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

Technical Source
2 min readJul 17, 2024

--

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 ?

NOTE:-

Matlabsolutions.com provide latest MatLab Homework Help,MatLab Assignment Help , Finance Assignment Help for students, engineers and researchers in Multiple Branches like ECE, EEE, CSE, Mechanical, Civil with 100% output.Matlab Code for B.E, B.Tech,M.E,M.Tech, Ph.D. Scholars with 100% privacy guaranteed. Get MATLAB projects with source code for your learning and research.

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);

SEE COMPLETE ANSWER CLICK THE LINK

https://www.matlabsolutions.com/resources/how-to-use-fminunc-for-a-2d-function-composed-of-two-functions-.php

--

--

Technical Source
Technical Source

Written by Technical Source

Simple! That is me, a simple person. I am passionate about knowledge and reading. That’s why I have decided to write and share a bit of my life and thoughts to.

No responses yet