Solving a quadratic optimization problem subjected to linear constraints

I'm trying to recreate the code behind this picture. I have a function

where . I need to solve the following quadratic optimization problem subject to linear constraint:

subject to:

the input data are: = [0, 0.25, 0.5, 1, 1.2, 1.8, 2]; = [2, 0.8, 0.5, 0.1, 1, 0.5, 1];

And I need the following result:

I tried to use the function fmincon but it gives me always the same value for lambda. Can you help me find the error or explain to me what kind of function I need to use instead?

clc;
clear;
close all;

x_i = [0, 0.25, 0.5, 1, 1.2, 1.8, 2];
f_i = [2, 0.8, 0.5, 0.1, 1, 0.5, 1];

ottimizzazione_quadratica(x_i,f_i);

function risultato = ottimizzazione_quadratica(x_i, f_i)
    x0 = zeros(size(x_i));
    A = [];
    b = [];
    Aeq = [];
    beq = [];
    lb = zeros(size(x_i));  % lambda_i >= 0
    ub = [];
    
    lambda_ottimale = fmincon(@(lambda) funzione_obiettivo(lambda, x_i, f_i), x0, A, b, Aeq, beq, lb, ub);

    risultato = F(lambda_ottimale, x_i);
    
    % Plot F(x) and points (xi, fi)
    x_vals = linspace(min(x_i), max(x_i), 1000);
    F_vals = arrayfun(@(x) F(lambda_ottimale, x), x_vals);

    figure;
    plot(x_i, f_i, 'ro', 'MarkerSize', 10, 'MarkerFaceColor', 'r');  % Punti dati
    hold on;
    plot(x_vals, F_vals, 'b-', 'LineWidth', 2);  % Funzione F(x)
    xlabel('x');
    ylabel('F(x)');
    grid on;
    hold off;
end

function risultato = funzione_obiettivo(lambda, x_i, f_i)
    risultato = norm(F(lambda, x_i) - f_i)^2;
end

% F(x)
function risultato = F(lambda, x_i)
    risultato = sum(lambda .* phi(x_i));
end

% phi
function risultato = phi(x_i)
    % max(0, 1 - norm(x - xi)^4)*(4*norm(x-xi)+1)
    risultato = arrayfun(@(x) max(0, 1 - norm(x - x_i)^4)*(4*norm(x - x_i) + 1), x_i);
end

Answer 1

[Matt J]

You can use lsqnonneg,

xi=[0, 0.25, 0.5, 1, 1.2, 1.8, 2]';  
fi= [2, 0.8, 0.5, 0.1, 1, 0.5, 1]';

phi=@(r) max(0, 1 - r).^4.*(4*r + 1);

C=phi(abs(xi-xi'));

[lambda,fval]=lsqnonneg(C,fi)

lambda = 7×1

1.8083 0 0 0 0.6845 0 0.8566

fval = 0.4904

[arianna]

It's better than before, but I still don't get the same interpolation of the picture. Now I get this curve.

I don't understand what I'm missing.

[Torsten]

Your code uses a different function for phi than in your mathematical description.

In your code:

phi(r)=max(0, 1 - r.^4).*(4*r + 1)

In your mathematical description:

phi(r)=(max(0, 1 - r)).^4.*(4*r + 1)

@Matt J used your mathematical description.