portfolio.optim and error in solve.QP: matrix D not positive definite
Hello everybody, sorry for my delayed "thanks" note - I was travelling. @Arun: Debugging the underlying code is a little bit difficult since the optimizer was written in FORTRAN. I think going for the nearest PD (as Krishna also suggested) might be the best way. However, I honestly don't understand why it is not PD... Does anybody have an explanation for that? @Guy: The weird thing is that I got the error code without the t(x) in the first place. t(x) solved the problem (for some assets) and the result indicated that it took a look at the assets and not the observations... I am going to give it a try with the covariance matrix again and let you know if it worked out... strange though. @Krishna: Thanks for your link! I think that really helps... I am going to try it out! Do you have an explanation why this is a common problem with a large number of assets? Thank you! Have a nice weekend! Lui
On Fri, Jan 28, 2011 at 3:51 PM, krishna <kriskumar at earthlink.net> wrote:
Also the link below might help with a large number of assets this is a common problem. https://stat.ethz.ch/pipermail/r-sig-finance/2008q3/002854.html Cheers Krishna On Jan 27, 2011, at 12:03 PM, Guy Yollin <gyollin at r-programming.org> wrote:
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
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