GATE Overflow Test Series | Mock GATE | Test 5 | Question: 34

Question

A variation of an inode is called a linode (short for little node). Assume the size of a disk block is $4K$ bytes and a disk block can contain $2K$ disk block addresses. A linode has the following structure:

• entries $0-6$ are direct disk block pointers
• entry $7$ is a single indirect block pointer
• entry $8$ is a double indirect block pointer

The maximum size of a file (rounded to the nearest GB) using a linode will be _________

Comments

Disk block size = 4KB 

Disk block addresses= 2K = 2^11 = We need 11 bits to address any of 2K 

So 11 bits ie Disk block address size = 2B (Assuming byte addressable)

 

Number of addresses= Size of disk block/address size =4KB/2B =  2K

7 direct, 1 single indirect and 1 Double indirect

Max file size = (7∗4K)+(2K∗4K)+(2K∗2K∗4K) 

                     = 28K+8M+16G≈16GB=28K+8M+16G≈16GB.

16 is answer

    

Answer 1

It would be $(7 \ast 4K) + (2K \ast 4K) + (2K\ast 2K \ast 4K) $

$\quad = 28K + 8M + 16G  \approx 16 GB$.

So, the correct answer is $16$.