-Reference Pattern

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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.

D = WSSi total demand frames if D > m Thrashing Policy if D > m, then suspend one of the

processes.

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Working-set model

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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.

Why is this not completely accurate? Improvement = 10 bits and interrupt

every 1000 time units.

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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.

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Other Considerations

Prepaging

Page size selection fragmentation table size I/O overhead locality

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Other Considerations (Cont.)

TLB Reach - The amount of memory accessible from the TLB.

TLB Reach = (TLB Size) X (Page Size)

Ideally, the working set of each process is stored in the TLB. Otherwise there is a high degree of page faults.

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Increasing the Size of the TLB

Increase the Page Size. This may lead to an increase in fragmentation as not all applications require a large page size.

Provide Multiple Page Sizes. This allows applications that require larger page sizes the opportunity to use them without an increase in fragmentation.

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Other Considerations (Cont.)

Program structure int A[][] = new int[1024][1024]; Each row is stored in one page Program 1 for (j = 0; j < A.length; j++)

for (i = 0; i < A.length; i++)

A[i,j] = 0;1024 x 1024 page faults

Program 2 for (i = 0; i < A.length; i++)for (j = 0; j < A.length;

j++)A[i,j] = 0;

1024 page faults