Scipy optimization with matrix multiplication - python-3.x

Question

I've tried to use spicy.optimize.minimize to solve a matrix multiplication optimization problem, however, the result gives me a dimension error, can someone help me with it?
import numpy as np
from scipy.optimize import minimize
# define known variables, mu, sigma, rf
mu = np.matrix([[0.12],
[0.08],
[0.05]])
sigma = np.matrix([[0.5, 0.05, 0.03],
[0.05, 0.4, 0.01],
[0.03, 0.01, 0.2]])
rf = 0.02
def objective_fun(x):
'''
This is the objective function
'''
s = np.sqrt(x.T * sigma * x)/(mu.T * x - rf)
return s
def constraint(x):
con = 1
for i in np.arange(0,3):
con = con - x[i]
return con
# set up the boundaries for x
bound_i = (0, np.Inf)
bnds = (bound_i, bound_i, bound_i)
#set up the constraints for x
con = {'type':'eq', 'fun':constraint}
# initial guess for variable x
x = np.matrix([[0.5],
[0.3],
[0.2]])
sol = minimize(objective_fun, x, method = 'SLSQP', bounds = bnds, constraints = con)
The error gives me:
ValueError Traceback (most recent call last)
<ipython-input-31-b8901077b164> in <module>
----> 1 sol = minimize(objective_fun, x, method = 'SLSQP', bounds = bnds, constraints = con)
e:\Anaconda3\lib\site-packages\scipy\optimize\_minimize.py in minimize(fun, x0, args, method, jac, hess, hessp, bounds, constraints, tol, callback, options)
606 elif meth == 'slsqp':
607 return _minimize_slsqp(fun, x0, args, jac, bounds,
--> 608 constraints, callback=callback, **options)
609 elif meth == 'trust-constr':
610 return _minimize_trustregion_constr(fun, x0, args, jac, hess, hessp,
e:\Anaconda3\lib\site-packages\scipy\optimize\slsqp.py in _minimize_slsqp(func, x0, args, jac, bounds, constraints, maxiter, ftol, iprint, disp, eps, callback, **unknown_options)
397
398 # Compute objective function
--> 399 fx = func(x)
400 try:
401 fx = float(np.asarray(fx))
e:\Anaconda3\lib\site-packages\scipy\optimize\optimize.py in function_wrapper(*wrapper_args)
324 def function_wrapper(*wrapper_args):
325 ncalls[0] += 1
--> 326 return function(*(wrapper_args + args))
327
328 return ncalls, function_wrapper
<ipython-input-28-b1fb2386a380> in objective_fun(x)
3 This is the objective function
4 '''
----> 5 s = np.sqrt(x.T * sigma * x)/(mu.T * x - rf)
6 return s
e:\Anaconda3\lib\site-packages\numpy\matrixlib\defmatrix.py in __mul__(self, other)
218 if isinstance(other, (N.ndarray, list, tuple)) :
219 # This promotes 1-D vectors to row vectors
--> 220 return N.dot(self, asmatrix(other))
221 if isscalar(other) or not hasattr(other, '__rmul__') :
222 return N.dot(self, other)
ValueError: shapes (1,3) and (1,3) not aligned: 3 (dim 1) != 1 (dim 0)
However, I tried individually every function I wrote, they all have no errors in the end, like, if after defining the x matrix as shown in the code, I simply run objective_fun(x) in the console, and I immediately get an answer:
optimize_fun(x)
matrix([[5.90897598]])
That means that my function can do the matrix multiplication correctly, so what is wrong with the code here?

Answer 1

The docs for minimize() says that x0 should be an (n,) shaped array, but you are trying to treat it like a (3,1) array. I'm not sure on the inner workings of minimize() but I suspect when it steps over different values of the fit parameters it converts to the format that it thinks it wants. Anyways, the following minor corrections make it so the code works.
import numpy as np
from scipy.optimize import minimize
# define known variables, mu, sigma, rf
mu = np.matrix([[0.12],
[0.08],
[0.05]])
sigma = np.matrix([[0.5, 0.05, 0.03],
[0.05, 0.4, 0.01],
[0.03, 0.01, 0.2]])
rf = 0.02
def objective_fun(x):
'''
This is the objective function
'''
x = np.expand_dims(x, 1) # convert the (3,) shape to (3,1). Then we can do our normal matrix math on it
s = np.sqrt(x.T * sigma * x)/(mu.T * x - rf) # Transposes so the shapes are correct
return s
def constraint(x):
con = 1
for i in np.arange(0,3):
con = con - x[i]
return con
# set up the boundaries for x
bound_i = (0, np.Inf)
bnds = (bound_i, bound_i, bound_i)
#set up the constraints for x
con = {'type':'eq', 'fun':constraint}
# initial guess for variable x
x = np.array([0.5, 0.3, 0.2]) # Defining the initial guess as an (3,) array)
sol = minimize(objective_fun, x, method = 'SLSQP', bounds = bnds, constraints = con)
print(sol) # and the solution looks reasonable
Output
fun: 5.86953830952583
jac: array([-1.70555401, -1.70578796, -1.70573896])
message: 'Optimization terminated successfully.'
nfev: 32
nit: 6
njev: 6
status: 0
success: True
x: array([0.42809911, 0.29522438, 0.27667651])
Take a look at the comments I put in for an explanation on what you need to do.