[SciPy-User] how to use properly the function fmin () to scipy.optimize

Warren Weckesser warren.weckesser@enthought....
Tue Mar 13 12:35:25 CDT 2012


On Tue, Mar 13, 2012 at 8:02 AM, javi <fralosal@ei.upv.es> wrote:

> Hello, I have been trying to find the right way to use the function fmin
> () to
> use downhill simplex.
>
> Mainly I have a problem with that is that the algorithm converges to good
> effect, ie as a solution with a value next to zero.
>
> To test the performance of the algorithm I used the following example:
>
> def minimize (x):
>
>         min = x [0] + x [1] + x [2] + x [3]
>         return min
>
> In which given a vector x would want to obtain the values of its elements
> that
> when added give the minimum possible value.
>
> To do this use the following function call:
>
> solution = fmin (minimize, x0 = array ([1, 2, 3, 4]), args = "1", xtol =
> 0.21, =
> 0.21 ftol, full_output = 1)
>
> print "value parameters", solution [0], "\ n"
>
> and I get the following results:
>
>       Optimization terminated successfully.
>                Current function value: 10.000000
>                Iterations: 1
>                Function evaluations: 5
>
>       value of the parameters: [1. 2. 3. 4.]
>
> As you can see the solution is VERY BAD, and I understand that due to large
> values of ftol and xtol that I gave it converges very quickly and gives a
> small value.
>
> Now, for that is a better result, ie, better than the 10 found understand
> that I
> must decrease and ftol xtol values​​, but in doing so I get:
>
>
> "Warning: Maximum number of function evaluations exceeded Has Been."
>
> Where I understand the algorithm before converging has made excessive
> calls to
> the function "minimize".
>
> Could you tell me what the correct use of the parameters ftol and  xtol to
> find
> a good minimum next to 0?. Sshould generally be used in subsequent cases
> of ftol
> and xtol values​​?, They differ?.
>
> A greeting and thank you very much.
>
>

It looks like you want to solve a *constrained* minimization problem, in
which all the components of x remain positive.  The function fmin() is for
unconstrained optimization, and your objective function has no
(unconstrained) minimum.

You can try fmin_cobyla or fmin_slsqp.   Here's a short demonstration:

-----
from scipy.optimize import fmin_slsqp, fmin_cobyla


def objective(x):
    """The objective function to be minized."""
    return x.sum()

def all_positive_constr(x):
    """Component constraint function for fmin_slsqp."""
    return x


# The following are the component constraint functions for fmin_cobyla.

def x0_positive(x):
    return x[0]

def x1_positive(x):
    return x[1]

def x2_positive(x):
    return x[2]

def x3_positive(x):
    return x[3]


if __name__ == "__main__":

    print "Using fmin_slsqp"
    result = fmin_slsqp(objective, [1,2,3,4], f_ieqcons=all_positive_constr)
    print result
    print

    print "Using fmin_cobyla"
    result = fmin_cobyla(objective, [1,2,3,4], [x0_positive, x1_positive,
x2_positive, x3_positive])
    print result
    print
-----

Warren

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