Round Robin MCQ Quiz in தமிழ் - Objective Question with Answer for Round Robin - இலவச PDF ஐப் பதிவிறக்கவும்

Concept:

The round-robin (RR) scheduling algorithm is designed especially for timesharing systems.

It is similar to FCFS scheduling, but preemption is added to enable the system to switch between processes.

A small unit of time, called a time quantum or time slice, is defined.

The ready queue is treated as a circular queue. The CPU scheduler goes around the ready queue, allocating the CPU to each process for a time interval of up to 1-time​ quantum.

Time quantum = 5 ms

Gantt Chart:

0       5        10     15       20      25      30      35     40       45      50      55      60      65

Calculation

Average turn around time = \(\frac{20 + 35 + 50 + 60 + 65}{5} = 46\)

Since s< q < r, so number of switches required = r/q

Time wasting during switching = s. r / q

Thus,  r + sr/ q = (qr + sr) /q = r(q + s) / q  is the actual time during which CPU is busy, where r is the actual time used for the process by CPU

Concept:

Time quantum is defined in round robin scheduling algorithm.

Explanation:

• The period of time for which a process is allowed to run in a pre-emptive multitasking system is generally called the time slice or quantum.
• Each process is assigned a fixed time (Time Quantum/Time Slice) in cyclic way. It is designed especially for the time-sharing system. The ready queue is treated as a circular queue.
• The CPU scheduler goes around the ready queue, allocating the CPU to each process for a time interval of up to 1-time quantum.
• If time quantum for Round Robin scheduling is very large, then it behaves same as FCFS scheduling.

Concept:

• By logic If we execute process with least burst time, algorithm will give minimum average waiting time hence P3 should be execute first then P2 and then P1.

Calculation:

Queue :

Process table:

\({\rm{Average\;waiting\;time}} = {\rm{\;}}\frac{{5\; + \;4\; + \;0}}{3} = 3{\rm{\;unit}}\) 

Gang chart:

Hence option 3 is the correct answer.

Gantt Chart with time slice of 2ms

0        2          4          5         7          9         11       12

Process Table: (time in milliseconds)

Average TAT = \(\frac{{17}}{3}\) = 5.66

Average WT = \(\frac{5}{3}\) = 1.66

Concept:

The round-robin (RR) scheduling algorithm is designed especially for timesharing systems.

It is similar to FCFS scheduling, but preemption is added to enable the system to switch between processes.

A small unit of time, called a time quantum or time slice, is defined.

The ready queue is treated as a circular queue. The CPU scheduler goes around the ready queue, allocating the CPU to each process for a time interval of up to 1-time​ quantum.

Gantt Chart:

Calculation

Turn around time = completion time – arrival time

Turn around time for

P1 = 11 – 0 = 11

P2 = 10 – 3 = 7

P3 = 9 – 2 = 7

For SJF, Gantt chart is

Here, the order of completion is  -> P1 – P3 – P2

For Round Robin, Gantt chart is

The correct answer is : Option 3

Key Points

Scheduling Algorithm: First Come First Serve (FCFS)

Gantt chart:

0 5 8 10 13 14

Process Table:

Average Turnaround Time: \({{5+6+8+8+8} \over 5} = 7.0\)

 Important Points:

TAT = CT - AT

The correct answer is option 2.

CONCEPT:

Highest Response Ratio Next (HRN):

HRN is a non-preemptive scheduling algorithm where the process with the highest response ratio is selected next. The response ratio is calculated as:

Response Ratio = (Waiting Time + Burst Time) / Burst Time

Formula:

Turnaround Time (TAT) = Completion Time (CT) - Arrival Time (AT)

Weighted TAT = Turnaround Time / Burst Time

Steps:

1. Select the process with the highest response ratio among all arrived processes.
2. Compute completion time, TAT, and weighted TAT.
3. Repeat until all processes are scheduled.

Calculated weighted turnaround times:

Average Weighted Turnaround Time = \({{1+ 2.5+2.67+2.67+8} \over 5 } = 2.834\)

∴ Hence, the correct answer is 2.834 ms

The correct answer is 76 ms

Key Points

• In round-robin scheduling, every process gets an equal share of CPU time, which is defined by the time quantum or time slice. Assuming there's no context switching time, each process will get to use the CPU for the duration of the time quantum then get placed at the end of the ready queue.
• In this case, the time quantum is 4 ms and there are 20 processes in total.
• Since one cycle for all the processes to get CPU once equals to "number of processes" times "time quantum", this means one cycle will take 20*4 = 80 ms.
• Given that CPU is free now, the first process doesn't have to wait, so it gets the CPU immediately. But the last process, i.e., the 20th process has to wait for 19 processes (not full 20, because the first process didn't wait) to complete their quanta before getting CPU. That means
• Maximum wait time = (Number of processes - 1) * Time quantum
• Maximum waiting time for a process is 19*4 = 76 ms.

Therefore, option 2), 76ms is correct.