GATE CS 2004 Q49 — I/O Scheduling

To find the maximum possible file size, we calculate the maximum number of disk blocks that can be addressed by the inode structure and multiply it by the disk block size.

Step 1: Calculate the number of block pointers per disk block

• Disk block size: $1 \text{ KB} = 1024 \text{ bytes}$
• Disk block address size: $32 \text{ bits} = 4 \text{ bytes}$

$$\text{Number of pointers per block} = \frac{1024 \text{ bytes}}{4 \text{ bytes}} = 256 = 2^8$$

Step 2: Calculate the total number of blocks addressable by the inode

• Direct pointers: $10$ blocks
• Single indirect pointer: $1 \times 2^8 = 2^8$ blocks
• Double indirect pointer: $1 \times 2^8 \times 2^8 = 2^{16}$ blocks
• Triple indirect pointer: $1 \times 2^8 \times 2^8 \times 2^8 = 2^{24}$ blocks

$$\text{Total blocks} = 10 + 2^8 + 2^{16} + 2^{24} \approx 2^{24} \text{ blocks}$$

Note: Since $2^{24}$ is significantly larger than the other terms, the smaller terms can be neglected for the approximation.

Step 3: Calculate the maximum file size

$$\text{Maximum file size} \approx 2^{24} \times 1 \text{ KB} = 2^{24} \times 2^{10} \text{ bytes} = 2^{34} \text{ bytes}$$

Step 4: Verify with the integer size constraint

The problem mentions that $48$-bit integers are used (likely for representing file sizes or offsets). A $48$-bit integer can address up to $2^{48}$ bytes. Since $2^{34} < 2^{48}$, the file size is limited by the inode structure, which is approximately $2^{34}$ bytes.

Thus, the maximum possible file size is $2^{34}$ bytes.