Solution2 - Technische Informatik

Folie a: Name 

Operating Systems and Computer Networks 

Universität Duisburg-Essen 

Institut für Medientechnik and Software Engineering 

& Data Processing 2 

Exercise 2: File Systems 

Dipl.-Ing. Bogdan Marin 


Fakultät für Ingenieurwissenschaften 

Abteilung Elektro-und Informationstechnik 

-Technische Informatik- 

Exercise 


& Datenverarbeitung 2

Objectives 

• File System Concept 

• File Management 

• Mass Storage 

• Physical Concept 

• File Allocation Table & Unix File System 

• Management of Free Blocks 



Exercise 



File System Concepts 

What are Files? 

– Named collection of information stored on mass 

storage media 

– Logical storage units 

Two aspects: 

– User‘s view – User Interface 

– System Designer‘s view – Implementation 

User Interface results in two distinct notions: 

– Files: storing a collection of data 

– Directory Structure: organising & providing info abt 

files in system 



Exercise 



File Management 

Administration of mass storage media and the devices 

which control them 

• Creation and deletion of files 

• To map and find files onto the physical media 

• File Access: WHO can and HOW to 



Exercise 



Mass Storage 

Refers to various techniques and devices for storing 

large amounts of data 

Three essential requirements are: 

• Able to store large amount of information 

• Storage must be non-volatile or persistent 

• Stored information can be accessed concurrently 

Mass Storage vs. Memory 

• Memory is volatile 



Exercise 



Mass Storage Media 

Magnetic tapes: tape streamer 

• + : cheap, large storage capacity, backup purposes 

• - : Disadv: slow, sequential access 

Magnetic disks type 1: harddisk 

• + : fast, higher storage capacity 

• - : more expensive, not necessay portable, bulky 

Magnetic disks type 2: floppy disks, zip 

• + : portable disk, universal, cheap, direct access 

• - : slow, small storage capacity 

Optical disks: CD, DVD 

• + : very large storage capacity, portable disk, read-only disk/drives cheap 

• - : slower than harddisk, write disk drives expensive 



Exercise 



Physical Concept (1/3) 

Hard-disk drive 

Multiple platters 



Example : Inside the Hard Disk 

Controller electronic Connection & Vent 

hole 

Multiple arms & heads 

Platters, Arms & 

Heads 

High-speed linear 

motor 

Exercise 




• One or more metal platters rotate at 5400, 7200 or 10 

800 rpm 

• A mechanical arm pivots over the platters from the 

corner 

• At any given arm position, each of the heads can 

read/write a circular region called a track 

• All the tracks for a given arm position form a cylinder 

• Each track is divided into sectors (typically 512 bytes 

per sector) 



Exercise 




Storing the Data 

NoH : Number of heads to read/write 

NoT : Number of tracks 

NoS : Number of sectors per track 

NBS : Number of bytes per sector 



0 

1 

NoH-2 

NoH-1 

Heads 0..NoH-1 

Sectors 

0... NoS-1 

Tracks 

0 ... NoT-1 

Exercise 



Storage Capacity 

The storage capacity of a mass storage media (BV) 

BV = NoH * NoT * NoS * NBS 

NOTE : NBS is by default 512 bytes ! 



Exercise 



Storage Capacity 

Given a standard format 8-inch floppy disk with 2 

heads, 80 tracks, 18 sectors per track, four 128byte 

units per sector (512 B), 

what is ist storage capacity in kilobytes [KB]? 

• BV = 2 * 80 * 18 * (4*128) [B] 

• BV = 2 * 80 * 18 * 512 [B] 

• BV = 1474560 [B] 

• BV = 1440 [KB] 



Exercise 



File Allocation Table (FAT) 

•FAT = a table that OS uses to locate files on the disk. 

•FAT12 is the oldest type of FAT, which uses a 12-bit binary number 

to hold the cluster number. 

•cluster number is not the same as the sector number (the system 

designer can define the number of physical sectors in a logical 

cluster on a disk). 

•A cluster = group of sectors on the disk that have information in 

them. 

•A 4K cluster has 8 physical sectors (NBS) in it (512*8=4096). 

•Each Cluster has an entry in the FAT table. 



Exercise 



File Allocation Table 

The FAT file system comes in three versions for MS 

DOS: FAT12, FAT16 and FAT32. Calculate the: 

• Size of maximum disk partition for each of the FAT version 

in Table 1 

• Size of FAT12 





Exercise 



File Allocation Table 

Cluster Size [KB] 

0.5 

1 

2 

4 

8 

16 

32 

FAT12 



? 

? 

? 

? 

FAT16 

? 

? 

? 

? 

? 

FAT32 

Exercise 


& Datenverarbeitung 2 

? 

? 

? 

?

Max Disk Partition (1) 

• Given Cluster Size = 0.5 [KB] with FAT12 

Max disk partition 

= 2 12 * 512 [B] 

= 2 [MB] 

• Given Cluster Size = 2 [KB] with FAT16 


= 2 16 * 2 [KB] 

= 128 [MB] 

• Given Cluster Size = 4 [KB] with FAT32 


= 2 28 * 4 [KB] 

= 2 10 * 2 10 * 2 8 * 2 2 [KB] 

= 1 [TB] 



Exercise 



Max Disk Partition (2) 

Cluster FAT12 

Size [KB] [MB] 

0.5 0.5*212 =2 

1 1* 2 12 =4 



FAT16 

[MB] 

2 2* 2 12 =8 128 

FAT32 [TB] 

4 2 2 *2 12 =16 2 2 *2 16 =256 1 

8 2 3 *2 16 =512 2 

16 2 4 *2 16 =1024 2 

32 2 5 *2 16 =2048 2 

Partitions are 

limited to 2 TB 

because internally, 

the system keeps 

track of the 

partition size in 

512 Byte sectors 

using a 32 bit 

number: 512 Bytes 

* 2 32 = 2 TB 

Exercise 



Unix File System 

i-node 



A Typical Unix partition 

Directory entry 

Exercise 




• Assume disc blocks are 8K bytes and that disc addresses are 32 bits: 

– What size can be directly addressed from the information in the i-node? 

– What size of file requires a double indirect block? 

– What is the largest possible file? 

Answer: 

a) Disk blocks= 8K bytes 

i-node => 10 disk block adress 

8K*10= 80K can be accessed directly 



Exercise 




b) Single indirect: (address of disk block which itself contains the disk block addresses) 

One 8K block can address : 2 3 *2 8 =2 11 = 2048 addresses of 8K blocks 

Can access file size: 2048*8k=16M 

Double indirect: (address of disk block with the addresses of single indirect blocks) 

2048*2048= 4 194 304 (2 22 ) addresses of 8K blocks 

Can access file size: 4194304 *8k=32G 

c) Triple indirect: (address of disk block with the addresses of double indirect blocks) 

2048*2 22 =2 33 addresses of 8K blocks 

Can access file size: 2 33 *8k=64TB 

Largest file: 64TB + 32G + 16M + 80K 



Exercise 



Free Space Management 

How to keep track of free blocks? 

•Method 1: Linked List 

•Linked lists of disk blocks, each block has as many free disk 

block no. as will fit in 

•One slot needed for the pointer to the next block 

•Method 2: Bitmap 

•A disk with n blocks requires a bitmap with n bits 

•Free blocks 1, used blocks 0 (or viceversa) 



Bitmap vs. Linked List 

1 bit/block vs. 32bits/block 

Exercise 




The beginning of a free space bitmap looks like this after the disk 

partition is first formatted: 1000 0000 0000 0000 (the first block 

is used by the root directory). The system always searches for 

free blocks starting at the lowest numbered block, so after 

writing file A, which uses 6 blocks the bitmap looks like this: 

1111 1110 0000 0000. 

Show the bitmap after each of the following additional actions: 

File B is written using 5 blocks 

File A is deleted 

File C is written, using 8 blocks 

File B is deleted 



Exercise 




1000 0000 0000 0000 (begining) 

1111 1110 0000 0000 (file A is written ) 

1111 1111 1111 0000 (file B is written ) 

1000 0001 1111 0000 (file A is deleted ) 

1111 1111 1111 1100 (file C is written ) 

1111 1110 0000 1100 (file B is deleted ) 



Exercise 



Next Week: 

Memory Management .. 

Check the webpage for exercises! 

http://www.fb9dv.uni-duisburg.de/ti/en/education/ss06/dv2/main.html 



Exercise

Solution2 - Technische Informatik

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