Optimization options parameter 'findiffrelstep'

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Pappu Murthy 2014-10-4

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编辑： Matt J 2014-10-8

I want to know exactly what this parameter is. i have two variables one is in the range 10 to 25 and other in the range 1000 to 2000... I am wondering whether setting findiffrelstep will improve my solution performance. Thanks in advance. Pappu murthy

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Matt J 2014-10-5

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The description at the FMINUNC webpage is pretty detailed.

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Matt J 2014-10-7

编辑：Matt J 2014-10-7

在 MATLAB Online 中打开

I think it is best if you show your objective function and your code and start talking in more detail about why you think you need to play with FinDiffRelStep.

I set [0.05 0.05] as FindiffRelStep. My variables are like [10,1100] so my step should be [.5 55]

I don't really follow that logic. The magnitude needed for delta is not related to the size of your variables. It is related to the size of the derivatives of your objective function. For example, if you're approximating the derivative of a 1D function f(x), the second order term in the Taylor remainder theorem gives you the error,

(f(x+delta)-f(x))/delta=f'(x)+f''(y)*delta

So, here the error term in the finite difference approximation of f'(x) is f''(y)*delta, where y is a point on the line between x and x+delta. You need to choose delta small enough to make the error term negligible. This requires some pre-analysis and some knowledge of a global bound on the second derivative magnitudes |f''(x)|.

Pappu Murthy 2014-10-8

Thanks for the valueable insights you are giving. I am sure my function is not differentiable in closed form. It involves a lot of computations which are done using a complicated fortran program as I mentioned earlie. That said, I have no reasons to believe there are 'undifferentiable' aspects to the compuations. I do have actually 5 parameters. The two parameter case is a sub case for testing purposes. I can try "fminsearch" if that can do the job. However, by scaling I am able to move properly all the 5 parameters now. I just divied one parameter by 100 and that seemed to do that job. Not sure though, whether I am doing this correctly. Now the question is i dod have sqrt in my least squares operation. But why should it matter? It is not doing a closed form differentiation anyway. So for finite difference it should not matter at all. Right?

Matt J 2014-10-8

编辑：Matt J 2014-10-8

It is not doing a closed form differentiation anyway. So for finite difference it should not matter at all. Right?

No, the theory of the algorithms assumes that the function is twice continuously differentiable. I'm not sure what distinction you're drawing between "closed-form differentiable" and other kinds. Either the function is continuously differentiable, or it's not. If it is not, the assumptions of the algorithm are violated, and that's the end of the story.

Finite difference operations are just a way of approximating the true derivative. The algorithm performs better if you provide an analytical gradient calculation (whether or not it is closed form). If the true derivative doesn't exist, the finite difference will not produce anything more useful. It will produce some unstable value greatly depending on the delta step size you use.

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