GATE CS 2012 Q41 — I/O Scheduling

To find the maximum possible file size, we first determine how many disk block addresses (pointers) can fit into a single disk block.

Step 1: Calculate the number of addresses per block

$$\text{Number of addresses per block } (N) = \frac{\text{Disk Block Size}}{\text{Disk Block Address Size}} = \frac{128 \text{ Bytes}}{8 \text{ Bytes}} = 16$$

Step 2: Calculate the maximum number of blocks a file can use

The file descriptor contains:

• Direct blocks: $8$ blocks
• Single indirect block: $1 \times N = 16$ blocks
• Doubly indirect block: $1 \times N^2 = 16^2 = 256$ blocks

$$\text{Total Blocks} = 8 + 16 + 256 = 280 \text{ blocks}$$

Step 3: Calculate the maximum file size

$$\text{Maximum File Size} = \text{Total Blocks} \times \text{Block Size}$$ $$\text{Maximum File Size} = 280 \times 128 \text{ Bytes} = 35,840 \text{ Bytes}$$

Converting this to KBytes ($1 \text{ KByte} = 1024 \text{ Bytes}$):

$$\text{Maximum File Size} = \frac{35,840}{1024} \text{ KBytes} = 35 \text{ KBytes}$$

Thus, the maximum possible file size is 35 KBytes, which corresponds to option B.