Computer Science Engineering (CSE) Exam > Computer Science Engineering (CSE) Questions > Consider three processes P1, P2 and P3 arrivi...
Start Learning for Free
Consider three processes P1, P2 and P3 arriving at 0ms, 4ms and 10ms respectively and have burst times 8ms, 4ms and 1ms respectively. Which of the following will have the highest average waiting time ?
- a)First Come First Serve (FCFS)
- b)Pre – emptive Shortest Job First
- c)Non Pre – emptive Shortest Job First
- d)Round Robin with quantum 1
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
Consider three processes P1, P2 and P3 arriving at 0ms, 4ms and 10ms r...
Round Robin with time quantum 1 will have the highest average waiting time of 3ms. Rest all will have average waiting time less than 3ms.
Most Upvoted Answer
Consider three processes P1, P2 and P3 arriving at 0ms, 4ms and 10ms r...
Answer:
To determine which scheduling algorithm will have the highest average waiting time, we need to consider the arrival times and burst times of the given processes. Let's analyze each scheduling algorithm and calculate the average waiting time for each.
First Come First Serve (FCFS):
In FCFS scheduling, the processes are executed in the order of their arrival times. Therefore, the average waiting time is calculated by summing up the waiting times of all processes and dividing it by the number of processes.
In this case, the waiting time for P1 is 0ms, for P2 is 4ms, and for P3 is 12ms. So the average waiting time for FCFS can be calculated as:
(0 + 4 + 12) / 3 = 16 / 3 = 5.33ms
Preemptive Shortest Job First:
In preemptive SJF scheduling, the process with the shortest remaining burst time is executed first. However, in this case, P2 arrives before P1 completes its execution. Therefore, P2 will start executing before P1, resulting in a lower average waiting time compared to FCFS.
Non-Preemptive Shortest Job First:
In non-preemptive SJF scheduling, the process with the shortest burst time is executed first. As the name suggests, the process will not be preempted until it completes its execution. Since P1 has the longest burst time, it will be executed first, resulting in a higher average waiting time compared to preemptive SJF and FCFS.
Round Robin with quantum 1:
In Round Robin scheduling, each process is given a fixed time quantum to execute. If a process doesn't complete within the time quantum, it is preempted and the next process is executed. In this case, all processes have burst times less than the time quantum, so each process will complete its execution in a single time quantum. As a result, the average waiting time for Round Robin with quantum 1 will be the same as the average waiting time for FCFS.
Therefore, the scheduling algorithm with the highest average waiting time among the given options is Round Robin with quantum 1.
To determine which scheduling algorithm will have the highest average waiting time, we need to consider the arrival times and burst times of the given processes. Let's analyze each scheduling algorithm and calculate the average waiting time for each.
First Come First Serve (FCFS):
In FCFS scheduling, the processes are executed in the order of their arrival times. Therefore, the average waiting time is calculated by summing up the waiting times of all processes and dividing it by the number of processes.
In this case, the waiting time for P1 is 0ms, for P2 is 4ms, and for P3 is 12ms. So the average waiting time for FCFS can be calculated as:
(0 + 4 + 12) / 3 = 16 / 3 = 5.33ms
Preemptive Shortest Job First:
In preemptive SJF scheduling, the process with the shortest remaining burst time is executed first. However, in this case, P2 arrives before P1 completes its execution. Therefore, P2 will start executing before P1, resulting in a lower average waiting time compared to FCFS.
Non-Preemptive Shortest Job First:
In non-preemptive SJF scheduling, the process with the shortest burst time is executed first. As the name suggests, the process will not be preempted until it completes its execution. Since P1 has the longest burst time, it will be executed first, resulting in a higher average waiting time compared to preemptive SJF and FCFS.
Round Robin with quantum 1:
In Round Robin scheduling, each process is given a fixed time quantum to execute. If a process doesn't complete within the time quantum, it is preempted and the next process is executed. In this case, all processes have burst times less than the time quantum, so each process will complete its execution in a single time quantum. As a result, the average waiting time for Round Robin with quantum 1 will be the same as the average waiting time for FCFS.
Therefore, the scheduling algorithm with the highest average waiting time among the given options is Round Robin with quantum 1.
|
Explore Courses for Computer Science Engineering (CSE) exam
|
|
Similar Computer Science Engineering (CSE) Doubts
Top Courses for Computer Science Engineering (CSE)View all
Consider three processes P1, P2 and P3 arriving at 0ms, 4ms and 10ms respectively and have burst times 8ms, 4ms and 1ms respectively. Which of the following will have the highest average waiting time ?a)First Come First Serve (FCFS)b)Pre – emptive Shortest Job Firstc)Non Pre – emptive Shortest Job Firstd)Round Robin with quantum 1Correct answer is option 'D'. Can you explain this answer?
Question Description
Consider three processes P1, P2 and P3 arriving at 0ms, 4ms and 10ms respectively and have burst times 8ms, 4ms and 1ms respectively. Which of the following will have the highest average waiting time ?a)First Come First Serve (FCFS)b)Pre – emptive Shortest Job Firstc)Non Pre – emptive Shortest Job Firstd)Round Robin with quantum 1Correct answer is option 'D'. Can you explain this answer? for Computer Science Engineering (CSE) 2025 is part of Computer Science Engineering (CSE) preparation. The Question and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus. Information about Consider three processes P1, P2 and P3 arriving at 0ms, 4ms and 10ms respectively and have burst times 8ms, 4ms and 1ms respectively. Which of the following will have the highest average waiting time ?a)First Come First Serve (FCFS)b)Pre – emptive Shortest Job Firstc)Non Pre – emptive Shortest Job Firstd)Round Robin with quantum 1Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Computer Science Engineering (CSE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Consider three processes P1, P2 and P3 arriving at 0ms, 4ms and 10ms respectively and have burst times 8ms, 4ms and 1ms respectively. Which of the following will have the highest average waiting time ?a)First Come First Serve (FCFS)b)Pre – emptive Shortest Job Firstc)Non Pre – emptive Shortest Job Firstd)Round Robin with quantum 1Correct answer is option 'D'. Can you explain this answer?.
Consider three processes P1, P2 and P3 arriving at 0ms, 4ms and 10ms respectively and have burst times 8ms, 4ms and 1ms respectively. Which of the following will have the highest average waiting time ?a)First Come First Serve (FCFS)b)Pre – emptive Shortest Job Firstc)Non Pre – emptive Shortest Job Firstd)Round Robin with quantum 1Correct answer is option 'D'. Can you explain this answer? for Computer Science Engineering (CSE) 2025 is part of Computer Science Engineering (CSE) preparation. The Question and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus. Information about Consider three processes P1, P2 and P3 arriving at 0ms, 4ms and 10ms respectively and have burst times 8ms, 4ms and 1ms respectively. Which of the following will have the highest average waiting time ?a)First Come First Serve (FCFS)b)Pre – emptive Shortest Job Firstc)Non Pre – emptive Shortest Job Firstd)Round Robin with quantum 1Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Computer Science Engineering (CSE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Consider three processes P1, P2 and P3 arriving at 0ms, 4ms and 10ms respectively and have burst times 8ms, 4ms and 1ms respectively. Which of the following will have the highest average waiting time ?a)First Come First Serve (FCFS)b)Pre – emptive Shortest Job Firstc)Non Pre – emptive Shortest Job Firstd)Round Robin with quantum 1Correct answer is option 'D'. Can you explain this answer?.
Solutions for Consider three processes P1, P2 and P3 arriving at 0ms, 4ms and 10ms respectively and have burst times 8ms, 4ms and 1ms respectively. Which of the following will have the highest average waiting time ?a)First Come First Serve (FCFS)b)Pre – emptive Shortest Job Firstc)Non Pre – emptive Shortest Job Firstd)Round Robin with quantum 1Correct answer is option 'D'. Can you explain this answer? in English & in Hindi are available as part of our courses for Computer Science Engineering (CSE).
Download more important topics, notes, lectures and mock test series for Computer Science Engineering (CSE) Exam by signing up for free.
Here you can find the meaning of Consider three processes P1, P2 and P3 arriving at 0ms, 4ms and 10ms respectively and have burst times 8ms, 4ms and 1ms respectively. Which of the following will have the highest average waiting time ?a)First Come First Serve (FCFS)b)Pre – emptive Shortest Job Firstc)Non Pre – emptive Shortest Job Firstd)Round Robin with quantum 1Correct answer is option 'D'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
Consider three processes P1, P2 and P3 arriving at 0ms, 4ms and 10ms respectively and have burst times 8ms, 4ms and 1ms respectively. Which of the following will have the highest average waiting time ?a)First Come First Serve (FCFS)b)Pre – emptive Shortest Job Firstc)Non Pre – emptive Shortest Job Firstd)Round Robin with quantum 1Correct answer is option 'D'. Can you explain this answer?, a detailed solution for Consider three processes P1, P2 and P3 arriving at 0ms, 4ms and 10ms respectively and have burst times 8ms, 4ms and 1ms respectively. Which of the following will have the highest average waiting time ?a)First Come First Serve (FCFS)b)Pre – emptive Shortest Job Firstc)Non Pre – emptive Shortest Job Firstd)Round Robin with quantum 1Correct answer is option 'D'. Can you explain this answer? has been provided alongside types of Consider three processes P1, P2 and P3 arriving at 0ms, 4ms and 10ms respectively and have burst times 8ms, 4ms and 1ms respectively. Which of the following will have the highest average waiting time ?a)First Come First Serve (FCFS)b)Pre – emptive Shortest Job Firstc)Non Pre – emptive Shortest Job Firstd)Round Robin with quantum 1Correct answer is option 'D'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice Consider three processes P1, P2 and P3 arriving at 0ms, 4ms and 10ms respectively and have burst times 8ms, 4ms and 1ms respectively. Which of the following will have the highest average waiting time ?a)First Come First Serve (FCFS)b)Pre – emptive Shortest Job Firstc)Non Pre – emptive Shortest Job Firstd)Round Robin with quantum 1Correct answer is option 'D'. Can you explain this answer? tests, examples and also practice Computer Science Engineering (CSE) tests.
|
|
Explore Courses for Computer Science Engineering (CSE) exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup