ITPEC FE Subject A October 2025 Question 9
ITPEC FE Subject A October 2025 — Question 9 of 60
HDD Average Access Time — calculate total access time from seek, rotation, overhead, and transfer.
Given HDD specs: seek time 5 ms, rotation 6000 RPM, transfer rate 1 MB/s, controller overhead 0.1 ms, data size 16 kB.
Formula: Access time = Seek + Controller overhead + Rotational latency + Transfer time
- •Seek time =
5 ms(given) - •Controller overhead =
0.1 ms(given) - •Rotational latency = half rotation =
(60 / 6000) / 2 = 5 ms - •Transfer time =
16 kB / 1024 kB/s = 15.625 ms
Total = 5 + 0.1 + 5 + 15.625 = `25.725 ms`
Why not others:
- (a) 20.1 — incorrect transfer time calculation
- (c) 30.725 — used full rotation (10 ms) instead of half
- (d) 74.1 — completely wrong calculation
Key rules:
- Average rotational latency = half of one full rotation
- Full rotation time = 60 / RPM (in seconds)
- Convert units carefully: 1 MB = 1024 kB
AI-generated — may contain errors
The original exam layout is preserved in the image so diagrams, formulas, tables, and code remain accurate.
This question comes from an official ITPEC past paper. ITPEC Practice is an independent study tool and is not affiliated with ITPEC. See the official FE past-paper collection or Report an issue.