Syntax for solving linear equations

I have found out that I have phased my problem less complicated then I think it is, so I'll try again. I have a set of data points on my y-axis which has the expression: a-x*b-c*y

(a,b,c are constants, x and y are the variables to be optimized)

In an ideal case of x and y, this expression equals a straight line

a-x*b-c*y=d*z+e

(d and e are constant),

I want x and y to fit this straight line the best, so I want to optimize:

abs( a-x*b-c*y- (d*z+e) ) = 0 with the conditions (x,y >= 0 and x,y =< 1)

Answer 1

[John D'Errico]

Hmm. I'm not sure what the other solvers are seeing in your question that I don't. Here is your question:

a-x*b-c*y (where a,b and c are constants).

Is minimized under the conditions that: x,y >= 0 and x,y =< 1

so, for given KNOWN constants, you want to solve for x and y. This is just a basic linear programming problem.

Note that a is irrelevant, since if you find the min of -b*x-c*y, the location does not change regardless of the value of a. Just set up linprog, assuming that you have the optimization toolbox.

TRY IT!

xy = linprog([-b,-c],[],[],[],[],[0 0],[1 1]);

[Alexander Kjærsgaard]

Thank you! Can I add a variant of the question? if I now wish to optimize the same expression but in absolute/numeric value, so:

abs(a-x*b-c*y)

under the same conditions, does this change the optimization code?

[John D'Errico]

Yes, completely so, because the objective is no longer linear. So linprog is now invalid for use directly. Depending on the values for a,b,c, and the ranges for x and y, that objective will now be piecewise linear, with a derivative discontinuity along a line. So most optimizers may have problems if you just fed them that problem.

So instead, use linprog, but solve it two times, with a linear inequality constraint, once with

a-x*b-c*y >= 0

and a second time with

a-x*b-c*y <= 0

Then take the better of the two solutions.

[Torsten]

The problem remains linear if you formulate it as

min: eps

a-x*b-y*c <= eps
a-x*b-y*c >= -eps
x,y >= 0
x,y <= 1

using linprog.

It's data fitting in the l_oo norm.

Best wishes

Torsten.

Answer 2

[Torsten]

x=(b>0);
y=(c>0);

Best wishes

Torsten.

Answer 3

[Alexandra Harkai]

https://uk.mathworks.com/help/matlab/ref/linsolve.html