Signup/Sign In

Shortest Job First(SJF) Scheduling

Shortest Job First scheduling works on the process with the shortest burst time or duration first.

  • This is the best approach to minimize waiting time.
  • This is used in Batch Systems.
  • It is of two types:
    1. Non Pre-emptive
    2. Pre-emptive
  • To successfully implement it, the burst time/duration time of the processes should be known to the processor in advance, which is practically not feasible all the time.
  • This scheduling algorithm is optimal if all the jobs/processes are available at the same time. (either Arrival time is 0 for all, or Arrival time is same for all)

Non Pre-emptive Shortest Job First

Consider the below processes available in the ready queue for execution, with arrival time as 0 for all and given burst times.

Non Pre-emptive Shortest Job First Scheduling

As you can see in the GANTT chart above, the process P4 will be picked up first as it has the shortest burst time, then P2, followed by P3 and at last P1.

We scheduled the same set of processes using the First come first serve algorithm in the previous tutorial, and got average waiting time to be 18.75 ms, whereas with SJF, the average waiting time comes out 4.5 ms.

Problem with Non Pre-emptive SJF

If the arrival time for processes are different, which means all the processes are not available in the ready queue at time 0, and some jobs arrive after some time, in such situation, sometimes process with short burst time have to wait for the current process's execution to finish, because in Non Pre-emptive SJF, on arrival of a process with short duration, the existing job/process's execution is not halted/stopped to execute the short job first.

This leads to the problem of Starvation, where a shorter process has to wait for a long time until the current longer process gets executed. This happens if shorter jobs keep coming, but this can be solved using the concept of aging.

Pre-emptive Shortest Job First

In Preemptive Shortest Job First Scheduling, jobs are put into ready queue as they arrive, but as a process with short burst time arrives, the existing process is preempted or removed from execution, and the shorter job is executed first.

Pre-emptive Shortest Job First Scheduling

The average waiting time will be,((5-3)+(6-2)+(12-1))/4=8.75

The average waiting time for preemptive shortest job first scheduling is less than both,non preemptive SJF scheduling and FCFS scheduling

As you can see in the GANTT chart above, as P1 arrives first, hence it's execution starts immediately, but just after 1 ms, process P2 arrives with a burst time of 3 ms which is less than the burst time of P1, hence the process P1(1 ms done, 20 ms left) is preemptied and process P2 is executed.

As P2 is getting executed, after 1 ms, P3 arrives, but it has a burst time greater than that of P2, hence execution of P2 continues. But after another millisecond, P4 arrives with a burst time of 2 ms, as a result P2(2 ms done, 1 ms left) is preemptied and P4 is executed.

After the completion of P4, process P2 is picked up and finishes, then P2 will get executed and at last P1.

The Pre-emptive SJF is also known as Shortest Remaining Time First, because at any given point of time, the job with the shortest remaining time is executed first.

Program for SJF Scheduling

In the below program, we consider the arrival time of all the jobs to be 0.

Also, in the program, we will sort all the jobs based on their burst time and then execute them one by one, just like we did in FCFS scheduling program.

// c++ program to implement Shortest Job first


using namespace std;
struct Process
   int pid;     // process ID
   int bt;      // burst Time
    this function is used for sorting all
    processes in increasing order of burst time
bool comparison(Process a, Process b)
    return ( <;
// function to find the waiting time for all processes
void findWaitingTime(Process proc[], int n, int wt[])
    // waiting time for first process is 0
    wt[0] = 0;
    // calculating waiting time
    for (int i = 1; i < n ; i++)
        wt[i] = proc[i-1].bt + wt[i-1] ;
// function to calculate turn around time
void findTurnAroundTime(Process proc[], int n, int wt[], int tat[])
    // calculating turnaround time by adding bt[i] + wt[i]
    for (int i = 0; i < n ; i++)
        tat[i] = proc[i].bt + wt[i];
// function to calculate average time
void findAverageTime(Process proc[], int n)
    int wt[n], tat[n], total_wt = 0, total_tat = 0;
    // function to find waiting time of all processes
    findWaitingTime(proc, n, wt);
    // function to find turn around time for all processes
    findTurnAroundTime(proc, n, wt, tat);
    // display processes along with all details
    cout << "\nProcesses "<< " Burst time "
         << " Waiting time " << " Turn around time\n";
    // calculate total waiting time and total turn around time
    for (int i = 0; i < n; i++)
        total_wt = total_wt + wt[i];
        total_tat = total_tat + tat[i];
        cout << " " << proc[i].pid << "\t\t"
             << proc[i].bt << "\t " << wt[i]
             << "\t\t " << tat[i] <<endl;
    cout << "Average waiting time = "
         << (float)total_wt / (float)n;
    cout << "\nAverage turn around time = "
         << (float)total_tat / (float)n;
// main function
int main()
    Process proc[] = {{1, 21}, {2, 3}, {3, 6}, {4, 2}};
    int n = sizeof proc / sizeof proc[0];
    // sorting processes by burst time.
    sort(proc, proc + n, comparison);
    cout << "Order in which process gets executed\n";
    for (int i = 0 ; i < n; i++)
        cout << proc[i].pid <<" ";
    findAverageTime(proc, n);
    return 0;

Order in which process gets executed 4 2 3 1 Processes Burst time Waiting time Turn around time 4 2 0 2 2 3 2 5 3 6 5 11 1 21 11 32 Average waiting time = 4.5 Average turn around time = 12.5

Try implementing the program for SJF with variable arrival time for different jobs, yourself.