Question
Consider three processes P1, P2, and P3 with burst times
10, 5, and 8 respectively. All arrive at time 0. Using the Shortest Job First (SJF) non-preemptive scheduling algorithm, what is the average waiting time?Solution
Processes: P1 (10), P2 (5), P3 (8) Â Â Â Â Â Â Order of execution (SJF non-preemptive): P2, P3, P1 Â Â Â Â Â Â P2: Starts at 0, finishes at 5. Waiting time = 0. Â Â Â Â Â Â P3: Starts at 5, finishes at 13. Waiting time = 5. Â Â Â Â Â Â P1: Starts at 13, finishes at 23. Waiting time = 13. Â Â Â Â Â Â Total waiting time = 0 + 5 + 13 = 18 Â Â Â Â Â Â Average waiting time = 18 / 3 = 6.0
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