The Weibull Distribution

The Weibull distribution is simply a lifetime model flexible enough to describe parts that die young, parts that fail at random and parts that wear out, all with one formula. It's named after Waloddi Weibull, the Swedish engineer who championed it for material strength and fatigue. The formula $R(t) = e^{-(t/\eta)^\beta}$ is the exponential from the last lesson with one twist: $t/\eta$ (age measured in units of the characteristic life) is raised to the power $\beta$ (the shape, which decides whether failures speed up or slow down with age). The other formulas, the hazard, percentile lives, the Weibull plot and the MTTF, all come straight out of that one line, so it's less than it looks.

Parts that wear out. A bearing in a conveyor gearbox doesn't fail at random. Every revolution adds a little fatigue damage under the rollers, so a bearing at 20,000 hours is far more likely to fail this week than a new one. The exponential model, with its constant failure rate, can't express that. The Weibull model can, with one extra parameter.

The model. $R(t) = e^{-(t/\eta)^\beta}$. With $\beta = 1$ it's the exponential with MTTF $\eta$. Two facts fall straight out:

• $\eta$ is the 63.2% life: $F(\eta) = 1 - e^{-1} = 0.632$ for every $\beta$, since $(\eta/\eta)^\beta = 1$. All Weibull CDFs with the same $\eta$ cross at that point.
• Percentile lives: solving $1 - e^{-(t/\eta)^\beta} = p$ gives $(t/\eta)^\beta = -\ln(1 - p)$, so $t_p = \eta[-\ln(1 - p)]^{1/\beta}$. The B10 life, the bearing industry's L10 rating, is $\eta(0.10536)^{1/\beta}$.

What β means: the failure rate of the survivors. The hazard $h(t) = f(t)/R(t)$ is the failure rate among parts still running at age $t$; reliability and hazard functions develops it in general. For the Weibull, $f = -R'$, and the chain rule gives

So the failure rate grows like $t^{\beta - 1}$:

• $\beta < 1$, a falling rate: infant mortality. Weak units, with a bad solder joint or a contaminated chip, die early, and the survivors are the robust ones. The fix is burn-in: run units for a while before they ship.
• $\beta = 1$, a constant rate: random failures, the memoryless world of the previous lesson.
• $\beta > 1$, a rising rate: wear-out by fatigue, corrosion or abrasion. Around $\beta \approx 3.5$ the density looks nearly normal. The fix is preventive replacement before the steep part of the curve (MTBF, repair and availability).

Over a product's whole life the hazard often traces a bathtub curve: infant mortality, then a long flat stretch of random failures, then wear-out. Each stretch is a Weibull with its own $\beta$.

Mean life. Substitute $u = (t/\eta)^\beta$ in $\int_0^\infty t f(t)\,dt$. Then $t = \eta u^{1/\beta}$ and $f(t)\,dt = e^{-u}\,du$, so

where $\Gamma(a) = \int_0^\infty u^{a - 1}e^{-u}\,du$ is the gamma function, an improper integral. For $\beta = 1$ that's $\eta$; for $\beta = 2$ it's $0.886\,\eta$. The mean is not $\eta$.

Weibull plots: making the model a straight line. Take logs twice. From $1 - F = e^{-(t/\eta)^\beta}$, $-\ln(1 - F) = (t/\eta)^\beta$, and then

Plot $y = \ln[-\ln(1 - \hat F_i)]$ against $x = \ln t_{(i)}$ with plotting positions $\hat F_i = (i - 0.5)/n$, as in normal probability plots. (Reliability software usually uses median ranks, $(i - 0.3)/(n + 0.4)$, which give nearly the same line.) If the points line up, the data are Weibull, the slope is $\beta$, and the line crosses $y = 0$, the 63.2% level, at $x = \ln\eta$. A least-squares line makes it precise: $\hat\beta$ is the slope $b_1$, and the intercept $b_0 = -\beta\ln\eta$ gives $\hat\eta = e^{-b_0/\hat\beta}$.

A common mistake. Since $\eta$ is called the characteristic life, it's tempting to treat it as the mean life. It isn't: the mean is $\eta\,\Gamma(1 + 1/\beta)$, which is $0.886\,\eta$ for $\beta = 2$. On a Weibull plot, watch for the related slip of reading the line's crossing point as $\eta$ itself rather than $\ln\eta$, and for putting $F$ into the logs as a percent. And if the points form a dog-leg with two different slopes, you're probably looking at two failure modes mixed together, which one Weibull can't fit.

Bearing catalogs rate every bearing by its L10 life, the B10 of this lesson. Wind farm operators fit Weibull models to gearbox and blade-bearing failures to schedule crane visits before the wear-out region, and warranty analysts plot early returns on Weibull paper to forecast how many units will fail before the warranty runs out. The fitted β is the first number a failure analyst reports, because it says whether to hunt for manufacturing defects ($\beta < 1$) or wear ($\beta > 1$). Wind engineers also describe wind speeds at a site with a Weibull distribution, with $\beta$ near 2.