SJF CPU Scheduling Program in C++

What is the quickest scheduling of jobs?

The job or process scheduling method that adheres to the non-preemptive scheduling discipline is called shortest job first scheduling. In this case, the scheduler chooses the job or process from the waiting list with the shortest completion time and assigns the CPU to it. Because SJF is more optimal than FIFO and decreases average wait times, which will enhance throughput, it is preferred over FIFO.

Both preemptive and non-preemptive SJF algorithms are possible. Shortest-remaining-time-first scheduling is another name for preemptive scheduling. The new process emerges in the preemptive method while the existing process is still running. The scheduler will prevent the execution of the process with the shorter burst time if the burst of the newly incoming process is smaller than the burst time of the current process.

What is Turn Around, Wait, and Completion time?

Completion Time is the amount of time needed for the process to finish running.
Turnaround Time is the period of time between starting a procedure and finishing it.
Turnaround time is equal to the time it takes to complete a procedure and submit it.
The delay between turnaround time and burst time is called waiting time.
Turnaround time minus burst time equals waiting time.

Algorithm

Start
Step 1-> Declare a struct Process
   Declare pid , bt , art
Step 2-> In function findTurnAroundTime ( Process proc [ ] , int n , int wt [ ] , int tat [ ] )
   Loop For i = 0 and i < n and i++
      Set tat [ i ] = proc [ i ].bt + wt [ i ]
Step 3-> In function findWaitingTime ( Process proc [ ] , int n , int wt [ ] )
   Declare rt [ n ]
   Loop For i = 0 and i < n and i++
      Set rt [ i ] = proc [ i ].bt
      Set complete = 0 , t = 0 , minm = INT_MAX
      Set shortest = 0 , finish_time
      Set bool check = false
      Loop While ( complete != n )
         Loop For j = 0 and j < n and j++
            If (proc [ j ].art <= t ) && ( rt [ j ] < minm ) && rt [ j ] > 0 then ,
               Set minm = rt [ j ]
               Set shortest = j
               Set check = true
            If check == false then ,
               Increment t by 1
               Continue
               Decrement the value of rt [ shortest ] by 1
               Set minm = rt [ shortest ]
            If minm == 0 then ,
               Set minm = INT_MAX
               If rt [ shortest ] == 0 then ,
               Increment complete by 1
               Set check = false
               Set finish_time = t + 1
               Set wt [ shortest ] = finish_time - proc [ shortest ].bt - proc [ shortest ].art
            If wt [ shortest ] < 0
               Set wt [ shortest ] = 0
               Increment t by 1
Step 4-> In function findavgTime ( Process proc [ ] , int n )
   Declare and set wt [ n ] , tat [ n ] , total_wt = 0 , total_tat = 0
   Call findWaitingTime ( proc , n , wt )
   Call findTurnAroundTime ( proc , n , wt , tat )
   Loop For i = 0 and i < n and i++
      Set total_wt = total_wt + wt [ i ]
      Set total_tat = total_tat + tat [ i ]
      Print proc [ i ].pid , proc [ i ].bt , wt [ i ] , tat [ i ]
      Print Average waiting time i.e. , total_wt / n
      Print Average turn around time i.e. , total_tat / n
Step 5-> In function int main ( )
   Declare and set Process proc [ ] = { { 1 , 5 , 1 } , { 2 , 3 , 1 } , { 3 , 6 , 2 } , { 4 , 5 , 3 } }
   Set n = sizeof ( proc ) / sizeof ( proc [ 0 ] )
   Call findavgTime ( proc , n )
Stop

Program for SJF Scheduling in C++

#include <bits/stdc++.h>
using namespace std;
//structure for every process
struct Process {
   int pid; // Process ID
   int bt; // Burst Time
   int art; // Arrival Time
};
void findTurnAroundTime(Process proc[], int n, int wt[], int tat[]) {
   for (int i = 0; i < n; i++)
   tat[i] = proc[i].bt + wt[i];
}
//waiting time of all process
void findWaitingTime(Process proc[], int n, int wt[]) {
   int rt[n];
   for (int i = 0; i < n; i++)
   rt[i] = proc[i].bt;
   int complete = 0, t = 0, minm = INT_MAX;
   int shortest = 0, finish_time;
   bool check = false;
   while (complete != n) {
      for (int j = 0; j < n; j++) {
         if ((proc[j].art <= t) && (rt[j] < minm) && rt[j] > 0) {
            minm = rt[j];
            shortest = j;
            check = true;
         }
      }
      if (check == false) {
         t++;
         continue;
      }
      // decrementing the remaining time
      rt[shortest]--;
      minm = rt[shortest];
      if (minm == 0)
         minm = INT_MAX;
         // If a process gets completely
         // executed
         if (rt[shortest] == 0) {
            complete++;
            check = false;
            finish_time = t + 1;
            // Calculate waiting time
            wt[shortest] = finish_time -
            proc[shortest].bt -
            proc[shortest].art;
            if (wt[shortest] < 0)
               wt[shortest] = 0;
         }
         // Increment time
         t++;
   }
}
// Function to calculate average time
void findavgTime(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);
   cout << "Processes " << " Burst time " << " Waiting time " << " Turn around time\n";
   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\t " << wt[i] << "\t\t " << tat[i] << endl;
   }
   cout << "\nAverage waiting time = " << (float)total_wt / (float)n; cout << "\nAverage turn around time = " << (float)total_tat / (float)n;
}
// main function
int main() {
   Process proc[] = { { 1, 5, 1 }, { 2, 3, 1 }, { 3, 6, 2 }, { 4, 5, 3 } };
   int n = sizeof(proc) / sizeof(proc[0]);
   findavgTime(proc, n);
   return 0;
}

OUTPUT:

Processes  Burst Time	Waiting Time  Turn Around Time
   1	        5	        3	            8
   2	        3	        0	            3
   3	        6	        12	            18
   4	        5	        6	            11

Average waiting time = 5.25
Average turn around time = 10
........................................................
Process executed in 1.33 seconds
Press any key to continue.

Whenever there is a tie, FCFS Scheduling decides who wins.
It is possible to employ SJF Scheduling in both preemptive and non-preemptive modes
3-The Shortest Remaining Time First is the preemptive mode of Shortest Job First (SRTF).

Next TopicVirtual functions and runtime polymorphism

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SJF CPU Scheduling Program in C++

What is the quickest scheduling of jobs?

What is Turn Around, Wait, and Completion time?

Algorithm

Program for SJF Scheduling in C++

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SJF CPU Scheduling Program in C++

What is the quickest scheduling of jobs?

What is Turn Around, Wait, and Completion time?

Algorithm

Program for SJF Scheduling in C++

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Contact info

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