Mean Time Between Failures and Mission Reliability: 22-Mec-B4 Integrated Manufacturing Systems, December 2018, Q3

Question 3: Mean Time Between Failures and Mission Reliability (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. Two reliability problems: a component population with constant failure rates, and a three-subsystem mission profile in which each subsystem must survive its own duty period.

Part (a) — component failure-rate data
Component	Quantity $n_i$	Failure rate $\lambda_i$ (per hour)
Silicon transistor	40	$74.0 \times 10^{-6}$
Film resistor	100	$3.0 \times 10^{-6}$
Paper capacitor	50	$10.0 \times 10^{-6}$

Part (b) — mission profile, total mission length 8 hours
Subsystem	Required operating time	Life distribution	Reliability information
A	8 hours	exponential	50 per cent last at least 14 hours
B	3 hours	normal	mean life 6 hours, standard deviation 1.5 hours
C	4 hours	Weibull, $\beta = 1.0$	mean life 40 hours

Find. (a) the mean time between failures of the electronic system; (b) the reliability of the three-subsystem system over the 8-hour mission.

Check: what part (b) asks for. The paper prints the mission data and the table but never states the quantity to be computed. The only question the data support — a required operating time and a life distribution for each subsystem, with nothing said about redundancy or repair — is the reliability of the system for the stated mission, and that is what is answered here, with each subsystem's reliability reported on the way so that partial credit is visible. Stated as an assumption under Note 1: the three subsystems are functionally in series (all three are needed) and their failures are statistically independent.

Approach. Part (a): for a series population of exponential components the system hazard is the quantity-weighted sum of the component hazards, and the mean time between failures is its reciprocal. Part (b): convert each subsystem's reliability information into the parameter of its own distribution, evaluate that distribution at that subsystem's required operating time, and multiply.

Figure 3.1 — Reliability block diagram for part (b). The three subsystems are in series, but each is evaluated at its own required operating time, not at the 8-hour mission length.

Part (a) — sum the hazard contributions. With every component critical, the system is a series of $40 + 100 + 50 = 190$ items, and constant hazards add: $$\lambda_s = \sum n_i \lambda_i = 40(74.0) + 100(3.0) + 50(10.0) \ \text{per } 10^{6} \text{ h}$$ $$\lambda_s = (2960 + 300 + 500) \times 10^{-6} = 3760 \times 10^{-6} = 3.760 \times 10^{-3} \text{ per hour}$$ The transistors alone account for 2960 of the 3760, that is 78.7 per cent of the total hazard, which is where any reliability-improvement effort should start.
Part (a) — invert to get the mean time between failures. For an exponential life the mean is the reciprocal of the constant hazard: $$\text{MTBF} = \frac{1}{\lambda_s} = \frac{1}{3.760 \times 10^{-3}} = 265.96 \text{ hours}$$ $$\boxed{\;\text{MTBF} \approx 266 \text{ hours}\;}$$ As a sanity figure, the reliability of that system over an 8-hour mission would be $e^{-\lambda_s(8)} = e^{-0.03008} = 0.9704$.
Part (b) — subsystem A from its median life. The statement that 50 per cent survive at least 14 hours makes 14 hours the median, not the mean. For an exponential life the median is $\ln 2/\lambda$, so $$\lambda_A = \frac{\ln 2}{14} = \frac{0.69315}{14} = 0.049511 \text{ per hour}, \qquad \text{MTTF}_A = \frac{1}{\lambda_A} = 20.20 \text{ hours}$$ $$R_A(8) = e^{-\lambda_A (8)} = e^{-0.39608} = 0.6730$$
Part (b) — subsystem B from the normal life model. Subsystem B is only needed for 3 hours of the 8-hour mission, so it is evaluated at 3 hours: $$z = \frac{t - \mu}{\sigma} = \frac{3 - 6}{1.5} = -2.00, \qquad R_B(3) = 1 - \Phi(-2.00) = \Phi(2.00) = 0.9772$$
Part (b) — subsystem C from the Weibull model. A Weibull life with shape $\beta = 1.0$ is exponential, and for $\beta = 1$ the mean life equals the characteristic life, so $\theta = 40$ hours and $$R_C(4) = \exp\!\left[-\left(\frac{t}{\theta}\right)^{\beta}\right] = e^{-(4/40)^{1.0}} = e^{-0.100} = 0.9048$$
Part (b) — combine in series. The three are independent and all are required, so the mission reliability is the product: $$R_{\text{sys}} = R_A R_B R_C = (0.6730)(0.9772)(0.9048)$$ $$\boxed{\;R_{\text{sys}} = 0.5951, \text{ that is about } 59.5 \text{ per cent}\;}$$ Subsystem A contributes essentially all of the shortfall: it is the only one asked to run the full 8 hours, and it does so on a mean life of only 20 hours. Even a perfect A would cap the mission at $R_B R_C = 0.8843$. A cold-standby unit for A alone, giving $R_A = e^{-\lambda_A t}(1 + \lambda_A t) = 0.9395$, would lift the mission reliability to 0.8308 without touching B or C; doubling A's mean life is not enough on its own, since it raises $R_A$ only to 0.8203 and the mission to 0.7254.

Question 3 — results
Quantity	Value
(a) Transistor hazard contribution, $40 \times 74.0 \times 10^{-6}$	$2960 \times 10^{-6}$ per hour
(a) Resistor hazard contribution, $100 \times 3.0 \times 10^{-6}$	$300 \times 10^{-6}$ per hour
(a) Capacitor hazard contribution, $50 \times 10.0 \times 10^{-6}$	$500 \times 10^{-6}$ per hour
(a) System failure rate $\lambda_s$	$3.760 \times 10^{-3}$ per hour
(a) Mean time between failures	266 hours
(b) $\lambda_A$ from a 14 h median	0.04951 per hour (mean life 20.2 h)
(b) $R_A$ at 8 hours	0.6730
(b) $R_B$ at 3 hours	0.9772
(b) $R_C$ at 4 hours	0.9048
(b) System reliability for the 8-hour mission	0.5951 (59.5 per cent)

Topic. Parts-count reliability prediction, the exponential, normal and Weibull life models, and the series combination rule for a mission profile.

Key relations. $\lambda_s = \sum n_i \lambda_i$; $\text{MTBF} = 1/\lambda_s$; $R(t) = e^{-\lambda t}$ with median $t_{0.5} = \ln 2/\lambda$; $R(t) = 1 - \Phi[(t - \mu)/\sigma]$ for a normal life; $R(t) = \exp[-(t/\theta)^{\beta}]$ for a Weibull life, which reduces to the exponential at $\beta = 1$; and $R_{\text{sys}} = \prod R_i$ for independent series elements.

Why this works. Hazards add in series because the probability that a system survives is the product of the probabilities that each element survives, and the logarithm of a product is a sum; for constant hazards those logarithms are simply $-\lambda_i t$. The parts-count method exploits this: a bill of material with a failure rate against each line item becomes a single system hazard by arithmetic alone, which is why it is the standard first-cut prediction in electronics.

Common pitfall. Two dominate. Reading “50 per cent will last at least 14 hours” as a mean life of 14 hours — for the exponential the mean is 1.44 times the median, and taking 14 as the mean gives $R_A = 0.565$ instead of 0.673. And evaluating every subsystem at the 8-hour mission length: the table gives each subsystem its own required operating time precisely because B and C are duty-cycled, and using 8 hours throughout would give 0.6730 × 0.0912 × 0.8187 = 0.050, an order of magnitude wrong.

Source. C. E. Ebeling, An Introduction to Reliability and Maintainability Engineering, 3rd ed., Ch. 4–5 (constant-failure-rate and time-dependent models, the Weibull distribution) and Ch. 6 (system reliability, series and parallel structures, parts-count prediction).

Question 3 of 7: Mean Time Between Failures and Mission Reliability

Question 3: Mean Time Between Failures and Mission Reliability (20 marks)