Shortest remaining time first the name itself says that it is actually we are going to take the shortest jobs but in a different way and it is actually based on bursts times criterium and the mode is preemptive which means even when the process is running we are going to stop it and then I showed you the other one if the situations.

T equal to zero and at that time only one process is available right so which is nothing but P 1 so at time 0 only P 1 is available I have no other option therefore I were to schedule P 1 fine and now once I should get P 1 here I’ll just be careful I’ll not directly completely run it to execution I’m in completion then what I do is I stop it at every one unit of time and then checked in meanwhile with mesh means during this time if I got any other process right which means. You could choose any one of them right but then depending on the shortest job shortest path time and supposed to choose P 2, therefore, the next one which is chosen is P 2 right and now for P 2 what is the best time it is actually 5 but then we don’t run it to completion which runs till one unit and then check it and then stops and sees so now I have no I have done this P 2 for 1 I 0 P 2 for 1 unit, therefore, the remaining time will be for now the time is 2 by the time.

The arrival time by the time – and see what are the process available so one more process got available which is P 3 now I have P 1 P 2 and P 3 available and the P 3 is burst time is 3, therefore, the shortest one among all these is 3 P 3, therefore, I will take P 3 and I should do it for how long just for one unit and an ambit so since I exceeded for one unit therefore what is the time here so the remaining time is two and now the time is three so when the time is three I have one more process arriving which is before therefore I have P 1 P 2 P 3 P 4.

One of them so which one is the minimum P 1 P 2 P 3 P 4 P 4 is a minimum so P 4 is going to have any no here scheduled here and anyway I will turn it for one unit I don’t have any other option I have to run it for all then I told it because the first time is one unit and then stop it anyway it is going to be stopped now the time is 4 and one of the processes which are available P 1 P 2 P 3 P 4 and then P file but P 4 is over, therefore, you need not worry about it before is not there right so among this P 1 P 2 P 3 P 5 you just check which one is a 3d stuff least time so we shouldn’t have the least time here so we have to process which are having the least remaining time.

One is P 3 and the other is P 4 P 5 sorry P 5 yeah P 3 and P 5 but you had to pick the one which is having the least arrival time so we were shown inside the dish arrival time P 3, therefore, you are supposed to pick P 3 itself right and now P 3 has to be run for one unit which means still file and then you should stop it and check it since you are at p3 for one unit the remaining time will be 1 and now we just check it if any process other areas one more photo has arrived whose phosphide is p6 and whose time is 1 so I have now P 1 P 2.