Tag: Round Robin Scheduling Example

Round Robin | Round Robin Scheduling | Examples

Round Robin Scheduling-

 

In Round Robin Scheduling,

  • CPU is assigned to the process on the basis of FCFS for a fixed amount of time.
  • This fixed amount of time is called as time quantum or time slice.
  • 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.
  • It is always preemptive in nature.

 

Round Robin Scheduling is FCFS Scheduling with preemptive mode.

 

 

Advantages-

 

  • It gives the best performance in terms of average response time.
  • It is best suited for time sharing system, client server architecture and interactive system.

 

Disadvantages-

 

  • It leads to starvation for processes with larger burst time as they have to repeat the cycle many times.
  • Its performance heavily depends on time quantum.
  • Priorities can not be set for the processes.

 

Important Notes-

 

Note-01:

 

With decreasing value of time quantum,

  • Number of context switch increases
  • Response time decreases
  • Chances of starvation decreases

 

Thus, smaller value of time quantum is better in terms of response time.

 

Note-02:

 

With increasing value of time quantum,

  • Number of context switch decreases
  • Response time increases
  • Chances of starvation increases

 

Thus, higher value of time quantum is better in terms of number of context switch.

 

Note-03:

 

  • With increasing value of time quantum, Round Robin Scheduling tends to become FCFS Scheduling.
  • When time quantum tends to infinity, Round Robin Scheduling becomes FCFS Scheduling.

 

Also read- FCFS Scheduling

 

Note-04:

 

  • The performance of Round Robin scheduling heavily depends on the value of time quantum.
  • The value of time quantum should be such that it is neither too big nor too small.

 

PRACTICE PROBLEMS BASED ON ROUND ROBIN SCHEDULING-

 

Problem-01:

 

Consider the set of 5 processes whose arrival time and burst time are given below-

 

Process Id Arrival time Burst time
P1 0 5
P2 1 3
P3 2 1
P4 3 2
P5 4 3

 

If the CPU scheduling policy is Round Robin with time quantum = 2 unit, calculate the average waiting time and average turn around time.

 

Solution-

 

Gantt Chart-

 

Ready Queue-

P5, P1, P2, P5, P4, P1, P3, P2, P1

 

 

Now, we know-

  • Turn Around time = Exit time – Arrival time
  • Waiting time = Turn Around time – Burst time

 

Also read- Various Times of Process

 

Process Id Exit time Turn Around time Waiting time
P1 13 13 – 0 = 13 13 – 5 = 8
P2 12 12 – 1 = 11 11 – 3 = 8
P3 5 5 – 2 = 3 3 – 1 = 2
P4 9 9 – 3 = 6 6 – 2 = 4
P5 14 14 – 4 = 10 10 – 3 = 7

 

Now,

  • Average Turn Around time = (13 + 11 + 3 + 6 + 10) / 5 = 43 / 5 = 8.6 unit
  • Average waiting time = (8 + 8 + 2 + 4 + 7) / 5 = 29 / 5 = 5.8 unit

 

Problem-02:

 

Consider the set of 6 processes whose arrival time and burst time are given below-

 

Process Id Arrival time Burst time
P1 0 4
P2 1 5
P3 2 2
P4 3 1
P5 4 6
P6 6 3

 

If the CPU scheduling policy is Round Robin with time quantum = 2, calculate the average waiting time and average turn around time.

 

Solution-

 

Gantt chart-

 

Ready Queue-

P5, P6, P2, P5, P6, P2, P5, P4, P1, P3, P2, P1

 

 

Now, we know-

  • Turn Around time = Exit time – Arrival time
  • Waiting time = Turn Around time – Burst time

 

Process Id Exit time Turn Around time Waiting time
P1 8 8 – 0 = 8 8 – 4 = 4
P2 18 18 – 1 = 17 17 – 5 = 12
P3 6 6 – 2 = 4 4 – 2 = 2
P4 9 9 – 3 = 6 6 – 1 = 5
P5 21 21 – 4 = 17 17 – 6 = 11
P6 19 19 – 6 = 13 13 – 3 = 10

 

Now,

  • Average Turn Around time = (8 + 17 + 4 + 6 + 17 + 13) / 6 = 65 / 6 = 10.84 unit
  • Average waiting time = (4 + 12 + 2 + 5 + 11 + 10) / 6 = 44 / 6 = 7.33 unit

 

Problem-03:

 

Consider the set of 6 processes whose arrival time and burst time are given below-

 

Process Id Arrival time Burst time
P1 5 5
P2 4 6
P3 3 7
P4 1 9
P5 2 2
P6 6 3

 

If the CPU scheduling policy is Round Robin with time quantum = 3, calculate the average waiting time and average turn around time.

 

Solution-

 

Gantt chart-

 

Ready Queue-

P3, P1, P4, P2, P3, P6, P1, P4, P2, P3, P5, P4

 

 

Now, we know-

  • Turn Around time = Exit time – Arrival time
  • Waiting time = Turn Around time – Burst time

 

Process Id Exit time Turn Around time Waiting time
P1 32 32 – 5 = 27 27 – 5 = 22
P2 27 27 – 4 = 23 23 – 6 = 17
P3 33 33 – 3 = 30 30 – 7 = 23
P4 30 30 – 1 = 29 29 – 9 = 20
P5 6 6 – 2 = 4 4 – 2 = 2
P6 21 21 – 6 = 15 15 – 3 = 12

 

Now,

  • Average Turn Around time = (27 + 23 + 30 + 29 + 4 + 15) / 6 = 128 / 6 = 21.33 unit
  • Average waiting time = (22 + 17 + 23 + 20 + 2 + 12) / 6 = 96 / 6 = 16 unit

 

Problem-04:

 

Four jobs to be executed on a single processor system arrive at time 0 in the order A, B, C, D. Their burst CPU time requirements are 4, 1, 8, 1 time units respectively. The completion time of A under round robin scheduling with time slice of one time unit is-

  1. 10
  2. 4
  3. 8
  4. 9

 

Solution-

 

Process Id Arrival time Burst time
A 0 4
B 0 1
C 0 8
D 0 1

 

Gantt chart-

 

Ready Queue-

C, A, C, A, C, A, D, C, B, A

 

 

Clearly, completion time of process A = 9 unit.

Thus, Option (D) is correct.

 

To gain better understanding about Round Robin Scheduling,

Watch this Video Lecture

 

Next Article- Priority Scheduling

 

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