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 _________
Answer
29.5 to 30.5
Explanation
At 6000 RPM the disk makes 100 rotations per second, so one rotation takes 10 ms.
- Seek time: 5 ms.
- Average rotational latency: half a rotation, 5 ms.
- Transfer time of one sector: \(10 / 500 = 0.02\) ms.
Because the sectors lie at random places, each one needs its own seek and rotational wait: \(5 + 5 + 0.02 = 10.02\) ms per sector. For 3000 sectors the total is \(3000 \times 10.02 = 30060\) ms, that is 30.06 s. The sector size (512 bytes) is not needed. Treating the file as stored on consecutive sectors would give a much smaller time.
Answer: 30.06