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Explain in detail

Explain in detail

1Comment
jayendra @jayendra
17 Dec 2014 09:12 pm

Disk block Size (DBS) = 4K bytes

here disk block 2 and disk block 3 will be used.

therefore 4KB+4KB=8KB

a)

1 file with 32 bit entry (4B entry)

so i can store 8KB/4B file pointers, which is maximum possible files = 2K

b)

for maximum possible file size:

Disk block address is 8 bits so maximum 2^8 disk blocks are possible = 256

out of 256 data blocks 0,1,2,3 are used for administrative purpose, therefore 256-4=252

possible max file size = no of data blocks * size of data block = 252 * 4KB