How to calculate Average Waiting Time and average Turn-around time in SJF Scheduling

Question

In SJF  Shortest Job First  Scheduling method      How to calculate Average Waiting Time and average Turn-around time   Is Gannt Chart correct

User · Answer

it is wrong  correct will be  P3   P2   P4   P5  P1 0     3   4    6    10   as the correct difference are these  Waiting Time  0 3 4 6 10  5     4 6  Ref  http   www it uu se edu course homepage oskomp vt07 lectures scheduling algorithms handout pdf

User · Answer

Gantt chart is wrong    First process P3 has arrived so it will execute first  Since the burst time of P3 is 3sec after the completion of P3  processes P2 P4  and P5 has been arrived  Among P2 P4  and P5 the shortest burst time is 1sec for P2  so P2 will execute next  Then P4 and P5  At last P1 will be executed   Gantt chart for this ques will be     P3   P2   P4   P5   P1    1    4    5    7   11   14   Average waiting time  0 2 2 3 3  5 2   Average Turnaround time  3 3 4 7 6  5 4 6

User · Answer

SJF are two type - i  non preemptive SJF   ii pre-emptive SJF  I have re-arranged the processes according to Arrival time   here is the non preemptive SJF  A T  Arrival Time  B T  Burst Time  C T  Completion Time  T T   Turn around Time   C T - A T  W T   Waiting Time   T T - B T    Here is the preemptive SJF Note  each process will preempt at time a new process arrives Then it will compare the burst times and will allocate the process which have shortest burst time  But if two process have same burst time then the process which came first that will be allocated first just like FCFS

User · Answer

The Gantt charts given by Hifzan and Raja are for FCFS algorithms     With an SJF algorithm  processes can be interrupted   That is  every process doesn t necessarily execute straight through their given burst time   P3 P2 P4 P3 P5 P1 P5  1 2 3 5 7 8 11 14  P3 arrives at 1ms  then is interrupted by P2 and P4 since they both have smaller burst times  and then P3 resumes   P5 starts executing next  then is interrupted by P1 since P1 s burst time is smaller than P5 s   You must note the arrival times and be careful   These problems can be trickier than how they appear at-first-glance     EDIT  This applies only to Preemptive SJF algorithms  A plain SJF algorithm is non-preemptive  meaning it does not interrupt a process

[scheduling] How to calculate Average Waiting Time and average Turn-around time in SJF Scheduling?

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