Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
981 views
in Technique[技术] by (71.8m points)

r - Error in nonlinear optimization problem : infinite or missing values in 'x'

I have to consider optimization problem in simulation study. An instance is given below:

library(mvtnorm)
library(alabama)

n = 200
q = 0.5
X <- matrix(0, nrow = n, ncol = 2)
X[,1:2] <- rmvnorm(n = n, mean = c(0,0), sigma = matrix(c(1,1,1,4),
                                                          ncol = 2))
x0 = matrix(c(X[1,1:2]), nrow = 1)
y0 = x0 - 0.5 * log(n) * (colMeans(X) - x0)
X = rbind(X, y0)

x01 = y0[1]
x02 = y0[2]
x1 = X[,1]
x2 = X[,2]

pInit = matrix(rep(1/(n + 1), n + 1), nrow = n + 1) 

f1 <- function(p) mean(((n + 1) * p ) ^ q)

heq1 <- function(p)
  c(sum(x1 * p) - x01, sum(x2 * p) - x02, sum(p) - 1)

sol <- alabama::auglag(pInit, fn = function(p) -f1(p), heq = heq1)
cat("The maximum objective value is:", -sol$value, '
')

This gives error:

Error in eigen(a$hessian, symmetric = TRUE, only.values = TRUE) : 
  infinite or missing values in 'x'

I am not sure how to point out and overcome this problem. If this occurs due to mis-specification of initial point, how one can specify it in simulation work so that the program can itself sets suitable initial point and gives the right solution? Otherwise, why this error occurs and how to get rid of it? Can someone please help. Thanks!

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

This answer is an ADDENDUM to the first answer, especially targeting your second question about significantly speeding up the whole process.

To make the run time estimation reproducible, we will fix the seed; all other definitions are yours.

set.seed(4789)
n = 200
q = 0.5
X <- mvtnorm::rmvnorm(n = n, mean = c(0,0),
                      sigma = matrix(c(1,1,1,4), ncol = 2))
x0 = matrix(c(X[1,1:2]), nrow = 1)
y0 = x0 - 0.5 * log(n) * (colMeans(X) - x0)
X = rbind(X, y0)
x01 = y0[1]; x02 = y0[2]
x1 = X[,1]; x2 = X[,2]
pInit = matrix(rep(1/(n + 1), n + 1), nrow = n + 1) 

First, let's do it with augmented Lagrangian and optim() as inner solver.

f1 <- function(p) sum(sqrt(pmax(0, p)))
heq1 <- function(p) c(sum(x1 * p) - x01, sum(x2 * p) - x02, sum(p) - 1)
hin1 <- function(p) p - 1e-06
system.time( sol <- alabama::auglag(pInit, fn = function(p) -f1(p), 
                           heq = heq1, hin = hin1) )
##    user  system elapsed 
##  24.631   0.054  12.324 
-1 * sol$value; heq1(sol$par)
## [1] 7.741285
## [1] 1.386921e-09 3.431108e-10 4.793488e-10

This problem is convex with linear constraints. Therefore we can apply an efficient convex solver such as ECOS. For modeling we will make use of the CVXR package.

# install.packages(c("ECOSolveR", "CVXR"))
library(CVXR)

p <- Variable(201)
obj <- Maximize(sum(sqrt(p)))
cons <- list(p >= 0, sum(p) == 1,
             sum(x1*p)==x01, sum(x2*p)==x02)
prbl <- Problem(obj, cons)
system.time( sol <- solve(prbl, solver="ECOS") )
##    user  system elapsed 
##   0.044   0.000   0.044 

ps <- sol$getValue(p)
cat("The maximum value is:", sum(sqrt(pmax(0, ps))))
## The maximum value is: 7.74226
c(sum(ps), sum(x1*ps) - x01, sum(x2*ps) - x02)
## [1]  1.000000e+00 -1.018896e-11  9.167819e-12

We see that the convex solver is about 500 times faster (!) than the first approach with a standard nonlinear solver. IMPORTANT: We do not need a starting value because a convex problem has only one optimum.


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...