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In a particular Unix Os, each data block is 256 bits, each node has 4 direct data block address and 3 additional addresses: one for single indirect block, one for double indirect block and one fot triple  indirect block. Each block is addressed with 64-bit. the total size of a file possible in the file system (in k bits ) is _________.

 

what is k bits here? what will the exact answer??

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Number of disk block address possible to store in 1 disk block=DB size/DBA.

DB size=disk block size.

DBA=disk block address.

Total size of file system=[# direct DBA's+DB size/DBA+$[\frac{DB size}{DBA}]^{2}$+.......] DB size.

=4+256/64 +$[\frac{256}{64}]^{2}$+$[\frac{256}{64}]^{3}$ ] *256 bits.

=21,504 bits.

=21,504/1024 Kbits

=21.00 Kbits.

Hence size of total file system=21 Kbits

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