Non preemptive scheduling – Solved Gate%20 Question Papers
Non preemptive scheduling – Solved Gate%20 Question Papers
Absolutely! Here’s a complete breakdown of Non-Preemptive Scheduling with a solved GATE-style question, including explanation, table, and Gantt chart — tailored for GATE CSE/IT aspirants.
Contents
🎯 Topic: Non-Preemptive Scheduling
Subject: Operating System
GATE Relevance: High
Type: Numerical-based Questions (Process Scheduling)
📘 What is Non-Preemptive Scheduling?
In Non-Preemptive Scheduling, once a process starts executing on the CPU, it cannot be stopped until it completes.
✅ The CPU is assigned only after the current process finishes.
This contrasts with Preemptive scheduling, where the CPU can be taken away by a higher-priority task.
✅ Common Non-Preemptive Scheduling Algorithms:
| Algorithm | Description |
|---|---|
| FCFS (First Come First Serve) | Processed in order of arrival |
| SJF (Shortest Job First) | Process with shortest burst time first |
| Priority Scheduling (Non-preemptive) | Highest priority (lowest number) runs first |
📝 Solved GATE-Style Question: SJF (Non-Preemptive)
📌 Question:
Consider the following processes:
| Process | Arrival Time | Burst Time |
|---|---|---|
| P1 | 0 ms | 8 ms |
| P2 | 1 ms | 4 ms |
| P3 | 2 ms | 9 ms |
| P4 | 3 ms | 5 ms |
Use Non-Preemptive SJF Scheduling.
Find the Average Waiting Time and Average Turnaround Time.
🧮 Step 1: Gantt Chart (SJF – Shortest Job First)
Initial time = 0, only P1 has arrived ⇒ Execute P1
At time 8, P2, P3, P4 have arrived
Shortest burst time among them = P2 (4ms)
Next = P4 (5ms)
Then = P3 (9ms)
📊 Gantt Chart:
📋 Step 2: Calculate Completion, Turnaround, and Waiting Times
| Process | Arrival Time | Burst Time | Completion Time | Turnaround Time (CT – AT) | Waiting Time (TAT – BT) |
|---|---|---|---|---|---|
| P1 | 0 | 8 | 8 | 8 | 0 |
| P2 | 1 | 4 | 12 | 11 | 7 |
| P3 | 2 | 9 | 26 | 24 | 15 |
| P4 | 3 | 5 | 17 | 14 | 9 |
🧠 Final Answers:
-
Average Waiting Time = (0 + 7 + 15 + 9) / 4 = 7.75 ms
-
Average Turnaround Time = (8 + 11 + 24 + 14) / 4 = 14.25 ms
📦 Summary:
| Metric | Value |
|---|---|
| Avg Waiting Time | 7.75 ms |
| Avg Turnaround Time | 14.25 ms |
💡 Key Concepts to Remember:
-
In Non-preemptive SJF, CPU waits for a process to finish before giving it to the next.
-
Always update available processes at each CPU idle time.
-
For equal burst times, use arrival time as tie-breaker.
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