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----- Original Message ----- From: Jim Langston <jlangston@conexfreight.com> > 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. Jim, The different lengths of transactions (I assume that you mean the service times) is folded into the standard deviation sd. To quote: --------------------- The basic result of single-server queuing theory was developed by Khintchine and Pollaczek and can be expressed: <w> = (u**2)/(2*(1-u))*(1 + 1/Rs) where the parameter Rs is given by Rs = (<ts>/sd)**2 --------------------- In your case the standard deviation becomes large (about 3.7 <ts>), so 1/Rs is 3.7**2 = 13.7 and the factor (1+1/Rs) is almot 15, which explains that the number of items waiting <w> is large (almost 15 times the "normal" case). +--- | 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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