Question

I m looking for a way to occupy exactly 80% (or any other number) of a single CPU in a consistent manner.
I need this for some unit test that tests a component that triggers under specific CPU utilization conditions
For this purpose I can assume that the machine is otherwise idle.

What s a robust and possibly OS independent way to do this?

Answer 1

没有任何东西,比如占用了80%的时间。万国邮联经常被使用或闲置。在一段时间内,你的平均使用率可以达到80%。是否有一个具体的时间期限,你是否希望这一期限得到平均? 这种假体编码应当跨平台运行,而超过1个平台使用率达到80%:

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

Answer 2

Its pretty easy to write a program that alternately spins and sleeps to get any particular load level you want. I threw this together in a couple of minutes:

#include <stdlib.h>
#include <signal.h>
#include <string.h>
#include <time.h>
#include <sys/time.h>

#define INTERVAL    500000
volatile sig_atomic_t   flag;
void setflag(int sig) { flag = 1; }

int main(int ac, char **av) {
    int load = 80;
    struct sigaction sigact;
    struct itimerval interval = { { 0, INTERVAL }, { 0, INTERVAL } };
    struct timespec pausetime = { 0, 0 };
    memset(&sigact, 0, sizeof(sigact));
    sigact.sa_handler = setflag;
    sigaction(SIGALRM, &sigact, 0);
    setitimer(ITIMER_REAL, &interval, 0);
    if (ac == 2) load = atoi(av[1]);
    pausetime.tv_nsec = INTERVAL*(100 - load)*10;
    while (1) {
        flag = 0;
        nanosleep(&pausetime, 0);
        while (!flag) { /* spin */ } }
    return 0;
}

Answer 3

The trick is that if you want to occupy 80% of CPU, keep the processor busy for 0.8 seconds (or 80% of any time interval. Here I take it to be 1 second), then let it sleep for 0.2 seconds, though it is advisable not to utilize the CPU too much, or all your processes will start running slow. You could try around 20% or so. Here is an example done in Python:

import time
import math
time_of_run = 0.1
percent_cpu = 80 # Should ideally replace this with a smaller number e.g. 20
cpu_time_utilisation = float(percent_cpu)/100
on_time = time_of_run * cpu_time_utilisation
off_time = time_of_run * (1-cpu_time_utilisation)
while True:
    start_time = time.clock()
    while time.clock() - start_time < on_time:
        math.factorial(100) #Do any computation here
    time.sleep(off_time)

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