written 8.6 years ago by
snehalb
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a) FCFS scheduling:
Here the processes are scheduled as per the order in which they arrive.
Gantt chart:
Waiting time for P1=0
Waiting time for P2=10
Waiting time for P3=11
Waiting time for P4=13
Waiting time for P5=14
Average waiting time$=\frac{Sum of waiting time for individual process}{No. of processes}$
$=\frac{0+10+11+13+14}{5} = 9.6 $
Average execution time$=\frac{Sum of execution time for individual process}{No. of processes}$ $=\frac{10+1+2+1+5}{5} = 3.8$
Average turn-around time = Average waiting time + Average execution time
= 9.6 + 3.8
= 13.4
b) SJF non-preemptive scheduling:
Here the processes are scheduled as per their burst time with the least timed-process being given the maximum priority.
Waiting time for P1=9
Waiting time for P2=0
Waiting time for P3=2
Waiting time for P4=1
Waiting time for P5=4
Average waiting time$=\frac{Sum of waiting time for individual process}{No. of processes}$ $=\frac{9+0+2+1+4}{5} = 3.2$
Average execution time$=\frac{Sum of execution time for individual process}{No. of processes}$ $=\frac{10+1+2+1+5}{5} = 3.8$
Average turn-around time= Average waiting time + Average execution time
= 3.2 + 3.8
= 7
c) Priority scheduling :
Here the scheduling happens on the basis of priority number assigned to each process. (Here we would be considering Priority=1 as highest priority)
Waiting time for P1=6
Waiting time for P2=0
Waiting time for P3=16
Waiting time for P4=18
Waiting time for P5=1
Average waiting time$=\frac{Sum of waiting time for individual process}{No. of processes}$ $=\frac{6+0+16+18+1}{5} = 8.2$
Average execution time$=\frac{Sum of execution time for individual process }{No. of processes}$
$=\frac{10+1+2+1+5}{5} = 3.8$
Average turn-around time= Average waiting time + Average execution time
= 8.2 + 3.8
= 12
d) Round Robin scheduling:
Here every process executes in the FCFS for the given time quantum. This is a pre-emptive method of scheduling. Here time quantum =1
Waiting time for $P1=(0+(5-1)+(8-6)+(10-9)+(12-11)+(14-13))=9$
Waiting time for P2=1
Waiting time for $P3=2+(6-3)=5$
Waiting time for P4=3
Waiting time for $P5=4+(7-5)+(9-8)+(11-10)+(13-12)=9$
Average waiting time$=\frac{Sum of waiting time for individual process}{No. of processes} $ $=\frac{9+1+5+3+9}{5} = 5.4$
Average execution time$=\frac{Sum of execution time for individual process}{No. of processes}$ $=\frac{10+1+2+1+5}{5} = 3.8 $
Average turn-around time= Average waiting time + Average execution time
= 5.4 + 3.8
= 9.2