Question 4 of 6: Reliability, Constant Failure Rate, and Normal Life-Data Analysis

Question 4: Reliability, Constant Failure Rate, and Normal 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) Reliability, the bathtub curve, and phase-appropriate distributions

Reliability $R(t)$ is the probability that an item performs its intended function without failure, under stated operating conditions, for a stated period of time $t$: $R(t)=P(T>t)=1-F(t)$, where $T$ is the (random) time to failure and $F$ is its cumulative distribution function. The life-cycle failure-rate (hazard-rate) curve, commonly called the "bathtub curve," plots the instantaneous hazard rate $h(t)=f(t)/R(t)$ against age and typically shows three distinct phases. The infant-mortality (early-life, DFR) phase shows a decreasing failure rate as manufacturing defects, weak components and installation errors are weeded out; the useful-life (CFR) phase shows an approximately constant, low failure rate driven by random, memoryless external stresses rather than aging; and the wear-out (IFR) phase shows an increasing failure rate as fatigue, corrosion, and cumulative wear mechanisms dominate.

Each phase is conventionally modelled with a distribution whose hazard-rate shape matches it. The infant-mortality phase is typically modelled with a Weibull distribution with shape parameter $\beta<1$ (decreasing hazard), sometimes a gamma distribution with shape $<1$. The useful-life phase, with its constant hazard, is modelled by the exponential distribution ($h(t)=\lambda$, constant, the unique memoryless continuous distribution) — this is the phase used in part (b). The wear-out phase, with its increasing hazard, is typically modelled by a Weibull distribution with shape parameter $\beta>1$, or by the normal or lognormal distribution when failures cluster fairly tightly around a characteristic wear-out life — this is the phase used in part (c), where the failure times are stated to come from a normal population.

(b) Reliability and MTTF for a constant-failure-rate component

Given. Constant failure rate $\lambda=0.00008\ \text{h}^{-1}$; operating time $t=5000$ h.

Find. $R(5000\,\text{h})$ and $MTTF$.

1. Reliability at $t=5000$ h (exponential life distribution, $R(t)=e^{-\lambda t}$): $$R(5000)=e^{-(0.00008)(5000)}=e^{-0.4}=\boxed{0.6703\ (67.03\%)}.$$
2. Mean time to failure. For an exponential (constant-hazard) life distribution, $MTTF=1/\lambda$: $$MTTF=\frac{1}{0.00008}=\boxed{12{,}500\ \text{h}}.$$

(c) Normal life-data: mean, standard deviation, and a conditional failure probability

Given. 8 failure times (h): 430, 400, 460, 435, 460, 530, 520, 490, verified to come from a normal population.

Find. $\hat\mu$, $\hat\sigma$ of the life distribution, and $P(T<500\mid T>400)$ — the probability that an item still operating at 400 h fails before it reaches 500 h.

1. Sample mean and standard deviation. $$\hat\mu=\frac{\sum t_i}{8}=\frac{430+400+460+435+460+530+520+490}{8}=\boxed{465.6\ \text{h}}.$$ $$\hat\sigma=\sqrt{\frac{\sum(t_i-\hat\mu)^2}{7}}=\boxed{45.15\ \text{h}}.$$
2. Conditional failure probability. Using the standard normal with $Z=(t-\hat\mu)/\hat\sigma$: $Z_{400}=(400-465.6)/45.15=-1.453$, $Z_{500}=(500-465.6)/45.15=0.761$, so $F(400)=\Phi(-1.453)=0.0731$, $F(500)=\Phi(0.761)=0.7768$. By the definition of conditional probability, $$P(T<500\mid T>400)=\frac{F(500)-F(400)}{1-F(400)}=\frac{0.7768-0.0731}{1-0.0731}=\boxed{0.759\ (75.9\%)}.$$