How to use fminunc for a 2D function composed of two functions?

Prashant Kumar answered . 2024-12-21 08:29:00

f = @(x) [x(1) x(2)]*[10 2;2 6]*[x(1);x(2)]-[5 5]*[x(1);x(2)]+(cos(30*x(1))+cos(30*x(2)))/40+(cos(80*x(1)^2)+cos(80*x(2)^2))/50;
options = optimoptions(@fmincon);
%options = optimoptions(@fmincon,'Display','iter','OptimalityTolerance',1e-12);
lb = [0 0];% lower bounds
ub = [1 1];% upper bounds
x0 = [.5 .5]; % initial guess
[xSol,fval,exitflag,output] = fmincon(f,x0,[],[],[],[],lb,ub,[],options); % solve optimization

Feasible point with lower objective function value found, but optimality criteria not satisfied. See output.bestfeasible..


Local minimum found that satisfies the constraints.

Optimization completed because the objective function is non-decreasing in 
feasible directions, to within the value of the optimality tolerance,
and constraints are satisfied to within the value of the constraint tolerance.

% brute force searching the entire space for min function value
x1 = 0:0.001:1;
x2 = 0:0.001:1;
[X1,X2] = meshgrid(x1,x2);
fValBrute = arrayfun(@(x1,x2)f([x1 x2]),X1,X2);
idxMin = find(fValBrute == min(fValBrute,[],'all')); % find minimum function value 
% plot
contourf(X1,X2,fValBrute,150)
hold on 
plot(xSol(1),xSol(2),'k*') % black star is the result of fmincon
plot(X1(idxMin),X2(idxMin),'rx') % red x is the brute force result
title({sprintf('Min function value found via fmincon: %.4f at [%.4f %.4f]',fval,xSol(1),xSol(2));
   sprintf('Min function value found via brute force: %.4f at [%.4f %.4f]',fValBrute(idxMin),X1(idxMin),X2(idxMin))})
hold off

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