Round Robin Algorithm.pdf
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The Round Robin algorithm schedules processes by allowing each one to run for a fixed time quantum before being preempted. This document provides an example of using Round Robin scheduling with 6 processes (P1-P6) and a time quantum of 4 units. It calculates the Gantt chart, ready queue, completion times, turnaround times, waiting times and average waiting and turnaround times. The average waiting time is 11.33 ms and average turnaround time is 15.33 ms.
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- 1. Round Robin Algorithm Scheduling Algorithm RR Scheduling Example: There are six processes named as P1, P2, P3, P4, P5 and P6. Their arrival time and burst time are given below in the table. The time quantum of the system is 4 units. Calculate Average waiting time and average turn around time. Process ID Arrival Time Burst Time 1 0 5 2 1 6 3 2 3 4 3 1 5 4 5 6 6 4 Answer: According to the algorithm, we have to maintain the ready queue and the Gantt chart. The structure of both the data structures will be changed after every scheduling. Ready Queue: Initially, at time 0, process P1 arrives which will be scheduled for the time slice 4 units. Hence in the ready queue, there will be only one process P1 at starting with CPU burst time 5 units. P1 5 GANTT chart: The P1 will be executed for 4 units first. P1 0 4 Ready Queue Meanwhile the execution of P1, four more processes P2, P3, P4 and P5 arrives in the ready queue. P1 has not completed yet, it needs another 1 unit of time hence it will also be added back to the ready queue.
- 2. Process P2 P3 P4 P5 P1 Burst Time 6 3 1 5 1 GANTT chart After P1, P2 will be executed for 4 units of time which is shown in the Gantt chart. P1 P2 0 4 8 Ready Queue During the execution of P2, one more process P6 is arrived in the ready queue. Since P2 has not completed yet hence, P2 will also be added back to the ready queue with the remaining burst time 2 units. P3 P4 P5 P1 P6 P2 3 1 5 1 4 2 GANTT chart After P1 and P2, P3 will get executed for 3 units of time since its CPU burst time is only 3 seconds. P1 P2 P3 0 4 8 12 Ready Queue Since P3 has been completed, hence it will be terminated and not be added to the ready queue. The next process will be executed is P4. P4 P5 P1 P6 P2 1 5 1 4 2 GANTT chart After, P1, P2 and P3, P4 will get executed. Its burst time is only 1 unit which is lesser then the time quantum hence it will be completed.
- 3. P1 P2 P3 P4 0 4 8 11 12 Ready Queue The next process in the ready queue is P5 with 5 units of burst time. Since P4 is completed hence it will not be added back to the queue. P5 P1 P6 P2 5 1 4 2 GANTT chart P5 will be executed for the whole time slice because it requires 5 units of burst time which is higher than the time slice. P1 P2 P3 P4 P5 0 4 8 11 12 16 Ready Queue P5 has not been completed yet; it will be added back to the queue with the remaining burst time of 1 unit. P1 P6 P2 P5 1 4 2 1 GANTT Chart The process P1 will be given the next turn to complete its execution. Since it only requires 1 unit of burst time hence it will be completed. P1 P2 P3 P4 P5 P1 0 4 8 11 12 16 17 Ready Queue P1 is completed and will not be added back to the ready queue. The next process P6 requires only 4 units of burst time and it will be executed next.
- 4. P6 P2 P5 4 2 1 GANTT chart P6 will be executed for 4 units of time till completion. P1 P2 P3 P4 P5 P1 P6 0 4 8 11 12 16 17 21 Ready Queue Since P6 is completed, hence it will not be added again to the queue. There are only two processes present in the ready queue. The Next process P2 requires only 2 units of time. P2 P5 2 1 GANTT Chart P2 will get executed again, since it only requires only 2 units of time hence this will be completed. P1 P2 P3 P4 P5 P1 P6 P2 0 4 8 11 12 16 17 21 23 Ready Queue Now, the only available process in the queue is P5 which requires 1 unit of burst time. Since the time slice is of 4 units hence it will be completed in the next burst. P5 1 GANTT chart P5 will get executed till completion. P1 P2 P3 P4 P5 P1 P6 P2 P5 0 4 8 11 12 16 17 21 23 24
- 5. As, we know to calculate: Turn Around Time = Completion Time - Arrival Time Waiting Time = Turn Around Time - Burst Time Process ID Arrival Time Burst Time Completion Time Turn Around Time (Completion Time – Arrival Time) Waiting Time (Turn Around Time – Arrival Time) 1 0 5 17 17 12 2 1 6 23 22 16 3 2 3 11 9 6 4 3 1 12 9 8 5 4 5 24 20 15 6 6 4 21 15 11 Avg Waiting Time = ∑ =(12+16+6+8+15+11)/6 =68/6 = 11.33 ms Avg Turn Around Time = ∑ =(17+22+9+9+20+15)/6 =92/6 = 15.33 ms Turn Around time of All Process Waiting Time of All Process No of Process No of Process