operating system (file system)

Question

Suppose a computer that has a file system for a 256GB disk, where each disk block is 16KB .If the OS for this computer uses a FAT, what is the smallest amount of memory that could possibly be used for the FAT (assuming the entire FAT is in memory)?

Answer 1

Disk size= 256GB= $2^{38}$B and Block size= 16KB= $2^{14}$B.

The FAT will have one entry for each disk block. No of disk blocks= Disk size/Block Size=$\frac{2^{38}}{2^{14}}$ = $2^{24}$ blocks.

So no of entries in FAT=$2^{24}$.

Smallest amount of memory used for the FAT = No of entries in FAT x Size of each entry.

Each entry will contain the block number. With 2^24 blocks, bits required to identify a block is 24 bits.

So size=$2^{24}$ x 24 bits = $2^{24}$ x 3 Bytes = 48MB.