pythonrcvxpycvxcvxr

Problems with DCP rules in CVXR

I am using the CVXR modelling package to solve a convex optimization problem. I know for sure that the problem is convex and that it follows the DCP rules, but if I check the DCP rules using CVXR it returns False. However, if I take the exact same problem and check it using CVXPY it returns True (as expected)

What is happening here? I attach a minimal reproducible example of this behavior in R and Python:

R code using CVXR

library(splines2)
library(CVXR)
deriv_basis = splines2::dbs(seq(0, 1, length.out=100), degree=3, intercept=T, df=30, derivs=2)
R = t(deriv_basis) %*% deriv_basis
beta_var = CVXR::::Variable(nrow(R))
q = CVXR::quad_form(beta_var, R)
CVXR::is_dcp(q)

[1] FALSE

write.table(x=R, file='R.csv'), row.names=F, sep=';')

Python code using CVXPY

import cvxpy
import pandas as pd

R = pd.read_csv('R.csv', sep=';').values
beta_var = cvxpy.Variable(R.shape[1])
q = cvxpy.quad_form(beta_var, R)
q.is_dcp()

Out[1]: True

Can someone explain what is happening here and how to solve it so I can use CVXR?

Solution

  • The problem is the negative eigenvalue in the R matrix. If you fix that by setting it to zero, say, then it satisfies the dcp condition. I have also fixed the syntax errors in the code in the question and removed the redundant :: . Another possibility (not shown) is to use nearest_spd in the pracma package to adjust the R matrix.

    library(splines2)
library(CVXR)

deriv_basis <- dbs(seq(0, 1, length.out=100), degree = 3, 
  intercept = TRUE, df = 30, derivs = 2)
R <- t(deriv_basis) %*% deriv_basis
e <- eigen(R)

# check decomposition
all.equal(R, e$vectors %*% diag(e$values) %*% t(e$vectors), 
 check.attributes = FALSE)
## [1] TRUE

e$values  # note negative value
##  [1]  1.095213e+08  1.095213e+08  1.056490e+07  1.055430e+07  1.052481e+07
##  [6]  1.046063e+07  1.034247e+07  1.015017e+07  9.866358e+06  9.485145e+06
## [11]  8.643220e+06  8.280963e+06  7.549803e+06  6.731472e+06  5.853402e+06
## [16]  4.949804e+06  4.056714e+06  3.209045e+06  2.437320e+06  1.759963e+06
## [21]  1.214976e+06  7.785251e+05  4.590441e+05  2.428199e+05  1.107300e+05
## [26]  4.060476e+04  1.040537e+04  1.320942e+03  7.239578e-09 -5.019224e-09

# zap negative eigenvalues making them zero
R <- with(e, vectors %*% diag(pmax(values, 0)) %*% t(vectors))

beta_var <- Variable(nrow(R))
q <- quad_form(beta_var, R)
is_dcp(q)
## [1] TRUE