CODEWITHSUNDEEP
pythonscipyminimizeboundary
ToTo10
ToTo10

Reputation: 73

Boundary in Powell method Scipy

Let's minimise the function

f =lambda x: (x+1)**2

using Powell method in scipy

If we use

scipy.optimize.minimize(f, 1, method='Powell', bounds=None)

the return is

   direc: array([[1.]])
     fun: array(0.)
 message: 'Optimization terminated successfully.'
    nfev: 20
     nit: 2
  status: 0
 success: True
       x: array(-1.)

i.e. the minimum is at -1 as it should. If we provide the bounds

scipy.optimize.minimize(f, 1, method='Powell', bounds=[(0,2)])

the return is again

   direc: array([[1.]])
     fun: array(0.)
 message: 'Optimization terminated successfully.'
    nfev: 20
     nit: 2
  status: 0
 success: True
       x: array(-1.)

which is now WRONG! The correct answer should be 0. It is like if the bounds are not taken int account. I am using scipy '1.4.1' and python 3.7.6. Someone has any clue?

Upvotes: 1

Views: 3315

Answers (2)

Valentin Formont
Valentin Formont

Reputation: 1

I want to add that I have noticed as well that the Powell method tends to explore out of bounds parameters.

Upvotes: 0

joni
joni

Reputation: 7157

With scipy 1.4.x the Powell method cannot handle constraints nor bounds, as you can see here. Updating to scipy 1.5.x, it can handle bounds, see here:

In [11]: scipy.optimize.minimize(f, x0=1.0, method='Powell', bounds=[(0.0,2.0)])
Out[11]:
   direc: array([[1.64428414e-08]])
     fun: array(1.)
 message: 'Optimization terminated successfully.'
    nfev: 103
     nit: 2
  status: 0
 success: True
       x: array([2.44756652e-12])

Upvotes: 5

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