Tiểu luận môn Giải thuật nâng cao chủ đề Virtual Memory - Pdf 23

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Virtual Memory
Nhóm 5
11070459 - Thái Tiểu Minh
11070460 - Nguyễn Kim Ngân
13070249 - Lê Minh Nam
13070250 - Trần Đức Nghĩa
13070251 - Phạm Ích Trí Nhân
13070247 - Trần Thị Mi
1
Contents
E. Examples
D. Protection
C. Segmentation
B. Replacement
A. Introduction
2
A. Introduction

Reasons for using virtual memory

Share main memory

Simplify memory management

Provide protection

What is a virtual memory?

Memory Hierarchy


 Simplify memory management
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1. Reason 3

Virtual memory :

One process can’t interfere with another

Because they operate in different address spaces

User process cannot access privileged information

Different sections of address spaces have different permissions
 Provide protection
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2. Memory Hierarchy
7
CPU
CPU
regs
regs
Cache
Memory
Memory
Disk
Disk
size:
speed:
$/Mbyte:
line size:

Controlled by operation system

Auto overlay

(+) Virtual continuous address space

(+) Size ~ disk size, speed ~ main mem speed

(+) Programmers don’t care about mem size  run more larger processes.
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3. Virtual Memory vs Cache
10
Contents
E. Examples
D. Protection
C. Segmentation
B. Replacement
A. Introduction
11
B. Replacement

Organization  Fully assosiative

Replacement  LRU

Write policy?  Write back

TLB

What is TLB?

Benefit for mem access time?
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4. TLB
15
Processor
TLB Cache
Main
Memory
miss
hit
data
hit
miss
Disk
Memory
OS Fault
Handler
page fault/
protection violation
Page
Table
data
virtual
addr.
physical
addr.
Page hit
4. Speeding up Translation with a TLB

“Translation Lookaside Buffer” (TLB)

TLB hit
TLB
Cache
.

4. Address Translation With a TLB
Contents
E. Examples
D. Protection
C. Segmentation
B. Replacement
A. Introduction
19
C. Segmentation

Classes :

Paging

Segmentation

Paged segmentation

Looking a block

Paging

Segmentation

Paged segmentation

Disk
Virtual
Addresses
Physical
Addresses
1. Page Table
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Valid bit :
-
1 : page in main mem
-
0 : page in disk
8 pages
4 page frames
= 4 pages
2. Segmentation

Virtual memory = a set of segments

Identify : segment number, size

Logical address :

Segment offset ≤ Segment size

Main mem = segments (+ holes)
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Segment number Seg offset
disk
mem

 10, 4, 8, 18, 7, 9, 12  10, 4, 8, 8, 7, 9, 12  10, 4, 8, 8, 7, 9, 3

Next fit : 20KB – 18KB – 9KB
 10, 4, 8, 18, 7, 9, 12  10, 4, 8, 8, 7, 9, 12  10, 4, 8, 8, 7, 12
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