如何求解一个线性规划方程组

Question

农民有一块农田，比方说L km2，种植小麦或大麦或两者的某种组合。农民有有限的肥料，F公斤，和杀虫剂，P公斤。每平方公里的小麦需要F1公斤的肥料和P1公斤的杀虫剂，而每平方公里的大麦需要F2公斤的肥料和P2公斤的杀虫剂。设S1为每平方公里小麦的销售价格，S2为大麦的销售价格。如果我们用x1和x2分别表示小麦和大麦的种植面积，那么可以通过选择x1和x2的最佳值来最大化利润。这个问题可以用以下标准形式的线性规划问题来表示:请参阅此页面，其中给出了约束。http://en.wikipedia.org/wiki/Linear_programming

解决这类问题的程序是什么？

Answer 1

你的问题就是所谓的线性规划问题。有各种算法和不同的实现。我用过一个叫GLPK的程序，它工作得很好。您使用一种特定于领域的编程语言来说明您的问题，GLPK将处理该程序以找到解决方案。在web上搜索GLPK就可以找到它。

Answer 2

import java.text.DecimalFormat;

public class CandidateCode {
    public String get_total_profit(String input1) {
        String[] inputs = input1.split(",");
        // Piece of farm land in square kilometer
        float L = Float.valueOf(inputs[0]);
        // Fertilizer in kg
        float F = Float.valueOf(inputs[1]);
        // Insecticide in kg
        float P = Float.valueOf(inputs[2]);
        // Fertilizer required in kg for square kilometer of Wheat
        float F1 = Float.valueOf(inputs[3]);
        // Insecticide required in kg for square kilometer of Wheat
        float P1 = Float.valueOf(inputs[4]);
        // Fertilizer required in kg for square kilometer of Rice
        float F2 = Float.valueOf(inputs[5]);
        // Insecticide required in kg for square kilometer of Rice
        float P2 = Float.valueOf(inputs[6]);
        // Selling price of wheat per square kilometer
        float S1 = Float.valueOf(inputs[7]);
        // Selling price of rice per square kilometer
        float S2 = Float.valueOf(inputs[8]);

        // Result Variables
        float totalRiceInsecUsed = 0f;
        float totalRiceFertUsed = 0f;
        float totalWheatInsecUsed = 0f;
        float totalWheatFertUsed = 0f;
        float areaOfWheat = 0.00f;
        float areaOfRice = 0.00f;
        float amount = 0.00f;

        while (true) {
            if ((L == areaOfRice + areaOfWheat)
                    || P == totalRiceInsecUsed + totalWheatInsecUsed
                    || F == totalRiceFertUsed + totalWheatFertUsed || F2 == 0
                    || F1 == 0 || P2 == 0 || P1 == 0) {
                break;
            }

            float calRiceProfit = Math.min(F / F2, P / P2) * S2;
            float calWheatProfit = Math.min(F / F1, P / P1) * S1;

            if (calRiceProfit > calWheatProfit) {
                float areaInsecUsed = P / P2;
                float areaFertUsed = F / F2;
                if (areaInsecUsed > areaFertUsed) {
                    L = L - areaFertUsed;
                    F2 = 0;
                    totalRiceFertUsed = totalRiceFertUsed + F2;
                    areaOfRice = areaOfRice + areaFertUsed;
                    amount = amount + areaFertUsed * S2;
                } else if (areaInsecUsed < areaFertUsed) {
                    L = L - areaInsecUsed;
                    P2 = 0;
                    totalRiceInsecUsed = totalRiceInsecUsed + areaInsecUsed;
                    areaOfRice = areaOfRice + areaInsecUsed;
                    amount = amount + areaInsecUsed * S2;
                }
            } else {
                float areaInsecUsed = P / P1;
                float areaFertUsed = F / F1;
                if (areaInsecUsed > areaFertUsed) {
                    L = L - areaFertUsed;
                    F1 = 0;
                    totalWheatFertUsed = totalWheatFertUsed + F1;
                    areaOfWheat = areaOfWheat + areaFertUsed;
                    amount = amount + areaFertUsed * S1;
                } else if (areaInsecUsed < areaFertUsed) {
                    L = L - areaInsecUsed;
                    P1 = 0;
                    totalWheatInsecUsed = totalWheatInsecUsed + areaInsecUsed;
                    areaOfWheat = areaOfWheat + areaInsecUsed;
                    amount = amount + areaInsecUsed * S1;
                }
            }

        }
        DecimalFormat df = new DecimalFormat();
        df.setMaximumFractionDigits(2);
        df.setMinimumFractionDigits(2);
        return df.format(amount) + "," + df.format(areaOfWheat) + ","
                + df.format(areaOfRice);
    }
}