Why Can CPU Utilization Exceed 100%? Exploring Time‑Slice Scheduling with C Examples

CPU Utilization and Time‑Slice Scheduling

In a multitasking operating system the scheduler assigns each runnable process a time slice T1 . The actual CPU time the process consumes during that slice is T2 . The ratio T2/T1 defines the process’s CPU utilization.

Low‑Utilization Example (sleep loop)

#include <sys/types.h>
#include <unistd.h>
#include <stdio.h>
#include <stdlib.h>

int main(void) {
    printf("%d: Cpu start...
", getpid());
    while (1) {
        sleep(1);
    }
    return 0;
}

This program spends most of its time sleeping, so a monitoring tool shows a very low CPU usage.

Higher Utilization with Busy Loop

#include <sys/types.h>
#include <unistd.h>
#include <stdio.h>
#include <stdlib.h>

int main(void) {
    int i = 0;
    printf("%d: Cpu start...
", getpid());
    while (1) {
        for (i = 0; i < 100000000; i++) {
            /* busy work */
        }
        sleep(1);
    }
    return 0;
}

The tight for‑loop consumes CPU cycles before each sleep, raising the measured utilization. The following screenshot shows the increase.

Maximum Utilization (no sleep)

#include <sys/types.h>
#include <unistd.h>
#include <stdio.h>
#include <stdlib.h>

int main(void) {
    int i = 0;
    printf("%d: Cpu start...
", getpid());
    while (1) {
        for (i = 0; i < 100000000; i++) {
            /* busy work */
        }
        /* sleep(1); */
    }
    return 0;
}

Removing the sleep call makes the process consume almost all CPU cycles. On a multi‑core machine the reported utilization can exceed 100 % because the metric is summed across cores.

Key Observation

On a single‑core system the utilization of any process cannot exceed 100 % because only one core is available. On multi‑core systems the sum of utilizations across cores can be greater than 100 %.