Re: Newbie question: big matrix calculations

*To*: mathgroup at smc.vnet.net*Subject*: [mg9296] Re: [mg9253] Newbie question: big matrix calculations*From*: Luci Ellis <elisha at dot.net.au>*Date*: Mon, 27 Oct 1997 02:47:06 -0500*Sender*: owner-wri-mathgroup at wolfram.com

>This is the first time I've ever been in this newsgroup, so please try >to excuse anything I do/say that's silly to the experts... > >I have three 100x100 sparse (diagonal) matrices (matrxD, matrxU, and >matrxL) and three 100x1 matrices (b, x2, and x1), and I'm performing >several operation on them: > >x2 = -Inverse[matrxD].(matrxL+matrxU).x1+Inverse[matrxD].b; One thing you can do is to calculate Inverse[matrxD] separately and call that object, instead of calculating the Inverse twice: x2 = With[{matrxDinv=Inverse[matrxD]},-matrxDinv.(matrxL+matrxU).x1+matrxDinv.b] Another thing is that you may be able to get the "optimal" ordering of the dot-product calculations with judicious use of parentheses. Roman Maeder's book "Programming in Mathematica" has, I believe, some discussion of this. I've just had a look at your code and I am certain there are faster ways to do what you are doing. I will have a closer look later -- there may be some built in functions that will speed things up. Luci > >I have to do it MANY, MANY times... with a 5x5 matrix it takes a few >minutes, but with a 100x100 it takes longer than eight hours (and still >running). Does anyone (offhandedly) know of a way to reduce the time >this takes for Mathematica to solve? > >I'm using a P133 and Mathematica 2.2. > >The entire hunk of code is viewable at: >http://www.cen.uiuc.edu/~dscott/PartB.gif solved for a 5x5 matrix >"size". > >PS-I've already tried decreasing the "accrcy" and changing my initial >"guess"... > >Thanks! > >-- >Dennis W. Scott, Jr. >University of Illinois at Urbana dscott at ews.uiuc.edu > ---------------------------------------------------------------- >Aspiring Electrical Engineer "I want to know God's >thoughts... -------------- Luci Ellis: elisha at dot.net.au http://www.dot.net.au/~elisha