File system :

Question

Consider a UNIX-like file system implemented with i-nodes that resides on a disk of size $512\text GB$. Each i-node has a total of $15$block addresses consisting of direct and indirect lock addresses.

Suppose we configure the file system to use a block size of $32$KB. How many bytes are needed to store all $15$ block addresses in an I-node?

a). $15$ Bytes                                                b). $29$ Bytes

c). $45$ Bytes                                                d). $75$ Bytes

Answer 1

Ans is C

Total Number of Blocks in Disk is = Size of Disk / Size of each block

Total Number of Blocks in Disk is =2³⁹/2¹⁵ =2²⁴

So, to address Each Block we need 24bits  ==>3 bytes

So space required by 15 address to block is 15*3=  45 Bytes

Comments

I think i need to study Inodes one more time, and then try these.

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Well you can.

But, no such formula is used here .

Just by using the concept that how much storage is needed to store 15 BA. If you know the size of BA , we just have to X by 15. Suppose, 32 blocks are there, a particular block only needs  only 5 bit address .

Also, we can find #blocks as BS is given and as well as DS.

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instead of blocks can we say total number of inodes present is disk