[R-SIG-Finance] portfolio.optim and error in solve.QP: matrix D not positive definite
Lui ##
lui.r.project at googlemail.com
Sat Jan 29 21:52:57 CET 2011
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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>>
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