GATE CS 2012 Q.41 (CS Q.41): File Systems, I/O and Disk Scheduling

A file system with 300 GByte disk uses a file descriptor with 8 direct block addresses, 1 indirect block address and 1 doubly indirect block address. The size of each disk block is 128 Bytes and the size of each disk block address is 8 Bytes. The maximum possible file size in this file system is

3 KBytes

35 KBytes

280 KBytes

dependent on the size of the disk

Official answer and solution

Official answer: B
From: Official GATE 2012 answer key

Solution

Calculate the maximum file size addressable via 8 direct, 1 single indirect, and 1 double indirect block pointers.

Given disk parameters:

Block size = 128 Bytes
Disk block address pointer size = 8 Bytes

Number of block pointers that fit in one disk block:

\text{Pointers per block} = \frac{128 \text{ Bytes}}{8 \text{ Bytes}} = 16

Calculate addressable data blocks:

Direct blocks: $8 \text{ blocks}$
Single indirect block: $1 \times 16 = 16 \text{ blocks}$
Doubly indirect block: $1 \times 16 \times 16 = 256 \text{ blocks}$

Total addressable data blocks:

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

Maximum file size:

\text{Maximum file size} = 280 \times 128 \text{ Bytes} = 35840 \text{ Bytes}

\text{In KBytes} = \frac{35840}{1024} \text{ KBytes} = 35 \text{ KBytes}

This matches option (B).

Final answer

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2012 Computer Science and Information Technology (CS)
Question: 41 of 65, printed as CS Q.41
Section: Computer Science and Information Technology

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