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Two linear equation with absolute value equation

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Murat YAPICI
Murat YAPICI on 13 Jan 2021
Commented: Murat YAPICI on 13 Jan 2021
Hello,
I have two linear equation and one absolute value equation. Is there a easy way to obtain minimum norm solution ?

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Accepted Answer

Bruno Luong
Bruno Luong on 13 Jan 2021
Edited: Bruno Luong on 13 Jan 2021
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))

More Answers (1)

Alan Stevens
Alan Stevens on 13 Jan 2021
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

  1 Comment

Murat YAPICI
Murat YAPICI on 13 Jan 2021
Thank you for your answer.
I mean something like this but I want to minimize (Like pseudo inverse). Not .

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