Hi Lui,
Without seeing the data this is just speculation but...
Are you sure you want t(x)? If you're mixing up your observations
versus your assets this may explain the error.
The first parameter of portfolio.optim (in the tseries package) is a
returns matrix, one column for each asset and one row for each day
(assuming daily returns). If you have this wrong then for your
small datasets you'd have more columns than rows and this could
produce that error.
Also, you don't have to pass the entire returns matrix to
portfolio.optim, you could pass just the covariance matrix you
calculate yourself and a vector (1-row matrix) of mean returns as
follows:
library(tseries)
set.seed(2)
R <- matrix(rnorm(100*10),nrow=100,ncol=10) # 10 assets, 100
observations
averet <- matrix(apply(R,2,mean),nrow=1)
rcov <- cov(R)
current_er <- 0.05
(op <- portfolio.optim(x=averet,pm=current_er,covmat=rcov,riskless =
FALSE,shorts = FALSE, rf = 0.0))
Hope this helps.
Best,
Guy
On 1/26/2011 7:51 PM, Lui ## wrote:
Dear Group,
I have a large set of stocks and want to determine the efficient
frontier. The data set covers approx. 1.5 years and S&P 500 companies
(nothing weird). portfolio.optim from the PerformanceAnalytics
package
works very well and fast. However, whenever I decrease the number of
stocks in the portfolio (to 10 or 400), I receive an error message:
"solve.QP(Dmat, dvec, Amat, bvec = b0, meq = 2) :
matrix D in quadratic function is not positive definite!"
My command settings for portfolio.optim were:
seed<- portfolio.optim(t(x), pm = current_er, riskless = FALSE,
shorts = FALSE, rf = 0.0)
Even when I tried it with shorts = TRUE the error would still remain.
x is the set of stocks (stocks in columns, time in rows), current_er
is the target return (lies between the minimal mean and the maximum
mean of a long only portfolio).
I can not post the stock data here - so maybe you have some general
suggestions for me of what could have gone wrong... The covariance
matrix is positive definite. What could cause the problem? It works
fine with the large data set but does not work at all with the small
one...
Thanks a lot for your suggestions!
Lui