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CRE (Hong Kong Civil Service Exams): Reliability, Maintainability, and Safety Body of Knowledge (CRE BoK)

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Reliability, Maintainability, and Safety Body of Knowledge (CRE BoK)

What this domain tests

The CRE Body of Knowledge bundles the three "-ilities" into one domain because the exam expects you to move fluently between them: the definitional precision of reliability, the probability-not-time nature of maintainability, and the risk logic behind safety analysis. Items are typically short quantitative or definitional questions where the qualifiers in the sentence decide the answer.

Core rules

  • Reliability is the probability that an item performs a required function, for a stated period, under stated conditions. All three qualifiers matter: drop "stated conditions" and the statement is meaningless.
  • Maintainability is the probability that a failed item is restored to a specified condition within a stated downtime, using prescribed procedures and resources. It is a probability, not a duration.
  • Availability: A = MTBF / (MTBF + MTTR). Inherent availability excludes logistics and administrative delay; operational availability includes them, so operational A is always the lower number.
  • Safety is freedom from unacceptable risk. The tool kit is hazard analysis, FMEA (bottom-up, component to system), FTA (top-down, deductive from the undesired top event), and a severity-by-probability risk matrix.
  • Life pattern: the bathtub curve has a decreasing-hazard infant-mortality region, a constant-hazard useful life (exponential, R(t) = e^(-λt)), and an increasing-hazard wear-out region (Weibull shape β > 1).
  • Structures: series R = ΠRᵢ and, for constant independent failure rates, λ_system = Σλᵢ. Active parallel redundancy gives R = 1 − Π(1 − Rᵢ).

Worked example

A series system has three subsystems with failure rates of 2, 3, and 5 failures per 10⁶ hours. Sum the rates: λ_s = 10 per 10⁶ h = 1 × 10⁻⁵ per hour, so MTBF = 1/λ_s = 100,000 h. Reliability for a 1,000-hour mission is R = e^(−0.01) = 0.9900. If the same equipment is repairable with MTTR = 5 h at that MTBF, inherent availability = 100,000/(100,005) = 0.99995; if logistics delay averages 20 h, operational availability falls to 100,000/100,025 = 0.99975.

Traps that cost marks

  • Treating MTBF and MTTF as synonyms - MTTF applies to non-repairable items, MTBF to repairable ones.
  • Using the linear approximation R ≈ 1 − λt when λt is not small; at λt = 0.1 the error is already about 0.5%.
  • Adding failure rates across redundant (parallel) units - rates add only in series.
  • Answering "maintainability" with an MTTR value, or confusing maintainability with maintenance.
  • Computing availability with MTTR alone after a question specifies logistics delay.
  • Applying the exponential model to wear-out data, or forgetting that RPN = Severity × Occurrence × Detection and that the three scales are ordinal.