How to use fminunc for a 2D function composed of two functions? (2024)

Sergio asked . 2024-07-17

I have:

and I want to minimize it on the unit square without using differentials.

I use the following code,

close all; clear; clc;options = optimoptions(@fminunc,'Display','iter','Algorithm','quasi-newton');xy_guess = [0,0];[xy_opt,fval] = fminunc(@quadratic,xy_guess,options)function f = quadratic(in)x = in(1);y = in(2);f = -5.*x - 5.*y + 10.*x.^2 + 2.*x.*yf = 1/200.*(-1000.*x - 1000.*y + 400.*x.*y + 1200.*y.^2 + 5.*cos(30.*x) + 4.*cos(80.*x.^2) + 5.*cos(30.*y) + 4.*cos(80.*y^2))

But declaring f twice does not work. How should I declare this double-function as an input for fminunc ?

Mathematics and Optimization , Optimization Toolbox , Optimization Results , Solver Outputs and Ite

Prashant Kumar answered . 2024-07-19 11:33:21

You dont seem to have a double function.

So here is a code to solve your problem. You have local minimas so fmincon and brute force approach gives you different results, since brute force approach is a global optimization.

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 boundsub = [1 1];% upper boundsx0 = [.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 valuex1 = 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 % plotcontourf(X1,X2,fValBrute,150)hold on plot(xSol(1),xSol(2),'k*') % black star is the result of fminconplot(X1(idxMin),X2(idxMin),'rx') % red x is the brute force resulttitle({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

Not satisfied with the answer ?? ASK NOW

MOST VIEWED QUESTIONS

Recent Matlab Projects