I have a question about the output of "show proce cpu"
Router#sh proc c
CPU utilization for five seconds: 8%/3%; one minute: 3%; five minutes: 2%
1 128 193 663 0.00% 0.00% 0.00% 0 Chunk Manager
2 4400 586915 7 0.00% 0.00% 0.00% 0 Load Meter
3 382504 7811772 48 0.00% 0.00% 0.00% 0 OSPF Hello
in above output
first line : five seconds: 8%/3%
The first number indicates the total,
the second number indicates the percent of CPU time spent at the interrupt level.
I don't know about the percent of cpu time spent for interrupt level exactly.
how the router use a cpu for interrupt level?
I have no idea about interrupt level?
as i known,
when router receive a packet in some interface,
the interface interrupt a cpu to process the packet.
cpu usage for interrupt is in proportion to received packets?
is there anyone explain about this concept more detail?
Interrupt Driven CPU Utilization
- CEF and fast packet switching
- ACL processing
- QoS processing
- Alignment or spurious memory access errors on MIPS and PPC based platforms (not on 68K based platforms). Check the command show align
- Spurious Interrupts, check the output of the show stacks command.
•Often a function of traffic rate and configured features in the switching path.
•Get more information with CPU profiling.
6500 - Interrupt Driven CPU utilization
Incorrect / Inappropriate switching path.
---CEF Disabled – globally or on specific interface.
---ACL Merge Algorithms / logging.
Router / Switch overloaded with traffic.
Memory issues like Spurious access, alignment errors etc.