负载的统计，必然是由内核完成的，因此在内核源码中找答案是再好不过的事情了，找来2.6.21的内核源码，开始探索。
节选部分源码
//kernel/timer.c
1254 active_tasks = count_active_tasks();
1256 CALC_LOAD(avenrun[0], EXP_1, active_tasks);
1257 CALC_LOAD(avenrun[1], EXP_5, active_tasks);
1258 CALC_LOAD(avenrun[2], EXP_15, active_tasks);
//include/linux/sched.h
110 #define FSHIFT 11 /* nr of bits of precision */
111 #define FIXED_1 (1<
113 #define EXP_1 1884 /* 1/exp(5sec/1min) as fixed-point */
114 #define EXP_5 2014 /* 1/exp(5sec/5min) */
115 #define EXP_15 2037 /* 1/exp(5sec/15min) */
117 #define CALC_LOAD(load,exp,n) \
118 load *= exp; \
119 load += n*(FIXED_1-exp); \
120 load >>= FSHIFT;
load(t) = ( load(t-1)*exp(i) + n(t)*(2048-exp(i)) ) / 2048
load(t-1)为上次计算出的结果
n(t)为t时刻的活动进程数
计算方式是累加各个CPU的运行队列中running和uninterruptible的值 再乘以2048
计算方式如下
1946 unsigned long nr_active(void)
1947 {
1948 unsigned long i, running = 0, uninterruptible = 0;
1949
1950 for_each_online_cpu(i) {
1951 running += cpu_rq(i)->nr_running;
1952 uninterruptible += cpu_rq(i)->nr_uninterruptible;
1953 }
1954
1955 if (unlikely((long)uninterruptible < 0))
1956 uninterruptible = 0;
1957
1958 return running + uninterruptible;
1959 }
1226 static unsigned long count_active_tasks(void)
1227 {
1228 return nr_active() * FIXED_1;
1229 }
exp(1) = 1884
exp(5) = 2014
exp(15) = 2037
exp(i) = 2048 * e^(-1/12/i)
从本质上看负载是完全由过去的一段时间里每个CPU上的活动进程数决定的但并不是在数值上等同于每秒钟需要进行调度的进程数具体的计算过程是个比较复杂的过程。