Memory Management

Memory Management

Chapter 7

Memory Management

Subdividing memory to accommodate multiple processes

Memory needs to allocated efficiently to pack as many processes into memory as possible

Memory Management Requirements (1)

RelocationProtectionSharingLogical organizationPhysical organization

Memory Management Requirements (2)Relocation

programmer does not know where the program will be placed in memory when it is executed

while the program is executing, it may be swapped to disk and returned to main memory at a different location

the processor hardware and operating system software must be able to translate the memory references found in the code of the program into actual physical memory address,reflecting the current location of the program in main memory

Memory Management Requirements (3)Protection

processes should not be able to reference memory locations in another process without permission

because the location of a program in main memory is unknown, it is impossible to check absolute addresses in programs since the program could be relocated

all memory references generated by a process must be checked during execution to ensure that they refer only to the memory space allocated to that process.


Sharing any protection mechanism must have

the flexibility to allow several processes to access the same portion of memory

If a number of process are executing the same program, it is advantageous to allow each process (person) access to the same copy of the program rather than have their own separate copy

Memory Management Requirements (5)Logical Organization

memory is organized as a linear address space. While this organization closely mirrors the actual machine hardware, it does not correspond t the way in which programs are typically constructed.

most programs are written in modules different degrees of protection given to

modules (read-only, execute-only) Segmentation is a tool to satisfies these

requirement.


Physical Organization memory available for a program plus its

data may be insufficientoverlaying allows various modules to be

assigned the same region of memory

secondary memory cheaper, larger capacity, and permanent

Memory PartitioningFixed PartitioningDynamic PartitioningSimple PagingSimple SegmentationVirtual-Memory PagingVirtual-Memory Segmentation

Fixed PartitioningPartition available memory into regions with

fixed boundariesEqual-size partitions

any process whose size is less than or equal to the partition size can be loaded into an available partition

if all partitions are full, the operating system can swap a process out of a partition

a program may not fit in a partition. The programmer must design the program with overlays

Fixed Partitioning

Main memory use is inefficient. Any program, no matter how small, occupies an entire partition. This is called internal fragmentation.

8 M

8 M

8 M

8 M

8 MOperating System

Fixed Partitioning

Unequal-size partitions lessens the problem with equal-size

partitions Operating System8 M

12 M

8 M

8 M

6 M

4 M

2 M

Placement Algorithm with Partitions

Equal-size partitions because all partitions are of equal size, it

does not matter which partition is usedUnequal-size partitions

can assign each process to the smallest partition within which it will fit

queue for each partition processes are assigned in such a way as to

minimize wasted memory within a partition

One Process Queue per Partition

NewProcesses

OperatingSystem

One Process Queue per Partition

When its time to load a process into main memory the smallest available partition that will hold the process is selected Operating

System

NewProcesses

Dynamic Partitioning

Partitions are of variable length and number

Process is allocated exactly as much memory as required

Eventually get holes in the memory. This is called external fragmentation

Must use compaction to shift processes so they are contiguous and all free memory is in one block

Example Dynamic Partitioning

Operating System

128 K

896 K

Operating System

Process 1 320 K

576 K

Operating System

Process 1 320 K

Process 2 224 K

352 K


Operating System

Process 1 320 K

Process 2

Process 3

224 K

288 K

64 K

Operating System

Process 1 320 K

Process 3

224 K

288 K

64 K

Operating System

Process 1 320 K

Process 3 288 K

64 K

Process 4 128 K

96 K


Operating System

320 K

Process 3 288 K

64 K

Process 4 128 K

96 K

Operating System

Process 3 288 K

64 K

Process 4 128 K

96 K

Process 2 224 k

96 K

Dynamic Partitioning Placement Algorithm

Operating system must decide which free block to allocate to a process

Best-fit algorithm chooses the block that is closest in size to the

request worst performer overall since smallest block is found for process, the

smallest amount of fragmentation is left memory compaction must be done more often


First-fit algorithm starts scanning memory from the

beginning and chooses the first available block that is large enough.

fastest may have many process loaded in the

front end of memory that must be searched over when trying to find a free block

Dynamic Partitioning Placement AlgorithmNext-fit

starts scanning memory from the location of the last placement and chooses the next available block that is large enough

more often allocate a block of memory at the end of memory where the largest block is found

the largest block of memory is broken up into smaller blocks

compaction is required to obtain a large block at the end of memory


Lastallocatedblock (14K)

Before After

8K 8K

12K 12K

22K

18K

6K 6K

8K 8K

14K 14K

6K

2K

36K20K

Next Fit

Free block

Allocated block

Best Fit

First Fit

Why Buddy System? - Buddy Systems (1)

A fixed partitioning scheme limits the number of active processes and may use space inefficiently if there is a poor match between available partition size and process size

A dynamic partitioning scheme is more complex to maintain and includes the overhead of compaction.

Memory Blocks in a Buddy System - Buddy Systems (2)

In a buddy system, memory blocks are available of size 2k, L <= K <= U,

where:2L = smallest block that is allocated.2U = largest size block that is allocated.Generally, 2U is the size of the entire

memory available for allocation.

Allocation Method - Buddy Systems (3)

To begin, the entire space available for allocation is treated as a single block of size 2U. If a request of size that 2 U-1 < s <= 2U is made, then the entire block is allocated.

Otherwise, the block is split into two equal buddies of size 2 U-1. If 2 U-2 < s<= 2U-1, then the request is allocated to one of the two buddies.

Otherwise, one of the buddies is split in half again.This process continues until the smallest block greater

than or equal to s is generated and allocated to the request.

Remove Unallocated Blocks - Buddy Systems (4)

At any time, the buddy system maintains a list of holes (unallocated blocks) of each size 2 i.

A hole may be removed from the (i + 1) list by splitting it in half to create two buddies of size 2i in the i list.

Whenever a pair of buddies on the i list both become unallocated, they are removed from that list and coalesced into a single block on the

(i + 1) list.

Example of Buddy System- Buddy Systems (5)

1 Megabyte Block 1MRequest 100K (A) A=128K 128K 256K 512K

Request 240K (B) A=128K 128K B=256K 512K

Request 256K (D) A=128K 128K B=256K D=256K 256K

Release B A=128K 128K 256K D=256K 256K

Release A 512K D=256K 256K

Request 75K (E) E=128K 128K 256K D=256K 256K

Release E 512K D=256K 256K

Release D 1M

Tree Representation of Buddy System - Buddy Systems (6)

1M

512K

256K

128K

A=128K 128K B=256K D=256K 256K

Fibonacci Buddy System (1)

The Fibonacci sequence is defined as follow:

F0 = 0, F1 = 1, F n+2 = F n+1 + Fn (n>=0)

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, ….


Show the results of the following sequence in a figure using Fibonacci Buddy system:Request 60 (A);

Request 135 (B); Request 70 (C);

Return A; Request 20 (D);

Return B; Return C;

Return D.


377 Byte Block 377(144+233)Request 60 (A) 55 A= 89 233

Request 135 (B) 55 A=89 89 B=144

Request 70 (C) 55 A=89 C=89 B=144

Return A 144 C=89 B=144

Request 60 (D) 55 D=89 C=89 B=144

Return B 55 D=89 C=89 144

Return D 144 C=89 144

Return C 377


377

144 + 233

(55+89)+ (89+144)

128K

64K 55 D=89 C=89 144

Relocation

When program loaded into memory the actual (absolute) memory locations are determined

A process may occupy different partitions which means different absolute memory locations during execution (from swapping)

Compaction will also cause a program to occupy a different partition which means different absolute memory locations

AddressesLogical

reference to a memory location independent of the current assignment of data to memory

translation must be made to the physical address

Relative address expressed as a location relative

to some known point, usually the beginning of the program

Physical the absolute address or actual location

Hardware Support for Relocation

Interrupt tooperating system

Process image inmain memory

Relative address

Absoluteaddress

Process Control Block

Program

Data

Stack

Adder

Comparator

Base Register

Bounds Register

Registers Used during Execution

Base register starting address for the process

Bounds register ending location of the process

These values are set when the process is loaded and when the process is swapped in

Registers Used during Execution

The value of the base register is added to a relative address to produce an absolute address

The resulting address is compared with the value in the bounds register

If the address is not within bounds, an interrupt is generated to the operating system

Paging

Partition memory into small equal-size chunks and divide each process into the same size chunks

The chunks of a process are called pages and chunks of memory are called frames

Operating system maintains a page table for each process contains the frame location for each page in the

process memory address consist of a page number and

offset within the page

PagingFrameNumber

01

2

3

4

5

6

7

8

9

10

11

12

13

14

01

2

3

4

5

6

7

8

9

10

11

12

13

14

A.0

A.1

A.2

A.3

01

2

3

4

5

6

7

8

9

10

11

12

13

14

A.0

A.1

A.2

A.3

B.0

B.1

B.2

(a) (c)(b)

Paging

01

2

3

4

5

6

7

8

9

10

11

12

13

14

A.0

A.1

A.2

A.3

B.0

B.1

B.2

C.0

C.1

C.2

C.3

01

2

3

4

5

6

7

8

9

10

11

12

13

14

A.0

A.1

A.2

A.3

C.0

C.1

C.2

C.3

01

2

3

4

5

6

7

8

9

10

11

12

13

14

A.0

A.1

A.2

A.3

C.0

C.1

C.2

C.3

D.0

D.1

D.2

D.3

D.4

(d) (e) (f)

Page Tables for Example

0

1

2

3

Process A

0

1

2

3

0

1

2

Process B

---

---

---

0

1

2

3

4

0

1

2

3

Process C

7

8

9

10

4

5

6

11

12

Process D

Free Frame List

13

14

Data Structures for the Example at Time Epoch (f)

Logical Address of PagingRelative Address = 1502 Logical Address =User Process (2700 Bytes) Page# =1, Offset = 4780000010111011110 0000010111011110

Page 0 1024

Page 1 478

Page 3 Internal

Fragmentation(a) Partitioning (b) Paging (Page Size =

1K)

Segmentation

All segments of all programs do not have to be of the same length

There is a maximum segment lengthAddressing consist of two parts - a

segment number and an offsetSince segments are not equal,

segmentation is similar to dynamic partitioning

Logical Address of Segmentation

Relative Address = 1502 Logical Address =User Process (2700 Bytes) Segment# =1, Offset

= 7520000010111011110 0001001011110000

Segment 0 750 Bytes 752 Segment 1 1950 Bytes (a) Partitioning (b) Segmentation

Question 1 - Exercise/Home Work (1)

A 1 Megabyte block of memory is allocated using buddy system.

A. Show the results of the following sequence in a figure:Request 70K (A); Request 35K (B); Request 80K (C); Return A; Request 60K (D); Return B; Return C; Return D.

B. Show the binary tree representation following Return B.


The Fibonacci sequence is defined as follow:

F0 = 0, F1 = 1, F n+2 = F n+1 + Fn (n>=0)

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, ….

A. Could this sequence be used to establish a buddy system?

B. What would be the advantage of this system over the binary buddy system?


C. Show the results of the following sequence in a figure using Fibonacci buddy system:Request 70 (A); Request 35 (B); Request 80 (C); Return A; Request 60 (D); Return B; Return C; Return D.

D. Show the binary tree representation following Return B.

Memory Management

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