One-Dimensional Semi-Infinite Constraints

Find values of x that minimize

f(x) = (x₁ – 0.5)² + (x₂– 0.5)² + (x₃– 0.5)²

where

$\begin{array}{l} K_{1} (x, w_{1}) = \sin (w_{1} x_{1}) \cos (w_{1} x_{2}) - \frac{1}{1000} {(w_{1} - 50)}^{2} - \sin (w_{1} x_{3}) - x_{3} \leq 1, \\ K_{2} (x, w_{2}) = \sin (w_{2} x_{2}) \cos (w_{2} x_{1}) - \frac{1}{1000} {(w_{2} - 50)}^{2} - \sin (w_{2} x_{3}) - x_{3} \leq 1, \end{array}$

for all values of w₁ and w₂ over the ranges

1 ≤ w₁ ≤ 100,
1 ≤ w₂ ≤ 100.

Note that the semi-infinite constraints are one-dimensional, that is, vectors. Because the constraints must be in the form K_i(x,w_i) ≤ 0, you need to compute the constraints as

$\begin{array}{l} K_{1} (x, w_{1}) = \sin (w_{1} x_{1}) \cos (w_{1} x_{2}) - \frac{1}{1000} {(w_{1} - 50)}^{2} - \sin (w_{1} x_{3}) - x_{3} - 1 \leq 0, \\ K_{2} (x, w_{2}) = \sin (w_{2} x_{2}) \cos (w_{2} x_{1}) - \frac{1}{1000} {(w_{2} - 50)}^{2} - \sin (w_{2} x_{3}) - x_{3} - 1 \leq 0. \end{array}$

First, write a file that computes the objective function.

function f = myfun(x,s)
% Objective function
f = sum((x-0.5).^2);

Second, write a file mycon.m that computes the nonlinear equality and inequality constraints and the semi-infinite constraints.

function [c,ceq,K1,K2,s] = mycon(X,s)
% Initial sampling interval
if isnan(s(1,1)),
   s = [0.2 0; 0.2 0];
end
% Sample set
w1 = 1:s(1,1):100;
w2 = 1:s(2,1):100;

% Semi-infinite constraints 
K1 = sin(w1*X(1)).*cos(w1*X(2)) - 1/1000*(w1-50).^2 -...
       sin(w1*X(3))-X(3)-1;
K2 = sin(w2*X(2)).*cos(w2*X(1)) - 1/1000*(w2-50).^2 -...
       sin(w2*X(3))-X(3)-1;

% No finite nonlinear constraints
c = []; ceq=[];

% Plot a graph of semi-infinite constraints
plot(w1,K1,'-',w2,K2,':')
title('Semi-infinite constraints')
drawnow

Then, invoke an optimization routine.

x0 = [0.5; 0.2; 0.3];      % Starting guess
[x,fval] = fseminf(@myfun,x0,2,@mycon);

After eight iterations, the solution is

The function value and the maximum values of the semi-infinite constraints at the solution x are

fval
fval =
    0.0771

[c,ceq,K1,K2] = mycon(x,NaN); % Initial sampling interval
max(K1)
ans =
   -0.0077
max(K2)
ans =
   -0.0812

A plot of the semi-infinite constraints is produced.

This plot shows how peaks in both constraints are on the constraint boundary.

The plot command inside mycon.m slows down the computation. Remove this line to improve the speed.

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