Using TORA Program please answer the question TORA program l
Using TORA Program please answer the question.
TORA program link is provided below link. It is a free program. http://media.pearsoncmg.com/ph/esm/ecs_taha_orai_9e/software_support/tora_system.zip/tora_system.zip
3-59. Consider the LP: Maximize z = 20m + 10x2 + x3 subject to 3x1-3x2 + 5r3 50 + a 10 By inspecting the constraints, determine the direction (x1,x2, or r) in which the solution space is unbounded. Without further computations, what can you conclude regarding the optimum objective value? (a) (b)Solution
#include<iostream.h>
#include<conio.h>
struct process
{
int no;
int at,et,wt,tt;
int tet;
int t;
};
void main()
{
process p[99];
int i,j,k;
cout<<\"\ Enter No of Processes:\";
int np;
cin>>np;
for (i=0;i<np;i++)
{
cout<<\"\ Enter Execution time of process\"<<i+1<<\":\";
cin>>p[i].et;
p[i].tet=p[i].et;
p[i].at=p[i].t=p[i].tt=p[i].wt=0;
p[i].no=i+1;
}
cout<<\"\ Enter Time Quantum:\";
int q;
cin>>q;
cout<<\"\ Entered Data\";
cout<<\"\ Process\\tET\";
for(i=0;i<np;i++)
{
cout<<\"\ \"<<p[i].no<<\"\\t\"<<p[i].et;
}
int totaltime=0;
for(i=0;i<np;i++)
{
totaltime+=p[i].et;
}
i=0;
k=0;
int rrg[99];
for(j=0;j<totaltime;j++)
{
if((k==0)&&(p[i].et!=0))
{
p[i].wt=j;
if((p[i].t!=0))
{
p[i].wt-=q*p[i].t;
}
}
if((p[i].et!=0)&&(k!=q))
{
rrg[j]=p[i].no;
p[i].et-=1;
k++;
}
else
{
if((k==q)&&(p[i].et!=0))
{
p[i].t+=1;
}
i=i+1;
if(i==np)
{
i=0;
}
k=0;
j=j-1;
}
}
/*
for(j=0;j<totaltime;j++)
{
cout<<\"\ \"<<rrg[j];
}
*/
int twt=0;
int ttt=0;
cout<<\"\ Result Of Round Robin\";
cout<<\"\ PNo\\tET\\tWT\\tTT\";
for(i=0;i<np;i++)
{
p[i].tt=p[i].wt+p[i].tet;
ttt+=p[i].tt;
twt+=p[i].wt;
cout<<\"\ \"<<p[i].no<<\"\\t\"<<\"\\t\"<<p[i].tet<<\"\\t\"<<p[i].wt<<\"\\t\"<<p[i].tt;
}
cout<<\"\ Average Waiting Time:\"<<(float)twt/np;
cout<<\"\ Average Turn Around Time:\"<<(float)ttt/np;
getch();
}
/* Output
Enter No of Processes:5
Enter Execution time of process1:10
Enter Execution time of process2:29
Enter Execution time of process3:3
Enter Execution time of process4:7
Enter Execution time of process5:12
Enter Time Quantum:10
Entered Data
Process ET
1 10
2 29
3 3
4 7
5 12
Result Of Round Robin
PNo ET WT TT
1 10 0 10
2 29 32 61
3 3 20 23
4 7 23 30
5 12 40 52
Average Waiting Time:23
Average Turn Around Time:35.2
*/


