ITPEC FE Subject A April 2024 Question 9
ITPEC FE Subject A April 2024 — Question 9 of 60
HDD Average Access Time — calculate total access time from disk specs.
Formula: access_time = seek_time + controller_overhead + rotational_latency + transfer_time
Given specs:
- Average seek time: 4 ms
- Controller overhead: 1.25 ms
- Rotation speed: 5,000 rpm
- Transfer rate: 60 MB/s
- Data size: 1 MB
Step-by-step:
- Rotational latency = half of one full rotation = (60,000 / 5,000) / 2 = 12 / 2 = 6 ms
- Transfer time = 1 MB / 60 MB/s = 1000 / 60 ≈ 16.67 ms
- Total = 4 + 1.25 + 6 + 16.67 = 27.92 ms
Why not others:
- (a) 21.93 — arithmetic error in one of the components
- (b) 26.67 — forgot controller overhead (4 + 6 + 16.67)
- (d) 33.92 — used full rotation (12 ms) instead of half
Key rule: Rotational latency = half of one full rotation time. Full rotation = 60,000 ms / rpm.
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.