Thus spake "Steven W. Raymond" <steven_raymond@eli.net>
> How come when issuing 'sho proc cpu' one gets a 5 sec, 1 min and 5 min
> processor utilization which will be some large number, but that the list
> of processes & their corresponding proc time does not add up to the time
> interval average?
> See below. The 5 minute average is 31%. I have filtered out the zero
> values as can be seen. The resulting non-zero time processes add up to
> less than say, 5%, but the 5 minute average is 31%. What's taking up
> the rest of the CPU? Kernel processes?
>
> Router#sh proc cpu | e 0.00%
> CPU utilization for five seconds: 31%/20%; one minute: 32%; five
> minutes: 31%
The key to understanding is right in that line... Note how the 5sec
utilization is listed as "31%/20%". If you go and add up the CPU time
consumed by the processes in the table below, you should come to
approximately 11%.
Turbo switching paths, such as fast, netflow, optimum, and CEF, take place
in interrupt handlers and not in any particular process. This means that
IOS cannot directly measure how much CPU time they are consuming.
The solution is to measure the total CPU load (e.g. 31%), and the CPU load
we can acccount for (e.g.11%). The difference (e.g. 20%) is attributable to
switching packets. IOS reports this last number for the current interval
only, since the interrupt load is almost always constant over time.
S
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