# Unix node I file system

**Ruturaj**#2

Really simple questionâ€¦

See the image above for clear understanding of the organisation.

First we need to calculate number of block address a single block can hold

= Block Size/ Block address size

= 4KB/4B

= 1K

= 1024

Now just calculate the total number if blocks in the system

= 8 direct pointers + 2 single indirect pointer + 4 double indirect pointer + 1 triple indirect pointer

= 8 + 2 * 1024 + 4 * 1024 * 1024 + 1 * 1024 * 1024 * 1024

Now we have total number of blocks. Just calculate total size by multiplying it with block size

= (8 + 2 * 1024 + 4 * 1024 * 1024 + 1 * 1024 * 1024 * 1024) * 4KB

**Ruturaj**#5

Yes 4TB to be approx. Rest factors wont be that significant. But if exact is asked then you have to evaluate it.