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我在这里找到有关 Simplex 方法的主题Alter Simplex Algorithm to Minimize on objective function NOT 最大化 但答案没有帮助。当我从

double[] variables = {  13.0,  23.0 };

至

double[] variables = { -13.0, -23.0 };

该程序不计算（无异常），它打印第一步，仅此而已。有人可以帮助我改变单纯形法从最大化到最小化吗？

代码：

导入 java.util.*；

public class Simplex
{
private static final double EPSILON = 1.0E-10;
private double[][] tableaux;
private int numOfConstraints;
private int numOfVariables;

private int[] basis;
/**
 * Constructor for objects of class Simplex
 */
public Simplex()
{


    double[][] thisTableaux = {
        {  5.0, 15.0 },
        {  4.0,  4.0 },
        { 35.0, 20.0 },
    };

    double[] constraints = { 480.0, 160.0, 1190.0 };

    double[] variables = {  -13.0,  -23.0 };

    numOfConstraints = constraints.length;
    numOfVariables = variables.length;

    tableaux = new double[numOfConstraints+1][numOfVariables+numOfConstraints+1];

    //adds all elements from thisTableaux to tableaux
    for(int i=0; i < numOfConstraints; i++)
    {
        for(int j=0; j < numOfVariables; j++)
        {
            tableaux[i][j] = thisTableaux[i][j];
        }
    } 


    //adds a slack variable for each variable there is and sets it to 1.0
    for(int i=0; i < numOfConstraints; i++)
    {
        tableaux[i][numOfVariables+i] = 1.0;
    }


    //adds variables into the second [] of tableux
    for(int j=0; j < numOfVariables; j++)
    {
        tableaux[numOfConstraints][j] = variables[j];
    }



    //adds constraints to first [] of tableaux
    for(int k=0; k < numOfConstraints; k++)
    {
        tableaux[k][numOfConstraints+numOfVariables] = constraints[k];
    }



    basis = new int[numOfConstraints];

    for(int i=0; i < numOfConstraints; i++)
    {
        basis[i] = numOfVariables + i;
    }

    show();

    optimise();

    assert check(thisTableaux, constraints, variables);


}

public void optimise() {
    while(true) {

        int q = findLowestNonBasicCol();

        if(q == -1) {
            break;
        }

        int p = getPivotRow(q);
        if(p == -1) throw new ArithmeticException("Linear Program Unbounded");

        pivot(p, q);

        basis[p] = q;
    }

}

public int findLowestNonBasicCol() {

    for(int i=0; i < numOfConstraints + numOfVariables; i++)
    {
        if(tableaux[numOfConstraints][i] > 0) {


            return i;
        }
    }

    return -1;


}

public int findIndexOfLowestNonBasicCol() {

    int q = 0;
    for(int i=1; i < numOfConstraints + numOfVariables; i++)
    {
        if(tableaux[numOfConstraints][i] > tableaux[numOfConstraints][q]) {
            q = i;
        }
    }

    if(tableaux[numOfConstraints][q] <= 0) {
        return -1;
    }

    else {
        return q;
    }
}

/**
 * Finds row p which will be the pivot row using the minimum ratio rule.
 * -1 if there is no pivot row
 */
public int getPivotRow(int q) {

    int p = -1;

    for(int i=0; i < numOfConstraints; i++) {

        if (tableaux[i][q] <=0) {
            continue;
        }

        else if (p == -1) {
            p = i;
        }

        else if((tableaux[i][numOfConstraints+numOfVariables] / tableaux[i][q] < tableaux[p][numOfConstraints+numOfVariables] / tableaux[p][q])) {
            p = i;
        }
    }
4

2 回答 2

1

您可能想研究对偶单纯形法（或对偶理论）。如果原始问题的标准形式是：

Maximize = 13*X1 + 23*X2;

有约束：

5*X1    +   15*X2   <= 480;
4*X1    +   4*X2    <= 160;
35*X1   +   20*X2   <= 1190;
X1 >= 0;
X2 >= 0;

那么对偶问题是：

Minimize = 480*Y1 + 160*Y2 + 1190*Y3;

有约束：

5*Y1    +   4*Y2    +   35*Y3   >= 13;
15*Y1 +     4*Y2    +   20*Y3   >= 23;
Y1 >= 0;
Y2 >= 0;
Y3 >= 0;

我在LINGO中测试了这两个问题并得到了相同的答案（Z = 800，X1 = 12，X2 = 28 -- Y1 = 1，Y2 = 2，Y3 = 0）。

于 2016-02-02T21:28:05.857 回答
-1

我猜该程序什么也没做，因为初始解决方案是最佳解决方案。

于 2014-05-18T19:19:12.740 回答