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A, b, Aeq, beq): solves the problem with the equality constraints Aeqx = beq. Set A=[ ] and b=[ ] if no inequalities exist;

       – x = linprog(f, A, b, Aeq, beq, lb, ub): defines a set of lower and upper bounds on the design variables, x, so that the solution is always in the range lb ≤ x ≤ ub. Set Aeq=[ ] and beq=[ ] if no equalities exist;

       – x = linprog(f, A, b, Aeq, beq, lb, ub, x0): defines the starting point at x0 (only used for the active-set algorithm);

       – x = linprog(f, A, b, Aeq, beq, lb, ub, x0, options): minimizes with the specified options;

       – [x, fval] = linprog(...): returns the value of the objective function at the solution x;

       – [x, fval, exitflag] = linprog(...): returns a value exitflag that describes the exit condition;

       – [x, fval, exitflag, output] = linprog(...): returns a structure output that contains information about the optimization process;

       – [x, fval, exitflag, output, lambda] = linprog(...): returns a structure output whose fields contain the Lagrange multipliers λ at the solution x.

      EXAMPLE 1.14.– Consider the linear program:

      [1.25]

      The following code can be executed to compute this example.

      1 1 variables = {'x1', 'x2', 'x3', 'x4', 'x5'}; % construct the vector x

      2 2 n = length(variables);

      3 3 for i = 1:n % create the indices of x

      4 4 eval([variables{i}, ' = ', num2str(i), ';']);

      5 5 end

      6 6 lb = zeros(1, n);

      7 7 lb([x1, x2]) = [5, 1]; % add the bounds of the variables x1,x2

      8 8 ub = Inf(1, n);

      9 9 A = zeros(3, n); % create the matrix A

      10 10 b = zeros(3, 1); % create the vector b

      11 11 % define the constraints

      12 12 A(1, [x1, x2, x3, x4]) = [1, 4, -2, -1]; b(1) = 8;

      13 13 A(2, [x1, x2, x3, x4]) = [1, 3, 2, -1]; b(2) = 10;

      14 14 A(3, [x1, x2, x3, x4, x5]) = [2, 1, 2, 3, -1]; b(3) = 20;

      15 15 Aeq = zeros(1, n); % create the matrix Aeq

      16 16 beq = zeros(1, 1); % create the vector beq

      17 17 % define the equality constraints

      18 18 Aeq(1, [x1, x2, x3, x4, x5]) = [1, 3, -4, -1, 1]; beq(1) = 7;

      19 19 c = zeros(n, 1); % create the vector c

      20 20 c([x1 x2 x3 x4 x5]) = [-2; -3; 1; 4; 1];

      21 21 % call linprog solver

      22 22 [x, objVal] = linprog(c, A, b, Aeq, beq, lb, ub);

      23 23 for i = 1:n

      24 24 fprintf('%s \t %20.4f\n', variables{i}, x(i))

      25 25 end

      26 26 fprintf(['The value of the objective function is' '%20.4f\n'],

      27 27 objVal)

       For a color version of this code, see www.iste.co.uk/radi/optimizations.zip

      Executing this code produces the following results:

      Конец ознакомительного фрагмента.

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