# Why trust-region-reflective algorithm in fmincon require a gradient as input?

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Hello

According to Matlab fmincon documentation, the 'trust-region-reflective' algorithm needs to specify the objective gradient.

1- Why does it need gradient as input?

2- Does the objective function have to be analytical? since my objective function, is function handle that calculates some scalar as output

Thanks for your help in advance

##### 12 Comments

Walter Roberson
on 26 Feb 2022

### Accepted Answer

Matt J
on 25 Feb 2022

Edited: Matt J
on 25 Feb 2022

Does the objective function have to be analytical? since my objective function, is function handle that calculates some scalar as output

I'm not sure what difference you see between a function being "analytical" and "specified by a function handle". If you mean, does fmincon require the objective function be specified as a single-line mathematical formula, the answer is no.

Why does it need gradient as input?

I believe it is because the trust-region algorithm is intended for problems of large dimension. If so, it would create a pitfall for users if finite difference approximations to the gradient were used by default, since for high dimensional problems, that would be very expensive computationally.

Regardless, even though the trust-region algorithm requires you to supply your own gradient calculation, you are of course free to do that calculation with your own finite difference implementation, or with one from the File Exchange (Example). However, as Torsten says, it might also be that the algorithm is sensitive to errors in the derivative calculations, so I'm not sure how well that would work.

### More Answers (2)

Zaikun Zhang
on 26 Feb 2022

##### 0 Comments

Bruno Luong
on 26 Feb 2022

Edited: Bruno Luong
on 26 Feb 2022

I don't think Matt's answer is correct, rather Torsen answer is better. The Trust region algorithm must approximate the objective function in a current "trust region". It is usually use the cost function and gradient in many points and create a quadratic function approximating. That requires both cost function and gradient calculation are continuous with respect to the points at which the gradients (and the objectiive) are evaluated. Therefore the numerical finite differences with adaptive step is not suitable, since there is no known method to estimate the step continuously with the point.

Therefore TR algorithm requires user to supply the "exact" gradient. Otherwise it won't converge.

Better still, if user provides the Hessian x vector calculation, it will help the TR to work even more efficiently.

##### 4 Comments

Bruno Luong
on 26 Feb 2022

OK, I miss that.

But my point does not change, if they compute H by finite differences on g, it requres the g to be exact so as H is meaningfuly estimatedl. Bottom line is the TR requires more accuracy on g, not because of the computation time.

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