GATE CS 2024 Set 2 Q.53: File Systems, I/O and Disk Scheduling

Consider a disk with the following specifications: rotation speed of 6000 RPM, average seek time of 5 milliseconds, 500 sectors/track, 512-byte sectors. A file has content stored in 3000 sectors located randomly on the disk. Assuming average rotational latency, the total time (in seconds, rounded off to 2 decimal places) to read the entire file from the disk is _________

Your answer

Official answer and solution

Accepted answers: 29.5 to 30.5
From: Official GATE 2024 answer key

Solution

Compute the seek time, average rotational latency, and sector transfer time for a single random sector, then multiply by 3000 sectors.

Time for one full disk rotation: $T_{\text{rot}} = \frac{60\text{ seconds}}{6000\text{ RPM}} = 0.01\text{ s} = 10\text{ ms}$
Average rotational latency: $T_{\text{latency}} = \frac{1}{2} T_{\text{rot}} = 5\text{ ms}$
Transfer time per sector ( $500$ sectors per track): $T_{\text{transfer}} = \frac{10\text{ ms}}{500} = 0.02\text{ ms}$
Total time for 3000 randomly located sectors:
Since each of the $3000$ sectors is located randomly on the disk, reading each sector incurs average seek time ( $5\text{ ms}$ ) + average rotational latency ( $5\text{ ms}$ ) + sector transfer time ( $0.02\text{ ms}$ ): $T_{\text{total}} = 3000 \times (5 + 5 + 0.02)\text{ ms} = 3000 \times 10.02\text{ ms} = 30,060\text{ ms} = 30.06\text{ seconds}$

Final answer

30.06

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

Source

Paper: GATE 2024 Computer Science and Information Technology - Set 2 (CS2)
Question: 53 of 65
Section: Computer Science and Information Technology

Report an error in this question

GATE questions are the work of the organising institutes. This site is independent and is not affiliated with them.

Question

Solution

Source