Two linear equation with absolute value equation

Hello,

I have two linear equation and one absolute value equation. Is there a easy way to obtain minimum norm solution ?

Answer 1

[Bruno Luong]

Correct minimum norm solution is

xmin =

   90.0000
  -40.0000
    5.0000
    5.0000


normxmin =

   98.7421

obtained with this code

s = cell(1,4);
[s{:}] = ndgrid([-1 1]);
s = reshape(cat(5,s{:}),[],4);
fmin = Inf;
xmin = nan(4,1);
for k=1:size(s,1)
    sk = s(k,:);
    Aeq = [1 1 -1 -1;
           1 1 1 1;
           sk.*[1 1 -1 -1]];
    beq = [40; 60; 120];
    A = -diag(sk);
    b = zeros(4,1);
    [x,f,flag] = quadprog(eye(4), zeros(4,1), ...
        A, b, ...
        Aeq, beq, ...
        [], []);
    if flag > 0 && f < fmin
        fmin = f;
        xmin = x;
    end
end
xmin
normxmin = norm(xmin,2)

% Check the constraints
xmin(1)+xmin(2)-xmin(3)-xmin(4)
xmin(1)+xmin(2)+xmin(3)+xmin(4)
abs(xmin(1))+abs(xmin(2))-abs(xmin(3))-abs(xmin(4))

[Murat YAPICI]

Thank you

Answer 2

[Alan Stevens]

Do you mean something like this

X0 = [-50 -5];
[X, Fval] = fminsearch(@(X) fn(X),X0);
x2 = X(1); x1 = 50-x2;
x4 = X(2); x3 = 10-x4;
disp([x1 x2 x3 x4])
disp(x1+x2+x3+x4)
disp(x1+x2-x3-x4)
disp(abs(x1)+abs(x2)-abs(x3)-abs(x4))

function F = fn(X)

         x2 = X(1); x1 = 50-x2;
         x4 = X(2); x3 = 10-x4;
         F = norm(abs(x1)+abs(x2)-abs(x3)-abs(x4)-120);
         
end

[Murat YAPICI]

Thank you for your answer.

I mean something like this but I want to minimize (Like pseudo inverse). Not .