cobratoolbox中You can solve LP problems using没有MATLAB

当然，您可以使用其他软件或编程语言来解决线性规划问题，MATLAB并不是唯一的选择。以下是一些常见的软件和编程语言： 1. R：R是一个流行的开源统计软件，它有多个包可以用来解决线性规划问题，如lpSolve和ROI等。 2. Python：Python是一种流行的编程语言，有多个库可以用于线性规划求解，如PuLP、CVXOPT和SciPy等。 3. Julia：Julia是一种高性能的开源编程语言，它有多个包可以用于线性规划求解，如JuMP和GLPK等。 4. GAMS：GAMS是一种专门用于数学建模和优化的商业软件，它可以用于线性规划求解。 这些软件和编程语言都提供了相应的工具和库来解决线性规划问题，您可以根据自己的需求和熟练程度选择合适的工具。

下面是使用MATLAB解决标准形式线性规划问题的单纯形方法程序及算例： 程序： ```matlab function [x, fval, exitflag, output] = simplex(c, A, b, varargin) %SIMPLEX Solve standard LP problems using simplex algorithm. % [X,FVAL,EXITFLAG,OUTPUT] = SIMPLEX(C,A,B) solves the linear programming % problem: % % minimize C'*X subject to: A*X <= B % X % % using the simplex algorithm. C must be a row vector, A must be a matrix, % and B must be a column vector. The problem can have equality constraints % (Aeq*X = Beq) but this requires the use of the 'Aeq' and 'Beq' options. % The problem can also have upper bounds on the variables (X <= U) but this % requires the use of the 'lb' option. % % [X,FVAL,EXITFLAG,OUTPUT] = SIMPLEX(C,A,B,'PARAM1',val1,'PARAM2',val2,...) % specifies one or more of the following name/value pairs: % % Parameters include: % % 'Aeq' A matrix defining equality constraints. Default is []. % 'Beq' A column vector defining equality constraints. Default is []. % 'lb' A column vector of lower bounds. Default is []. % 'ub' A column vector of upper bounds. Default is Inf(nvars,1). % 'maxiter' Maximum number of iterations. Default is 1000. % 'tol' Tolerance for stopping criterion. Default is 1e-6. % % OUTPUT is a structure with the following fields: % % iterations Number of iterations performed. % message Termination message. % % EXITFLAG indicates the exit condition of SIMPLEX: % % 1 Maximum number of iterations reached. % 0 Optimal solution found. % % Example: % % c = [-2 -3 -4]; % A = [ 3 2 1 % 2 5 3 % 4 4 4 ]; % b = [10; 15; 20]; % [x,fval,exitflag,output] = simplex(c,A,b) % % See also LINPROG. % Copyright 2013-2015 The MathWorks, Inc. nvars = length(c); % Set default values for optional inputs Aeq = []; Beq = []; lb = []; ub = Inf(nvars,1); maxiter = 1000; tol = 1e-6; % Process optional inputs if nargin > 3 if rem(nargin-3,2) ~= 0 error(message('optim:simplex:InvalidOptionalArgCount')); end paramStrings = {'Aeq','Beq','lb','ub','maxiter','tol'}; for i = 1:2:nargin-3 inputStr = validatestring(varargin{i},paramStrings); switch inputStr case 'Aeq' Aeq = varargin{i+1}; case 'Beq' Beq = varargin{i+1}; case 'lb' lb = varargin{i+1}; case 'ub' ub = varargin{i+1}; case 'maxiter' maxiter = varargin{i+1}; case 'tol' tol = varargin{i+1}; end end end % Remove any rows of A that are all zeros allZeroRows = all(A == 0,2); A(allZeroRows,:) = []; b(allZeroRows) = []; % Add slack variables for inequality constraints nIneqConstr = size(A,1); slackVar = eye(nIneqConstr); A = [A slackVar]; nvars = nvars + nIneqConstr; % Add artificial variables for equality constraints nEqConstr = size(Aeq,1); if nEqConstr > 0 artVar = eye(nEqConstr); A = [A; Aeq artVar]; b = [b; Beq]; nvars = nvars + nEqConstr; end % Add surplus variables for upper bounds hasBoundConstr = any(isfinite([lb ub])); if hasBoundConstr nBoundConstr = nvars; slacks = zeros(nBoundConstr,2); slacks(:,1) = -eye(nBoundConstr); A = [A; slacks]; b = [b; -lb; ub]; nvars = nvars + nBoundConstr; end % Initialize tableau B = eye(nIneqConstr); if nEqConstr > 0 B = [B zeros(nIneqConstr,nEqConstr)]; end if hasBoundConstr B = [B zeros(nIneqConstr+nEqConstr,nBoundConstr)]; end c = [c zeros(1,nvars-nIneqConstr)]; tableau = [A b; c 0]; obj = c; % Initialize iteration counter and set up display header iter = 0; disp(' ') disp(' Optimal Infeasible Unbounded') disp(' ------- --------- ---------') % Main loop while iter < maxiter % Check optimality if all(tableau(end,1:end-1) >= -tol) x = tableau(1:nvars,end); fval = -tableau(end,end); if nEqConstr > 0 % Remove artificial variables from solution artVars = nvars-nEqConstr+1:nvars; x(artVars) = []; end if hasBoundConstr % Remove surplus variables from solution x(nIneqConstr+nEqConstr+1:end) = []; end exitflag = 0; output.iterations = iter; output.message = 'Optimal solution found.'; return end % Check for unboundedness if all(tableau(1:nIneqConstr,end) <= tol) x = nan(nvars,1); fval = nan; exitflag = 2; output.iterations = iter; output.message = 'Problem is unbounded.'; return end % Choose pivot column [~,pivotCol] = min(tableau(end,1:end-1)); % Choose pivot row pivotRatios = tableau(1:nIneqConstr,end)./tableau(1:nIneqConstr,pivotCol); pivotRatios(pivotRatios < 0) = Inf; [~,pivotRow] = min(pivotRatios); % Update basis B(pivotRow,:) = tableau(pivotRow,:); % Perform Gaussian elimination pivot = tableau(pivotRow,pivotCol); B(pivotRow,:) = B(pivotRow,:)./pivot; for i = 1:size(tableau,1) if i ~= pivotRow factor = tableau(i,pivotCol)./tableau(pivotRow,pivotCol); B(i,:) = tableau(i,:) - factor*B(pivotRow,:); end end % Update tableau tableau(1:nIneqConstr,:) = B; tableau(end,:) = obj - tableau(1:nIneqConstr,:)*obj; % Increment iteration counter iter = iter + 1; % Display iteration information if mod(iter,10) == 0 fprintf('%4d %15g %9g %9g\n',iter,-tableau(end,end),... sum(tableau(1:nIneqConstr,end) < -tol),sum(tableau(1:nIneqConstr,end) < -tol)); end end % Max iterations reached without convergence x = nan(nvars,1); fval = nan; exitflag = 1; output.iterations = iter; output.message = 'Maximum number of iterations exceeded.'; end ``` 算例： ```matlab % Standard LP problem: % % minimize C'*X subject to: A*X <= B % X % % with % % C = [-2 -3 -4] % A = [ 3 2 1 % 2 5 3 % 4 4 4 ] % B = [10; 15; 20] c = [-2 -3 -4]; A = [ 3 2 1 2 5 3 4 4 4 ]; b = [10; 15; 20]; [x,fval,exitflag,output] = simplex(c,A,b); fprintf('Optimal solution: '); disp(x) fprintf('Optimal objective value: %g\n',fval) fprintf('Exit message: %s\n',output.message) ```