Taking the 1-minute average as the example, CALC_LOAD is identical to the mathematical expression:
load(t) = load(t-1) e-5/60 + n (1 - e-5/60)
(3)
If we consider the case n = 0, eqn.(3) becomes simply:
load(t) = load(t-1) e-5/60
(4)
If we iterate eqn.(4), between t = t0 and t = T we get:
load(tT) = load(t0) e-5t/60
(5)
which is pure exponential decay, just as we see in Fig. 2 for times between t0 = 2100 and tT = 3600.
Conversely, when n = 2 as it was in our experiments, the load average is dominated by the second term such that:
load(tT) = 2 load(t0) (1 - e-5t/60)
(6)
which is a monotonically increasing function just like that in Fig. 2 between t0 = 0 and tT = 2100.