Microcomputer Operating 1. Suppose there are 5 jobs in the ready queue. Processe
ID: 672632 • Letter: M
Question
Microcomputer Operating
1. Suppose there are 5 jobs in the ready queue. Processes’ arrival time and burst time are shown as below:
Process Arrival Time Burst Time
P1 0.0 11
P2 4.0 8
P3 5.0 5
P4 7.0 6
P5 9.0 1
All the processes are non-preemptive. Show your show solution in Gantt chart for SJF Scheduling and calculate the average waiting time.
2. Suppose there are 5 jobs in the ready queue. Processes’ arrival time and burst time are shown as below:
Process Arrival Time Burst Time
P1 0.0 11
P2 4.0 8
P3 5.0 5
P4 7.0 6
P5 9.0 1
All the processes are preemptive. Show your show solution in Gantt chart for SJF Scheduling and calculate the average waiting time.
Explanation / Answer
Microcomputer Operating
1. Suppose there are 5 jobs in the ready queue. Processes’ arrival time and burst time are shown as below:
Process Arrival Time Burst Time
P1 0.0 11
P2 4.0 8
P3 5.0 5
P4 7.0 6
P5 9.0 1
All the processes are non-preemptive. Show your show solution in Gantt chart for SJF Scheduling and calculate the average waiting time.
Solution :
----------------------------------------------------------------------------------------
| P1 | p5 | p3 | p4 | P2 | |
| | | | | |
----------------------------------------------------------------------------------------
0 11 12 17 23 31
Avg Time is :
P1 Wait Time : 0
P2 Wait Time : 23 - 8 = 15
P3 Wait Time : 12 - 5 = 7
P4 Wait Time : 17 - 6 = 11
P5 Wait Time : 11 - 1 = 10
Total Avg Time is : (0 + 15 + 7 + 11 +10) / 5 = 8.6
2. Suppose there are 5 jobs in the ready queue. Processes’ arrival time and burst time are shown as below:
Process Arrival Time Burst Time
P1 0.0 11 11-4 = 7
P2 4.0 8 8- 1 = 7
P3 5.0 5 5 - 2 = 3
P4 7.0 6 7 - 2 = 5
P5 9.0 1 0
All the processes are preemptive. Show your show solution in Gantt chart for SJF Scheduling and calculate the average waiting time.
Solution :
Remaining Time Wait Time
-----------------------------------------------
| P1 | P2 | P3 | P3 | P5 | P4 | P2 | P1 |
| | | | | | | | | |
-----------------------------------------------
0 4 5 7 9 10 15 22 29
Waiting Time :
P1 Waiting Time : End Of P1 Process time Complete - Beginning Time of P1 Process Next time In chat = 22 - 4 = 18
P2 Waiting Time : End Of P2 Process time Complete - Beginning Time of P2 Process Next time In chat = 15 - 5 = 10
P3 Waiting Time : End Of P3 Process time Complete - Beginning Time of P3 Process Next time In chat = 2
P4 Waiting Time : 5
P5 Waiting Time : 0
Total Avg Time is : ( 18 + 10 + 2 + 5+ 0) / 5 = 7