Thread: Simplex algorithm problem View Single Post
 11-29-2012, 07:54 AM PM User | #1 musicawlatch New to the CF scene   Join Date: Nov 2012 Posts: 1 Thanks: 0 Thanked 0 Times in 0 Posts Simplex algorithm problem Hey guys! I am trying to write a Simplex algorithm program that reads input from a file. It runs but every time gives a no solution output ('No'). I have spent several days working on this program and still didn't find a bug. Code: ```import java.util.*; import java.io.*; import java.math.*; public class first { public static void main (String[]args) throws IOException { /* Reading input from a file*/ File inputFile = new File("input.txt"); Scanner in = new Scanner(inputFile); // Reading the number of free variables int n = in.nextInt(); // Reading the number of equations int m = in.nextInt(); // Defining indexes int i=0, j=0, k=0, l=0; double f = 0; //Creating a simplex tableau double[][] S = new double[m+1][n+m+1]; // Helping variable double x=0; // A very big number int M=10000; // Filling up the coefficients of function z for (i = 0; i < n; i++) S[m][i] = in.nextDouble(); // Filling up the coefficients of constraint equations for (i = 0; i < m; i++) for (j = 0; j < n; j++) S[i][j] = in.nextDouble(); // Filling up the RHS for (i = 0; i < m; i++) { S[i][n+m] = in.nextDouble(); // if RHS is negative, change the sign if (S[i][n+m]<0) { S[i][n+m]=-S[i][n+m]; for (j = 0; j < n; j++) S[i][j]=-S[i][j]; } x+=S[i][n+m]; //Adding basis variables to the simplex table S[m][n+i]=0; S[i][n+i]=1; } S[m][m+n] = x*M; x=0; // Changing the last row/column of the tableau for (j = 0; j < n; j++) { for (i = 0; i < m; i++) x+=S[i][j]; S[m][j]=x*M - S[m][j]; x=0; } // Now the tableu is filled up // Creating an array for the basis indexes int[] num = new int[n]; for(i=0; i0) { x=10000; for (i = 0; i < m; i++) if (S[i][k] > 0) // For the positive values we count (S[i][n+m]/S[i][k]) if ((S[i][n+m]/S[i][k]) < x) { //Find the minimum among them x = S[i][n+m]/S[i][k]; l=i; } //If all elements of the column are <= 0, there are no solutions if (((x-10000)<1.e-009)||((x-10000)>-(1.e-009))) { bw.write ("No"); bw.flush(); bw.close(); return; } // We found k and l indexes, so now we can perform the Gauss-Jordan method x=S[l][k]; for (j = 0; j <= m+n; j++) S[l][j]/=x; for (i=0; i<=m; i++) { x=S[i][k]; if(i!=l) for (j = 0; j <= m+n; j++) S[i][j]=S[i][j]-(S[l][j]*x); } // Finding the largest maximum element in the last row x=S[m][0]; for (j = 1; j < n; j++) if (S[m][j]>x) { x=S[m][j]; k=j; } num[ni]=k; // Record the index in the array of indexes of basis variables ni++; // After the first iteration we return to Step 1 } /* for(i=0;i<=m;i++){ for(j=0;j<=m+n;j++){ String str3 = new String(); str3 = Double.toString(S[i][j]); bw.write(str3); bw.write(" "); } bw.write("\n"); } bw.write("\n"); */ // At this point, we got a solution. Need to check it for (j = n; j < n+m-1; j++) {if ((S[m][j]<1.e-009)&&(S[m][j]>-(1.e-009))) if ((S[m][m+n]<1.e-009)&&(S[m][m+n]>-(1.e-009))) { bw.write ("Unbounded"); bw.flush(); bw.close(); return; } else { bw.write ("No"); bw.flush(); bw.close(); return; } } // If the solution is not unbounded, then the solution is found. bw.write("Yes\n"); double f1; f1 = S[m][n+m]; new BigDecimal(f1).setScale(6, BigDecimal.ROUND_HALF_EVEN); String str1 = new String(); str1 = Double.toString(f1); str1 = String.format("%8.6f", f1).replace(',','.'); bw.write(str1.replaceAll(" ","") + "\n"); //bw.write(Double.toString(S[m][n+m])+"\n"); for(i=0;i-(1.e-009))) { for (j=0; j-(1.e-009))) { f = S[j][n+m]; new BigDecimal(f).setScale(6, BigDecimal.ROUND_HALF_EVEN); String str = new String(); str = Double.toString(f); str = String.format("%8.6f", f).replace(',','.'); bw.write(str.replaceAll(" ","")); //bw.write(Double.toString(S[j][n+m])); bw.write(" "); } } else bw.write("0.000000 "); } bw.flush(); bw.close(); return; } }``` Last edited by musicawlatch; 11-29-2012 at 08:00 AM..