I was going through the various scheduling algorithm and came across the shortest remaining time Job fist algorithm.
What I understand: Shortest remaining time job first algorithm is the preemptive version of the shortest time job algorithm. Here when a new process with lower burst time has arrived,it prempts the current process and the the new process is executed.
I came across a website while doing my study. The link for the site :
https://www.studytonight.com/operating-system/shortest-job-first
What I Don't understand is that the author of the article says that when a process P4 with a CPU burst time of 2ms arrives and process P2 with a remaining burst time of 1ms is executed,P2 is pre-empted. Why is that so? Shouldn't P4 be put to the ready queue since it has more burst time than the remaining burst time of P2?
I understand the fact that burst times are compared and in this way the cpu burst time of P2 is (3ms) more than the CPU burst time of p4(2ms).But since 2 burst time have already been executed, the remaining burst time of p2 is less than that of p4.Why is it preempted?