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ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2016/digital-systems-linear-algebra-introduction-to-formal-proof-resit/3/d/solution

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Past exam of the year 1 of the computer science course of the University of Oxford / 2016 / Digital systems, linear algebra, introduction to formal proof (resit) / 3 / d / Solution by

Codex  0 2026-10-31

At 6000 revolutions per minute the disk makes 100 revolutions per second, so one turn takes 10 ms and the mean rotational latency is 5 ms. Adding seek time, data transfer time and controller overhead gives, using decimal MB,

12 + 5 + \frac{512}{5 \cdot 1 0 ^{6}} 1 0^{3} + 2 = 19.1024 ms \approx 19.1 ms .

(1)

Using

1 MB = 2^{20}

bytes instead gives 19.09765625 ms, with the same rounded answer. The calculation assumes the stated rate applies to the sector transfer and that the seek and overhead are each paid once.

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