Optimization not working because fitness function must be a function handle

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xu
xu 2023-6-4
编辑： Matt J 2023-6-4
% Define the objective functions
objFun1 = @(x) -0.0365*x(1) + 0.18*x(2) - 0.389; % F(1)
objFun2 = @(x) 0.0435*x(1) - 0.266*x(2) + 4.2*x(3) + 0.019*x(1)*x(2) - 0.3*x(1)*x(3) - 0.2485; % F(2)
% Define the constraint functions
constraintFun = @(x) deal([], []); % No constraints in this case
% Define the lower and upper bounds
lb = [6; 1; 0.05]; % Lower bounds
ub = [22; 5; 0.45]; % Upper bounds
% Define the problem structure
problem.objective = {@(x) -objFun1(x), @(x) objFun2(x)}; % Multi-objective function to minimize F(1) and maximize -F(2)
problem.x0 = [6; 1; 0.05]; % Initial guess
problem.lb = lb; % Lower bounds
problem.ub = ub; % Upper bounds
problem.nonlcon = constraintFun; % Nonlinear constraints
% Set the options for gamultiobj
options = optimoptions('gamultiobj', 'Display', 'iter', 'PlotFcn', @gaplotpareto);
% Solve the multi-objective optimization problem
[x, fval, exitflag, output] = gamultiobj(problem, options);
% Display the Pareto front solutions
fprintf('Pareto Front Solutions:\n');
for i = 1:numel(fval)
fprintf('Solution %d:\n', i);
fprintf('F(1): %.4f\n', -fval(i, 1)); % Multiply by -1 to get the maximum value of F(1)
fprintf('F(2): %.4f\n', fval(i, 2));
fprintf('x1: %.4f\n', x(i, 1));
fprintf('x2: %.4f\n', x(i, 2));
fprintf('x3: %.4f\n', x(i, 3));
fprintf('-------------------------\n');
end
% Find the optimal solution based on preferences
optimalIndex = find(fval(:, 1) == max(fval(:, 1))); % Find the solution with the maximum value of F(1)
optimalSolution = x(optimalIndex, :);
fprintf('Optimal Solution:\n');
fprintf('F(1): %.4f\n', -fval(optimalIndex, 1)); % Multiply by -1 to get the maximum value of F(1)
fprintf('F(2): %.4f\n', fval(optimalIndex, 2));
fprintf('x1: %.4f\n', optimalSolution(1));
fprintf('x2: %.4f\n', optimalSolution(2));
fprintf('x3: %.4f\n', optimalSolution(3));

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回答（1 个）

Matt J
Matt J 2023-6-4
编辑：Matt J 2023-6-4
problem.objective = ...
@(x) [+0.0365*x(1) - 0.18*x(2) + 0.389;
0.0435*x(1) - 0.266*x(2) + 4.2*x(3) + 0.019*x(1)*x(2) - 0.3*x(1)*x(3) - 0.2485];
  1 个评论
Steven Lord
Steven Lord 2023-6-4
Ran in:
Or to reuse the previous definition of the two objective functions:
% Define the objective functions
objFun1 = @(x) -0.0365*x(1) + 0.18*x(2) - 0.389; % F(1)
objFun2 = @(x) 0.0435*x(1) - 0.266*x(2) + 4.2*x(3) + 0.019*x(1)*x(2) - 0.3*x(1)*x(3) - 0.2485; % F(2)
objective = @(x) [-objFun1(x); objFun2(x)];
To check:
objective([1, 2, 3])
ans = 2×1
0.0655 11.0010
problem.objective = ...
@(x) [-0.0365*x(1) + 0.18*x(2) - 0.389;
0.0435*x(1) - 0.266*x(2) + 4.2*x(3) + 0.019*x(1)*x(2) - 0.3*x(1)*x(3) - 0.2485];
problem.objective([1, 2, 3])
ans = 2×1
-0.0655 11.0010

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