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Re: Performance Predictions - How to?

From: EscVector <>
Date: 22 Dec 2006 09:57:32 -0800
Message-ID: <>

Mark D Powell wrote:
> On Dec 22, 10:23 am, "EscVector" <> wrote:
> > I have a multiple part process and I'm able to predict runtime based on
> > record count within +-30 minutes. The prediction is made by taking the
> > total to-be-processed record count from the longest single process and
> > dividing it by the runtime total of all processes.
> > How is this mathematically possible/sound? This isn't even a linear
> > projection. Just dumb luck?
> >
> > I'm looking for ways to project performance based on record counts, so
> > any suggestions are welcome.
> Table A and B both have 1 million rows. Table A has a row length of 20
> bytes. Table B has a row length of 2000 bytes. Do you think the time
> to process 500,000 rows of Table A will be the same as to process
> 500,000 rows of Table B? It could be if the 500k access were by PK in
> randon order. But if full table scans are used I would expect the 20
> byte row task to run in less time.
> A + or - estimate of 30 minutes is not very good for a job that runs
> one hour but is pretty good for a job that runs 24 hours so your
> current method may or may not be working very well. You did not
> provide any information to how accurate 30 minutes is.
> If you compare prior runs of a job to current runs where the row count
> varied then using row count will probably work fairly well, but using
> an average row count number for all jobs is likely to, at least in a
> few circumstances, miss the mark badly. How badly will depend on how
> similar your jobs are.
> You are likely to have better luck estimating the number of blocks of
> data that each job has to read and basing the estimate off of this.
> HTH -- Mark D Powell --

This typical run is over 5 days . New client with more data had projected at 40 days, now down to 8. I usually predict with linear and quadratic equations and hand management a nice chart. An example is in Niemiec's Oracle 9i Perf Tune book (p353-377) In this case, I stumbled upon a "magic" number that always foots with runtime.

Even Odder:
Typical runtime is 5 days. 30 minutes is a small percentage of 5 days. In fact rounded up it comes to 0.42% (very odd)

3600	seconds per hour
1800	30 minutes
86400	seconds per day
432000	seconds in 5 days
0.416666666666667%	Percentage that 1800 is of 432k
0.42%	Rounded
Received on Fri Dec 22 2006 - 11:57:32 CST

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