How to use ~= operator as constraint function?

I have a function called "error" that I want to minimize with initial X0 = 1.34. The definition is:

function error = objectiveFcn(X,F)
D = 1.33;
M = 0.2;
error = (abs((((F)*(X/((F)*(1/M-1)+X))+(1-(X/((F)*(1/M-1)+X)))*X)-D)/D))*100;
end

The most straightforward answer is X=F=1.33 for which Error = 0 but I want X~=F.

I don't seem to be able to correctly include this under constraintFcn... Any suggestions?

UPDATE:

I have actually solved this in Excel, but I want to automate the process. This is how it looks like:

The exact solution is X=F=1.33 but this is not an accpetable result for me because X and F in reality can never be equal. The next best answer is X=1.265 and F=1.675 with an error of 0.01%.

F can be further defined as

F = (D-(1-V)*X)/V;
V=0.158;

Update 2: I realized this can be reduced to a single-variable problem. Here: but how do I extract the X for which X~=F. Is there a function for this?

D = 1.33;
M = 0.2;
V = 0.158;
err = @(X) (abs((((((D-(1-V)*X)/V))*(X/((((D-(1-V)*X)/V))*(1/M-1)+X))+(1-(X/((((D-(1-V)*X)/V))*(1/M-1)+X)))*X)-D)/D))*100;
fplot(err,[1.22,1.36])

[Matt J]

The problem is ill-defined. Consider the simpler case of the 1D objective function f(x)=x.^2. What would be the minimum of this function subject to the constraint that x~=0?

[Jan]

Do not call a function "error", because error() is a very important built-in function. Shadowing it will cause serious troubles. By the way, your function is not called "error", but "objectiveFunction".

A nicer version of the formula:

err = 100 * abs((X / (1 / M - 1 + X / F) + (1 - X / (F / M - F + X)) * X - D) / D)

How do you search for a solution? With a local or global optimizer?

[Pelajar UM]

Thanks. I am using fmincon and it returns 1.33 as expected because whenever I add X=~F in the constraint, it won't run...

Answer 1

[Matt J]

Here: but how do I extract the X for which X~=F. Is there a function for this?

Yes, but clearly you are not doing continuous optimization, so you should use min() rather than fmincon().

D = 1.33;
M = 0.2;
V = 0.158;

X=setdiff( 1.24:0.005:1.34  ,D);

err = (abs((((((D-(1-V).*X)./V)).*(X./((((D-(1-V).*X)./V)).*(1./M-1)+X))+(1-(X./((((D-(1-V).*X)./V)).*(1./M-1)+X))).*X)-D)./D)).*100;

[~,imin]=min(err);

X(imin)

ans = 1.2650

[Pelajar UM]

Thanks, but the answer shouldn't be 1.34. It should be 1.265 or something in that vicinity.

[Matt J]

No, that was your original function. The function as of your updates looks different:

D = 1.33;
M = 0.2;
V = 0.158;

X=1.24:0.05:1.34;

err = (abs((((((D-(1-V)*X)/V))*(X/((((D-(1-V)*X)/V))*(1/M-1)+X))+(1-(X/((((D-(1-V)*X)/V))*(1/M-1)+X)))*X)-D)/D))*100;

plot(X,err)

[Pelajar UM]

Hmm... When I plot, this is what I get:

D = 1.33;
M = 0.2;
V = 0.158;
err = @(X) (abs((((((D-(1-V)*X)/V))*(X/((((D-(1-V)*X)/V))*(1/M-1)+X))+(1-(X/((((D-(1-V)*X)/V))*(1/M-1)+X)))*X)-D)/D))*100;
fplot(err,[1.22,1.36])

Warning: Function behaves unexpectedly on array inputs. To improve performance, properly vectorize your function to return an output with the same size and shape as the input arguments.

[Matt J]

Never mind. I fixed my original answer. It now gives X=1.265

D = 1.33;
M = 0.2;
V = 0.158;

X=setdiff( 1.24:0.005:1.34  ,D);

err = (abs((((((D-(1-V).*X)./V)).*(X./((((D-(1-V).*X)./V)).*(1./M-1)+X))+(1-(X./((((D-(1-V).*X)./V)).*(1./M-1)+X))).*X)-D)./D)).*100;

[~,imin]=min(err);

X(imin)

ans = 1.2650

[Pelajar UM]

Perfect! Thanks a lot.

Answer 2

[John D'Errico]

For which solver? This is a 1-variable problem. If you want help, it does help if you explain more than you have done.

Note that ~= is not something you can specify for ANY solver, since that allows your solution to be infinitely close to equality, yet not exactly equal. And that makes your problem ill-posed.

Anyway, is this problem even meaningful?

D = 1.33;
M = 0.2;
F = 1.33;
err = @(X) (abs((((F)*(X/((F).*(1/M-1)+X))+(1-(X./((F)*(1/M-1)+X))).*X)-D)/D))*100;
fplot(err,[-3,50])

Warning: Function behaves unexpectedly on array inputs. To improve performance, properly vectorize your function to return an output with the same size and shape as the input arguments.

It looks like I missed a dot in there to vectorize it. Regardless, there seems to be only a single minimizer, at X == F. We can probably prove that to be the case with a little effort, merely by showing the behavior of this simple function as X grows large in both directions, that there are no points where the function has a zero derivative.

syms x
err(x)

ans = 

We could play with that and now look at the derivatives.

Or, is F also an unknown? If it is, then your objective is not formulated in a form that any solver in MATLAB will directly take. You seem to be treating F as a constant.

[Pelajar UM]

Thanks. Please see the details I have added to the original post. You are right, there's only one perfect solution and that is when X=F but since that's not acceptable for me, I'm searching for the next best combination of X and F that gives the lowest error.

F is not known. In general the problem is some constant X + some constant F should be as close as possible to D which is 1.33. But X cannot be equal to F.

Answer 3

[Fabio Freschi]

You may have a look here or here