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Free disk space can be used to keep track of using a free list or a bit map. Disk addresses require $d$ bits. For a disk with $B$ blocks, $F$ of which are free, state the condition under which the free list uses less space than the bit map.
asked in Operating System by Veteran (59.5k points)
edited by | 995 views

2 Answers

+22 votes
Best answer

Bit map maintains one bit for each block, If it is free then bit will be "$0$" if occupied then bit will be "$1$".
For space purpose, it doesn't matter what bit we are using, only matters that how many blocks are there.
For $B$ blocks, Bit map takes space of "$B$" bits.

Free list is a list that maintains addresses of free blocks only. If we have $3$ free blocks then it maintains $3$ addresses in a list, if $4$ free blocks then $4$ address in a list and like that.


Given that we have $F$ free blocks, therefore $F$ addresses in a list, and each address size is d bits therefore Free list takes space of "$Fd$".

condition under which the free list uses less space than the bit map: $Fd<B$

answered by Boss (15.9k points)
edited by
0
what is tree list ?? or is it free list ??
0
its free list. tree list typo
+10 votes
Solution for Part a :-

Assume that size of each block is S bits.

Then no of bits required for free list is = Fd, No of blocks required = Fd/S

No of bits required for Bit map = B (No of blocks ) , No of block required is =  B /S

Condition under which free list uses less space than the bit map.

Fd / S < B / S
answered by Boss (42.6k points)
0
why divide by S ??

in the above selected answer there is no such division and condition FD<B is sufficient i think ?

explain your approach please .


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