Explanation : Scheduling order will be
P2 , P3 , P1 , P5 , P1, P4
Waiting time of processes will be
P2 = 0
P3= 5-2=3
P1=10+2=12
P5=0
P4= 15
Average waiting time will be = (0+3+12+0+15)/5= 30/5=6ms
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Process Management Questions for UGC NET Computer Science