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Well, I must admit, I read only half of it, and scanned the other half. But, I did not notice anything about the length of the transactions. Consider... With the same number of transactions, if one set takes x amount of time and the other set takes x*6 amount of time, either CPU utilization will go up or the jobs will start to queue, or in our case, both. We have quite a number of batch jobs that are submitted throughout the day, and things are fine. I would guesstimate 20 an hour. We never even notice them. But, when some hugh jobs get submitted the CPU usage goes up with one slightly, barely noticeable. If two get submitted at the same time there is noticeable delay in response time by the system. If 3 or 4 get submitted the system practically crawls, and jobs queue up. There is a limit of 4 batch jobs running through QBATCH at a time, because if 5 ever got submitted we wouldn't be able to do anything. >From what I read, it seems you are discussing transactions of a fixed length in duration. Consider the effects of transactions that are twice the length in duration, for they will throw calculations. If, on average, I submit 10 transactions a second (to use your time frame, for us it would be minutes or decs of minutes) and each transaction took 1/20th of a second to process, I should be fairly stable. But, if I submit 10 transactions a second and each transaction took 1/10th of a second there would be a very noticeable response time lag (from transaction overlap). As you could imagine, if I submitted 10 jobs a second and each job took 1/9th of a second there would be a definite problem, as I would be submitting more jobs than can be processed. How are you calculating the length of the transaction in your formulas? You may have and I just missed it, as I said, I skimmed the second half. Regards, Jim Langston leif@ibm.net wrote: > Since there seems to be some interest in this, I'll chip in with something > I wrote for a project once: > > System Configuration Considerations Using Simple Queuing Theory > --------------------------------------------------------------- <SNIP> > I.e. the standard deviation is of the same size as the length itself. > The probability that a value is more than three standard deviations > removed from the mean is very low (0.001) so that the maximum queue > length could be set to > > w(max) = <w> + 3 * sd(w) = 4 * <w> > > You could add an extra one for added margin and use w(max) = 5 * <w> > with confidence. > > +--- > | This is the Midrange System Mailing List! > | To submit a new message, send your mail to MIDRANGE-L@midrange.com. > | To subscribe to this list send email to MIDRANGE-L-SUB@midrange.com. > | To unsubscribe from this list send email to MIDRANGE-L-UNSUB@midrange.com. > | Questions should be directed to the list owner/operator: david@midrange.com > +--- +--- | This is the Midrange System Mailing List! | To submit a new message, send your mail to MIDRANGE-L@midrange.com. | To subscribe to this list send email to MIDRANGE-L-SUB@midrange.com. | To unsubscribe from this list send email to MIDRANGE-L-UNSUB@midrange.com. | Questions should be directed to the list owner/operator: david@midrange.com +---
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