Updated: Solved my first question about sorting, but now I can't figure out how to show the correct diagram for the earliest deadline first algorithm with idle times.
Here is my code so far:
import java.util.*;
class deadlineprototype
{
public static void main(String args[])
{
Scanner sc = new Scanner(System.in);
System.out.println("enter no. of processes : ");
int n=sc.nextInt();
int job[]=new int[n+1];
int burst[]=new int[n+1];
int newburst[]=new int[n+1];
int arrival[]=new int[n+1];
int deadline[]=new int[n+1];
int wt[]=new int[n+1];
int turn[]=new int[n+1];
int tot_turn=0;
int tot_wait=0;
float avg_turn=0;
float avg_wait=0;
int j;
for(int m=1;m<=n;m++)
{
arrival[m]=m;
}
for(int m=1;m<=n;m++)
{
job[m]=m;
}
for(int m=1;m<=n;m++)
{
System.out.println("enter arrival time, burst time and deadline of process "+(m)+"(0 for none):");
arrival[m]=sc.nextInt();
burst[m]=sc.nextInt();
deadline[m]=sc.nextInt();
if (deadline[m]==0){
deadline[m]=1000;
}
}
int temp;
for(int i=0;i<n;i++)
{
for(j=i;j<n;j++)
{
if(arrival[j+1] == arrival[j])
{
if(deadline[j+1] < deadline[j])
{
temp=deadline[j+1];
deadline[j+1]=deadline[j];
deadline[j]=temp;
temp=job[j+1];
job[j+1]=job[j];
job[j]=temp;
temp=burst[j+1];
burst[j+1]=burst[j];
burst[j]=temp;
temp=arrival[j+1];
arrival[j+1]=arrival[j];
arrival[j]=temp;
}
}
}
}
turn[1]=burst[1];
for(int i=2;i<=n;i++)
{
turn[i]=burst[i]+turn[i-1];
wt[i]=turn[i]-burst[i];
}
for(int i=1;i<=n;i++)
{
tot_turn+=(wt[i]+burst[i])-arrival[i];
avg_turn=(float)tot_turn/n;
tot_wait+=wt[i]-arrival[i];
avg_wait=(float)tot_wait/n;
}
System.out.println("----------Earliest Deadline Scheduling Diagram----------");
for(int m=1;m<=n;m++)
{
if(deadline[m]==1000){
deadline[m]=0;
}
if(wt[m]==0){
System.out.println("0"+wt[m]+" _____");
}
else{
System.out.println(wt[m]+" _____");
}
System.out.println(" | |");
System.out.println(" |job "+job[m]+"|");
System.out.println(" |_____|");
try
{
//newburst[m]=(burst[m]*1000);
Thread.sleep(1000);
}catch (InterruptedException ie)
{
System.out.println(ie.getMessage());
}
}
System.out.println((wt[wt.length-1]+burst[burst.length-1]));
If I input 2 processes without idle then it will show correct output:
enter no. of processes :
2
enter arrival time, burst time and deadline of process 1(0 for none):
0 17 0
enter arrival time, burst time and deadline of process 2(0 for none):
0 13 10
----------Earliest Deadline Scheduling Diagram----------
00 _____
| |
|job 2|
|_____|
13 _____
| |
|job 1|
|_____|
30
But if it has idle time then it will output:
enter no. of processes(5-10):
2
enter arrival time, burst time and deadline of process 1(0 for none):
0 5 0
enter arrival time, burst time and deadline of process 2(0 for none):
10 10 10
----------Earliest Deadline Scheduling Diagram----------
00 _____
| |
|job 1|
|_____|
5 _____
| |
|job 2|
|_____|
15
First sort the Processes based on the arrival time as follows,
for(i=0;i<n;i++)
{
for(j=i;j<n;j++)
{
if(arrival[j+1] < arrival[j])
{
temp=deadline[j+1];
deadline[j+1]=deadline[j];
deadline[j]=temp;
temp=job[j+1];
job[j+1]=job[j];
job[j]=temp;
temp=burst[j+1];
burst[j+1]=burst[j];
burst[j]=temp;
temp=arrival[j+1];
arrival[j+1]=arrival[j];
arrival[j]=temp;
}
}
}
Afterwards if the arrival time of processes are equal sort them based on the deadline as follows,
for(i=0;i<n;i++)
{
for(j=i;j<n;j++)
{
if(arrival[j+1] == arrival[j])
{
if(deadline[j+1] < deadline[j])
{
temp=deadline[j+1];
deadline[j+1]=deadline[j];
deadline[j]=temp;
temp=job[j+1];
job[j+1]=job[j];
job[j]=temp;
temp=burst[j+1];
burst[j+1]=burst[j];
burst[j]=temp;
temp=arrival[j+1];
arrival[j+1]=arrival[j];
arrival[j]=temp;
}
}
}
}