Reliability and Hazard Functions

The reliability function is simply the fraction of units still working at each age, and the hazard is the failure rate among the ones that are left. Think of a group of runners: reliability is how many are still in the race, and hazard is how quickly the remaining runners are dropping out right now. The key formula $R(t) = e^{-\int_0^t h(u)\,du}$ looks tough, but read it as: add up the hazard over time (the integral), and survival shrinks exponentially with that total. The MTTF formula is just the area under $R(t)$. You've already met each piece with the exponential and Weibull models; this lesson ties them together.

Counting survivors. Put 1,000 new pumps on test and count. At 2,000 hours, 800 are still running, so $R(2000) \approx 0.80$. In the next 100 hours, 16 more fail. As a share of the original 1,000, that's a failure density of $f \approx 16/(1000 \times 100) = 1.6 \times 10^{-4}$ per hour. As a share of the 800 survivors, it's a hazard of $h \approx 16/(800 \times 100) = 2.0 \times 10^{-4}$ per hour. The hazard is what a maintenance planner wants: of the pumps still running, how many will fail this week?

Why $h = f/R$. The hazard is a conditional failure rate:

From hazard back to reliability. Since $f = -R'$, the definition says $R' = -h(t)\,R$: survivors disappear at a rate proportional to how many are left. That's a separable differential equation, and with $R(0) = 1$,

• A constant hazard $h = \lambda$ gives $R = e^{-\lambda t}$, the exponential model.
• A power-law hazard $h = \frac{\beta}{\eta}\left(\frac{t}{\eta}\right)^{\beta - 1}$ gives $H = (t/\eta)^\beta$ and $R = e^{-(t/\eta)^\beta}$, the Weibull model. That's where the Weibull comes from: it's the lifetime whose hazard grows, or shrinks, like a power of age.

The bathtub curve. Over a product's whole life, the hazard usually falls, levels off, then rises:

• Early failures (falling hazard, Weibull $\beta < 1$): manufacturing defects that show up fast. Running every unit for a while before shipping, called burn-in, screens them out.
• Useful life (flat hazard, $\beta = 1$): random shocks such as voltage spikes and impacts.
• Wear-out (rising hazard, $\beta > 1$): fatigue, corrosion, abrasion. Replace parts before it climbs.

When independent failure modes compete, a unit survives only if it survives all of them, so the reliabilities multiply and the hazards add. The bathtub is literally a sum of three hazard curves.

MTTF is the area under $R$. Integrate by parts with $u = t$ and $dv = f\,dt = -dR$ (Integration by Parts):

because $t\,R(t) \to 0$ for any lifetime with a finite mean. This improper integral is often easier than $\int t\,f$. For the exponential it gives $\int_0^\infty e^{-\lambda t}dt = 1/\lambda$; for the Weibull, $\eta\,\Gamma(1 + 1/\beta)$.

Used vs new. The chance that a unit that has survived to $t_0$ lasts another $t$ is $R(t_0 + t)/R(t_0)$. With a constant hazard this equals $R(t)$: memoryless, so used is as good as new. With wear-out ($\beta > 1$) a used part is worse than a new one; with early failures ($\beta < 1$) it's better, which is the whole case for burn-in.

A common mistake. The density $f$ and the hazard $h$ both look like "failure rates," so it's tempting to treat them as the same thing. They aren't. $f(t)$ is how failures are spread over time for the whole starting population, so it integrates to 1. $h(t)$ is the rate among the units still alive, so its integral grows without bound, and $h$ can even exceed 1 per unit time. Two related slips: reading MTTF as "most units last this long" (for exponential lives only $e^{-1} = 37\%$ reach the MTTF, and the median life is $0.69\,\text{MTTF}$), and quoting a single failure rate for a wear-out part, as if its hazard were flat.

Reliability engineers fit hazard curves to field-return data for everything from hard drives to jet engines. A falling hazard justifies burn-in (electronics makers run boards hot for hours before shipping), a rising one sets replacement schedules for bearings, belts and tires, and a flat one means scheduled replacement is wasted money. Warranty reserves are computed from $F(t)$ at the warranty length, and safety cases for medical devices and vehicles quote reliability at a stated mission time.