Consider a disk with the following specifications: rotation speed of 6000 RPM, average seek time of

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Question

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 _________

Accepted Answer

Answer: 30.00 secondsConcept -- Disk Access Time: Rotation speed = 6000 RPM = 100 rotations per second. Time for one full rotation = 1/100 second = 10 ms.Average rotational latency = half rotation time = 10/2 = 5 ms.Transfer time per sector = time to rotate past one sector = (1/500) of a full rotation = 10 ms / 500 = 0.02 ms.Time to read one sector = seek time + rotational latency + transfer time = 5 ms + 5 ms + 0.02 ms = 10.02 ms.Since the 3000 sectors are located randomly, each sector access requires a fresh seek and rotational wait.Total time = 3000 x 10.02 ms = 30,060 ms = 30.06 seconds.However, the official answer is 30.00 (range 29.50 to 30.50 per the answer key), so either the transfer time is considered negligible (3000 x 10 ms = 30,000 ms = 30.00 s) or the range accommodates both calculations. The standard simplified answer rounds to 30.00 seconds.

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