GATE Overflow Test Series | Mock GATE | Test 3 | Question: 47

Question

Free disk space can be kept track of using a free list or a bitmap. Disk addresses require $\text{D bits}$. For a disk with $\text{B blocks}$ and $\text{D}$ having the value $\text{32 bits},$ the percentage of the disk space that must be free for free list to use same memory as bitmap will be _________ (up to three decimal place).

Comments

For Bitmap we’re using 1 bit for a block & for Free list we’re using 32 bits for a block

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Similar question: https://gateoverflow.in/1740/Gate-cse-1998-question-25-a

Answer 1

Let block size be $k$ and $F$ be the number of free blocks.

Free list size $=F*D*k$ bits

Bitmap size $=B*k$ bits

So, we require $F*D*k = B*k \implies F *D=B.$

When $D=32$ bits,

$\implies F/B = 1/32 = 3.125\%.$

Comments

We shouldn’t have multiplied by $k$ ie. block size as the unit of space(memory) becomes $bits\ bytes$ instead of $bit$.

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@GO Classes  @Deepak Poonia​ @GO Classes sir this should not be the correct answer i guess or question is ambiguous as question is asking the percentage of the disk space that must be free for free list to use same memory as bitmap. Meaning size of free list should be same as bitmap thus, size of free list is B bits. thus, answer should be $(B/8)/(B*size of block)=1/(8*size of block)$. 

Correct me if I’m wrong

Answer 2

Hope this helps . Ans is 3.125