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Date:   Mon, 17 Aug 2009 13:29:03 -0700
Reply-To:   Dale McLerran <stringplayer_2@YAHOO.COM>
Sender:   "SAS(r) Discussion" <SAS-L@LISTSERV.UGA.EDU>
From:   Dale McLerran <stringplayer_2@YAHOO.COM>
Subject:   Re: IML Matrix Inversion - correct size of matrix
Comments:   To: Howard Schreier <hs@howles.com>
Content-Type:   text/plain; charset=iso-8859-1

Howard,

Oh my! How did I manage that? You are absolutely correct!

In theory, a 64-bit computer (not running a Window's OS) could be built with memory that allowed 800+ Gb of ram. Of course, that is the amount of ram necessary for just storage of the matrix. Additional ram of at least 1600 Gb or so would be needed for computing the inverse. Thus, one would need a 64-be computer with at least 2400 Gb or ram to invert such a large matrix.

Assuming that such a computer could be found, there would be some further question as to whether such a computer ran SAS. Even then, there might be question as to whether SAS IML could operate in this environment and address all of the memory.

I am guessing that right now, it is extremely unlikely that there is any system in the world running SAS which would allow operations on this size a data matrix.

Dale

--------------------------------------- Dale McLerran Fred Hutchinson Cancer Research Center mailto: dmclerra@NO_SPAMfhcrc.org Ph: (206) 667-2926 Fax: (206) 667-5977 ---------------------------------------

--- On Mon, 8/17/09, Howard Schreier <hs@howles.com> wrote:

> From: Howard Schreier <hs@howles.com> > Subject: Re: IML Matrix Inversion > To: "Dale McLerran" <stringplayer_2@YAHOO.COM> > Date: Monday, August 17, 2009, 12:44 PM > 1) The number > of elements in the matrix is E = 330000*330000 > = 1.089E+11 2) The number of bytes in the matrix > is > Bytes = 8*E = 8.712E+11 > 3) The number of kilobytes is > Kb = Bytes / (1024^2) = 830841 > 4) The number of megabytes is > Mb = Kb / (1024^2) = 0.79235Where do the ^2 > factors come from?Seems to me the footprint is about > 8e2 GIGA bytes. >


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