Round Robin CPU Scheduling Problem

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written 2.1 years ago by

binitamayekar ★ 6.4k

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Round Robin (RR) CPU Scheduling

In RR scheduling CPU is assigned to the process based on FCFS for a fixed amount of time called time quantum.
After the time quantum expires, the running process is preempted and sent to the ready queue.
Then, the processor is assigned to the next arrived process.
This process continues until all the processes complete their execution.

In the given problem total of 5 jobs named P0, P1, P2, P3, and P4 arrived in the sequence.

The system works based on Round Robin (RR) CPU Scheduling with Time Quantum = 15

Gantt Chart for RR

General explanation of how to create this Gantt Chart (only for Understanding Purpose)

Step 1 -

At time 0, process P0 arrives in the ready queue and executes its tasks for time quantum 15 units.
During 15 units of the time slice, another process P1 and P2 arrived in the ready queue.
After that, process P0 will return to the end of the ready queue and await its execution.

Step 2 -

Now, process P1 starts its execution for a time quantum of 15 units.
Then process P2 starts its execution for a time slot of 10 units because the Burst Time is 10, and it does not go to the ready queue for further execution.

Step 3 -

Now, process P0 will start its execution again for 15 units because no new process arrived till the time of 40 ms or unit.
Then process P1 will start its execution again for the remaining 5 units and completes its execution.

Step 4 -

Now, process P0 will start its execution again for 15 units because no new process arrived till the time of 60 ms or unit.
After completing this execution process P0 will start its execution again for the next 15 units because no new process arrived till the time of 75 ms or unit.
But, during this 15 units of the time slice, another process P3 and P4 arrived in the ready queue.

Step 5 -

Step 6 -

Now, process P3 will start its execution again for the remaining 5 units and completes its execution.
After that process, P4 starts its execution for a time quantum of 15 units.
Finally, process P0 will start its execution again for the remaining 5 units and completes its execution.

Step 7 -

Now, process P4 is the only remaining process to complete its execution.
The process P4 completes its remaining execution in 15 unit and 5 unit time slices.

Formulae -

$$Turn\ Around\ Time = Process\ Completion\ Time – Process\ Arrival\ Time$$

$$Waiting\ Time = Turn\ Around\ Time – Burst\ Time$$

Calculate the Turnaround Time and waiting Time for every process in the tabular format as follows:

From the Gantt Chart shown above process completion time for each process can be confirmed.
- Process P0 completes its execution at 160 ms or units
- Process P1 completes its execution at 60 ms or units
- Process P2 completes its execution at 40 ms or units
- Process P3 completes its execution at 140 ms or units
- Process P4 completes its execution at 180 ms or units
That means process completes their execution in the sequence of P2 -> P1 -> P3 -> P0 -> P4

Process	Burst Time	Arrival Time	Process Completion Time	Turn Around Time	Waiting Time
P0	80	0	160	160 – 0 = 160	160 – 80 = 80
P1	20	10	60	60 – 10 = 50	50 – 20 = 30
P2	10	10	40	40 – 10 = 30	30 – 10 = 20
P3	20	80	140	140 – 80 = 60	60 – 20 = 40
P4	50	85	180	180 – 85 = 95	95 – 50 = 45

Now,

Calculate the Average Waiting Time

Average Waiting time = (80 + 30 + 20 + 40 + 45) / 5 = 43 ms or units

Similarly we can also compute Average Turnaround Time based on the above data

Average Turn Around Time = (160 + 50 + 30 + 60 + 95) / 5 = 79 ms or units

i] Gantt Chart illustrated that all processes finishes its execution in the sequence of P2 -> P1 -> P3 -> P0 -> P4

ii] Turnaround Time for process P3 = 60 ms or units

iii] Average Wait Time = 43 ms or units

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