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trust-region reflective algorithm in lsqnonlin and fmincon: same behavior for "unconstrained" optimization"?

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Say I provide constraints to lsqnonlin and fmincon. For both, I will choose trust-region-reflective algorithm
Do both fmincon and lsqnonlin behave identical in case the constraints are not active (unconstrained optimization)?
By identical, I mean is the same variant of trust-region-reflective algorithm implemented in fmincon and lsqnonlin, or are they different?

Accepted Answer

Matt J
Matt J on 15 Feb 2023
Edited: Matt J on 15 Feb 2023
No, they are not the same. They both take steps in a 2D subspace, but for fmincon, the basis vectors for the subspace are determined from the exact Hessian (see equations (3,4) here), whereas for lsqnonlin, they are determined from the Jacobian (see equation (6) here).
  4 Comments
SA-W
SA-W on 16 Feb 2023
Interestingly, the three solutions differ only by a constant. Would you consider the solution "FMINCON - ONLY GRADIENT" to be the worst of the three solutions?
I obtained very good results with lsqnonlin (trust-region-reflective), but now my problem has both bounds and inequality constraints, which is why I have to switch to fmincon. If both bounds and inequality constraints are present, I can not use trust-region-reflective algorithm anymore. Which of the algorithm in fmincon resembles the trust-region-reflective algorithm most closely?
Matt J
Matt J on 16 Feb 2023
Edited: Matt J on 16 Feb 2023
Interestingly, the three solutions differ only by a constant.
Because [1;1;1] is in the null space of A.
Would you consider the solution "FMINCON - ONLY GRADIENT" to be the worst of the three solutions?
They all achieve the same resnorm so how could it be worse than any of the others?
Which of the algorithm in fmincon resembles the trust-region-reflective algorithm most closely?
I don't think a comparison can be drawn. Inequality constraints are going to change the algorithm a lot.

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