File Systems Lab

8/13/2019 File Systems Lab

1/19

Operating Systems

6 File SystemsLab notes

Course lectured by Prof. Gabriel Kuper

Lab assist. Ilya Zaihrayeu

http://www.dit.unitn.it/~ilya/os.htm


2/19


3/19

Ilya Zaihrayeu Operating Systems Course 3

Recall: file allocation strategies

contiguous

linked

indexed


4/19


File allocation strategies, contd

A:Operations: read blocks + write blocks + write the file control

block

0r + 1w = 199r + 2w = 1010r + 1w = 1f)

0r + 1w = 150r + 1w = 51

51r + 1w = 52

49r + 50w = 99

50r + 51w = 101

e)

0r + 1w = 11r + 1w = 299r + 100w =199d)

0r + 2w = 21r + 3w = 40r + 2w = 2c)0r + 2w = 250r + 2w = 5250r + 52w = 102b)

0r + 2w = 20r + 2w = 2100r + 102w=202a)

IndexedLinkedContiguous


5/19


File allocation strategies-2

Q: One way to use contiguous allocation of the diskand not suffer from holes is to compact the diskevery time a file is removed.

Since all files are contiguous, copying a file requires aseek and rotational delay to read the file, followedby the transfer at full speed. Writing the file backrequires the same work.

a) Assuming a seek time of 5 msec, a rotational delayof 4 msec, a transfer rate of 8 MB/sec, and anaverage file size of 8 KB, how long does it take toread a file into main memory then write it back to the

disk at a new location?b) Using these numbers, how long would it take to

compact half of a 16-GB disk?


6/19


File allocation strategies-2, contd

A:

a) Time to read a file = seek time + latency time

+ read time = 5 msec + 4 msec + 8 KB / (8 *1024 KB / 1 * 1000 msec) 0.98 msec

Therefore read + write = 2 * 0.98 msec = 1.96

msecb) Half of a 16-GB disk is 8 GB = 8 388 608 KB

That is approximately 1 048 576 files. Each file

takes 1.96 msec, thus the whole processwill take about 2055 sec, i.e. about 34 min


7/19


Free disk space management

Q: Free disk space can be kept track of using a

free list or a bit map. Disk addresses require

D bits. For a disk with B blocks, F of whichare free, state the condition under which the

free list uses less space than the bit map.

For D having the value 16 bits, express your

answer as a percentage of the disk spacethat must be free.


8/19


Free disk space management, contd

A:

The bit map requires B blocks, whereas the

free list requires F*D bits. Thus the statewhen the free list uses less space than the bitmap is F*D < B.

The percentage of disk space free is given bythe fraction F/B. F*D less than B becomes

F/B less than 1/D. If D is 16, then this fractionis 1/16 or 6.25 percent


9/19


File system consistency

Q: A file system checker has built up its counters as

shown bellow

Are there any errors? If so, are they serious? Why?

How to fix them?

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

In use 1 1 0 1 0 1 2 1 1 0 0 1 1 1 0 0

free 0 0 0 0 2 0 0 0 0 1 1 0 1 0 1 1


10/19


Recall: file system consistency

Initially:

Checking i-nodes: if a blockis referenced, its number Inuse is incremented

Checking free lists or bitmaps: if a block is

referenced, its number freeis incremented

If the system is consistent,each block will have 1 either

in In use or in free (in thisexample the system isinconsistent)

0 1 2 3 4 5 6

In use 0 0 0 0 0 0 0

free 0 0 0 0 0 0 0

0 1 2 3 4 5 6

In use 0 2 0 1 0 0 0

free 0 0 0 0 0 0 0

0 1 2 3 4 5 6

In use 0 2 0 1 0 0 0

free 1 0 1 0 1 1 1

0 1 2 3 4 5 6

In use 0 2 0 1 0 0 0

free 1 0 1 0 1 1 1


11/19


File system consistency, contd

A:

Block 2 is not reported in either table (missing block). Not

serious, but wastes space. Add it to the free list Block 4 is twice in a free list. Not serious at the moment. Just

fix the free list

Block 6 is present in 2 files at the same time. If a file is

removed then the block will appear in the both lists. Thesolution is to allocate a free block, copy the contents of 6

there, and insert the copy onto one of the files

Block 12 is present in the both lists. Remove it from the free

list

0 1 2 3 4 5 6 7 8 9 10 111213 14 15

In use 1 1 0 1 0 1 2 1 1 0 0 1 1 1 0 0

free 0 0 0 0 2 0 0 0 0 1 1 0 1 0 1 1


12/19


Free memory space management

Q: The beginning of a free space bitmap looks like thisafter 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 startingat 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 thefollowing additional actions:

a) File B us written, using 5 blocks.

b) File A is deleted.

c) File C is written, using 8 blocks.d) File B is deleted


13/19


Free memory space management, contd

A:

a) 1111 1111 1111 0000

b) 1000 0001 1111 0000

c) 1111 1111 1111 1100

d) 1111 1110 0000 1100

A B

B

BC C

C C


14/19


Free memory space management-2

Q: A certain file system uses 2 KB disk blocks.

The median file size is 1 KB. What fraction of

the disk space is wasted for 1 KB files? Howmight you justify wasting the disk space?


15/19


Free memory space management-2, contd

A: The 1 KB median file size consumes only half of the 2 KB

disk block, and thus leads to 50% wasted disk space

Because disks generally read large disk blocks more efficiently,

and realistic file systems include large files, disk performance

can justify the wasted disk space

2K


16/19


Directories and files

Q: How many disk operations are needed to

fetch the i-node for the file

/usr/ast/courses/os/handout.t?

Assume that the i-node for the root directory is

in memory, but nothing else along the path is

in memory. Also assume that all directories fit

in one disk block.


17/19


Recall: i-nodes

I-node stores the attributes

and disk addresses of a

files blocks

The first few addresses are

stored in the i-node itself

Additional addresses are

available via single, doubleand triple indirect blocks

reference

In UNIX directories are

implemented as i-nodes


18/19


Directories and files, contd

A: Since all directories fit in one disk block then all required data ona directory is stored in the i-node itself

To open /usr/ast/courses/os/handout.t, assuming the root i-node is

already in memory, we do the following reads:

1: directory /2: i-node for /usr3: directory /usr4: i-node for /usr/ast5: directory /usr/ast6: i-node for /usr/ast/courses7: directory /usr/ast/courses

8: i-node for /usr/ast/courses/os/9: directory /usr/ast/courses/os/10: i-node for /usr/ast/courses/os/handout.t


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References

Andrew S. Tanenbaum, Modern Operating

Systems

Silberschatz Galvin, Operating System

Concepts

Documents

File Systems Lab