0 votes 0 votes Can anyone please explain how this calculation is done? A 1.3 GB disk with 512 byte blocks would need a bit map of over 332 KB to track its free blocks . nandini gupta asked Aug 27, 2018 nandini gupta 596 views answer comment Share Follow See all 4 Comments See all 4 4 Comments reply Shaik Masthan commented Aug 27, 2018 reply Follow Share 1.3 GB disk with 512 byte blocks so total blocks = $\frac{1.3 \;X\; 2^{30} }{2^{9}}$ = ${1.3 \;X\; 2^{21} }$ Blocks ∴ ${1.3 \;X\; 2^{21} }$ Bits required to represent those blocks = ${1.3 \;X\; 2^{18} }$ Bytes = ${1.3 \;X\;256\;X\; 2^{10} }$ Bytes = ${1.3 \;X\;256}$ KB = 332 KB 1 votes 1 votes nandini gupta commented Aug 28, 2018 reply Follow Share Thanks a lot 0 votes 0 votes nandini gupta commented Aug 28, 2018 reply Follow Share There are 1.3 * (2)^21 blocks . But I think there should be log(1.3 * (2)^21) bits to represent those blocks. How 1.3 * (2)^21 blocks is equal to 1.3 * (2)^21 bits 0 votes 0 votes Shaik Masthan commented Aug 28, 2018 reply Follow Share one bit required for one block in bitmap, 1 is used to say that block is free, 0 is used to say that block is not free ( may be you can interchange ). 0 votes 0 votes Please log in or register to add a comment.