GATE CS 2019 Q.52 (CS Q.42): File Systems, I/O and Disk Scheduling

The index node (inode) of a Unix-like file system has 12 direct, one single-indirect and one double-indirect pointers. The disk block size is $4\text{ kB}$ , and the disk block address is $32$ -bits long. The maximum possible file size is (rounded off to 1 decimal place) _________ GB.

Your answer

Official answer and solution

Accepted answers: 3.7 to 3.8, or 4 to 4.1
From: Official GATE 2019 answer key

Note on this question

Official answer key accepted both 3.7 to 3.8 and 4.0 to 4.1. The calculation yields ~4.0 GB in binary GiB units (4.004 GB) or ~3.7 GiB if 4 GB is converted to binary gigabytes.

Solution

Compute the total number of blocks addressable via direct, single-indirect, and double-indirect pointers in the inode, then convert to GB.

We are given:

Disk block size = $4\text{ kB} = 4096\text{ bytes} = 2^{12}\text{ bytes}$
Disk block address = $32\text{ bits} = 4\text{ bytes}$

The number of block pointers that fit in a single disk block is:

\text{Pointers per block} = \frac{4096\text{ bytes}}{4\text{ bytes/pointer}} = 1024 = 2^{10}\text{ pointers}

The total number of data blocks addressable by the inode is:

Direct pointers ( $12$ ): $12\text{ blocks}$
Single-indirect pointer ( $1$ ): $2^{10} = 1024\text{ blocks}$
Double-indirect pointer ( $1$ ): $2^{10} \times 2^{10} = 2^{20} = 1048576\text{ blocks}$

Total blocks:

N_{\text{blocks}} = 12 + 1024 + 1048576 = 1049612\text{ blocks}

Total maximum file size in bytes:

\text{Size} = 1049612 \times 4096\text{ bytes} = 4,299,210,752\text{ bytes}

In binary gigabytes ( $1\text{ GiB} = 2^{30}\text{ bytes}$ ):

\text{Size} = \frac{1049612 \times 2^{12}}{2^{30}} = \frac{1049612}{2^{18}} \approx 4.003955\text{ GB} \approx 4.0\text{ GB}

In decimal gigabytes ( $1\text{ GB} = 10^9\text{ bytes}$ ), or converting $4\text{ GB}$ to GiB ( $4 \times 10^9 / 2^{30} \approx 3.725\text{ GiB}$ ), the GATE answer key officially accepted both 3.7 to 3.8 and 4.0 to 4.1.

Final answer

3.7 to 3.8

Common trap

Forgetting that indirect blocks contain block addresses (4 bytes each) rather than direct data.

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

Source

Paper: GATE 2019 Computer Science and Information Technology (CS)
Question: 52 of 65, printed as CS Q.42
Section: Computer Science and Information Technology

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