Question

A file system with 300 GByte disk uses a file descriptor with 8 direct block addresses, I indirect block address and 1 doubly indirect block address. The size of each disk block is 128 Bytes and the size of each disk block address is 8 Bytes. The maximum possible file size in this file system is

Answer 1

Maximum Possible size
[(address point by doubly inirect block)${}^2$ + (address pointed by single address) + Address points by single direct address ] $\times$ [ Block size]

$=[{\left(\frac{128}{8}\right)}^2+\left(\frac{128}{8}\right)+8]\times 128$ B

$=[2^8+2^4+2^3]\times 128$ B

= 32 kB + 2 kB + 1 kB = 35 kB