ITPEC FE Subject A October 2025 Question 9
When an HDD has the specifications shown in the table below, what is the average access time for the HDD to transfer 16 kB of data in milliseconds (ms)? The average access time is calculated as the sum of the average seek time, controller overhead, rotational latency, and transfer time. Other overheads can be ignored. Here, 1 MB = 1,024 kB.
Explanation
25.725 ms — add the four components, taking half a rotation as the average rotational latency.
- •Average seek time: 5 ms
- •Controller overhead: 0.1 ms
- •One rotation takes 60 ÷ 6,000 = 0.01 s = 10 ms, so the average rotational latency is half of it, 5 ms
- •Transfer rate 1 MB/s = 1,024 kB/s, so the transfer time is 16 ÷ 1,024 = 0.015625 s = 15.625 ms
- •Total: 5 + 0.1 + 5 + 15.625 = 25.725 ms
Answer (b)
Why not others:
- •(c) 30.725 — uses a full rotation, 10 ms, as the latency instead of half a rotation
- •(d) 74.1 — divides the wrong way round for the transfer time, 1,024 ÷ 16 = 64 ms
- •(a) 20.1 — has no .625 part, so it cannot contain the 15.625 ms transfer time of 16 kB at 1,024 kB/s
Key rule: Average rotational latency is half of one rotation, 60 ÷ RPM seconds, and transfer time is data size ÷ transfer rate in matching units.
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