Consider a typical disk that rotates at 15000 rotations per minute (RPM) and…
2015
Consider a typical disk that rotates at 15000 rotations per minute (RPM) and has a transfer rate of 50 × 10⁶ bytes/sec. If the average seek time of the disk is twice the average rotational delay and the controller’s transfer time is 10 times the disk transfer time, the average time (in milliseconds) to read or write a 512 byte sector of the disk is _________
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Correct answer: 6.1 to 6.2
Step-by-step calculation of the average time to read/write a 512-byte sector:
Rotation speed and average rotational delay: 15000 RPM = 15000/60 = 250 revolutions per second. One rotation = 1/250 = 0.004 s = 4 ms, so the average rotational delay (half a rotation) = 2 ms.
Average seek time: Given as twice the average rotational delay → 2 × 2 ms = 4 ms.
Disk transfer time for 512 bytes: 512 bytes / (50 × 10^6 bytes/s) = 512 / 50,000,000 = 1.024×10⁻⁵ s = 0.01024 ms.
Controller transfer time: Given as 10 × disk transfer time = 10 × 0.01024 ms = 0.1024 ms.
Total average time: seek (4 ms) + rotational delay (2 ms) + disk transfer (0.01024 ms) + controller transfer (0.1024 ms) = 6.11264 ms ≈ 6.11 ms.
This result (about 6.11 ms) falls within the provided answer range of 6.1 to 6.2 ms.
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