Hi All, Good Morning !!
We have IEFACTRT exit routine installed in our shop which shows the Job condition code and the CPU value. Here one of our Application developer is interested to know the value in MIPS for a Job. I checked one article which says like " http://itknowledgeexchange.techtarget.com/itanswers/convert-cpu-time-into-mips/ " This makes the assumption that the job is not multi-tasking either through > it's own code or with the built-in cpu parallelism of DB2. > > 1) divide the cpu time by the elapsed time > 2) divide the total processor MIPS by the number of engines on the > processor > 3) multiply the two above results together > > Example: > > 1) A job that ran for 60 minutes consumed 20 minutes of cpu time. So 20/60 > = .33 > > 2) The processor has a 6 engines and a total of 1200 MIPS. So 1200/6 = 200 > MIPS per engine. > > 3) 200*.33 = 66 MIPS > > Bill Can we really represent the Total CPU time taken for a Job in MIPS(average) ? Does it really makes sense to represent the CPU time in MIPS ? To be honest our shop does not own any SAS or MXG to process the RAW data on SMF Record 30 type, but yes I do have a unconventional way of Processing the SMF records but again it shows only CPU time taken. Could anyone please shed some light on the above. Jake ---------------------------------------------------------------------- For IBM-MAIN subscribe / signoff / archive access instructions, send email to lists...@bama.ua.edu with the message: INFO IBM-MAIN
