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Chapter 5 of Optimizing Oracle Performance

I am trying to grasp the concepts laid out in this chapter, but the numbers in my trace files don't seem to work.  Here are my questions:

1) First, do I have the correct understanding in the following points:
- {c} is total cpu for a call and its recursive progeny
- {e} is the elapsed time within a call and its recursive progeny
- {e} includes {c}, so the difference between {e} and {c} is accounted for in elapsed time between calls - {wait ela}

2) If this is not true and the difference between {e} and {c} is not accounted for by waits, then what does account for the difference between the two?

3) Now for the question about the trace file.  If {c} is accounted for within {e}, then how do I explain circumstances in the trace file where {c} is greater than {e}?  Here is a line from my trace file:

FETCH #1:c=150000,e=146737,p=0,cr=0,cu=0,mis=0,r=1,dep=0,og=4,tim=2734191821385


Thanks for your assistance.

Regards,
Jeremy Drobnick
Jeremy Drobnick Send private email
Monday, August 20, 2007
 
 
Jeremy,

Your understanding of {c} and {e} are correct. Your third statement in question #1 is partly true, but it's missing the concept of unaccounted-for time, which is also key to answering your question #2.

I think the key to your complete understanding lies in Chapter 7, where the unaccounted-for topic is convered in detail. Specifically, the answer to your question #3 is demonstrated in and around Figure 7-8 on page 164. There is another example of the problem you're asking about on page 166, in the example that discusses "FETCH #96".

Hope this helps...
Cary Millsap Send private email
Monday, August 20, 2007
 
 
Guess I need to keep reading...  :)  Thanks for the additional insight.  Great job on the book.

Regards,
JD
Jeremy Drobnick Send private email
Tuesday, August 21, 2007
 
 

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