Respuesta :
Answer:
128 KB
Explanation:
Given data :
The block size = 512 bytes
Block pointer size = 2 bytes
In an indexed allocation, a special block named Index Block contains all the pointers to the blocks that the file had occupied.
Therefore, the maximum number of block pointer in 1 block :
[tex]$=\frac{\text{block size}}{\text{pointer size}}$[/tex]
[tex]$=\frac{512}{2}$[/tex]
= 256
Thus, a block can hold maximum 256 pointers.
It is given the level of indexing is one. So,
Maximum size of the file = maximum number of block pointers in 1 file x block size
= 256 x 512
[tex]$=2^8 \times 2^9$[/tex]
[tex]$=2^{8+9}$[/tex]
[tex]$=2^{17}$[/tex]
[tex]$=2^7 \times 2^{10}$[/tex]
[tex]$=128 \times 2^{10}$[/tex]
= 128 KB (∵ 1 KB = [tex]2^{10} \ B[/tex] )
Therefore, the largest file that 1 level if indexed allocation method can handle is 128 KB.
