Calculate CPU per process

Question

I'm trying to write a script that gives back the CPU usage (in %) for a specific process I need to use the /proc/PID/stat because ps aux is not present on the embedded system.

I tried this:

#!/usr/bin/env bash

PID=$1

PREV_TIME=0
PREV_TOTAL=0
while true;do
    TOTAL=$(grep '^cpu ' /proc/stat |awk '{sum=$2+$3+$4+$5+$6+$7+$8+$9+$10; print sum}')

    sfile=`cat /proc/$PID/stat`

    PROC_U_TIME=$(echo $sfile|awk '{print $14}')
    PROC_S_TIME=$(echo $sfile|awk '{print $15}')
    PROC_CU_TIME=$(echo  $sfile|awk '{print $16}')
    PROC_CS_TIME=$(echo $sfile|awk '{print $17}')

    let "PROC_TIME=$PROC_U_TIME+$PROC_CU_TIME+$PROC_S_TIME+$PROC_CS_TIME"

    CALC="scale=2 ;(($PROC_TIME-$PREV_TIME)/($TOTAL-$PREV_TOTAL)) *100"

    USER=`bc <<< $CALC`

    PREV_TIME="$PROC_TIME"
    PREV_TOTAL="$TOTAL"

    echo $USER
    sleep 1
done

But is doesn't give the correct value if i compare this to top. Do some of you know where I make a mistake?

Thanks

Answer 1

Under a normal invocation of top (no arguments), the %CPU column is the proportion of ticks used by the process against the total ticks provided by one CPU, over a period of time.

From the top.c source, the %CPU field is calculated as:

float u = (float)p->pcpu * Frame_tscale;

where pcpu for a process is the elapsed user time + system time since the last display:

hist_new[Frame_maxtask].tics = tics = (this->utime + this->stime);
...
   if(ptr) tics -= ptr->tics;
...
// we're just saving elapsed tics, to be converted into %cpu if
// this task wins it's displayable screen row lottery... */
this->pcpu = tics;

and:

et = (timev.tv_sec - oldtimev.tv_sec)
     + (float)(timev.tv_usec - oldtimev.tv_usec) / 1000000.0;
Frame_tscale = 100.0f / ((float)Hertz * (float)et * (Rc.mode_irixps ? 1 : Cpu_tot));

Hertz is 100 ticks/second on most systems (grep 'define HZ' /usr/include/asm*/param.h), et is the elapsed time in seconds since the last displayed frame, and Cpu_tot is the numer of CPUs (but the 1 is what's used by default).

So, the equation on a system using 100 ticks per second for a process over T seconds is:

(curr_utime + curr_stime - (last_utime + last_stime)) / (100 * T) * 100

The script becomes:

#!/bin/bash
PID=$1
SLEEP_TIME=3 # seconds
HZ=100       # ticks/second
prev_ticks=0
while true; do
    sfile=$(cat /proc/$PID/stat)

    utime=$(awk '{print $14}' <<< "$sfile")
    stime=$(awk '{print $15}' <<< "$sfile")
    ticks=$(($utime + $stime))

    pcpu=$(bc <<< "scale=4 ; ($ticks - $prev_ticks) / ($HZ * $SLEEP_TIME) * 100")

    prev_ticks="$ticks"

    echo $pcpu
    sleep $SLEEP_TIME
done

The key differences between this approach and that of your original script is that top is computing its CPU time percentages against 1 CPU, whereas you were attempting to do so against the aggregate total for all CPUs. It's also true that you can compute the exact aggregate ticks over a period of time by doing Hertz * time * n_cpus, and that it may not necessarily be the case that the numbers in /proc/stat will sum correctly:

$ grep 'define HZ' /usr/include/asm*/param.h
/usr/include/asm-generic/param.h:#define HZ 100
$ grep ^processor /proc/cpuinfo | wc -l
16
$ t1=$(awk '/^cpu /{sum=$2+$3+$4+$5+$6+$7+$8+$9+$10; print sum}' /proc/stat) ; sleep 1 ; t2=$(awk '/^cpu /{sum=$2+$3+$4+$5+$6+$7+$8+$9+$10; print sum}' /proc/stat) ; echo $(($t2 - $t1))
1602