Bit Map

Question

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 .

Comments

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

---

Thanks a lot

---

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

---

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 ).