Answer:
a) 4100.00 GB
b) 4.0039 GB
c) 6291456 B
Explanation:
Given:
14 direct pointers
2 single indirect pointers =[tex]2^1^0[/tex]
1 double indirect pointer = [tex]2^2^0[/tex]
1 triple indirect pointer = [tex]2^3^0[/tex]
a) Let's first find the number of block pointers in one disk block.
Size of disk block / disk block address
[tex] = \frac{4*1024}{4B} = 1024 = 2^1^0[/tex]
A single indirect pointer can address up to 1024 or [tex]2^1^0[/tex] blocks
The maximum possible file size, will be:
[tex] [14 + 2^1^0 + 2^1^0 + 2^2^0 + 2^3^0] 4KB [/tex]
= 4100.00 GB
b) For overhead:
[tex] [14 + 2^1^0 + 2^1^0 + 2^2^0 + 2^3^0] 4B [/tex]
= 4.0039 GB
c) For overhead for a file of 6GB,
we need to check for the required number of block pointers :
[tex] \frac{6 * 2^3^0 }{4 * 2^1^0} = 1572864 B [/tex]
The overhead for a file of 6GB will be:
1572864 * 4 =
= 6291456 B
