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Article #3032

Re: Floating point performance

#3032
From: Dave Seaman
Date: Fri, 17 Oct 1986 02:57
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In article <8575@lanl.ARPA> jlg@a.UUCP (Jim Giles) writes:
>In article <8184@sun.uucp> dgh@sun.UUCP writes:
>>
>>                            Mflops Per MHz
>>
>>                             David Hough
>>			   dhough@sun.com
>>
>Mflops:(Millions of FLoating point OPerations per Second)
>MHz: (Millions of cycles per second)
>
>Therefore 'Mflops per MHz':(Millions^2 FLoating point OPeration cycles per
>			    sec^2)

You didn't divide properly.  "Mflops per MHz" means "Mflops divided by MHz",
which, by the "invert the denominator and multiply" rule, comes out to

	"Floating point operations per cycle"

after cancelling the "millions of ... per second" from numerator and
denominator.

I'm not claiming that this is a particularly useful measure, but that's
what it means.
-- 
Dave Seaman	  					
ags@h.cc.purdue.edu
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References: <340@euroies.UUCP> <1989@videovax.UUCP> <722@mips.UUCP> <8184@sun.uucp> <8575@lanl.ARPA>

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From: ags@h.cc.purdue.edu (Dave Seaman)
Newsgroups: net.arch
Subject: Re: Floating point performance
Message-ID: <3153@h.cc.purdue.edu>
Date: Thu, 16-Oct-86 22:57:38 EDT
Article-I.D.: h.3153
Posted: Thu Oct 16 22:57:38 1986
Date-Received: Sat, 18-Oct-86 00:33:05 EDT
References: <340@euroies.UUCP> <1989@videovax.UUCP> <722@mips.UUCP> <8184@sun.uucp> <8575@lanl.ARPA>
Reply-To: ags@h.cc.purdue.edu.UUCP (Dave Seaman)
Organization: Purdue University Computing Center
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