C/C++ implementation of simplex method [closed]

Question

I am unable to find an implemenation of simplex method.I have a set of points and want to minimize theie distance so i only need the method simplex I have google before posting this question and could nt find anything that I could use

Inhere

Is this what you're talking about? en.wikipedia.org/wiki/Simplex_algorithm – Loisloise

no..i need one in which we have a set of equations and slack variables..LP problem – Inhere

sorry i think this is the one i need..but a c/c++ function – Inhere

Answer 1

/*
  What: Simplex in C
  AUTHOR: GPL(C) moshahmed/at/gmail.

  What: Solves LP Problem with Simplex:
    { maximize cx : Ax <= b, x >= 0 }.
  Input: { m, n, Mat[m x n] }, where:
    b = mat[1..m,0] .. column 0 is b >= 0, so x=0 is a basic feasible solution.
    c = mat[0,1..n] .. row 0 is z to maximize, note c is negated in input.
    A = mat[1..m,1..n] .. constraints.
    x = [x1..xm] are the named variables in the problem.
    Slack variables are in columns [m+1..m+n]

  USAGE:
    1. Problem can be specified before main function in source code:
      c:\> vim mosplex.c  
      Tableau tab  = { m, n, {   // tableau size, row x columns.
          {  0 , -c1, -c2,  },  // Max: z = c1 x1 + c2 x2,
          { b1 , a11, a12,  },  //  b1 >= a11 x1 + a12 x2
          { b2 , a21, a22,  },  //  b2 >= a21 x1 + a22 x2
        }
      };
      c:\> cl /W4 mosplex.c  ... compile this file.
      c:\> mosplex.exe problem.txt > solution.txt

    2. OR Read the problem data from a file:
      $ cat problem.txt
            m n
            0  -c1 -c2
            b1 a11 a12
            b2 a21 a11 
      $ gcc -Wall -g mosplex.c  -o mosplex
      $ mosplex problem.txt > solution.txt
*/

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <string.h>
#include <assert.h>

#define M 20
#define N 20

static const double epsilon   = 1.0e-8;
int equal(double a, double b) { return fabs(a-b) < epsilon; }

typedef struct {
  int m, n; // m=rows, n=columns, mat[m x n]
  double mat[M][N];
} Tableau;

void nl(int k){ int j; for(j=0;j<k;j++) putchar('-'); putchar('\n'); }

void print_tableau(Tableau *tab, const char* mes) {
  static int counter=0;
  int i, j;
  printf("\n%d. Tableau %s:\n", ++counter, mes);
  nl(70);

  printf("%-6s%5s", "col:", "b[i]");
  for(j=1;j<tab->n; j++) { printf("    x%d,", j); } printf("\n");

  for(i=0;i<tab->m; i++) {
    if (i==0) printf("max:"); else
    printf("b%d: ", i);
    for(j=0;j<tab->n; j++) {
      if (equal((int)tab->mat[i][j], tab->mat[i][j]))
        printf(" %6d", (int)tab->mat[i][j]);
      else
        printf(" %6.2lf", tab->mat[i][j]);
      }
    printf("\n");
  }
  nl(70);
}

/* Example input file for read_tableau:
     4 5
      0   -0.5  -3 -1  -4 
     40    1     1  1   1 
     10   -2    -1  1   1 
     10    0     1  0  -1  
*/
void read_tableau(Tableau *tab, const char * filename) {
  int err, i, j;
  FILE * fp;

  fp  = fopen(filename, "r" );
  if( !fp ) {
    printf("Cannot read %s\n", filename); exit(1);
  }
  memset(tab, 0, sizeof(*tab));
  err = fscanf(fp, "%d %d", &tab->m, &tab->n);
  if (err == 0 || err == EOF) {
    printf("Cannot read m or n\n"); exit(1);
  }
  for(i=0;i<tab->m; i++) {
    for(j=0;j<tab->n; j++) {
      err = fscanf(fp, "%lf", &tab->mat[i][j]);
      if (err == 0 || err == EOF) {
        printf("Cannot read A[%d][%d]\n", i, j); exit(1);
      }
    }
  }
  printf("Read tableau [%d rows x %d columns] from file '%s'.\n",
    tab->m, tab->n, filename);
  fclose(fp);
}

void pivot_on(Tableau *tab, int row, int col) {
  int i, j;
  double pivot;

  pivot = tab->mat[row][col];
  assert(pivot>0);
  for(j=0;j<tab->n;j++)
    tab->mat[row][j] /= pivot;
  assert( equal(tab->mat[row][col], 1. ));

  for(i=0; i<tab->m; i++) { // foreach remaining row i do
    double multiplier = tab->mat[i][col];
    if(i==row) continue;
    for(j=0; j<tab->n; j++) { // r[i] = r[i] - z * r[row];
      tab->mat[i][j] -= multiplier * tab->mat[row][j];
    }
  }
}

// Find pivot_col = most negative column in mat[0][1..n]
int find_pivot_column(Tableau *tab) {
  int j, pivot_col = 1;
  double lowest = tab->mat[0][pivot_col];
  for(j=1; j<tab->n; j++) {
    if (tab->mat[0][j] < lowest) {
      lowest = tab->mat[0][j];
      pivot_col = j;
    }
  }
  printf("Most negative column in row[0] is col %d = %g.\n", pivot_col, lowest);
  if( lowest >= 0 ) {
    return -1; // All positive columns in row[0], this is optimal.
  }
  return pivot_col;
}

// Find the pivot_row, with smallest positive ratio = col[0] / col[pivot]
int find_pivot_row(Tableau *tab, int pivot_col) {
  int i, pivot_row = 0;
  double min_ratio = -1;
  printf("Ratios A[row_i,0]/A[row_i,%d] = [",pivot_col);
  for(i=1;i<tab->m;i++){
    double ratio = tab->mat[i][0] / tab->mat[i][pivot_col];
    printf("%3.2lf, ", ratio);
    if ( (ratio > 0  && ratio < min_ratio ) || min_ratio < 0 ) {
      min_ratio = ratio;
      pivot_row = i;
    }
  }
  printf("].\n");
  if (min_ratio == -1)
    return -1; // Unbounded.
  printf("Found pivot A[%d,%d], min positive ratio=%g in row=%d.\n",
      pivot_row, pivot_col, min_ratio, pivot_row);
  return pivot_row;
}

void add_slack_variables(Tableau *tab) {
  int i, j;
  for(i=1; i<tab->m; i++) {
    for(j=1; j<tab->m; j++)
      tab->mat[i][j + tab->n -1] = (i==j);
  }
  tab->n += tab->m -1;
}

void check_b_positive(Tableau *tab) {
  int i;
  for(i=1; i<tab->m; i++)
    assert(tab->mat[i][0] >= 0);
}

// Given a column of identity matrix, find the row containing 1.
// return -1, if the column as not from an identity matrix.
int find_basis_variable(Tableau *tab, int col) {
  int i, xi=-1;
  for(i=1; i < tab->m; i++) {
    if (equal( tab->mat[i][col],1) ) {
      if (xi == -1)
        xi=i;   // found first '1', save this row number.
      else
        return -1; // found second '1', not an identity matrix.

    } else if (!equal( tab->mat[i][col],0) ) {
      return -1; // not an identity matrix column.
    }
  }
  return xi;
}

void print_optimal_vector(Tableau *tab, char *message) {
  int j, xi;
  printf("%s at ", message);
  for(j=1;j<tab->n;j++) { // for each column.
    xi = find_basis_variable(tab, j);
    if (xi != -1)
      printf("x%d=%3.2lf, ", j, tab->mat[xi][0] );
    else
      printf("x%d=0, ", j);
  }
  printf("\n");
} 

void simplex(Tableau *tab) {
  int loop=0;
  add_slack_variables(tab);
  check_b_positive(tab);
  print_tableau(tab,"Padded with slack variables");
  while( ++loop ) {
    int pivot_col, pivot_row;

    pivot_col = find_pivot_column(tab);
    if( pivot_col < 0 ) {
      printf("Found optimal value=A[0,0]=%3.2lf (no negatives in row 0).\n",
        tab->mat[0][0]);
      print_optimal_vector(tab, "Optimal vector");
      break;
    }
    printf("Entering variable x%d to be made basic, so pivot_col=%d.\n",
      pivot_col, pivot_col);

    pivot_row = find_pivot_row(tab, pivot_col);
    if (pivot_row < 0) {
      printf("unbounded (no pivot_row).\n");
      break;
    }
    printf("Leaving variable x%d, so pivot_row=%d\n", pivot_row, pivot_row);

    pivot_on(tab, pivot_row, pivot_col);
    print_tableau(tab,"After pivoting");
    print_optimal_vector(tab, "Basic feasible solution");

    if(loop > 20) {
      printf("Too many iterations > %d.\n", loop);
      break;
    }
  }
}

Tableau tab  = { 4, 5, {                     // Size of tableau [4 rows x 5 columns ]
    {  0.0 , -0.5 , -3.0 ,-1.0 , -4.0,   },  // Max: z = 0.5*x + 3*y + z + 4*w,
    { 40.0 ,  1.0 ,  1.0 , 1.0 ,  1.0,   },  //    x + y + z + w <= 40 .. b1
    { 10.0 , -2.0 , -1.0 , 1.0 ,  1.0,   },  //  -2x - y + z + w <= 10 .. b2
    { 10.0 ,  0.0 ,  1.0 , 0.0 , -1.0,   },  //        y     - w <= 10 .. b3
  }
};

int main(int argc, char *argv[]){
  if (argc > 1) { // usage: cmd datafile
    read_tableau(&tab, argv[1]);
  }
  print_tableau(&tab,"Initial");
  simplex(&tab);
  return 0;
} 

Malt

in find_pivot_row function, min_ratio can still be negative and not equal to -1 – Capillaceous