ITPEC FE Subject A October 2025 Question 8
ITPEC FE Subject A October 2025 — Question 8 of 60
Preemptive Interrupt Scheduling — calculate main process CPU time with multiplex interrupts.
Given interrupt A (High, 0.5ms) and B (Low, 1.5ms), multiplex interrupts enabled:
- •Multiplex interrupts = higher-priority interrupt preempts lower-priority
- •Preempted process resumes after higher-priority finishes
- •Overhead is ignored
Method: simulate the timeline left to right
- 0.0–1.0 → B runs (arrived at t=0, needs 1.5ms, done 1.0ms)
- 1.0–1.5 → A preempts B (A arrived at t=1), B waits
- 1.5–2.0 → B resumes (0.5ms remaining) → B done
- 2.0–2.5 → Main process (0.5ms)
- 2.5–3.0 → A runs (arrived at t=2.5)
- 3.0–3.5 → Main process (0.5ms)
- 3.5–4.0 → A runs (arrived at t=3.5)
- 4.0–5.0 → Main process (1.0ms)
Main total = 0.5 + 0.5 + 1.0 = `2.0 ms`
Why not others:
- (b) 2.5 — miscounts interrupt occurrences or timing
- (c) 3.5 — overcounts free CPU time
- (d) 5 — ignores all interrupts
Key rule: Draw a Gantt chart. Fill each time slot with the highest-priority active interrupt. Main gets only the leftover slots. Preemption pauses low-priority work but doesn't lose progress.
AI-generated — may contain errors
The original exam layout is preserved in the image so diagrams, formulas, tables, and code remain accurate.
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