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
