Question

A file system with 300 GByte disk uses a file descriptor with 8 direct block addresses, 1 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 

(A) 3 KBytes 

(B) 35 KBytes 

(C) 280 KBytes 

(D) dependent on the size of the disk

Answer 1

(B) 35 KBytes

Each block size = 128 Bytes 

Disk block address = 8 Bytes 

∴ Each disk ca n contain = 128/8 = 16 addresses

Size due to 8 direct block addresses: 8 x 128 

Size due to 1 indirect block address: 16 x 128 

Size due to 1 doubly indirect block address : 16 x 16 x 128 

Size due to 1 doubly indirect block address: 16 x 16 x 128

 So, maximum possible file size: 

= 8 X 128 + 16 X 128 + 16 X 16 X 128= 1024 + 2048 + 32768 = 35840 Bytes = 35kBytes