A Unix – I node is maintaining 256 direct disk block address 4 triple indirect disk block, 2 double indirect disk block and one single indirect disk block. The DBA sequence is of 64 bit and database size is 2 KB then determine the file size.
(A) 131072 MB
(B) 128 MB
(C) 131329 MB
(D) 256 MB


Answer: (C)

Explanation: DB size = 2KB, DBA = 64 bit = 8 Byte
Let X = DB size / DBA = 2KB / 8B= 256
Maximum File size possible = [N₀ + N₁X + N₂*X² + N₃*X³] * DB size
N₀ = Direct disk block = 256
N₁= Single indirect disk block = 1
N₂ = Double indirect disk block = 2
N₃ = Triple indirect disk block = 4
Maximum file size = [ 256 + 1*256 + 2 * 256² + 4 * 256³ ] * 2KB
= [ 1/ 2 + 1/ 2 + 256 + 131072] * 1MB
= 131329 MB

So, option (C) is correct.


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