Cpu usage per Node

[ah...@reezocar.com]

Hello guys !

am trying to collect informations about the cpu usage per node i wrote this query sort(rate(node_cpu_seconds_total{mode="user"}[5m]) * 100 !=0)
but this query returns values for every cpu core per node . what i have to do to sum the values for every core in node cpu and devide it  on the number of cores and display the result for every node

[bakou....@gmail.com]

sort(sum(rate(node_cpu_seconds_total{mode="user"}[5m]) * 100)by("the variable of the node"))

[Ben Kochie]

Better, is to use `without (mode,cpu)`, allows all other labels to be passed through.

sum without (mode,cpu) (

  rate(node_cpu_seconds_total{mode!="idle"}[5m])

)

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[Chris Siebenmann]

> am trying to collect informations about the cpu usage per node i wrote

> this query *sort(rate(node_cpu_seconds_total{mode="user"}[5m]) * 100
> !=0)* but this query returns values for every cpu core per node . what

> i have to do to sum the values for every core in node cpu and devide
> it on the number of cores and display the result for every node

The simple and faster way to do this is to use avg(), instead of trying
to count how many CPUs your machine has in order to divide the sum() by
it. So:

(avg( rate(node_cpu_seconds_total{mode="user"}[5m]) ) without (cpu)) * 100

(Using avg() instead of sum() divided by CPU count is not original
to me. I got the idea from a commentator on my blog, who pointed out
the equivalence.)

- cks

[Ben Kochie]

Yup, this is a very good way to do it.

Another way to express the "whole CPU use" would be like this:

avg without (cpu) (

  sum without (mode) (

    rate(node_cpu_seconds_total{mode!="idle"}[5m])

  )

)

Maybe I should add this as a canonical example to the example rules[0].

[0]: https://github.com/prometheus/node_exporter/blob/master/example-rules.yml

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