Why does cpu load not change more than a few hundredths?

问题

I'm running this command

grep 'cpu ' /proc/stat | awk '{usage=($2+$4)*100/($2+$4+$5)} END {print usage "%"}'

Yet it only outputs something like 0.99xxxx%

If I do an apt-get upgrade or any process, I would imagine it would go above 1%. Even running stress -c 1 doesn't make it change any.

Is there a way to log CPU usage accurately? Server has 1 vCPU.

Need to have this log every 5 seconds.

while sleep 5; do "code" >> logfile; done

回答1:

Why does cpu load not change more than a few hundredths?

Because /proc/stat is returning aggregated CPU load statistics since the system last booted, not real-time ones. If you run your script just after a reboot, the reported load might significantly change as long as the CPU load itself changes. However, the longer the script runs the lesser load changes will impact the displayed value and after a while, the value will essentially stay constant.

If you want to compute the load from /proc/stat and not use the already available tools that do it, you need to compute the difference from two consecutive samples, eg :

while sleep 5; do grep -w cpu /proc/stat ; done | \
    awk '{
        print (o2+o4-$2-$4)*100/(o2+o4+o5-$2-$4-$5) "%"
        o2=$2;o4=$4;o5=$5}'

Otherwise, a simpler but less accurate way might be:

vmstat -n 5 | \
    awk '{used=$13+$14;total=used+$15
          if(total>0) print used*100/total "%"}'

回答2:

If you want to log the highest cpu percentage (this is, the process with the highest cpu usage in the moment of querying), you can use:

ps -e -o %cpu | tail -n +2 | sort -r | head -n 1 > cpu-usage.log

Explanation of the command:

ps -e -o %cpu gives you process information of every process in the output format that consist only in the cpu usage percentage
tail -n +2 filters the previous output starting from the second line (thus ignoring the header printed by ps)
sort -r sort the values in reverse order (highest first)
head -n 1 filters the data returned by sort so you discard all but the first line

回答3:

Slightly modified #3 for my own uses:

for x in $(seq 1 11);do sleep 5;grep -w cpu /proc/stat ; done | \
awk '{
    print (o2+o4-$2-$4)*100/(o2+o4+o5-$2-$4-$5) "%"
    o2=$2;o4=$4;o5=$5}'

Where seq is the number of samples taken, Sleep is still the length of time between samples

来源：https://stackoverflow.com/questions/32029015/why-does-cpu-load-not-change-more-than-a-few-hundredths

标签

bash

cpu

cpu-load