Yes -- this seems to be the case; the following example works as expected. Thank you!
R = matrix(rnorm(9),3,3) R[lower.tri(R)] = 0 R.inv = solve(R) Dmat = t(R) %*% R dvec = c(0,5,0) Amat = matrix(c(-4,-3,0,2,1,0,0,-2,1),3,3) bvec = c(-8,2,0) x1 = solve.QP(Dmat=Dmat, dvec=dvec, Amat=Amat, bvec=bvec, factorized=FALSE) x2 = solve.QP(Dmat=R.inv, dvec=dvec, Amat=Amat, bvec=bvec, factorized=TRUE) print(x1$solution) print(x2$solution) -----Original Message----- From: Peter Dalgaard BSA [mailto:[EMAIL PROTECTED] Sent: 02 June 2003 18:12 To: David S. Khabie-Zeitoune Cc: [EMAIL PROTECTED] Subject: Re: [R] Help with factorized argument in solve.QP "David S. Khabie-Zeitoune" <[EMAIL PROTECTED]> writes: > I modified the example in the helpfile slightly to test this out: > > R = matrix(rnorm(9),3,3) > R.inv = solve(R) > Dmat = t(R) %*% R > dvec = c(0,5,0) > Amat = matrix(c(-4,-3,0,2,1,0,0,-2,1),3,3) > bvec = c(-8,2,0) > > x1 = solve.QP(Dmat=Dmat, dvec=dvec, Amat=Amat, bvec=bvec, > factorized=FALSE) > x2 = solve.QP(Dmat=R.inv, dvec=dvec, Amat=Amat, bvec=bvec, > factorized=TRUE) > print(x1$solution) > print(x2$solution) > > I would have expected that x1$solution and x2$solution were the same > (or numerically similar); however they are typically very different. > Where am I going wrong...? Hmmm. Could it be that it is assuming a *triangular* square root of the matrix? -- O__ ---- Peter Dalgaard Blegdamsvej 3 c/ /'_ --- Dept. of Biostatistics 2200 Cph. N (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~~~~~~~~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907 ______________________________________________ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
