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Consider the processes P1, P2, P3, P4 given in the below table, arrives for execution in the same order, with Arrival Time 0, and given Burst Time.
written 5.7 years ago by | • modified 3.1 years ago |
Draw the Gantt chart and find the average waiting time using the FCFS and SJF (Non-Pre-emptive) scheduling algorithm.
Process | Burst Time |
---|---|
P0 | 21 |
P1 | 3 |
P2 | 6 |
P3 | 2 |
Subject: Operating System
Topic : Process And Process Scheduling
Difficulty: Hard
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