Implementation of simplex algorithm in C. (WITH OUTPUT IMAGE)

#include<stdio.h>
#include<conio.h>

void main()
{
float c[10],val[10][10],basis[10]={0},z[10]={0},del[10],b[10];
int i,j,k,eno,vno,og,in,bname[10];
float min,max,multiplier,z1=0,ba[10],key_el;
clrscr();

printf("Enter no. of variables");
scanf("%d",&vno);

printf("\n Enter coeffiients with variables in objective function Z::");
for(i=0;i<vno;i++)
scanf("%f",&c[i]);

printf("\n How many equations??");
scanf("%d",&eno);

printf("\n Enter coefficents of variables in each equation WITH SLACK VARIABLES");
for(i=0;i<eno;i++)
{ printf("\n Equaion %d::",i+1);
     for(j=0;j<vno;j++)
     {printf("coeff with var%d=",j+1);
      scanf("%f",&val[i][j]);
     }
     for(j=vno;j<eno+vno;j++)
     { printf("coeff with slack%d=",j-vno+1);
      scanf("%f",&val[i][j]);
     }
}

printf("\n Enter RHS values for each equation");
for(i=0;i<eno;i++)
{    printf("\n Equaion %d::",i+1);
     scanf("%f",&b[i]);

}

//now for slack we have to do cj=0 in Z;
for(i=vno;i<eno+vno;i++)
{c[i]=0;}

//basis are already set to zero bt to know which variable is in basis,bname
//we will slack=number of equation  i.e if 3 equations then 3 slack variables.
//so last number will be vno+eno=total variables in each equation containg slack.

for(i=0,j=vno;j<(vno+eno);i++,j++)
{bname[i]=j;
}



//start processing simplex
for(k=0;k<10;k++)
{
max=0;

//after each iteration make zj=0;
for(i=0;i<vno+eno;i++)
z[i]=0;

//find z to calculate delta
for(i=0;i<vno+eno;i++)
{ for(j=0;j<eno;j++)
  z[i]=z[i]+(basis[j]*val[j][i]);
}

// find delta and maximum among them
for(j=0;j<vno+eno;j++)
{ del[j]=0;

 del[j]=c[j]-z[j];
  if(del[j]>max)
  {max=del[j];
   in=j;
  }
  printf("del %f",del[j]);
}

//if max=0 it means all neagtive valuesin delta so stop.
if(max==0)
{break;
}



//printf("here1");
for(i=0;i<eno;i++)
{
   if(val[i][in]!=0)
   ba[i]=b[i]/val[i][in];
  //first of all take first value of bi/aij as smallest.
  if(i==0)
  {min=ba[i];
  }

  //we are finding outgoing variable so it will be minimum amongst bi/aij;
  if(ba[i]<=min)
  {min=ba[i];
  og=i;
  }
 //d printf("%f",ba[i]);
}


//we got o/g variable so rename it with i/g.
bname[og]=in;
printf("incomming=%d",in);
//basis will be changed according to cj value.
printf("outgoing=%d",og);
basis[og]=c[in];
//key element will be intersection of o/g variable's row & i/g variable's coloumn.
key_el=val[og][in];
printf("key %f",key_el);

//NOTE: here i/g shows key coloumn &
//           o/g shows key row.

//now do pivoting.

//firstly for key row which will be of 'o/g'.
// printf("here2");
 for(j=0;j<eno+vno;j++)
 {val[og][j]=val[og][j]/key_el;

 //at the end of loop do it for rhs also;
 //  printf("h3");
   if(j==(eno+vno)-1)
   b[og]=b[og]/key_el;
 }

//for rest of the rows;
//we have to set one multiplier which is above and below values of key_element.
for(i=0;i<eno;i++)
{
 //only process rows except key_row;
  if(i!=og)
  {multiplier=val[i][in];

     // multiply whole key row with negative multiplier
     //  and now add it in previous row and overwite value.
      for(j=0;j<eno+vno;j++)
      { val[i][j]=val[i][j]+(-1*multiplier*val[og][j]);
      }
     //after doing for variables do it for rhs;
     b[i]=b[i]+(-1*multiplier*b[og]);
  }
}

   //done
 //updated basis,basis_name,b(rhs),all coefficients.
   //so next loop will use updated value and do same;
}  //finish loop with 'k';

printf("\n Solution");
for(i=0;i<eno;i++)
{  if(bname[i]>=vno)
    printf("\n slack%d=%f",(bname[i]%vno)+1,b[i]);

    else
    printf("\n x%d=%f",bname[i]+1,b[i]);
}

//calculating z by multiplying rhs of variables and its corredsponding basis;
for(j=0;j<eno;j++)
{z1=z1+(basis[j]*b[j]);
}

printf("\n Z=%f",z1);

getch();
}