Question 4 of 6: Reliability, the Bathtub Curve, and Weibull Analysis

Question 4: Reliability, the Bathtub Curve, and Weibull 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 of failure-rate evolution and their distributions

The classic reliability "bathtub curve" plots the instantaneous failure (hazard) rate $h(t)$ against product age and shows three distinct phases. Phase 1 — infant mortality (burn-in): a DECREASING failure rate, driven by manufacturing defects, weak components and workmanship errors that fail early and are then "weeded out" of the surviving population; modelled by a distribution with a decreasing failure rate (DFR), typically a Weibull distribution with shape parameter $\beta<1$, or a mixture of exponentials. Phase 2 — useful life: an approximately CONSTANT failure rate, dominated by random external stresses/chance events rather than wear-out; modelled by the exponential distribution (constant failure rate, CFR), which is the Weibull special case $\beta=1$. Phase 3 — wear-out: an INCREASING failure rate as fatigue, corrosion, and cumulative wear mechanisms dominate; modelled by a distribution with increasing failure rate (IFR) — a Weibull with $\beta>1$, or a normal/lognormal distribution, are standard choices.

(b) Reliability function, failure rate, mean residual life, and the memoryless property

The reliability function $R(t)=P(T\gt t)=1-F(t)$ is the probability the item survives beyond age $t$. The failure rate (hazard function) $h(t)=f(t)/R(t)$ is the instantaneous conditional rate of failure given survival to $t$. The mean residual life $L(t)=E[T-t\mid T\gt t]$ is the expected remaining life given survival to age $t$. For the exponential distribution with rate $\lambda$: $R(t)=e^{-\lambda t}$, $h(t)=\lambda$ (constant, independent of $t$), and $L(t)=1/\lambda$ for every $t$ — the expected remaining life is the SAME regardless of the item's current age.

This is the memoryless property: $P(T\gt t+s\mid T\gt t)=P(T\gt s)$ for all $t,s\ge0$, i.e. the distribution of remaining life is independent of how long the item has already survived. A used, still-working exponential item is statistically indistinguishable from a brand-new one — it carries no "memory" of wear or age. This is precisely why the mean residual life is constant in $t$, and it is only exactly true for the exponential distribution among continuous distributions (any Weibull with $\beta\ne1$ has an age-dependent hazard, so surviving items ARE statistically different from new ones).

(c) Weibull analysis of the life-test data

Given. Ten failure times (hours): 305, 470, 960, 1150, 185, 520, 615, 720, 655, 885. The attached Weibull probability paper referenced by the question is not reproducible on this page; the graphical estimate is instead obtained by an algebraically equivalent least-squares fit on the linearized Weibull axes, which is the closed-form counterpart of reading the slope/intercept directly off Weibull paper.

Find. The Weibull shape parameter $\beta$, the characteristic life $\eta$, and the mean time to failure (MTTF).

Approach. Order the data, assign median-rank plotting positions $F_i\approx(i-0.3)/(N+0.4)$ (Benard's approximation), linearize $\ln\ln\!\left(\frac1{1-F}\right)=\beta\ln t-\beta\ln\eta$, and fit a straight line by least squares; recover $\eta$ from the intercept and MTTF from $\eta\,\Gamma(1+1/\beta)$.

1. Rank and assign median-rank plotting positions. Sorted (h): 185, 305, 470, 520, 615, 655, 720, 885, 960, 1150 with $F_i=(i-0.3)/10.4$ for $i=1,\dots,10$, giving $F_i=0.067,\,0.163,\,0.260,\,0.356,\,0.452,\,0.548,\,0.644,\,0.740,\,0.837,\,0.933$.
2. Linearize and fit. Regressing $y_i=\ln\ln\!\left(1/(1-F_i)\right)$ on $x_i=\ln t_i$ gives slope $=\hat\beta$ and intercept $=-\hat\beta\ln\hat\eta$: $$\hat\beta=\boxed{2.01},\qquad \text{intercept}=-13.27,\ \ (r=0.993,\text{ an excellent linear fit}).$$
3. Characteristic life. $$\hat\eta=\exp\!\left(\frac{-\text{intercept}}{\hat\beta}\right)=\exp\!\left(\frac{13.27}{2.006}\right)=\boxed{745.5}\ \text{h}.$$
4. Mean time to failure. $$MTTF=\hat\eta\,\Gamma\!\left(1+\frac1{\hat\beta}\right)=745.5\times\Gamma(1.499)=745.5\times0.8862=\boxed{660.6}\ \text{h}.$$