我被要求写下一个Matlab程序，以便使用修订的Simplex方法来解决LP。
我编写的代码运行没有输入数据的问题，虽然我已经意识到它没有正确解决问题，因为它没有更新基础B的逆（上述方法的真正核心思想）。 / p>
问题只与代码的一部分有关，脚本底部的代码是针对：
通过对[B ^ -1 u]执行基本行操作来计算新的逆基础B ^ -1（轴行索引为l_out）。向量u被转换为单位向量，其中u（l_out）= 1而u（i）= 0表示其他i。
这是我写的代码：
%% Implementation of the revised Simplex. Solves a linear
% programming problem of the form
%
% min c'*x
% s.t. Ax = b
% x >= 0
%
% The function input parameters are the following:
% A: The constraint matrix
% b: The rhs vector
% c: The vector of cost coefficients
% C: The indices of the basic variables corresponding to an
% initial basic feasible solution
%
% The function returns:
% x_opt: Decision variable values at the optimal solution
% f_opt: Objective function value at the optimal solution
%
% Usage: [x_opt, f_opt] = S12345X(A,b,c,C)
% NOTE: Replace 12345X with your own student number
% and rename the file accordingly
function [x_opt, f_opt] = SXXXXX(A,b,c,C)
%% Initialization phase
% Initialize the vector of decision variables
x = zeros(length(c),1);
% Create the initial Basis matrix, compute its inverse and
% compute the inital basic feasible solution
B=A(:,C);
invB = inv(B);
x(C) = invB*b;
%% Iteration phase
n_max = 10; % At most n_max iterations
for n = 1:n_max % Main loop
% Compute the vector of reduced costs c_r
c_B = c(C); % Basic variable costs
p = (c_B'*invB)'; % Dual variables
c_r = c' - p'*A; % Vector of reduced costs
% Check if the solution is optimal. If optimal, use
% 'return' to break from the function, e.g.
J = find(c_r < 0); % Find indices with negative reduced costs
if (isempty(J))
f_opt = c'*x;
x_opt = x;
return;
end
% Choose the entering variable
j_in = J(1);
% Compute the vector u (i.e., the reverse of the basic directions)
u = invB*A(:,j_in);
I = find(u > 0);
if (isempty(I))
f_opt = -inf; % Optimal objective function cost = -inf
x_opt = []; % Produce empty vector []
return % Break from the function
end
% Compute the optimal step length theta
theta = min(x(C(I))./u(I));
L = find(x(C)./u == theta); % Find all indices with ratio theta
% Select the exiting variable
l_out = L(1);
% Move to the adjacent solution
x(C) = x(C) - theta*u;
% Value of the entering variable is theta
x(j_in) = theta;
% Update the set of basic indices C
C(l_out) = j_in;
% Compute the new inverse basis B^-1 by performing elementary row
% operations on [B^-1 u] (pivot row index is l_out). The vector u is trans-
% formed into a unit vector with u(l_out) = 1 and u(i) = 0 for
% other i.
M=horzcat(invB,u);
[f g]=size(M);
R(l_out,:)=M(l_out,:)/M(l_out,j_in); % Copy row l_out, normalizing M(l_out,j_in) to 1
u(l_out)=1;
for k = 1:f % For all matrix rows
if (k ~= l_out) % Other then l_out
u(k)=0;
R(k,:)=M(k,:)-M(k,j_in)*R(l_out,:); % Set them equal to the original matrix Minus a multiple of normalized row l_out, making R(k,j_in)=0
end
end
invM=horzcat(u,invB);
% Check if too many iterations are performed (increase n_max to
% allow more iterations)
if(n == n_max)
fprintf('Max number of iterations performed!\n\n');
return
end
end % End for (the main iteration loop)
end % End function
%% Example 3.5 from the book (A small test problem)
% Data in standard form:
% A = [1 2 2 1 0 0;
% 2 1 2 0 1 0;
% 2 2 1 0 0 1];
% b = [20 20 20]';
% c = [-10 -12 -12 0 0 0]';
% C = [4 5 6]; % Indices of the basic variables of
% % the initial basic feasible solution
%
% The optimal solution
% x_opt = [4 4 4 0 0 0]' % Optimal decision variable values
% f_opt = -136 % Optimal objective function cost
答案 0 :(得分：1)
好吧，经过大量的时间花在密集使用printmat和disp从数学的角度来理解代码中发生了什么我意识到这是一个问题，索引j_in和规范化除以因此，我设法解决了以下问题。 现在它完美运行。欢呼声。
%% Implementation of the revised Simplex. Solves a linear
% programming problem of the form
%
% min c'*x
% s.t. Ax = b
% x >= 0
%
% The function input parameters are the following:
% A: The constraint matrix
% b: The rhs vector
% c: The vector of cost coefficients
% C: The indices of the basic variables corresponding to an
% initial basic feasible solution
%
% The function returns:
% x_opt: Decision variable values at the optimal solution
% f_opt: Objective function value at the optimal solution
%
% Usage: [x_opt, f_opt] = S12345X(A,b,c,C)
% NOTE: Replace 12345X with your own student number
% and rename the file accordingly
function [x_opt, f_opt] = S472366(A,b,c,C)
%% Initialization phase
% Initialize the vector of decision variables
x = zeros(length(c),1);
% Create the initial Basis matrix, compute its inverse and
% compute the inital basic feasible solution
B=A(:,C);
invB = inv(B);
x(C) = invB*b;
%% Iteration phase
n_max = 10; % At most n_max iterations
for n = 1:n_max % Main loop
% Compute the vector of reduced costs c_r
c_B = c(C); % Basic variable costs
p = (c_B'*invB)'; % Dual variables
c_r = c' - p'*A; % Vector of reduced costs
% Check if the solution is optimal. If optimal, use
% 'return' to break from the function, e.g.
J = find(c_r < 0); % Find indices with negative reduced costs
if (isempty(J))
f_opt = c'*x;
x_opt = x;
return;
end
% Choose the entering variable
j_in = J(1);
% Compute the vector u (i.e., the reverse of the basic directions)
u = invB*A(:,j_in);
I = find(u > 0);
if (isempty(I))
f_opt = -inf; % Optimal objective function cost = -inf
x_opt = []; % Produce empty vector []
return % Break from the function
end
% Compute the optimal step length theta
theta = min(x(C(I))./u(I));
L = find(x(C)./u == theta); % Find all indices with ratio theta
% Select the exiting variable
l_out = L(1);
% Move to the adjacent solution
x(C) = x(C) - theta*u;
% Value of the entering variable is theta
x(j_in) = theta;
% Update the set of basic indices C
C(l_out) = j_in;
% Compute the new inverse basis B^-1 by performing elementary row
% operations on [B^-1 u] (pivot row index is l_out). The vector u is trans-
% formed into a unit vector with u(l_out) = 1 and u(i) = 0 for
% other i.
M=horzcat(u, invB);
[f g]=size(M);
if (theta~=0)
M(l_out,:)=M(l_out,:)/M(l_out,1); % Copy row l_out, normalizing M(l_out,1) to 1
end
for k = 1:f % For all matrix rows
if (k ~= l_out) % Other then l_out
M(k,:)=M(k,:)-M(k,1)*M(l_out,:); % Set them equal to the original matrix Minus a multiple of normalized row l_out, making R(k,j_in)=0
end
end
invB=M(1:3,2:end);
% Check if too many iterations are performed (increase n_max to
% allow more iterations)
if(n == n_max)
fprintf('Max number of iterations performed!\n\n');
return
end
end % End for (the main iteration loop)
end % End function
%% Example 3.5 from the book (A small test problem)
% Data in standard form:
% A = [1 2 2 1 0 0;
% 2 1 2 0 1 0;
% 2 2 1 0 0 1];
% b = [20 20 20]';
% c = [-10 -12 -12 0 0 0]';
% C = [4 5 6]; % Indices of the basic variables of
% % the initial basic feasible solution
%
% The optimal solution
% x_opt = [4 4 4 0 0 0]' % Optimal decision variable values
% f_opt = -136 % Optimal objective function cost