Constrained optimization in python

Question

In Python, using SciPy, I need to find the maximum of a function Q*((1+y)*2-3*Q-0.1)-y**2 Given the restriction (1-z)*(Q*((1+y)*2-3*Q-0.1))-y**2=0 . For z I would like to input some values to find the arguments that maximize the function given that value of z .

I have tried many ways to use SciPy optimization functions but I cannot figure out how to go about doing this. I did succeed in doing it using WolframAlpha , but this does not provide me with an answer to questions that follow up on this one.

Attempt:

from scipy.optimize import minimize

def equilibrium(z):
    #Objective function
    min_prof = lambda(Q,y): -1*(Q*((1+y)*2-3*Q-0.1)-y**2)

    #initial guess
    x0 = (0.6,0.9)

    #Restriction function
    cons = ({'type': 'eq', 'fun': lambda (Q,y): (1-z)*(Q*((1+y)*2-3*Q-0.1))-y**2})

    #y between 0 and 1, Q between 0 and 4
    bnds = ((0,4),(0,1))

    res = minimize(min_prof,x0, method='SLSQP', bounds=bnds ,constraints=cons)

    return res.x

from numpy import arange

range_z = arange(0,1,0.001)

print equilibrium(range_z)

Error:

---------------------------------------------------------------------------
ValueError                                Traceback (most recent call last)
<ipython-input-20-527013574373> in <module>()
     21 range_z = arange(0,1,0.01)
     22 
---> 23 print equilibrium(range_z)

<ipython-input-20-527013574373> in equilibrium(z)
     14     bnds = ((0,4),(0,1))
     15 
---> 16     res = minimize(min_prof,x0, method='SLSQP', bounds=bnds ,constraints=cons)
     17 
     18     return res.x

/Users/Joost/anaconda/lib/python2.7/site-packages/scipy/optimize/_minimize.pyc in minimize(fun, x0, args, method, jac, hess, hessp, bounds, constraints, tol, callback, options)
    456     elif meth == 'slsqp':
    457         return _minimize_slsqp(fun, x0, args, jac, bounds,
--> 458                                constraints, callback=callback, **options)
    459     elif meth == 'dogleg':
    460         return _minimize_dogleg(fun, x0, args, jac, hess,

/Users/Joost/anaconda/lib/python2.7/site-packages/scipy/optimize/slsqp.pyc in _minimize_slsqp(func, x0, args, jac, bounds, constraints, maxiter, ftol, iprint, disp, eps, callback, **unknown_options)
    324             + 2*meq + n1 + ((n+1)*n)//2 + 2*m + 3*n + 3*n1 + 1
    325     len_jw = mineq
--> 326     w = zeros(len_w)
    327     jw = zeros(len_jw)
    328 

ValueError: negative dimensions are not allowed

Answer 1

You need to evaluate your function for one z at a time. A minimal modification to make work your code is as follows:

print [equilibrium(z) for z in z_range]

In your current code the function describing the constraints returns a vector instead of a scalar, which leads to the error message.

Instead of optimizing numerically you might note that your problem is solvable analytically:

a = 0.1
Q = (6-3*a+3**.5 *(4-4*a+a**2-4*z+4*a*z-a**2 *z)**.5)/(6*(2+z))
y = Q*(1-z)+(Q*(-1+z)*(-2+a+Q*(2+z)))**.5

You can test this and convince yourself that it gives the same result (up-to numerical precision) then the numerical optimization. (I've tested for z=0.745 - you need to check the 2nd derivative to select the correct maximum generally. But this is doable.)