Question

Write a program in Java that will simulate FCFS (First Come First Serve) considering context switching is 0. The program should compute waiting time and turnaround time of every job as well as the average waiting time and average turnaround time. Then change the context switching time to 0.4 milliseconds.
Arrival time CPU Burst (in milliseconds Process 0) P2 P3 4 100 P6

Answer 1

Dear student,

as per property of FCFS, the process coming first will be executed completely without any interruption or context switching.

The java program is provided below. It simulate FCFS, which sorts processes according to their arrival time and executes and computes waiting & arrival time. It also calculates their average waiting and average arrival time.

import java.util.Scanner;

public class FCFS {

   public static void main(String[] args) {
      
       Scanner sc = new Scanner(System.in);
       int n, i, j, temp, time=0;
       float avg_wt=0, avg_tt=0;
       System.out.println("Enter no. of processes");
       n = sc.nextInt();
      
       int at[] = new int[n+1]; // Arrival time
       int ft[] = new int[n+1]; // Finish time
       int tt[] = new int[n+1]; // Turn around Time
       int wt[] = new int[n+1]; // Waiting time
       int ct[] = new int[n+1]; // CPU/Burst time
       String p[] = new String[n+1]; // Process name
       String ts;
      
       System.out.println("Insert details of processes as (Arrival_time, CPU_time)");
      
       for(i=1;i<=n;i++)
       {
           System.out.println("Process "+i);
           at[i] = sc.nextInt();
           ct[i] = sc.nextInt();
           ft[i] = 0;
           wt[i] = 0;
           tt[i] = 0;
           p[i] = "p"+i;
       }
      
       // sorts processes according to their arrival times
       for(i=1;i<n;i++)
       {
           for(j=0;j<n;j++)
           {
               if(at[j]>at[j+1])
               {
                   temp = at[j];
                   at[j] = at[j+1];
                   at[j+1] = temp;
                  
                   temp = ct[j];
                   ct[j] = ct[j+1];
                   ct[j+1] = temp;  
                  
                   ts = p[j];
                   p[j] = p[j+1];
                   p[j+1] = ts;
               }
           }
       }
      
       // Process each task
       for(i=1;i<=n;i++)
       {
           if(at[i]<=time)
           {
               time += ct[i];
               ft[i] += time;
               tt[i] = ft[i]-at[i];
               wt[i] = tt[i]-ct[i];
               avg_tt += tt[i];
               avg_wt += wt[i];
           }
           else
           {
               while(at[i]>time)
                   time++;
               time += ct[i];
               ft[i] = time;
               tt[i] = ft[i]-at[i];
               wt[i] = tt[i]-ct[i];
               avg_tt += tt[i];
               avg_wt += wt[i];              
           }

       }
      
       avg_tt /= n;
       avg_wt /= n;
       System.out.println("\nProcess AT \tFT \tCT \tTT \tWT");
       for(i=1;i<=n;i++)
       {
           System.out.println(p[i]+"\t"+at[i]+"\t"+ft[i]+"\t"+ct[i]+"\t"+tt[i]+"\t"+wt[i]);
       }
      
       System.out.println("\nAverage turnaound time = "+avg_tt);
       System.out.println("Average waiting time = "+avg_wt);
      
       sc.close();
   }
}

The output of this code for example provided by you is given below.....

Enter no. of processes
6
Insert details of processes as (Arrival_time, CPU_time)
Process 1
0 6
Process 2
3 2
Process 3
5 1
Process 4
9 7
Process 5
10 4
Process 6
11 3

Process AT    FT    CT    TT    WT
p1   0   6   6   6   0
p2   3   8   2   5   3
p3   5   9   1   4   3
p4   9   16   7   7   0
p5   10   20   4   10   6
p6   11   23   3   12   9

Average turnaound time = 7.3333335
Average waiting time = 3.5