975 views
2 votes
2 votes
consider  a unix file system impleneted with i-node,that resides on a disk of size 512GB.Each i-node has atotal of 15 block addresses,consisting of direct and indirect block adresses.

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

a)15byte

b)29byte

c)45byte

d)75byte

1 Answer

Best answer
7 votes
7 votes

No of disk blocks = Disk size / Disk block size

                          = 239 / 215

                          = 224

So disk block address size  =  log 2(No of disk blocks)

                                         =  24 bits

                                         =  3 B

Given in inode we have 15 such entries..

Hence total size                  = 15 * 3

                                          = 45 B

Hence C) is correct answer..  

selected by

Related questions

0 votes
0 votes
1 answer
2
Aboveallplayer asked Dec 31, 2016
308 views
How large a file can be addressed by a double indirect block alone in a 64 bit system?minimum block entry size 8 byte and block size 8KB