Disk Allocation

Discussion of disk allocation algorithms from Silberschatz's 2012 Operating System textbook

Table of Contents:

03:27 - Contiguous Allocation
05:33 - Contiguous Allocation of Disk Space
06:39 - Linked Allocation
07:01 - Linked Allocation
07:17 - Linked Allocation
07:18 - Linked Allocation
07:18 - Allocation Methods - Linked
08:16 - Linked Allocation
08:16 - Linked Allocation
10:22 - Linked Allocation
11:10 - FAT Tables
11:24 - Linked Allocation
11:36 - FAT Tables
12:04 - File-Allocation Table (FAT)
14:07 - Extent-Based Systems
15:45 - Allocation Methods - Indexed
16:56 - Example of Indexed Allocation
17:44 - Indexed Allocation (Cont.)
19:28 - Indexed Allocation with Linking
20:53 - Indexed Allocation with Dual-Level
21:52 - Indexed Allocation – Mapping (Cont.)
22:20 - Indexed Allocation with Dual-Level
23:24 - Indexed Allocation – Mapping (Cont.)
23:25 - Combined Scheme: UNIX UFS (4K bytes per block, 32-bit addresses)
25:15 - Performance
27:31 - Performance (Cont.)
29:02 - Free-Space Management
30:35 - Free-Space Mgt. via Bit Map
31:29 - Free-Space Mgt. via Free List
32:19 - Linked Free Space List on Disk
32:38 - Protecting Free Management
33:38 - Free-Space Management (Cont.)
34:57 - Free-Space Management (Cont.)

[bsuryasaradhi]

I am trying to make a hardisk from old casette tapes, this helped a lot during my research time❤ thanks

[rishanmascarenhas]

Beautifully explained sir. Thankyou very much for your efforts. 😉

[life_of_cloudie]

Very helpful this video solved all my problems.Thank you, Sir :)

[atanuguin]

Very good discussion on Disk Allocation. Thank you.

[mevinimunaweera]

Nice explanation.I understand it very well

[tman]

thank you ! finally someone can explain it

[ulysses_grant]

Sir, I have a question:
In 5:19 you explain, "By the way, because 512 is a power of two, you don't need to make the integer division, you just look at the 9 LSB of the logical address and that'll be our value Q, the block number we're trying to access"
I tried it with an hipothetical Logical Address 0x1010 (4112 in decimal). In binary, we will have: 0001 0000 0001 0000. So, if we look to the 9 least significant bits, we will have 16.
If we calculate Q == 4112/512 (block size), we will have 8.
If we calculate R == 4112%512, THEN we will have 16.
So, the 9 least significant bits are the Q value or the R value? I got a little confused.
Thanks a lot for your video and for your explanation!

[anarchyat]

i dont get the divide and modulus part

[kaushikkumaran]

In linked allocation address translation. why is 508 taken to calculate the block number and offset. Even though four bytes are used for the pointer the complete block still requires 512 bytes, so to find the block number why are we not using integer divsion by 512. And awesome lecture sir. Thanks a lot:-)

[deepakkumawat]

sir please speak slow i cant understand some things