Charles, In theory, you divide the rated SU/second by the number of processors giving SUs/processor/second, adjusting for "MP effect" overhead. Similarly, you could use MIPS/processor such that:
273.8 (2064-2C3) divided by 426.1 (2094-722) equals 0.643. 0.146 seconds times 0.643 equals 0.0939 seconds Subtle factors render the ratio less than exact, especially with very small values, but your tests should prove to be in the ballpark. Test by averaging several runs and let us know how it turns out. db -----Original Message----- From: IBM Mainframe Discussion List [mailto:IBM-MAIN@LISTSERV.UA.EDU] On Behalf Of Charles Mills Sent: Tuesday, July 17, 2012 9:53 AM To: IBM-MAIN@LISTSERV.UA.EDU Subject: Help with elementary CPU speed question I have gotten dragged into a CPU performance question; a field I know little about. I run a test on a 2094-722. It is rated at 19778 SU/Second. The job consumes .146 CPU seconds total. I run the same job on a 2064-2C3. It is rated at 13378 SU/Second. All other things being roughly equal, should I expect that the job will consume 1.48 (19778/13378) times as much CPU time, or .216 CPU seconds? Is my logic right, or am I off somewhere? I'm not worried about a millisecond or two; just the broad strokes. Thanks, Charles ---------------------------------------------------------------------- For IBM-MAIN subscribe / signoff / archive access instructions, send email to lists...@listserv.ua.edu with the message: INFO IBM-MAIN ---------------------------------------------------------------------- For IBM-MAIN subscribe / signoff / archive access instructions, send email to lists...@listserv.ua.edu with the message: INFO IBM-MAIN
