Unconstrained optimization problem fminunc with modified least squares

2021 6 30
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更新时间：2021 6 30
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Hello, I am trying an unconstrained optimization problem with 5 design variables (each design variable is a vector). The function evaluation is of the form (written as a separate function file):
funcVal = sumsqr(Z-z_tilde) + (1/k0)*norm(b-b0).^2 + (1/k1)*norm(c-c0).^2
Here, "b0" and "c0" are the known values and I want the optimizer to optimize the values for b and c so that they are as close as possible to b0 and c0, respectively.
z_tilde is the function that takes the data point along with 8 Gaussians with parameters (a,b,c).
The optimization setup is as shown below:
x_ = rand(30,1);
x_ (1:8) = theta_ (1:3:end) -10;
x_ (9:16) = mus - 10;
x_ (17:24) = vars - 10;
funcVal = @(x_)evalObjFunc(X, Y, Z,...
x_ (1:8), reshape(x_ (9:16),1,8), reshape(x_ (17:24),1,8), reshape(x_(25:28),1,4), reshape(x_(29:30),1,2), ...
mus,vars);
opts = optimoptions(@fminunc,'MaxIterations',10000,'MaxFunctionEvaluations',50000,'CheckGradients',true);
opts.OptimalityTolerance = 1.000000e-16;
opts.StepTolerance = 1.000000e-11;
[theta,fval,grad] = fminunc(funcVal,x_, opts);
The error I'm getting is "fminunc stopped because it cannot decrease the objective function along the current search direction."
P.S: The function definition is correct, I have checked it several times and for different values of x_ the function returns different values.

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Matt J
Matt J 2021-6-30
编辑：Matt J 2021-6-30
Since it is an unconstrained least squares problem, you might see performance advantages if you use lsqcurvefit() or lsqnonlin() instead.

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回答（1 个）

Matt J
Matt J 2021-6-30

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The exit message looks fine. It's not an error.

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Vinay PRAKASH
Vinay PRAKASH 2021-6-30
Hey Matt, thanks for your reply. But the problem is I get the same exit message for any value initial input value. It means the optimizer is not even trying to optimize the parameters and at the end I get the same output that I initially provided. Perhaps, I am going wrong somewhere.
Matt J
Matt J 2021-6-30
编辑：Matt J 2021-6-30
Ran in:
Ran in:
The exit message means fminunc thinks it succeeded. If it is terminating with every inital point that you choose, it means that it thnks every initial point is optimum. This can happen in functions that are locally flat everywhere, for example,
x=fminunc(@floor,1.3)
Initial point is a local minimum. Optimization completed because the size of the gradient at the initial point is less than the value of the optimality tolerance.
x = 1.3000
x=fminunc(@floor,1.5)
Initial point is a local minimum. Optimization completed because the size of the gradient at the initial point is less than the value of the optimality tolerance.
x = 1.5000
x=fminunc(@floor,2.7)
Initial point is a local minimum. Optimization completed because the size of the gradient at the initial point is less than the value of the optimality tolerance.
x = 2.7000

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Matt J
Matt J
2021-6-30

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