Matlab is violating restrictions of fmincon

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Javier 2013-11-27

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https://ww2.mathworks.cn/matlabcentral/answers/107794-matlab-is-violating-restrictions-of-fmincon

评论： Javier 2013-12-2

采纳的回答： Matt J

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Hi

I want to solve a constrained maximization problem and i am using the function fmincon.

 [theta] = fmincon('gmmsimobj2',theta0,[-1 0 0;0 0 -1;0 1 0;0 6.85988 1;0 -5 -1],[0.001 0 0 0.5 0],[],[],[],[],[],foptions,simfunc,e,sp,beta,momfunc,y,mp,alg,W);

One of the restrictions I am including is that the first parameter cannot be negative, but when matlab solves the problem insists on setting it at -1.0000e-003. The problem has no solution for non-positive values. Why does matlab try to solve the problem with this value even violating the restriction? Am I including it wrong or it is a common issue?

Thank you!

Javier

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Matt J 2013-11-27

编辑：Matt J 2013-11-27

Incidentally, it is bad practice to pass more than 10 arguments to fmincon. I don't know how much longer this will be supported. There are newer and better ways to pass fixed parameters to your objective function.

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Answer 1

Matt J 2013-11-27

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https://ww2.mathworks.cn/matlabcentral/answers/107794-matlab-is-violating-restrictions-of-fmincon#answer_116559

编辑：Matt J 2013-11-27

In the A,b constraint data that you've shown, it doesn't appear that you've constrained x(1) to be non-negative. I see a lower bound of -1e-3, in perfect consistency with the output you're reporting.

Neverthelss, you should know that FMINCON can violate any constraints within foptions.TolCon, so you should still expect to see constraint violations by that amount even after fixing your b(1)=-1e-3 value. You should NOT try to avoid this by setting TolCon=0.

Finally, non-negativity and other bounds should really be specified using the lb,ub input arguments, not the A,b arguments. If you do it using ub, lb some algorithms like interior-point (with AlwaysHonorConstraints 'on') and sqp can enforce the bounds rigorously.

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Matt J 2013-11-28

编辑：Matt J 2013-11-28

If A*X<B and the first row of A is 1 0 0 then 1*x1+0*x2+0*x3 <0. Isn´t that correct?

It would be correct if A(1,:)=[-1 0 0] and B(1)=0. However, you do not have B(1)=0 in what you posted. You have B(1)=.001.

Javier 2013-12-2

You´re right, I had 0 previously but then I changed to make the restriction more binding.

Well, I am now using lower and upper bound restrictions as you suggested. The program is working I have understood better how Matlab works with this function.

Thank you!

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