idk if I dont have enough sleep or if im too fckng dumb that I am not able to fully understand the subject. anyways, after P3 why P1 go in next instead of P4?
I'm at video 51 of the Operating Systems playlist and I just have to stop and say how grateful I am for this. This man does not miss a beat in his explanations. THANK YOU !!!
because till time 2 there was no process to put in ready queue, therefore scheduler put the p1 after p3 to complete the job, however at time 3 p4 has arrived and it had to be put in behind the p1 and it kept going like that.
Because p1 gets preempted at time 2ms and at that time p4 hasn't arrived (p4 arrives at 3ms) so p1 enters the ready queue and is placed after p3 and then at 3ms p4 arrives and is placed after p1..
In a typical round-robin scheduling system, the decision of whether to place the preempted process or the newly arrived process first at the back of the queue depends on the specific implementation of the scheduling algorithm. There isn't a universally standardized rule for this, and different operating systems or schedulers may handle this situation differently. Here are two possible scenarios: Place Preempted Process First: The preempted process could be placed back in the queue before the newly arrived process. This means that the preempted process, which was already in the middle of its execution, gets another chance to continue from where it left off before the newly arrived process gets its turn. Place Newly Arrived Process First: The newly arrived process could be placed in the queue before the preempted process. This means that the system prioritizes the new arrival, allowing it to start execution before the preempted process gets another turn. in this lecture, they have taught for 2nd scenario only.
I'm struggling with process synchronisation and concurrency. I'm waiting for lectures on those topics. Not comprehensive but a little introductory with little explanation can really help me me out. By the way, in upcoming march, I will apply for interview at Neso Academy for Compiler Design Course.
Sir? I have a question? That I've done by myself first but without the ready queue. Means I've got arrangement as p1,p3,p4,p1,p1,p2,p5,p2, p5 Is it correct if we don't follow the ready queue.?
@Neso Academy 11:57 one thing I didn't understand that when p1 arrives the remaining time quantum was only 1unit due to process p3 voluntary releases itself from the CPU as it's execution is over. So it should execute for only 1 ms after it would again be preempted. Please clear my doubt
after executing p1 in the first time p1 get a ready queue at time 2 and p3 gets ready queue at time 2 so the arrival time of p1 and p3 is the same so we execute p1 because the pid of p1 is 1 less than the pid of 3 which is 3
Sir..thank you for this video..and i get knowledge on this problem But i have a question....that is. Assume the following workload in a system: Process Arrival Time Burst Time P1 5 5 P2 4 6 P3 3 7 P4 1 9 P5 2 2 P6 6 3 Draw a Gantt chart illustrating the execution of these jobs using Round robin scheduling algorithm and also Calculate the average waiting time and average turnaround time Plz..plz..solve the problem
Your videos are amazing for computer science students like myself taking courses in concurrent programming and operating systems. Thanks for all the help with your videos!
