fmincon does not fulfil constraints

Hi Matlab Community,

I am stuck at one point with fmincon and I am not getting anywhere. Your help would be very much appreciated.

I am trying to solve an optimization problem. With two main conditions:

x=fmincon(@(x)fAraucoMarkt(x,AraucoEnergy),x0,[],[],[],[],LB,RB,@(x)bedMarkt(x,AraucoEnergy));

But within the results these conditions are not fulfilled:

Here the conditions:

function [c,ceq] = bedMarkt(x,AraucoEnergy)
    minSpeicher=fminSpeich(x,AraucoEnergy);
    Restschuld=fRestschuld(x,AraucoEnergy);
    c=0.1-minSpeicher;
    ceq =Restschuld-0;
end

If I understood everything right than c means that the minSpeicher is always larger than 0.1 right? and that is most of the time not the case.

Maybe you can see already what I am doing wrong otherwise let me know if I could provide more information.

Thanks a lot

Erik

Answer 1

[Alan Weiss]

As documented, Iterations Can Violate Constraints. So I am not sure whether you are complaining about intermediate iterations (where the constraints can be violated) or the final result (which should have the constraints violated by no more than options.ConstraintTolerance).

Alan Weiss

MATLAB mathematical toolbox documentation

[Erik Beste]

Hi Alan thanks for your quick response. I am talking about the final results (I guess).

I would get a massage like this:

Local minimum found that satisfies the constraints.

Optimization completed because the objective function is non-decreasing in

feasible directions, to within the value of the optimality tolerance,

and constraints are satisfied to within the value of the constraint tolerance.

But if I use these Values my result for c would be nagative tho gave the constraint of 0,1 as a min value.

[Alan Weiss]

c is negative means the constraints are satisfied. Did you formulate your constraints so that negative c is what you want?

Alan Weiss

MATLAB mathematical toolbox documentation

[Erik Beste]

sorry I said it in a wrong way minSpeicher is negativ.

but acording to my constrain:

c=0.1-minSpeicher;

it should be at least 0.1 right?

[Alan Weiss]

Clearly, fmincon thinks that the constraints are satisfied. I think that you should check two things:

1. With x as the returned solution point, evaluate [c,ceq] = bedMarkt(x,AraucoEnergy) and check that c is indeed negative.
2. Once you have verified that c is negative at the solution, check how c relates to your desired constraint.

Alan Weiss

MATLAB mathematical toolbox documentation

[Erik Beste]

Ok I took the solution gained with fmincon and used it like you said in 1.

Unforanatly c in this case was positiv (3.33) I honestly dont know what kind of conclution I can draw from this.

[Alan Weiss]

Clearly, what you think you are passing to fmincon as a nonlinear constraint function is not what fmincon is calculating. You need to debug your fmincon call to ensure that the correct nonlinear constraint is being passed.

Alan Weiss

MATLAB mathematical toolbox documentation

[Erik Beste]

Thanks a lot Alan, I think the mistake was that what I added into c was a linear constrain. But now I wander what I added into c was linear but a formula that I created. Is there anyway to use formulas in A?

[Matt J]

If you mean that fminSpeich(x,AraucoEnergy) is a linear function of x, that would not explain the problem you've described. However, it is better to implement linear inequality constraints using the A,b matrix arguments. You can obtain the matrix represention of a function using this:

https://www.mathworks.com/matlabcentral/fileexchange/44669-convert-linear-functions-to-matrix-form

[Erik Beste]

Thanks Matt, so fminSpeich(x,AraucoEnergy) beaves somewhat linear. It simulates a production line over a year with supply and demand but it afterwards just shows if the stock of incoming material is still existing. As the production line must not run out on stock at any time the stock should always be bigger than 10% of the maximum capaxity. So fminSpeich(x,AraucoEnergy) just shows how full the stock is. But the beaviour in it is linear if I buy one more unit/machine (x(1,2)) use I y more of the stock ...

So it is a simulation of any hour of a given year depending on how many units/machines etc. I have.

[Erik Beste]

I will try to use your formula but I did not get it jet thanks anyway.

[Erik Beste]

@Matt J for your function to work how would I use it. So my imputs are x (1x4 double) and AraucoEnergy (8760x8 double) my output (lets call it y) is a 1X1 double.

[Alan Weiss]

Erik, I suspect that you genuinely have a nonlinear constraint, because you need to run a simulation to calculate its value. I think that Matt might have been misled by your comment that the constraint is linear. Yes, it is linear in the variable minSpeicher, but I suspect that it is not linear in the variable x, which contains your optimization parameters.

So I would abandon the idea of rewriting your constraints as linear, and concentrate on debugging the existing program. Is it true that, in some way, minSpeicher depends on x? My understanding is yes, it depends on x through the fminSpeich function. What you need to check is how things are being passed between functions so that fmincon thinks that the nonlinear constraints are satisfied, but when you calculate the value of c it is positive, meaning the constraint is violated.

Alan Weiss

MATLAB mathematical toolbox documentation

[Erik Beste]

so x is a (1x4 double) that contains the 4 variables that are supposed to be opimized:

1. How it is financed (so debt ratio of the project)--> only one without influence on minSpeicher
2. Amount of Machine one (produces a gas)
3. Size of the stock/tank
4. Amount of Machine Two (produces another gas --> uses the stock/tank as a feedstock)

minSpeicher calculates the min stock in the gastank. So the question is to optimze the projectlayout.

I will check my funcions lets see if that leeds me anywhere.

[Erik Beste]

Ok so I used the debugging now for a while an I noticed one thing. fmincon changes the values of x just very little from step to step. But apart from x (1,1) all the values are integers (but I just do this in the formula fminspeicher).

So in fminSpeich I have code like this:

AnSpeich=double(int8(x(1,3)));

Is there any way to increace the steps? at least for x(1,2:4). That fmincon knows that I want theses numbers to be integers so it changes like 1,2,3,... and not 4,05 --> 4,06?

[Alan Weiss]

Since the internal workings of your code have integer values of x, you cannot use fmincon. That is the unfortunate truth: fmincon operates only on differentiable functions. The small steps that you observe are from fmincon attempting to estimate gradients, which obviously doesn't work when you have an integer variable.

The only solvers that handle integers with truly nonlinear problems are in Global Optimization Toolbox: ga and surrogateopt. It is possible that you can adapt your problem to fit into their syntaxes.

However, it is entirely possible that your problem could be formulated to be linear and integer after all. See, for example, Maximize Long-Term Investments Using Linear Programming (no integer variables here): Solver-Based, Factory, Warehouse, Sales Allocation Model: Problem-Based, Optimal Dispatch of Power Generators: Problem-Based, and other examples in the Linear Programming and Mixed-Integer Linear Programming category.

Good luck,

Alan Weiss

MATLAB mathematical toolbox documentation

[Erik Beste]

Thanks a lot for this information. I tried it now without integers and it works perfectly fine. As it is for my master thesis I will tallk with my supervisor and see what he thinks what is the best solution.

Thanks anyway!

[Erik Beste]

Maybe it would be a good idea to include this into the description of the function. I could not find it there anywhere and I guess it is an important information.

[Matt J]

It's there, but I agree that it really should be at the top of the document, rather than a footnote.

https://www.mathworks.com/help/optim/ug/fmincon.html#busp5fq-7

"fmincon is a gradient-based method that is designed to work on problems where the objective and constraint functions are both continuous and have continuous first derivatives."

Answer 2

[Matt J]

so I used the debugging now for a while an I noticed one thing. fmincon changes the values of x just very little from step to step. But apart from x (1,1) all the values are integers (but I just do this in the formula fminspeicher)

fmincon does not handle integer-valued variables, and it also doesn't handle functionsthat involve rounding to integers, like with int8. How many combinations of of possible integer values are there for the 3 variables? Could you just loop over them all, solving for x(1) given fixed values for the other three?

[Erik Beste]

no I think that is not an option thanks anyway