How to Set CPU Load on a Red Hat Linux Box

How to set CPU load on a Red Hat Linux box?

This is exactly what you need (internet archive link):
https://web.archive.org/web/20120512025754/http://weather.ou.edu/~apw/projects/stress/stress-1.0.4.tar.gz

From the homepage:
"stress is a simple workload generator for POSIX systems. It imposes a configurable amount of CPU, memory, I/O, and disk stress on the system. It is written in C, and is free software licensed under the GPL."

How to set Load Generator on Red Hat server (Linux)

Install:

cd /media/Linux // the directory which contains the installer.sh
./installer.sh

Start/Stop LG:

cd /opt/HP/HP_LoadGenerator/bin
./m_daemon_setup -install
./m_daemon_setup -kill

Linux: How to put a load on system memory?

I didn't understand very well if you want to generate arbitrary CPU load or CPU utilization. Yes, they are different things indeed. I'll try to cover both problems.

First of all: load is the average number of processes in the running, runnable or waiting for CPU scheduler queues in a given amount of time, "the one that wants your CPU" so to speak.

So, if you want to generate arbitrary load (say 0.3) you have to run a process for 30% of the time and then remove it from the run queue for 70% of the time, moving it to the sleeping queue or killing it, for example.

You can try this script to do that:

export LOAD=0.3
while true
     do yes > /dev/null &
     sleep $LOAD
     killall yes
     sleep `echo "1 - $LOAD" | bc`
done

Note that you have to wait some time (1, 10 and 15 minutes) to get the respective numbers to come up, and it will be influenced by other processes in your system. The more busy your system is the more this numbers will float. The last number (15 minutes interval) tends to be the most accurate.

CPU usage is, instead, the amount of time for which CPU was used for processing instructions of a computer program.

So, if you want to generate arbitrary CPU usage (say 30%) you have to run a process that is CPU bound 30% of the time and sleeps 70% of it.

I wrote an example to show you that:

#include <stdlib.h>
#include <unistd.h>
#include <err.h>
#include <math.h>
#include <sys/time.h>
#include <stdarg.h>
#include <sys/wait.h>

#define CPUUSAGE 0.3      /* set it to a 0 < float < 1 */
#define PROCESSES 1       /* number of child worker processes */
#define CYCLETIME 50000   /* total cycle interval in microseconds */

#define WORKTIME (CYCLETIME * CPUUSAGE)
#define SLEEPTIME (CYCLETIME - WORKTIME)

/* returns t1-t2 in microseconds */
static inline long timediff(const struct timeval *t1, const struct timeval *t2)
{
  return (t1->tv_sec - t2->tv_sec) * 1000000 + (t1->tv_usec - t2->tv_usec);
}

static inline void gettime (struct timeval *t)
{
  if (gettimeofday(t, NULL) < 0)
  {
    err(1, "failed to acquire time");
  }
}

int hogcpu (void)
{
  struct timeval tWorkStart, tWorkCur, tSleepStart, tSleepStop;
  long usSleep, usWork, usWorkDelay = 0, usSleepDelay = 0;

  do
  {
    usWork = WORKTIME - usWorkDelay;
    gettime (&tWorkStart);
    do
    {
      sqrt (rand ());
      gettime (&tWorkCur);
    }
    while ((usWorkDelay = (timediff (&tWorkCur, &tWorkStart) - usWork)) < 0);

    if (usSleepDelay <= SLEEPTIME)
      usSleep = SLEEPTIME - usSleepDelay;
    else
      usSleep = SLEEPTIME;

    gettime (&tSleepStart);
    usleep (usSleep);
    gettime (&tSleepStop);
    usSleepDelay = timediff (&tSleepStop, &tSleepStart) - usSleep;
  }
  while (1);
  return 0;
}

int main (int argc, char const *argv[])
{
  pid_t pid;
  int i;
  for (i = 0; i < PROCESSES; i++)
  {
    switch (pid = fork ())
    {
    case 0:
      _exit (hogcpu ());
    case -1:
      err (1, "fork failed");
      break;
    default:
      warnx ("worker [%d] forked", pid);
    }
  }
  wait(NULL);
  return 0;
}

If you want to eat up a fixed amount of RAM you can use the program in the cgkanchi's answer.

How to create a CPU spike with a bash command

You can also do

dd if=/dev/zero of=/dev/null

To run more of those to put load on more cores, try to fork it:

fulload() { dd if=/dev/zero of=/dev/null | dd if=/dev/zero of=/dev/null | dd if=/dev/zero of=/dev/null | dd if=/dev/zero of=/dev/null & }; fulload; read; killall dd

Repeat the command in the curly brackets as many times as the number of threads you want to produce (here 4 threads).
Simple enter hit will stop it (just make sure no other dd is running on this user or you kill it too).

rsyslogd using 100% CPU Utilization on all RHEL EC2 Instances

Issue is due to cert expiry for rsyslog. Observed a lot of connection retry errors in system messages for all clients. Post renewing certs in rsyslog, cpu went down.

How to occupy 80% CPU consistently?

There is no such thing as occupying the CPU 80% of the time. The CPU is always either being used, or idle. Over some period of time, you can get average usage to be 80%. Is there a specific time period you want it to be averaged over? This pseudo-code should work across platforms and over 1 second have a CPU usage of 80%:

while True:
    startTime = time.now()
    while date.now() - startTime < 0.8:
        Math.factorial(100) // Or any other computation here
    time.sleep(0.2)

CentOS 6.5 spike

In the first place, you are comparing apples and oranges: CentOS 6 corresponds to RHEL 6. Very likely your code would behave the same on RHEL 6.5 as it does on CentOS 6.5, and the same on CentOS 5.8 as on RHEL 5.8. It is misleading to describe the issue as a difference between RHEL and CentOS.

In the second place, if your CPU utilization is that strongly affected by a few usleep() calls (executed, apparently, very many times), then your code is flawed and you should fix it. Building a custom kernel to mask the problem would be pretty backward. Nevertheless, if the objective is more to move over to CentOS than to move up to a (somewhat) more up-to-date environment, then switch to CentOS 5 instead of to CentOS 6.

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