Operating Systems - Round Robin Scheduling - Quantum & Turnaround Time
In a Round Robin scheduler with a 3ms time quantum, a process arrives with a CPU burst of 7ms. How many times will this process be preempted before completion?
In a Round Robin scheduler with a 3ms time quantum, a process arrives with a CPU burst of 7ms. How many times will this process be preempted before completion?
easy🧾 Trace Q3 of Q15
Step-by-Step Solution
Solution:
Step 1: Calculate number of quanta needed
The process needs 7ms total, with each quantum 3ms, so it requires 3 quanta (3+3+1).Step 2: Determine preemptions
Preemption occurs after each quantum except the last. So, preemptions = 3 - 1 = 2.Final Answer:
Option A -> Option AQuick Check:
7ms / 3ms = 2 full quanta + remainder, so 2 preemptions [OK]
Quick Trick: Preemptions = number of quanta minus one [OK]
Common Mistakes:
MISTAKES- Assuming preemption equals number of quanta
- Ignoring the last partial quantum
- Confusing burst time with quantum length
Trap Explanation:
PITFALL- Counting all quanta as preemptions ignores that the last quantum finishes the process.
Interviewer Note:
CONTEXT- Tests understanding of how Round Robin preemption relates to quantum and burst time.
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