![]() The process with the least remaining CPU burst time among the available processes will get executed. P3 is also available and requires 5 units of total CPU burst time. Since P2 also becomes available at time 4 because it has completed the IO operation and now it needs another 1 unit of CPU burst time. Processor becomes free for the execution of other jobs. At time unit 4, it will change its state from running to waiting. Hence, P1 will remain in the running state from time 3 to time 4. Since the remaining burst time of P1 is lesser then P3 hence CPU will continue its execution. The process P2 went to waiting state and the CPU becomes idol at this time.įrom time 1 to 3, since P2 is being in waiting state, and no other process is available in ready queue, hence the only available process P1 will be executed in this period of time.Īt time 3, the process P3 arrived with the total CPU burst time of 5 units. The following diagram illustrates the processes and states at Time 1. Since No other process is available at this point of time other than P1 so P1 will get executed. The processor becomes free to execute other jobs. After 1 unit of execution, P2 will change its state from running to waiting. P2 also needs some IO time in order to complete its execution. In this case, it is P2.įrom time 0 to time 1, P2 will be in running state. Since the algorithm we are using is SRTF hence, the process with the shortest burst time will be scheduled on the CPU. Process IdĪt time 0, the process P1 and P2 arrives. Their Arrival Time, and the CPU Burst time are given in the table below. In the Example, there are four jobs with process ID P1, P2, P3 and P4 are available. In this Example, we are considering, the IO bound processes. However, the process might need some IO operation or some resource to complete its execution. Till now, we were considering the CPU bound jobs only. Next → ← prev SRTF with Processes contains CPU and IO Time ![]()
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