代码线性规划行使手(Code a linear programming exercise by ha

我一直在做线性规划问题，在我的班级通过以图表他们，但我想知道如何编写一个程序来解决这个问题对我来说是特别的问题。如果有太多的变量或限制，我不可能通过以图表形式做到这一点。

实施例的问题，最大化5X + 3Y与约束：

5倍 - 2Y> = 0
X + Y <= 7
X <= 5
X> = 0
Y> = 0

我绘制这一点，得到了明显的区域有3角。 X = 5 Y = 2是最佳点。

我怎么变成这个代码？我知道单纯形法。而且很重要的是，将全部LP问题，在相同结构的编码？将蛮力的工作？

Answer 1:

有相当数量的单纯实现的，如果你搜索，你会发现。

除了在注释（数字食谱在C）提到的，你还可以发现：

谷歌自己的单纯，求解
再有就是COIN-OR
GNU有自己的GLPK
如果你想要一个C ++实现，这一个在谷歌代码实际上是访问。
有许多实现R中包括引导软件包。（在R，您可以通过不括号键入它看到一个函数的实现。）

为了解决您的另外两个问题：

将所有的LP进行编码以同样的方式？ 是的，一个普通的LP求解器可以被写入装载和解决任何LP。（有行业标准格式读取LP的像mps和.lp
将蛮力的工作？ 请记住，许多公司和大机构花微调求解很长一段时间。有LP的有有趣的特性，很多求解器将尝试利用。另外，某些计算可并行地解决。该算法是指数，因此在一些大量的变量/约束，蛮力将无法正常工作。

希望帮助。

Answer 2:

这个是我写的matlab昨天，这可以很容易地转录成C ++，如果您使用本征库或使用STD的一个std ::向量写自己的矩阵类::矢量

function [x, fval] = mySimplex(fun, A, B, lb, up)

%Examples paramters to show that the function actually works 

% sample set 1 (works for this data set)

% fun = [8 10 7];
% A = [1 3 2; 1 5 1];
% B = [10; 8];
% lb = [0; 0; 0];
% ub = [inf; inf; inf];

% sample set 2 (works for this data set)

fun = [7 8 10];
A = [2 3 2; 1 1 2];
B = [1000; 800];
lb = [0; 0; 0];
ub = [inf; inf; inf];


% generate a new slack variable for every row of A 

numSlackVars = size(A,1); % need a new slack variables for every row of A 

% Set up tableau to store algorithm data 
tableau = [A; -fun];

tableau = [tableau, eye(numSlackVars + 1)];

lastCol = [B;0];

tableau = [tableau, lastCol];

% for convienience sake, assign the following: 

numRows = size(tableau,1);
numCols = size(tableau,2);

% do simplex algorithm 

% step 0: find num of negative entries in bottom row of tableau 

numNeg = 0; % the number of negative entries in bottom row

for i=1:numCols 
    if(tableau(numRows,i) < 0)
        numNeg = numNeg + 1;
    end
end

% Remark: the number of negatives is exactly the number of iterations needed in the
% simplex algorithm 

for iterations = 1:numNeg 
    % step 1: find minimum value in last row 
    minVal = 10000; % some big number 
    minCol = 1; % start by assuming min value is the first element 
    for i=1:numCols
        if(tableau(numRows, i) < minVal)
            minVal = tableau(size(tableau,1), i);
            minCol = i; % update the index corresponding to the min element 
        end
    end 

    % step 2: Find corresponding ratio vector in pivot column 
    vectorRatio = zeros(numRows -1, 1);
    for i=1:(numRows-1) % the size of ratio vector is numCols - 1
        vectorRatio(i, 1) = tableau(i, numCols) ./ tableau(i, minCol);
    end 

    % step 3: Determine pivot element by finding minimum element in vector
    % ratio

    minVal = 10000; % some big number 
    minRatio = 1; % holds the element with the minimum ratio 

    for i=1:numRows-1
        if(vectorRatio(i,1) < minVal)
            minVal = vectorRatio(i,1);
            minRatio = i;
        end 
    end 

    % step 4: assign pivot element 

    pivotElement = tableau(minRatio, minCol);

    % step 5: perform pivot operation on tableau around the pivot element 

    tableau(minRatio, :) = tableau(minRatio, :) * (1/pivotElement);

    % step 6: perform pivot operation on rows (not including last row)

    for i=1:size(vectorRatio,1)+1 % do last row last 
        if(i ~= minRatio) % we skip over the minRatio'th element of the tableau here 
            tableau(i, :) = -tableau(i,minCol)*tableau(minRatio, :) +  tableau(i,:);
        end
    end
end 

% Now we can interpret the algo tableau 

numVars = size(A,2); % the number of cols of A is the number of variables 

x = zeros(size(size(tableau,1), 1)); % for efficiency 

% Check for basicity 
for col=1:numVars
    count_zero = 0;
    count_one = 0;
    for row = 1:size(tableau,1)
        if(tableau(row,col) < 1e-2)
            count_zero = count_zero + 1;
        elseif(tableau(row,col) - 1 < 1e-2)
            count_one = count_one + 1;
            stored_row = row; % we store this (like in memory) column for later use 
        end
    end
    if(count_zero == (size(tableau,1) -1) && count_one == 1) % this is the case where it is basic 
        x(col,1) = tableau(stored_row, numCols);
    else 
        x(col,1) = 0; % this is the base where it is not basic 
    end
end

% find function optimal value at optimal solution 
fval = x(1,1) * fun(1,1); % just needed for logic to work here 
for i=2:numVars 
    fval = fval + x(i,1) * fun(1,i);
end


end