Question 1 of 7: Parts-Count Reliability and Mission Reliability

Question 1: Parts-Count Reliability 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. Part (a): a series subsystem of 190 components in three part classes, with the quantities and constant hourly failure rates tabulated above. Part (b): three independent subsystems in series, each with its own required operating time within an 8-hour mission and its own life model.

Find. The mean time between failures of the electronic subsystem, and the probability that all three subsystems survive their required operating times on one mission.

Approach. Both parts rest on the same idea: for items in series the hazard rates add, so a parts-count prediction sums the population failure rates to get one system rate whose reciprocal is the MTBF, and a mission reliability multiplies the individual survival probabilities, each computed from the life model that applies to that subsystem.

1. Part (a) — write the parts-count model for a series subsystem. When every component is critical, the subsystem survives only if all components survive, and for constant hazard rates the system rate is the population-weighted sum $$\lambda_s = \sum_{i} n_i \lambda_i$$ where $n_i$ is the quantity of part class $i$ and $\lambda_i$ its failure rate per hour. No reliability is computed component by component; the rates simply add.
2. Accumulate the three contributions. Taking each row in turn, $$n_1\lambda_1 = 40 \times 74.0 \times 10^{-6} = 2960 \times 10^{-6}\ \text{h}^{-1}$$ $$n_2\lambda_2 = 100 \times 3.0 \times 10^{-6} = 300 \times 10^{-6}\ \text{h}^{-1}$$ $$n_3\lambda_3 = 50 \times 10.0 \times 10^{-6} = 500 \times 10^{-6}\ \text{h}^{-1}$$ Adding the three and re-adding them as a check gives $\lambda_s = 3760 \times 10^{-6} = 3.760 \times 10^{-3}\ \text{h}^{-1}$. The transistors carry 78.7 per cent of the system failure rate on 21.1 per cent of the component count, which is the first thing a reliability review would act on.
3. Invert the system rate to get the mean time between failures. For an exponential time to failure the mean is the reciprocal of the rate, $$\text{MTBF} = \frac{1}{\lambda_s} = \frac{1}{3.760\times10^{-3}}$$ $$\boxed{\ \text{MTBF} = 266\ \text{hours}\ }$$ (265.96 h before rounding). Two sanity figures follow immediately: over an 8-hour mission this subsystem alone has $R = e^{-3.760\times10^{-3}\times 8} = 0.9704$, and at $t = \text{MTBF}$ the reliability is $e^{-1} = 0.368$ — the exponential model always gives 36.8 per cent survival at the mean, never 50 per cent.
4. Part (b) — convert subsystem A's median life into a rate. The phrase "50 per cent of subsystems will last at least 14 hours" states the median, not the mean. For an exponential life the median $t_{0.5}$ satisfies $e^{-\lambda t_{0.5}} = 0.5$, so $$\lambda_A = \frac{\ln 2}{t_{0.5}} = \frac{0.6931}{14} = 0.049511\ \text{h}^{-1}$$ which corresponds to a mean life of $1/\lambda_A = 20.20$ h, not 14 h. Over its required 8 hours, $$R_A = e^{-0.049511 \times 8} = e^{-0.39609} = 0.6730$$
5. Take subsystem B from the normal life model at 3 hours. With $\mu = 6$ h and $\sigma = 1.5$ h the standardised required time is $$z = \frac{t - \mu}{\sigma} = \frac{3 - 6}{1.5} = -2.00$$ and the subsystem survives if its life exceeds 3 h, so $$R_B = 1 - \Phi(-2.00) = \Phi(2.00) = 0.9772$$
6. Recognise that subsystem C's Weibull collapses to an exponential. A Weibull with shape $\beta = 1.0$ is the exponential distribution, whose mean equals its characteristic life, so a mean life of 40 h means $\theta = 40$ h and $$R_C = e^{-(t/\theta)^{\beta}} = e^{-4/40} = e^{-0.1} = 0.9048$$
7. Multiply the three survival probabilities. The subsystems are independent and all three are needed, so $$R_{\text{mission}} = R_A R_B R_C = 0.6730 \times 0.9772 \times 0.9048$$ $$\boxed{\ R_{\text{mission}} = 0.595 \quad (59.5\ \text{per cent})\ }$$ Subsystem A is the weak link by a wide margin: it contributes 67.3 per cent where B and C contribute 97.7 and 90.5 per cent, so any reliability-growth effort belongs there.