Question

Consider a file system that uses inodes to represent files. Disk blocks are 2KB in size and a pointer to a disk block requires 4 bytes. Consider a file whose inode has 6 direct disk blocks, plus 1 single indirect, and 1 double indirect disk blocks. What is the maximum size of the file? If we use bitmap to manage the free space, how many disk blocks are required to hold the bitmap of 32 GB disk?

Answer 1

Answer:

a.  Maximum size of file = 8.003 GB

Explanation:

a. Maximum size of file

= (6 * 2 KB) + (2048 * 2 KB) + (2048 * 2048 * 2 KB)

= 12kb + 4096 Kb +  8388608 kb

= 8392716 kb

8392716/1024 mb = 8196.01 MB

8196.01 /1024 GB= 8.003 GB

b. According to above calculation

For 8 GB 6 direct, 1 single and 1 double indirect block requires

so

for 32 GB

24 direct Block

4 single and 4 double indirect block requires