07-14-2003 11:20 AM - edited 03-02-2019 08:51 AM
Can someone please explain the scheduler allocate numbers to me?...according to the following link http://www.cisco.com/univercd/cc/td/doc/product/software/ios121/121cgcr/fun_r/frprt3/frd3003.htm#1019339....by default 5% of CPU is avail for process tasks...i assume this number is reached by dividing the default process time(200ms) by the default interrupt time(4000ms)...ok...so far so good then an example is given later...where it says that to ensure that 20% of CPU is avail for process time type the following "scheduler allocate 2000 500"....huh..isn't 500/2000=25%? ....and don't worry i have no intention of modifying these values... ;-)................
john
07-14-2003 01:44 PM
the answer is that the denominator is the sum of the two allocate values....thanks....
john
09-08-2003 05:41 AM
hello john,
I am also looking for an explanation for these values.
The problem is, that we sometimes have a high cpu on our c7206 Internet-Router (100%) and with the show cpu command I can not see an high consuming process. I try to troubleshoot, but the telnet and console session is not usable.
What are cisco recommended values ?
Maybe I have to change these values to do be able for troubleshooting.
Thanks
Guenther
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