model:
max=13*A+ 23*B;
5*A + 15*B <480 ;
4*A + 4 *B <160 ;
35* A + 20 *B <1190 ;
end
Variable Value Reduced Cost
A 12.00000 0.000000
B 28.00000 0.000000
Row Slack or Surplus Dual Price
1 800.0000 1.000000
2 0.000000 1.000000
3 0.000000 2.000000
4 210.0000 0.000000
public static void test0() {
double[][] A = {
{ 5, 15 },
{ 4, 4 },
{ 35, 20 }
};
double[] c = { 13, 23 };
double[] b = { 480,160,1190};
test(A, b, c);
}
value = 800.0
x[0] = 11.999999999999998
x[1] = 28.0
y[0] = 1.0
y[1] = 2.0
y[2] = -0.0
package 线性规划; /** * Created by han on 2016/5/5. */ import java.io.PrintStream; /************************************************************************* * Compilation: javac Simplex.java * Execution: java Simplex * * Given an M-by-N matrix A, an M-length vector b, and an * N-length vector c, solve the LP { max cx : Ax <= b, x >= 0 }. * Assumes that b >= 0 so that x = 0 is a basic feasible solution. * * Creates an (M+1)-by-(N+M+1) simplex tableaux with the * RHS in column M+N, the objective function in row M, and * slack variables in columns M through M+N-1. * *************************************************************************/ public class Simplex { private static final double EPSILON = 1.0E-10; private double[][] a; // tableaux private int M; // number of constraints private int N; // number of original variables private int[] basis; // basis[i] = basic variable corresponding to row i private static PrintStream StdOut=System.out; // only needed to print out solution, not book // sets up the simplex tableaux public Simplex(double[][] A, double[] b, double[] c) { M = b.length; N = c.length; a = new double[M+1][N+M+1]; for (int i = 0; i < M; i++) for (int j = 0; j < N; j++) a[i][j] = A[i][j]; for (int i = 0; i < M; i++) a[i][N+i] = 1.0; for (int j = 0; j < N; j++) a[M][j] = c[j]; for (int i = 0; i < M; i++) a[i][M+N] = b[i]; basis = new int[M]; for (int i = 0; i < M; i++) basis[i] = N + i; solve(); // check optimality conditions assert check(A, b, c); } // run simplex algorithm starting from initial BFS private void solve() { while (true) { // find entering column q int q = bland(); if (q == -1) break; // optimal // find leaving row p int p = minRatioRule(q); if (p == -1) throw new RuntimeException("Linear program is unbounded"); // pivot pivot(p, q); // update basis basis[p] = q; } } // lowest index of a non-basic column with a positive cost private int bland() { for (int j = 0; j < M + N; j++) if (a[M][j] > 0) return j; return -1; // optimal } // index of a non-basic column with most positive cost private int dantzig() { int q = 0; for (int j = 1; j < M + N; j++) if (a[M][j] > a[M][q]) q = j; if (a[M][q] <= 0) return -1; // optimal else return q; } // find row p using min ratio rule (-1 if no such row) private int minRatioRule(int q) { int p = -1; for (int i = 0; i < M; i++) { if (a[i][q] <= 0) continue; else if (p == -1) p = i; else if ((a[i][M+N] / a[i][q]) < (a[p][M+N] / a[p][q])) p = i; } return p; } // pivot on entry (p, q) using Gauss-Jordan elimination private void pivot(int p, int q) { // everything but row p and column q for (int i = 0; i <= M; i++) for (int j = 0; j <= M + N; j++) if (i != p && j != q) a[i][j] -= a[p][j] * a[i][q] / a[p][q]; // zero out column q for (int i = 0; i <= M; i++) if (i != p) a[i][q] = 0.0; // scale row p for (int j = 0; j <= M + N; j++) if (j != q) a[p][j] /= a[p][q]; a[p][q] = 1.0; } // return optimal objective value public double value() { return -a[M][M+N]; } // return primal solution vector public double[] primal() { double[] x = new double[N]; for (int i = 0; i < M; i++) if (basis[i] < N) x[basis[i]] = a[i][M+N]; return x; } // return dual solution vector public double[] dual() { double[] y = new double[M]; for (int i = 0; i < M; i++) y[i] = -a[M][N+i]; return y; } // is the solution primal feasible? private boolean isPrimalFeasible(double[][] A, double[] b) { double[] x = primal(); // check that x >= 0 for (int j = 0; j < x.length; j++) { if (x[j] < 0.0) { StdOut.println("x[" + j + "] = " + x[j] + " is negative"); return false; } } // check that Ax <= b for (int i = 0; i < M; i++) { double sum = 0.0; for (int j = 0; j < N; j++) { sum += A[i][j] * x[j]; } if (sum > b[i] + EPSILON) { StdOut.println("not primal feasible"); StdOut.println("b[" + i + "] = " + b[i] + ", sum = " + sum); return false; } } return true; } // is the solution dual feasible? private boolean isDualFeasible(double[][] A, double[] c) { double[] y = dual(); // check that y >= 0 for (int i = 0; i < y.length; i++) { if (y[i] < 0.0) { StdOut.println("y[" + i + "] = " + y[i] + " is negative"); return false; } } // check that yA >= c for (int j = 0; j < N; j++) { double sum = 0.0; for (int i = 0; i < M; i++) { sum += A[i][j] * y[i]; } if (sum < c[j] - EPSILON) { StdOut.println("not dual feasible"); StdOut.println("c[" + j + "] = " + c[j] + ", sum = " + sum); return false; } } return true; } // check that optimal value = cx = yb private boolean isOptimal(double[] b, double[] c) { double[] x = primal(); double[] y = dual(); double value = value(); // check that value = cx = yb double value1 = 0.0; for (int j = 0; j < x.length; j++) value1 += c[j] * x[j]; double value2 = 0.0; for (int i = 0; i < y.length; i++) value2 += y[i] * b[i]; if (Math.abs(value - value1) > EPSILON || Math.abs(value - value2) > EPSILON) { StdOut.println("value = " + value + ", cx = " + value1 + ", yb = " + value2); return false; } return true; } private boolean check(double[][]A, double[] b, double[] c) { return isPrimalFeasible(A, b) && isDualFeasible(A, c) && isOptimal(b, c); } // print tableaux public void show() { StdOut.println("M = " + M); StdOut.println("N = " + N); for (int i = 0; i <= M; i++) { for (int j = 0; j <= M + N; j++) { StdOut.printf("%7.2f ", a[i][j]); } StdOut.println(); } StdOut.println("value = " + value()); for (int i = 0; i < M; i++) if (basis[i] < N) StdOut.println("x_" + basis[i] + " = " + a[i][M+N]); StdOut.println(); } public static void test(double[][] A, double[] b, double[] c) { Simplex lp = new Simplex(A, b, c); StdOut.println("value = " + lp.value()); double[] x = lp.primal(); for (int i = 0; i < x.length; i++) StdOut.println("x[" + i + "] = " + x[i]); double[] y = lp.dual(); for (int j = 0; j < y.length; j++) StdOut.println("y[" + j + "] = " + y[j]); } public static void test0() { double[][] A = { { 5, 15 }, { 4, 4 }, { 35, 20 } }; double[] c = { 13, 23 }; double[] b = { 480,160,1190}; test(A, b, c); } public static void test1() { double[][] A = { { -1, 1, 0 }, { 1, 4, 0 }, { 2, 1, 0 }, { 3, -4, 0 }, { 0, 0, 1 }, }; double[] c = { 1, 1, 1 }; double[] b = { 5, 45, 27, 24, 4 }; test(A, b, c); } // x0 = 12, x1 = 28, opt = 800 public static void test2() { double[] c = { 13.0, 23.0 }; double[] b = { 480.0, 160.0, 1190.0 }; double[][] A = { { 5.0, 15.0 }, { 4.0, 4.0 }, { 35.0, 20.0 }, }; test(A, b, c); } // unbounded public static void test3() { double[] c = { 2.0, 3.0, -1.0, -12.0 }; double[] b = { 3.0, 2.0 }; double[][] A = { { -2.0, -9.0, 1.0, 9.0 }, { 1.0, 1.0, -1.0, -2.0 }, }; test(A, b, c); } // degenerate - cycles if you choose most positive objective function coefficient public static void test4() { double[] c = { 10.0, -57.0, -9.0, -24.0 }; double[] b = { 0.0, 0.0, 1.0 }; double[][] A = { { 0.5, -5.5, -2.5, 9.0 }, { 0.5, -1.5, -0.5, 1.0 }, { 1.0, 0.0, 0.0, 0.0 }, }; test(A, b, c); } // test client public static void main(String[] args) { try { test0(); } catch (Exception e) { e.printStackTrace(); } StdOut.println("--------------------------------"); try { test1(); } catch (Exception e) { e.printStackTrace(); } StdOut.println("--------------------------------"); try { test2(); } catch (Exception e) { e.printStackTrace(); } StdOut.println("--------------------------------"); try { test3(); } catch (Exception e) { e.printStackTrace(); } StdOut.println("--------------------------------"); try { test4(); } catch (Exception e) { e.printStackTrace(); } StdOut.println("--------------------------------"); int M = Integer.parseInt(args[0]); int N = Integer.parseInt(args[1]); double[] c = new double[N]; double[] b = new double[M]; double[][] A = new double[M][N]; for (int j = 0; j < N; j++) c[j] = StdRandom.uniform(1000); for (int i = 0; i < M; i++) b[i] = StdRandom.uniform(1000); for (int i = 0; i < M; i++) for (int j = 0; j < N; j++) A[i][j] = StdRandom.uniform(100); Simplex lp = new Simplex(A, b, c); StdOut.println(lp.value()); } private static class StdRandom { public static double uniform(int i) { return Math.random()*i; } } }