#define T 26
#define J 8

continuous X[T][J];
int ready[J], due[J];
int dur[J];

data()
{
	ready = { 0,1,1,11,11,5,11,7 };
	due = { 5,5,8,26,15,10,24,22 };
	dur = { 2,3,3,1,4,2,2,5 };
}

network()
{
	and(int i=0;i<T;i++)						// At most one job per time unit
		sigma(int j=0;j<J;j++) X[i][j] <= 1;
	and(int j=0;j<J;j++) 						// Each job must be completed
		sigma(int k=ready[j];k<due[j];k++) X[k][j] == dur[j];
	min: sigma(int i=0;i<T;i++)				// Charge i for active unit i							
			sigma(int j=0;j<J;j++) i*X[i][j];
}

output()
{
int prev;
prev = -1;

	and(int t=0;t<T;t++)
		and(int j=0;j<J;j++) 
			if wp(X[t][j]) > 0; 
			then {
				if j == prev;
				then;
				else {
					if prev == -1;
					then;
					else printf(" run job %d \n",prev);
					}
				if j == prev;
				then printf(", %d",t);
				else printf("At time %d",t);
			prev = j;
			}
			printf (" run job %d \n",prev);
}

2lp_main()
{
	data();
	network();
	output();
}
 