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How do I use a minimization function in scipy with constraints

I need some help regarding optimisation functions in python(scipy) the problem is optimizing f(x) where x=[a,b,c...n]. the constraints are that values of a,b etc should be between 0 and 1, and sum(x)==1. The scipy.optimise.minimize function seems best as it requires no differential. How do I pass the arguments?

Creating an ndarray using permutation is too long. My present code as below:-

import itertools as iter
all=iter.permutations([0.0,.1,.2,.3,.4,.5,.6,.7,.8,.9,1.0],6) if sum==1
all_legal=[]
for i in all:
if np.sum(i)==1:
    #print np.sum(i)
    all_legal.append(i)
print len(all_legal)
lmax=0
sharpeMax=0
for i in all_legal:
    if sharpeMax<getSharpe(i):
        sharpeMax=getSharpe(i)
        lmax=i
like image 770
anand Avatar asked Sep 12 '13 14:09

anand


Video Answer


2 Answers

You can do a constrained optimization with COBYLA or SLSQP as it says in the docs.

from scipy.optimize import minimize

start_pos = np.ones(6)*(1/6.) #or whatever

#Says one minus the sum of all variables must be zero
cons = ({'type': 'eq', 'fun': lambda x:  1 - sum(x)})

#Required to have non negative values
bnds = tuple((0,1) for x in start_pos)

Combine these into the minimization function.

res = minimize(getSharpe, start_pos, method='SLSQP', bounds=bnds ,constraints=cons)
like image 66
Daniel Avatar answered Sep 24 '22 06:09

Daniel


Check .minimize docstring:

scipy.optimize.minimize(fun, x0, args=(), method='BFGS', jac=None, hess=None, hessp=None, \
              bounds=None, constraints=(), tol=None, callback=None, options=None)

What matters the most in your case will be the bounds. When you want to constrain your parameter in [0,1] (or (0,1)?) You need to define it for each variable, such as:

bounds=((0,1), (0,1).....)

Now, the other part, sum(x)==1. There may be more elegant ways to do it, but consider this: instead of minimizing f(x), you minimize h=lambda x: f(x)+g(x), a new function essential f(x)+g(x) where g(x) is a function reaches it minimum when sum(x)=1. Such as g=lambda x: (sum(x)-1)**2.

The minimum of h(x) is reached when both f(x) and g(x) are at their minimum. Sort of a case of Lagrange multiplier method http://en.wikipedia.org/wiki/Lagrange_multiplier

like image 37
CT Zhu Avatar answered Sep 21 '22 06:09

CT Zhu