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

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Sergio
Sergio 2024-6-28
编辑： Sergio 2024-6-28
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.*y
f = 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 ?
Thanks!
  2 个评论
Ashutosh Thakur
Ashutosh Thakur 2024-6-28
Hi Sergio,
With the last line of code the value of f is overwritten, and only this value is passed to the fminunc. Also I have observed that you have a constraint on the unit square. You can take advantage of the fmincon function, https://www.mathworks.com/help/optim/ug/fmincon.html, as fminunc does not have any constraint.
Can you try to use fmincon function?
Sergio
Sergio 2024-6-28
编辑：Sergio 2024-6-28
Thanks for the tip Ashutosh. While I will try fmincon, how do I represent f from that double map to a single input function?
With the new input, accordingly to your tip I get convergence, but I have no idea what f is considered as...
So I tried this (with the comma after + 2.*x.*y):
f = -5.*x - 5.*y + 10.*x.^2 + 2.*x.*y, 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))
options = optimoptions(@fmincon,'Display','iter','Algorithm','interior-point');
xy_guess = [0,0];
A = [0,1];
[xy_opt,fval] = fmincon(@quadratic,xy_guess,A,1)
function f = quadratic(in)
% Unpack inputs
x = in(1);
y = in(2);
% The Quadratic function in 2D
f = -5.*x - 5.*y + 10.*x.^2 + 2.*x.*y
f = 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))
end

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采纳的回答

Aquatris
Aquatris 2024-6-28
编辑：Aquatris 2024-6-28
Ran in:
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 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
  1 个评论
Sergio
Sergio 2024-6-28
编辑：Sergio 2024-6-28
Superb!! Thanks for this. I see the correct use with the semi-colonuse for the function representation, which is in fact a vector field rather than a function

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