Exercise5.4 (28.218.81608.81609.81613.222.81848.81877.81906)
FCFS: P1 P2 P3 P4 P5 0 10 11 13 14 19 SJF： P2 P4 P3 P5 P1 0 1 2 4 9 19 Nonpreemptive priproty： P2 P5 P1 P3 P4 0 1 6 16 18 19 RR: P1 P2 P3 P4 P5 P1 P3 P5 P1 P5 P1 P5 P1 P5 P1 P1 P1 P1 P1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 b.What is the turnaround time of each process for each of the scheduling algorithums in part a? Answer： FCFS RR SJF Priority P1 10 19 19 16 P2 11 2 1 1 P3 13 7 4 18 P4 14 4 2 19 P5 19 14 9 6 c.What is the waiting time of each process for each of the scheduling algorithms in part a? Answer： FCFS RR SJF Priority P1 0 9 9 6 P2 10 1 0 0 P3 11 5 2 16 P4 13 3 1 18 P5 14 9 4 1 d.Which of the algorithms in part a results in the minimum average waiting time (over all processes)? Answer：FCFS AVG=(0+10+11+13+14)/5=9.6 SJF AVG=(0+9+2+1+4)/5=3.2 RR AVG=(9+1+5+3+9)/5=5.4 PRIORITY AVG=(6+0+16+18+1)=8.2 So the SJF algorithms has the minimum average waiting time.
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