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Question

A hard disk has a capacity of 40MB.Disk block size is 2KB.Disk block address is 3 bytes.If the system uses a bit map approach to maintain free space utilization, what is the number of blocks required to store the bit map?

Comments

is it 2 or 3????

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can you explain it

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Blocks are ::$ 40 * 2^{20} / 2 * 2^{10} = 20*2^10  $

And block size in bits is :: $2*2^{10} * 8 bits$

So, required blocks should be :: $ceil(20*2^{10} / 16 * 2^{10} )$

                                         $ =>  ceil(1.25)  = 2$

Am i right ???

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sorry but i don't have the answer key

but i think your answer is correct

Answer 1

Size of Disk= 40 MB

Size of Disk Block= 2 KB

Size of DBA= 3 Byte

No. of Disk Blocks= 40MB/2KB= 20K

In Bit Map approach, every disk block will be mapped with one binary bit.

 No. of DiskBlocks we can map in 1 Disk Block= Disk Block Size in bits=2KB= 2K*8 bits=16 K bits

Therefore we can map 16K disk block's  using one disk block.

So to map 20 K disk block's, we need 20K/16K disk blocks= 1.25 Disk Block which is possible in not less than 2 Disk Blocks.