CODEWITHSUNDEEP
pythoncvxpyquadratic-programming
George
George

Reputation: 680

Why am I getting this DCPError?

I'm trying to optimize a binary portfolio vector to be greater than a benchmark using CVXPY.

import cvxpy as cp
import numpy as np

# Generate a random non-trivial quadratic program.

n = 10 # number of options

np.random.seed(1)
mu = np.random.randn(n) # expected means
var_covar = np.random.randn(n,n) # variance-covariance matrix
var_covar = var_covar.T.dot(var_covar) # cont'd
bench_cov = np.random.randn(n) # n-length vector of cov(benchmark, returns)

lamd = 0.01 # risk tolerance

# Define and solve the CVXPY problem.

x = cp.Variable(n, boolean=True)

prob = cp.Problem(cp.Maximize(mu.T@x + lamd * (cp.quad_form(x, var_covar) - (2 * bench_cov.T@x))), [cp.sum(x) == 4])

prob.solve()

I get this error using CVXPY version 1.1.0a0 (downloaded directly from github):

DCPError: Problem does not follow DCP rules. Specifically:

The objective is not DCP, even though each sub-expression is.

You are trying to maximize a function that is convex.

From what I've read maximizing a convex function is very difficult, but I got this equation from a paper. I figure I must be doing something wrong as I'm new to quadratic programming and CVXPY.

Thank you!

Upvotes: 2

Views: 3429

Answers (1)

Erwin Kalvelagen
Erwin Kalvelagen

Reputation: 16724

The problem with your model is that max x'Qx is non-convex. As we have binary variables x we can use a trick.

Define 

y(i,j) = x(i)*x(j)

as extra binary variable. Then we can write

sum((i,j), x(i)*Q(i,j)*x(j))

as

sum((i,j), y(i,j)*Q(i,j))

The binary multiplication y(i,j) = x(i)*x(j) can be linearized as:

 y(i,j) <= x(i)
 y(i,j) <= x(j)
 y(i,j) >= x(i)+x(j)-1

With this reformulation we have a completely linear model. It is a MIP as we have binary variables.

We can do this in CVXPY as:

import numpy as np
import cvxpy as cp

# Generate a random non-trivial quadratic program.

n = 10 # number of options

np.random.seed(1)
mu = np.random.randn(n) # expected means
var_covar = np.random.randn(n,n) # variance-covariance matrix
var_covar = var_covar.T.dot(var_covar) # cont'd
bench_cov = np.random.randn(n) # n-length vector of cov(benchmark, returns)

lamd = 0.01 # risk tolerance

e = np.ones((1,n))

x = cp.Variable((n,1), "x", boolean=True)
y = cp.Variable((n,n), "y", boolean=True)


prob = cp.Problem(cp.Maximize(mu.T@x + lamd * (cp.sum(cp.multiply(y,var_covar)) -2*bench_cov.T@x) ),
                  [y <= x@e, y <= (x@e).T, y >= x@e + (x@e).T - e.T@e, cp.sum(x)==4 ])

prob.solve(solver=cp.ECOS_BB)
print("status",prob.status)
print("obj",prob.value)
print("x",x.value)

This gives the result:

status optimal
obj 4.765120794509871
x [[1.00000000e+00]
 [3.52931931e-10]
 [3.80644178e-10]
 [2.53300872e-10]
 [9.99999999e-01]
 [1.79871537e-10]
 [1.00000000e+00]
 [3.46298454e-10]
 [9.99999999e-01]
 [1.00172269e-09]]

Notes:

  • You are encouraged to use a better MIP solver than ECOS_BB. For this model it gives the correct results, but it is somewhat of a toy solver and is known to give problems on more difficult data sets.
  • I don't understand the economics of the model. We are maximizing risk here. It may not be prudent to base your investment decisions on the results of this model.
  • Note that some high-end solvers (like Cplex and Gurobi) do this reformulation automatically. However CVXPY will not allow you to pass on the non-convex model to the solver.

Upvotes: 4

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