Question 3 of 6: The Bathtub Curve, Reliability Fundamentals, and Weibull Life-Data Analysis

Question 3: The Bathtub Curve, Reliability Fundamentals, and Weibull Life-Data Analysis (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

(a) The bathtub curve: three phases and their governing distributions

The bathtub curve plots a population's instantaneous failure (hazard) rate $\lambda(t)$ against age and shows three characteristic phases. (1) Infant mortality (burn-in): a DECREASING failure rate, caused by manufacturing defects, weak components, and workmanship errors that fail quickly under use — as these weak units are weeded out, the surviving population becomes progressively more reliable and $\lambda(t)$ falls. This region is well modelled by a Weibull distribution with shape parameter $\beta<1$ (a decreasing hazard function), and manufacturers often deliberately "burn in" units past this region before shipment to avoid delivering early failures to customers. (2) Useful life: an approximately CONSTANT failure rate, where failures are caused by random, memoryless external events (shocks, misuse, random component failures unrelated to age) rather than accumulated wear — this region is modelled by the exponential distribution (the special case of Weibull with $\beta=1$, constant hazard $\lambda(t)=\lambda$). Most of a well-designed product's service life, and most reliability calculations, fall in this region. (3) Wear-out: an INCREASING failure rate as fatigue, corrosion, and accumulated wear cause the population to age and fail at a growing rate — modelled by a Weibull distribution with $\beta>1$ (or a normal/lognormal distribution, which also captures the accelerating, clustered failures typical of fatigue-driven wear-out). Because Weibull's hazard function is $h(t)=(\beta/\eta)(t/\eta)^{\beta-1}$, a single distribution family (Weibull) with a varying shape parameter can represent all three phases, which is exactly why it is the workhorse distribution of part (c)'s life-data analysis.

(b) Reliability function, density, failure rate, MTTF; the exponential; the memoryless property

For a nonnegative failure time $T$ with cumulative distribution function $F(t)=P(T\le t)$:

• Reliability (survival) function: $R(t)=P(T>t)=1-F(t)$ — the probability the unit is STILL functioning at age $t$.
• Failure density function: $f(t)=\dfrac{dF(t)}{dt}=-\dfrac{dR(t)}{dt}$ — the unconditional probability density of failing at exactly age $t$.
• Failure (hazard) rate: $h(t)=\dfrac{f(t)}{R(t)}$ — the CONDITIONAL instantaneous failure rate GIVEN survival to age $t$ (failures per unit time, per surviving unit); it is $h(t)$, not $f(t)$, that traces the bathtub curve of part (a).
• Mean time to failure: $MTTF=E[T]=\displaystyle\int_0^\infty t\,f(t)\,dt=\int_0^\infty R(t)\,dt$ (the second form, integrating the survival function directly, is usually the easier computation).

For the exponential distribution with constant failure rate $\lambda$ (the useful-life region of part (a)): $$f(t)=\lambda e^{-\lambda t},\qquad R(t)=e^{-\lambda t},\qquad h(t)=\frac{f(t)}{R(t)}=\lambda\ (\text{constant, independent of }t),\qquad MTTF=\int_0^\infty e^{-\lambda t}\,dt=\frac{1}{\lambda}.$$

The exponential's memoryless property states $P(T>t+s\mid T>t)=P(T>s)$ for any $t,s\ge0$: given that a unit has already survived to age $t$, the probability it survives an ADDITIONAL $s$ hours is exactly the same as the probability a brand-new unit survives $s$ hours — a used-but-still-working exponential unit is statistically as good as new. This follows directly from the constant hazard rate: since $h(t)=\lambda$ never depends on how long the unit has already run, there is no accumulated wear effect to condition on. It means age-based preventive replacement (replacing a unit simply because it has run for a while) gives NO reliability benefit for a purely exponential (constant-hazard) failure mode — a policy that only makes sense once the failure mode enters the increasing-hazard (wear-out, $\beta>1$) region of part (a).

(c) Weibull analysis of the 10 life-test failure times

Given. $N=10$ failure times (hours): 175, 303, 467, 521, 617, 665, 685, 713, 961, 1153 (sorted). Median-rank plotting position (Bernard's approximation), $F_i=(i-0.3)/(N+0.4)$.

Find. The Weibull shape parameter $\beta$, characteristic life $\eta$, and $MTTF$.

Approach. The Weibull CDF $F(t)=1-e^{-(t/\eta)^\beta}$ linearizes to $\ln[-\ln(1-F(t))]=\beta\ln t-\beta\ln\eta$ — plotting $y_i=\ln[-\ln(1-F_i)]$ against $x_i=\ln t_i$ (i.e. on Weibull probability paper) gives a straight line whose slope is $\beta$ and whose $x$-intercept ($y=0$) is $\ln\eta$; fit by least-squares regression through the ranked data.

1. Least-squares fit of $y$ on $x$. Regressing the table's $(x_i,y_i)$ pairs, $$y=\beta x-\beta\ln\eta=1.998x-13.154.$$
2. Shape parameter. The slope of the fitted line is the Weibull shape parameter directly: $$\boxed{\hat\beta=1.998\approx2.0}.$$ Since $\hat\beta\approx2$ is close to (but above) 1, this insulation's failures sit near the boundary between random (constant-hazard) failures and a mild wear-out trend — a mildly increasing hazard rate, consistent with a life test intended to characterize an insulation's aging behaviour.
3. Characteristic life. From the intercept, $\ln\eta=-(\text{intercept})/\beta=13.154/1.998=6.584$, so $$\boxed{\hat\eta=e^{6.584}=723.5\ \text{h}}$$ ($\eta$ is the age at which $R(\eta)=e^{-1}=36.8\%$ of the population is expected to still be surviving.)
4. Mean time to failure. For the Weibull distribution, $MTTF=\eta\,\Gamma(1+1/\beta)$: $$MTTF=723.5\times\Gamma(1+1/1.998)=723.5\times\Gamma(1.501)=723.5\times0.8862=\boxed{641.2\ \text{h}}.$$ Cross-check: the raw sample mean of the 10 failure times is 626.0 h, reasonably close to the fitted $MTTF$ — the small gap is expected sampling noise from fitting a 2-parameter distribution to only 10 points via a linearized (rather than full maximum-likelihood) regression.