Mean-Variance Portfolio Constraints in CVXOPT Quadratic Programs

# How do I do a mean variance optimization with constraints?


# How do I do a mean variance optimization with constraints?












I am using python and the cvxopt library to calculate an efficient frontier, per the docs:

http://cvxopt.org/examples/book/portfolio.html

However, I cannot figure out how to add a constraint so that there is an upper bound on a particular asset's maximum allowed weight. Is that possible using cvxopt?

Here is my code so far that produces an efficient frontier with no constraints, except I believe b, which sets the max sum of weights to 1. I'm not sure what G, h, A, and mus do, and the docs don't really explain. Where does the 10**(5.0*t/N-1.0) in the formula for mus come from?

```
from math import sqrt
from cvxopt import matrix as cmatrix
from cvxopt.blas import dot as cdot 
from cvxopt.solvers import qp, options 

# Number of assets
n = 4
# Convariance matrix
S = cmatrix( [[ 4e-2,  6e-3, -4e-3,   0.0 ], 
             [ 6e-3,  1e-2,  0.0,    0.0 ],
             [-4e-3,  0.0,   2.5e-3, 0.0 ],
             [ 0.0,   0.0,   0.0,    0.0 ]] )
# Expected return
pbar = cmatrix([.12, .10, .07, .03])

# nxn matrix of 0s
G = cmatrix(0.0, (n,n))
# Convert G to negative identity matrix
G[::n+1] = -1.0
# nx1 matrix of 0s
h = cmatrix(0.0, (n,1))
# 1xn matrix of 1s
A = cmatrix(1.0, (1,n))
Aadd = cmatrix(0.0, (n,n))
# Convert Aadd to identity matrix
Aadd[::n+1] = 1.0
A = cmatrix(np.vstack((A,Aadd)))

# bounds 
b = cmatrix([1.0,1.0,1.0,1.0,1.0])

N = 100
mus = [ 10**(5.0*t/N-1.0) for t in range(N) ]
options['show_progress'] = False
xs = [ qp(mu*S, -pbar, G, h, A, b)['x'] for mu in mus ]
returns = [ cdot(pbar,x) for x in xs ]
risks = [ sqrt(cdot(x, S*x)) for x in xs ]

sharpes = [x/y for x,y in zip(returns,risks)]
max_sharpe = max(sharpes)
max_index = sharpes.index(max_sharpe)
print xs[max_index]
```

Output:

```
  File "<ipython-input-45-8e583df5adc5>", line 29, in <module>
    xs = [ qp(mu*S, -pbar, G, h, A, b)['x'] for mu in mus ]

  File "C:\Users\Anaconda\lib\site-packages\cvxopt\coneprog.py", line 4496, in qp
    return coneqp(P, q, G, h, None, A,  b, initvals)

  File "C:\Users\Anaconda\lib\site-packages\cvxopt\coneprog.py", line 1986, in coneqp
    raise ValueError("Rank(A) < p or Rank([P; G; A]) < n")

ValueError: Rank(A) < p or Rank([P; G; A]) < n
```

## Answer by Luke (score 1)

https://quant.stackexchange.com/a/11538

If you don't know the meaning of the other matrices, I'd look more at the docs and the definition of the quadratic program:

http://cvxopt.org/userguide/coneprog.html#quadratic-programming

This is also an example from the book:

http://www.ee.ucla.edu/~vandenbe/publications/mlbook.pdf

And there is a good deal of explanation there.

Finally, if you don't want to get into that I'd look at CVXPY which allow much simpler constraints (from their page):

```
# Construct the problem.
x = Variable(n)
objective = Minimize(sum_squares(A*x - b))
constraints = [0 <= x, x <= 1]
prob = Problem(objective, constraints)
```