2 equations, 2 unknowns - fminsearch(), fminunc() applied on numerical data

I have 2 equations and , which should deliver the same result:

All of them are matrices with [m=800, n=1300]; A and B are certainly known, I tried to calculate Cx and Cy, but my problem is that the outcome is not the same - the equation is not satisfied!

I'm trying to treat Cx and Dy as unknowns and use fminsearch & fminunc by saying: to find the values for Cx and Cy which satisfy the equations. The calculation is done for every element of the matrices.

Cy_new = zeros(size(A));

Cx_new = zeros(size(A));

f2min_val = zeros(size(A));

for i=1:size(A,1)

for j=1:size(A,2)

f2min = @(C_xy) DRx(i,j) + C_xy(1) - ( DRy(i,j) + C_xy(2) );

[ C_xy, f2min_val(i,j) ] = fminunc( F2min, [-1, 0.5]); % Cx(i,j), Cy(i,j)

Cy_new(i,j) = C_xy(1); Cx_new(i,j) = C_xy(2);

end

end

The data for previously calculated Cx(i,j) and Cy(i,j) lies in a range of [-1, 0.5]

But I get the message for fminunc:

Problem appears unbounded.

fminunc stopped because the objective function value is less than

or equal to the value of the objective function limit.

<stopping criteria details>

fminunc stopped because the objective function value, -2.461927e+20, is less than

or equal to options.ObjectiveLimit = -1.000000e+20.

C_xy =

1.0e+20 *

-1.2310 1.2310

and for fminsearch

Exiting: Maximum number of function evaluations has been exceeded

- increase MaxFunEvals option.

Current function value: -111848999850590921804554051032055239655030784.000000

C_xy =

1.0e+44 *

-1.0233 0.0952

no matter what I use as starting values. Someone has an idea about what I might improve? Or is it just impossible with my data?

Answer 1

[Bjorn Gustavsson]

From your description you have 800-by-1300 linear equations of the type:

a + x = b + y

and you want to solve for both x and y. Therein liest the problem. If you re-arrange the terms you will find:

x-y = b - a

and that is all you can achieve with this type of problem. You will have to accept that you can only determine the difference between x and y, and either settle for that or add more information to the problem. The results you get from the optimization-problems is due to the problem being underdetermined.

HTH

[Torsten]

Additionally to what @Bjorn Gustavsson said, you would have to define

f2min = @(C_xy) (DRx(i,j) + C_xy(1) - ( DRy(i,j) + C_xy(2) )).^2;

as objective function.

Only then you will get C_xy with DRx(i,j) + C_xy(1) - ( DRy(i,j) + C_xy(2) ) = 0.

If you don't square the expression, the optimizers try to make DRx(i,j) + C_xy(1) - ( DRy(i,j) + C_xy(2) ) = -Inf.

[Philipp Bisignano]

Thank you both for your answer!

By squaring the eq. fminsearch() works, although it take a tremendous amount of time.

But what exactly is the answer? Is it the answer to my problem or is it the difference Cx - Cy, how @Bjorn Gustavsson suggested?

[Torsten]

From fminsearch, e.g., you get one answer for C_xy(1) and C_xy(2) that satisfies

DRx(i,j) + C_xy(1) - ( DRy(i,j) + C_xy(2) ) = 0

or

C_xy(1) - C_xy(2) = DRy(i,j) - DRx(i,j)

You get all solutions of your equation if you add an arbitrary value s both to C_xy(1) and C_xy(2) because this way, the difference

(C_xy(1) + s) - (C_xy(2) + s) = C_xy(1) - C_xy(2)

remains the same.

[Philipp Bisignano]

Thank you for your answer @Torsten!

so do you suggest to implement the eq. to minimize like this: ?

s = 4; % arbitrary value

f2min = @(C_xy) (C_xy(1) + s) - (C_xy(2) + s) - ( DRy(i,j) + C_xy(2) ) = 0

[ C_xy, f2min_val(i,j) ] = fminunc( f2min, [-1, 0.5])

How do you mean exactly all solutions of the equations? I don't understand exactly the clue with the constant s and how I could integrate it in my equation.

As for now the solution, which I had calculated with

DRx(i,j) + C_xy(1) - ( DRy(i,j) + C_xy(2) ) = 0

is wrong.

[Torsten]

DRx = 3;
DRy = -22;
f2min = @(C_xy) DRx + C_xy(1) - ( DRy + C_xy(2) )

f2min = function_handle with value:
    @(C_xy)DRx+C_xy(1)-(DRy+C_xy(2))

sol = fminsearch(@(C_xy)f2min(C_xy).^2,[1,1])

sol = 1×2

-12.5968 12.4032

f2min(sol)

ans = 1.2971e-08

s = -pi;
sols = sol + s

sols = 1×2

-15.7384 9.2616

f2min(sols)

ans = 1.2971e-08

[Philipp Bisignano]

So if I get it right I miss only the right value for the constant s, to have the exact results for my problem?

(Which I may retrieve from some BC in my problem)

[Torsten]

In principle, you need a second equation because one equation cannot fix two unknowns.

[Philipp Bisignano]

I'd have another question, so I get a result for C_xy from Matlab by doing the minimization on f2min.

By applying some boundary conditions I am also able to find a value for the constant s, by which the single terms (C_xy(1) - C_xy(2)) could be shifted.

What do I have: -

• A possible solution out of many possible solutions for C_xy?
• A solution for C_xy out of 2 possible ones?

Or just nonsense?

Thank you in advance!

[Bjorn Gustavsson]

To my understanding you will have one solution for C_xy(1) and C_xy(2) that both gives the same difference and the correct average value that you have determined "from elsewhere". That boundary-condition selects one pair of C_xy-values out of an otherwise infinite number of possible pairs with the same difference.