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homework3 Version 0
👤 Author: by 1589080613qqcom 2019-10-22 00:29:00
Suppose the burst time is as follows:
P1:4  P2:5 P3:9 P4:6 P5:2
for FCFS: the average waiting time is:(4+9+18+24)/5=10
for Shortest-job-First:  the average waiting time is :7.2

then let us add arrival time for each process like:
P1:1 P2:2 P3:9 P4:0 P5:13
then for Shortest-job-First algorithm the average waiting time is:(1+5+10+0+0)/5=3.2

and for Priority Scheduling we add priorities for each processes:
P1:1 P2:2 P3:4 P4:3 P5:5
the average waiting time is :(4+9+15+24)/5=10.4

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