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Hazard Rate Ratio⚠ unverified

General Calculations / Reliability · Compute the ratio of Weibull hazard rates at two times

Parameters

InputSymbolUnitDefaultDescription
t1t11.0Reference time (denominator of the ratio), in time units
t2t21.0Comparison time (numerator of the ratio), in time units
shapeshape1.0Weibull shape parameter (beta), dimensionless
scalescale1.0Weibull scale parameter (eta), in the same time units as ``t1`` and ``t2``
OutputSymbolUnitDescription
resultratioRatio of the hazard rate at ``t2`` to the hazard rate at ``t1``, dimensionless. Returns positive infinity when the hazard rate at ``t1`` is zero

The science & history

Understanding the Parameters

Derivation (Approaching a Proof)

Take the Weibull hazard rate at each time, $h(t) = \frac{\beta}{\eta}(t/\eta)^{\beta-1}$ (Weibull Failure Rate), and form the ratio. Every factor except the time-power is common to numerator and denominator:

$$\frac{h(t_2)}{h(t_1)} = \frac{\frac{\beta}{\eta}(t_2/\eta)^{\beta-1}}{\frac{\beta}{\eta}(t_1/\eta)^{\beta-1}} = \frac{(t_2/\eta)^{\beta-1}}{(t_1/\eta)^{\beta-1}} = \left(\frac{t_2}{t_1}\right)^{\beta-1}. \qquad\blacksquare$$

The prefactor $\beta/\eta$ and the $\eta$ inside each power cancel, leaving a pure function of $t_2/t_1$ and $\beta - 1$. This is the mathematical statement that the Weibull hazard is a power law: its ratio over any two times is set by the exponent alone, the signature of scale-invariance in the failure process.

Dimensional check. $t_2/t_1$ is dimensionless, so its power $(\cdot)^{\beta-1}$ is dimensionless — a pure ratio. $\checkmark$

History and Development

Related Concepts: Weibull Failure Rate, Weibull Reliability, Weibull Distribution PDF, Bathtub Curve, Exponential Reliability, B10 Life

Notes: Registry calculator hazard-rate-ratio (unverified). $h(t_2)/h(t_1) = (t_2/t_1)^{\beta-1}$ — correct as shipped; independent of $\eta$ (which the registry keeps as an unused input). $\beta=1$ ⇒ ratio $1$ (no aging); $\beta>1$ ⇒ aging; $\beta<1$ ⇒ improving. Scale-free measure of aging. Returns $+\infty$ if $h(t_1)=0$.

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