Reliability and Maintainability of an Assembly: 22-Mec-B4 Integrated Manufacturing Systems, May 2016, Q3

Question 3: Reliability and Maintainability of an Assembly (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.

Given. One assembly whose time between failures is described first by an exponential and then by a normal model of the same mean, and one product whose repair time is exponential.

Given data — Question 3
Quantity	Symbol	Value
Mean time between failures, exponential model	θ	100 hours
Mean of the alternative normal model	μ	100 hours
Standard deviation of the normal model	σ	20 hours
Mission time to be matched	t	100 hours
Mean time to repair, exponential	MTTR	4 hours

Find. (a) the exponential mission time whose reliability equals that of a 100-hour mission under the normal model, and (b) the probability that one repair lasts between 3 and 5 hours.

Reliability against mission time for the two life models of the same mean. Equal reliability of 0.50 is reached at 100 hours under the normal model but at only 69.3 hours under the exponential.

Approach. Write the reliability function for each model, evaluate the normal one at the stated mission time to get the target reliability, then invert the exponential reliability function to recover the mission time that matches it; part (b) is a difference of two exponential survivor probabilities.

Part (a) — evaluate the reliability under the normal model. Reliability is the probability of surviving the mission, $$R_{N}(t) = 1-\Phi\!\left(\frac{t-\mu}{\sigma}\right) = 1-\Phi\!\left(\frac{100-100}{20}\right) = 1-\Phi(0)$$ $$\boxed{R_{N}(100) = 0.500}$$ The mission time equals the mean, so the standard deviation of 20 hours never enters the answer — half of a symmetric distribution lies above its mean whatever its spread.
Invert the exponential reliability function. For a constant failure rate the survivor function is $R_{E}(t)=e^{-t/\theta}$, so setting it equal to the target reliability and taking logarithms gives $$e^{-t/\theta} = 0.500 \quad\Longrightarrow\quad t = -\theta\ln(0.500) = \theta\ln 2$$
Evaluate the equivalent mission time. Substituting the mean of 100 hours, $$t = 100 \times 0.6931 = 69.31\ \text{hours}$$ $$\boxed{t_{E} = 69.3\ \text{hours}}$$ so an exponential assembly of the same average life may be trusted for only about 69 hours to give the same 50 per cent chance of success that the normal assembly gives over a full 100 hours.
Read the practical meaning of the gap. Run the comparison the other way and the point sharpens: over a 100-hour mission the exponential assembly has reliability $e^{-1}=0.368$ against 0.500 for the normal one. The exponential has no wear-out shoulder — its hazard rate is constant from the first hour — whereas the normal model concentrates its failures around the mean and is therefore very safe early in life. Two items with identical mean lives can thus carry quite different mission risks, which is why a mean time between failures should never be quoted on its own.
Part (b) — set up the repair-time probability. With repair time exponential of mean $MTTR = 4$ hours, the probability that a repair is still unfinished after t hours is $e^{-t/4}$, so the probability of finishing between 3 and 5 hours is the difference of the two survivor probabilities: $$P(3 < T < 5) = e^{-3/4} - e^{-5/4}$$
Evaluate it. The two exponentials are 0.4724 and 0.2865, so $$P(3 < T < 5) = 0.4724 - 0.2865 = 0.1859$$ $$\boxed{P(3 < T < 5) = 18.6\ \text{per cent}}$$ For context, 52.8 per cent of repairs are finished within 3 hours and 28.7 per cent are still running after 5, so the two-hour window either side of the mean captures fewer than one repair in five — the signature of a distribution whose standard deviation equals its mean.

Final results — Question 3
Quantity	Value
Reliability of a 100-hour mission, normal model	0.500
Equivalent exponential mission time	69.3 hours
Reliability of a 100-hour mission, exponential model	0.368
P(repair finished within 3 h)	52.8 per cent
P(repair still running after 5 h)	28.7 per cent
P(3 h < repair time < 5 h)	18.6 per cent

Topic. Life models in reliability and maintainability: the exponential (constant hazard) and the normal (wear-out) descriptions of the same mean life.

Key relations. Exponential reliability $R(t)=e^{-t/\theta}$ with hazard rate $\lambda = 1/\theta$ constant; normal reliability $R(t)=1-\Phi\!\left((t-\mu)/\sigma\right)$; equivalence is imposed by $e^{-t/\theta} = R_{N}$, giving $t=-\theta\ln R_{N}$; for maintainability $M(t) = 1-e^{-t/MTTR}$ and $P(a < T < b) = e^{-a/MTTR}-e^{-b/MTTR}$.

Why this works. The exponential is memoryless: an assembly that has run 90 hours is exactly as likely to survive the next ten as a new one, so reliability decays geometrically from the first hour and reaches one half well before the mean. The normal model has an increasing hazard rate, so it holds reliability near unity for most of its life and then collapses through 0.5 exactly at the mean. Equating the two at a single point, as this question asks, is the cleanest way to show that the mean life is not by itself a statement about mission risk.

Common pitfall. Assuming that a 100-hour mission on a 100-hour mean life must give 0.5 reliability under both models; under the exponential it gives 0.368. The second trap in part (b) is to reach for a density and integrate, or worse to multiply the density by the two-hour width; the difference of survivor functions is exact and needs no integration.

Source. Ebeling, An Introduction to Reliability and Maintainability Engineering, 3rd ed., ch. 2–3 and ch. 8; Chase and Jacobs, Operations and Supply Chain Management, 16th ed., reliability and maintenance material.

Question 3 of 6: Reliability and Maintainability of an Assembly

Question 3: Reliability and Maintainability of an Assembly (20 marks)