Round Robin scheduling in Operating Systems - Step-by-Step Execution

Concept Flow - Round Robin scheduling

Start with ready queue

↓

Pick first process

↓

Run process for time quantum

↓

Process finished?

Yes→Remove from queue

No↓

Move process to end of queue

↓

More processes in queue?

Yes→Pick next process

No↓

End

Round Robin scheduling cycles through processes, giving each a fixed time slice, then moves unfinished processes to the queue's end until all finish.

Execution Sample

Operating Systems

Processes = [P1(5), P2(3), P3(7)]
Time Quantum = 2
While processes remain:
  Run current process for 2 units
  If done, remove
  Else, move to end

This simulates running three processes with different burst times using a 2-unit time slice each cycle.

Analysis Table

Step	Current Process	Time Run	Remaining Time Before	Remaining Time After	Queue State	Action
1	P1	2	5	3	[P1, P2, P3]	Run P1 for 2, move to end
2	P2	2	3	1	[P2, P3, P1]	Run P2 for 2, move to end
3	P3	2	7	5	[P3, P1, P2]	Run P3 for 2, move to end
4	P1	2	3	1	[P1, P2, P3]	Run P1 for 2, move to end
5	P2	1	1	0	[P2, P3, P1]	Run P2 for 1, finish and remove
6	P3	2	5	3	[P3, P1]	Run P3 for 2, move to end
7	P1	1	1	0	[P1, P3]	Run P1 for 1, finish and remove
8	P3	2	3	1	[P3]	Run P3 for 2, move to end
9	P3	1	1	0	[P3]	Run P3 for 1, finish and remove
10	-	-	-	-	[]	All processes finished, stop

💡 All processes have zero remaining time, queue is empty, scheduling ends.

State Tracker

Variable	Start	After 1	After 2	After 3	After 4	After 5	After 6	After 7	After 8	After 9	Final
P1 Remaining Time	5	3	3	1	1	1	1	0	0	0	0
P2 Remaining Time	3	3	1	1	1	0	0	0	0	0	0
P3 Remaining Time	7	7	7	5	5	5	3	3	1	0	0
Queue State	[P1, P2, P3]	[P2, P3, P1]	[P3, P1, P2]	[P1, P2, P3]	[P2, P3, P1]	[P3, P1]	[P1, P3]	[P3]	[P3]	[]	[]

Key Insights - 3 Insights

Why does a process move to the end of the queue after its time quantum?

If the process is not finished after its time quantum (see steps 1, 2, 3), it must wait its turn again, so it moves to the queue's end to allow others to run.

What happens when a process finishes before using the full time quantum?

At step 5, P2 only needs 1 unit to finish, so it runs less than the quantum and is removed immediately from the queue.

How do we know when the scheduling ends?

When the queue is empty (step 10), all processes have finished, so the scheduler stops running.

Visual Quiz - 3 Questions

Look at the execution table, what is P1's remaining time after step 4?

A 3

B 0

C 1

D 5

💡 Hint

Check the 'Remaining Time After' column for P1 at step 4 in the execution_table.

At which step does P2 finish execution?

A Step 5

B Step 4

C Step 7

D Step 9

💡 Hint

Look for when P2's remaining time becomes 0 in the execution_table.

If the time quantum was increased to 3, how would the queue state after step 1 change?

A [P2, P3, P1]

B [P3, P1, P2]

C [P1, P2, P3]

D [P2, P1, P3]

💡 Hint

With a 3-unit quantum, P1 would run for 3 units, reducing its remaining time from 5 to 2, then move to the end; check how this affects queue order.

Concept Snapshot

Round Robin Scheduling:
- Each process runs for a fixed time slice (quantum).
- If unfinished, process moves to queue end.
- Continues cycling until all finish.
- Fair and simple for time-sharing systems.
- Prevents any process from starving.
- Time quantum size affects responsiveness and overhead.

Full Transcript

Round Robin scheduling works by cycling through all processes in the ready queue, giving each a fixed time slice to run. If a process finishes within its time slice, it is removed from the queue. If not, it is moved to the end of the queue to wait for its next turn. This continues until all processes have completed. The execution table shows step-by-step how processes run, their remaining times, and how the queue changes. Key moments include understanding why processes move to the queue's end, what happens when a process finishes early, and when the scheduling ends. The visual quiz tests understanding of remaining times, finishing steps, and effects of changing the time quantum.

Practice

1. What is the main idea behind Round Robin scheduling in operating systems?

easy

A. The shortest job runs first until completion.

B. Processes are run based on their priority levels.

C. Each process gets an equal fixed time slice to run in turns.

D. Processes run only when they request CPU time.

Solution

1. Step 1: Understand Round Robin scheduling basics

   Round Robin scheduling assigns each process a fixed time slice called a quantum, and processes run in a cyclic order.

2. Step 2: Compare options with the definition

   Only "Each process gets an equal fixed time slice to run in turns." correctly describes this fixed time slice and cyclic turn-taking approach.

3. Final Answer:

   Each process gets an equal fixed time slice to run in turns. -> Option C

4. Quick Check:

   Round Robin = fixed time slice per process [OK]

Hint: Round Robin means equal time slices in a cycle [OK]

Common Mistakes:

• Confusing Round Robin with priority scheduling
• Thinking shortest job runs first
• Assuming processes run only on request

2. Which of the following is the correct way to represent the time quantum in Round Robin scheduling?

easy

A. A fixed time interval each process runs before switching.

B. The total time a process needs to complete.

C. The priority level assigned to a process.

D. The time a process waits before starting.

Solution

1. Step 1: Define time quantum in Round Robin

   The time quantum is the fixed time interval given to each process to run before the CPU switches to the next process.

2. Step 2: Eliminate incorrect options

   Options B, C, and D describe other concepts like total burst time, priority, and waiting time, not the time quantum.

3. Final Answer:

   A fixed time interval each process runs before switching. -> Option A

4. Quick Check:

   Time quantum = fixed run time per process [OK]

Hint: Time quantum is the fixed run time slice [OK]

Common Mistakes:

• Mixing time quantum with total process time
• Confusing quantum with priority
• Thinking quantum is waiting time

3. Consider three processes P1, P2, and P3 with burst times 5, 3, and 8 units respectively. Using Round Robin scheduling with a time quantum of 3 units, what is the order of process execution in the first two cycles?

medium

A. P1, P3, P2, P1, P2, P3

B. P3, P1, P2, P3, P1, P2

C. P2, P1, P3, P2, P1, P3

D. P1, P2, P3, P1, P3, P3

Solution

1. Step 1: Calculate first cycle execution

   Each process runs for 3 units or less if burst time is less. P1 runs 3 (remaining 2), P2 runs 3 (done), P3 runs 3 (remaining 5).

2. Step 2: Calculate second cycle execution

   Next, P1 runs remaining 2 (done), P3 runs 3 (remaining 2), then P3 runs remaining 2 (done).

3. Final Answer:

   P1, P2, P3, P1, P3, P3 -> Option D

4. Quick Check:

   Round Robin cycles through processes with quantum 3 [OK]

Hint: Run each process max quantum, repeat until done [OK]

Common Mistakes:

• Not updating remaining burst times correctly
• Mixing process order in cycles
• Assuming processes finish in one quantum

4. A Round Robin scheduler has a time quantum of 4 units. A process with burst time 6 units is scheduled. The process runs for 6 units without interruption. What is the likely error in the scheduling?

medium

A. The process voluntarily gave up CPU before quantum ended.

B. The time quantum was ignored; process should have been preempted after 4 units.

C. The scheduler used priority instead of Round Robin.

D. The process was too short to be preempted.

Solution

1. Step 1: Understand expected Round Robin behavior

   With quantum 4, a process running longer than 4 units should be preempted after 4 units to allow others to run.

2. Step 2: Analyze the given scenario

   The process ran full 6 units without interruption, which means the scheduler did not preempt it as expected.

3. Final Answer:

   The time quantum was ignored; process should have been preempted after 4 units. -> Option B

4. Quick Check:

   Quantum ignored means no preemption [OK]

Hint: Process must be preempted after quantum expires [OK]

Common Mistakes:

• Assuming short processes don't get preempted
• Confusing voluntary yield with scheduler preemption
• Ignoring time quantum enforcement

5. In a Round Robin system, if the time quantum is set too large, what is the most likely effect on system performance?

hard

A. It behaves like First-Come-First-Served, causing longer wait times for some processes.

B. Processes switch too frequently, increasing overhead.

C. All processes finish faster due to longer CPU bursts.

D. The system becomes unfair by always running the shortest job first.

Solution

1. Step 1: Understand effect of large time quantum

   If the quantum is very large, each process runs almost to completion before switching, similar to First-Come-First-Served scheduling.

2. Step 2: Analyze performance impact

   This causes longer wait times for other processes and reduces the fairness and responsiveness of Round Robin.

3. Final Answer:

   It behaves like First-Come-First-Served, causing longer wait times for some processes. -> Option A

4. Quick Check:

   Large quantum = FCFS behavior, longer waits [OK]

Hint: Large quantum makes Round Robin act like FCFS [OK]

Common Mistakes:

• Thinking large quantum reduces overhead
• Assuming all processes finish faster
• Confusing with shortest job first scheduling