How do I get a positive solution from rref?

I want to solve a linear combination from a table and a vector, but after solving it (using rref())always gives me some negative numbers. How do i get only positive feedback from it?

Answer 1

[Shantanu Dixit]

Hi,

If I understand your query correctly, you want to solve an exact system 'Ax=b' (where 'A' is the table and 'b' is the vector) while ensuring 'x≥0'. The 'rref' function only computes the reduced row echelon form and does not enforce nonnegativity.

To achieve this, you can use 'linprog' which allows you to impose 'x≥0' by setting lower bounds as an input argument 'lb'. Here’s a simple example solving a 3×2 system using 'linprog':

f = zeros(size(A,2),1);      % Trivial objective function: minimize 0'*x
Aeq = A;                     % Equality constraint: A*x = b
beq = b;
lb = zeros(size(A,2),1);     % Lower bound: x >= 0
options = optimoptions('linprog','Display','none');
x = linprog(f, [], [], Aeq, beq, lb, [], options);

If such a solution exists then 'linprog' will return it as 'x' else an empty vector is returned.

You can refer to linprog for additional details: https://www.mathworks.com/help/optim/ug/linprog.html

Hope this helps!