调度器3—PELT算法下util_avg的增速和减速

这个程序用于 debug 若一个任务突然一直跑下去，其 util_avg 增加的速度。util_avg 是对一个正在运行的任务计算的，若其不允许了，对系统的 util 的影响应该是0，若是 n 个 periods(周期1024us) 后又开始运行了，其 uitl_avg 直接乘以 y^n 即可。

一、测试增速

1. 测试程序

#include <stdio.h>
#include <math.h>
#include <stdlib.h>

#define HALFLIFE_32 32
#define LOAD_AVG_MAX_32 47742

#define HALFLIFE_8 8
#define LOAD_AVG_MAX_8 12326

#define HALFLIFE_4 4
#define LOAD_AVG_MAX_4 6430

#define HALFLIFE_3 3
#define LOAD_AVG_MAX_3 4959

#define HALFLIFE_2 2
#define LOAD_AVG_MAX_2 3493


void calc_converged_max(double y)
{
    int n = -1;
    /* first period */
    long max = 1024;
    long last = 0, y_inv = ((1UL << 32) - 1) * y;

    for (; ; n++) {
        if (n > -1) {
            max = ((max * y_inv) >> 32) + 1024;
            /* This is the same as: max = max*y + 1024; */
        }
        if (last == max)
            break;

        last = max;
    }
    n--;

    printf("#define LOAD_AVG_MAX %ld
", max);
    printf("#define LOAD_AVG_MAX_N %d

", n);
}


int util_avg_from_sudden_running_all_time(double y, int periods, int load_avg_max) {
    int i;
    double util_avg;
    double util_sum = 0;
    
    for (i = 0; i < periods; i++) {
        util_sum += 1024 * pow(y, i);

        util_avg = util_sum / load_avg_max;
    
        printf("util_sum=%d, periods=%d, util_avg=%%%d
", (int)util_sum, i+1, (int)(util_avg * 100));
    }
    
    return 0;
}


void main(int argc, char *argv[])
{
    double y;
    int choose = 32, periods = 200;

    if (argc == 2) {
        choose = atoi(argv[1]);
    }

    if (argc == 3) {
        choose = atoi(argv[1]);
        periods = atoi(argv[2]);
    }
    
    printf("y^%d=0.5, periods=%d
", choose, periods);

    switch(choose) {
        case 32:
            y = pow(0.5, 1/(double)HALFLIFE_32);
            calc_converged_max(y); //47742
            util_avg_from_sudden_running_all_time(y, periods, LOAD_AVG_MAX_32);
            break;
        case 8:
            y = pow(0.5, 1/(double)HALFLIFE_8);
            calc_converged_max(y); //12326
            util_avg_from_sudden_running_all_time(y, periods, LOAD_AVG_MAX_8);
            break;
        case 4:
            y = pow(0.5, 1/(double)HALFLIFE_4);
            calc_converged_max(y); //6430
            util_avg_from_sudden_running_all_time(y, periods, LOAD_AVG_MAX_4);
            break;
        case 3:
            y = pow(0.5, 1/(double)HALFLIFE_3);
            calc_converged_max(y); //4959
            util_avg_from_sudden_running_all_time(y, periods, LOAD_AVG_MAX_3);
            break;
        case 2:
            y = pow(0.5, 1/(double)HALFLIFE_2);
            calc_converged_max(y); //3493
            util_avg_from_sudden_running_all_time(y, periods, LOAD_AVG_MAX_2);
            break;
        default: break;
    }
}

2. 测试结果

/*
$ ./pp 8 20
y^8=0.5, periods=20
#define LOAD_AVG_MAX 12326
#define LOAD_AVG_MAX_N 85

util_sum=1024, periods=1, util_avg=%8
util_sum=1963, periods=2, util_avg=%15
util_sum=2824, periods=3, util_avg=%22
util_sum=3613, periods=4, util_avg=%29
util_sum=4337, periods=5, util_avg=%35
util_sum=5001, periods=6, util_avg=%40
util_sum=5610, periods=7, util_avg=%45
util_sum=6168, periods=8, util_avg=%50
util_sum=6680, periods=9, util_avg=%54
util_sum=7150, periods=10, util_avg=%57
util_sum=7581, periods=11, util_avg=%61
util_sum=7975, periods=12, util_avg=%64
util_sum=8337, periods=13, util_avg=%67
util_sum=8669, periods=14, util_avg=%70
util_sum=8974, periods=15, util_avg=%72
util_sum=9253, periods=16, util_avg=%75
util_sum=9509, periods=17, util_avg=%77
util_sum=9744, periods=18, util_avg=%78
util_sum=9959, periods=19, util_avg=%80
util_sum=10156, periods=20, util_avg=%82
*/

二、测试递减

1. 测试程序

#include <stdio.h>
#include <math.h>
#include <stdlib.h>

#define HALFLIFE_32 32
#define HALFLIFE_8 8
#define HALFLIFE_4 4
#define HALFLIFE_3 3
#define HALFLIFE_2 2

int util_avg_from_sudden_sleep_all_time(double y, int periods) {
    int i;
    double util_avg;
    double util_sum = 0;
    
    for (i = 0; i < periods; i++) {
        util_avg = 1 * pow(y, i);
        printf("periods=%d, util_avg=%%%d
", i+1, (int)(util_avg * 100));
    }
    
    return 0;
}

void main(int argc, char *argv[])
{
    double y;
    int choose = 32, periods = 200;

    if (argc == 2) {
        choose = atoi(argv[1]);
    }

    if (argc == 3) {
        choose = atoi(argv[1]);
        periods = atoi(argv[2]);
    }
    
    printf("y^%d=0.5, periods=%d

", choose, periods);

    switch(choose) {
        case 32:
            y = pow(0.5, 1/(double)HALFLIFE_32);
            util_avg_from_sudden_sleep_all_time(y, periods);
            break;
        case 8:
            y = pow(0.5, 1/(double)HALFLIFE_8);
            util_avg_from_sudden_sleep_all_time(y, periods);
            break;
        case 4:
            y = pow(0.5, 1/(double)HALFLIFE_4);
            util_avg_from_sudden_sleep_all_time(y, periods);
            break;
        case 3:
            y = pow(0.5, 1/(double)HALFLIFE_3);
            util_avg_from_sudden_sleep_all_time(y, periods);
            break;
        case 2:
            y = pow(0.5, 1/(double)HALFLIFE_2);
            util_avg_from_sudden_sleep_all_time(y, periods);
            break;
        default: break;
    }
}

2. 执行结果

$ ./pp 8 10
y^8=0.5, periods=10

periods=1, util_avg=%100
periods=2, util_avg=%91
periods=3, util_avg=%84
periods=4, util_avg=%77
periods=5, util_avg=%70
periods=6, util_avg=%64
periods=7, util_avg=%59
periods=8, util_avg=%54
periods=9, util_avg=%49
periods=10, util_avg=%45

三、补充

1. 测试取 8 和 32，PELT 下连续慢跑 10 个周期 util_avg 的变化情况

               y^8=0.5      y^32=0.5                    
y              0.917004043  0.978572062                    
y^10           0.420448208  0.805245166                    
1-y^10         0.579551792  0.194754834                    
1-y            0.082995957  0.021427938                    
(1-q^10)/(1-y) 6.982891875  9.088827624                    
util_avg       0.698289188  0.908882762

y越大(指数N越小)，满跑的情况下，负载增加的越快。同理衰减的越慢，越利于性能越不利于功耗。

2. 满载增速：

q = (1/2)^32
LOAD_AVG_MAX = 47742
util_avg = (1-q^n)/(1-q) * 1024(1ms~=1024us) * 1024(scale)

一直满载后一直休眠的util_avg的减速：

util_avg = 1024(scale) * q^n