Linux: How to put a load on system memory?

Question

I'm working on a small function, that gives my users a picture of how occupied the CPU is. I'm using cat /proc/loadavg, wich returns the wellknown 3 numbers.

My problem is that the CPU doesn't do anything right now, while I'm developing.

Is there a good way to generate som load on the CPU, I was thinking something like makecpudosomething 30, for a load of 0.3 or similar. Does an application like this exist?

Also, are there any way to eat up RAM in a controlled fashion?

Thanks

Michael

Answer 1

Have you considered using prime95?

I'm not sure if you can limit it to a percentage like that tho...

Answer 2

while true;
    do openssl speed;
done

also the stress program will let you load the cpu/mem/disk to the levels you want to simulate:

• http://weather.ou.edu/~apw/projects/stress/

stress is a deliberately 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 GPLv2.

to maintain a particular level of cpu utilization, say 30%, try cpulimit:

• http://cpulimit.sourceforge.net/

it will adapt to the current system environment and adjust for any other activity on the system.

there's also a patch to the kernel for native cpu rate limits here: http://lwn.net/Articles/185489/

Answer 3

mark@localhost$ time pi 1048576 | egrep '.*total$'

Is a simple benchmarking command that will give your cpu a rousting, post your times :D

Answer 4

You can try http://code.google.com/p/stressapptest/

Answer 5

To eat up a fixed amount of RAM, you could just:

#include <stdlib.h>
#include <string.h>
#define UNIX 1

//remove the above line if running under Windows

#ifdef UNIX
    #include <unistd.h>
#else
    #include <windows.h>
#endif

int main(int argc, char** argv)
{
    unsigned long mem;
    if(argc==1)
        mem = 1024*1024*512; //512 mb
    else if(argc==2)
        mem = (unsigned) atol(argv[1]);
    else
    {
        printf("Usage: loadmem <memory in bytes>");
        exit(1);
    }

    char* ptr = malloc(mem);
    while(1)
    {
        memset(ptr, 0, mem);
        #ifdef UNIX
            sleep(120);
        #else
            Sleep(120*1000);
        #endif
    }
}

The memset() call seems to be required, because at least on OS X, the actual memory doesn't seem to get used until it is actually initialised.

EDIT: Fixed in response to comment

Answer 6

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.