Consider a disk with the following specifications: rotation speed of 6000 RPM,…
2024
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 _________
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Correct answer: 30.06
Given: rotation speed = 6000 RPM; average seek time = 5 ms; 500 sectors per track; sector size = 512 bytes; file size = 3000 sectors; assume average rotational latency.
Rotation time = 6000 RPM = 6000/60 = 100 revolutions per second, so one rotation = 1/100 s = 0.01 s = 10 ms.
Average rotational latency = half a rotation = 0.01/2 = 0.005 s = 5 ms.
Transfer time for one sector = rotation time / sectors per track = 0.01 s / 500 = 0.00002 s = 0.02 ms.
For randomly located sectors, each sector on average requires one seek and one average rotational latency before its transfer. So per-sector access time = average seek + avg rotational latency + sector transfer = 0.005 + 0.005 + 0.00002 = 0.01002 s.
Total time for 3000 sectors = 3000 * 0.01002 s = 30.06 s.
Answer: 30.06 seconds (rounded to 2 decimal places).
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