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V. Round Robin and SJF comparison 6 Process Arrival Time CPU Burst Time P1 0 9 P2 4 4 P3 6 P4 For each of the following algor

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Answer #1
process id P1 P1 P2 P3 P1 P4 P2 P4

0 3 6 9 10 13 17 18 20

process arrival time burst time completion time turn around time waiting time
P1 0 9 13 13 4
P2 4 4 18 14 10
P3 6 1 9 3 2
P4 8 6 20 12 6

Average waiting time=(4+10+2+6)/4=5.5

average turn around time= (13+14+3+12)/4=10.5

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