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i like moldy chess. cheese....shests
yes in the preemptive sjf where every time the system recieve job shorter than the current it stop the current and start the new one
no.
find primitive sjf? with avg waiting time and TAW? P Bust TIme Arivel time p1 5 1 p2 3 0 p3 2 2 p4 4 3 p5 8 2
The university solved slips for third year Bsc Computer Science are usually issued after the students satisfy the conditions set out by the examination body. The students also have to meet the conditions set out by the senate.
Shortest Job First (SJF) scheduling algorithm selects the process with the smallest burst time for execution next, which helps in reducing average waiting time. Shortest Remaining Time First (SRTF), on the other hand, preempts the currently running process if a new process with a smaller burst time arrives, resulting in even lower waiting times but with increased context switching.
Non pre-emptive means once CPU starts executing one process, it will not be taken out of the CPU until it is terminated or it has to wait for some event. In preemptive SJF scheduling, current running process is moved to the ready queue when a new process with a shorter CPU burst joins the ready queue.
short job first (SJF) is a job scheduling algorithm where shorter jobs are favored for being serviced first. this increases the overall system's number of processed jobs but has the disadvantage that long jobs can face starvation in waiting state irrespective to their wait time..
It is SJF.
in pre emptive scheduling. a limited time period is fixed for every process in the CPU. no matter whether the process is completed or not ... the resource assinged to it will be taken back back abd will be given to the next process in the queue.while in non preemptive the resorces are with a process untill it finishes completely others wait for their turn till then. this kind of scheduling has a high probability of going into a deadlock.
In that work system, the shortest job gets higher priority because more gets accomplished in any amount of time. If you have two hours to do 4 tasks and one task will take five minutes, one will take 30 minutes, one will take an hour and one will take an hour and a half, by using the shortest job first method, you will accomplish three tasks in those two hours and part of the fourth task. If you do the longest job first, you will accomplish the task that takes an hour and a half, then only part of the one hour task, but nothing on the 5 minute or 30 minute tasks. That means that by doing the shortest jobs first, you will accomplish three times as many tasks as you will by doing the longest jobs first.
IN QUEueing theory,it is like how we serve people in waiting line --- 1.fcfs 2.round robin 3.sjf 4. siro 5.fifo