CRE (ASQ) common mistakes
CRE · Understand the traps and how to avoid them.
When computing system reliability from a block diagram, the candidate multiplies every component reliability together regardless of whether the blocks are in series or parallel, or treats a redundant (parallel) subsystem as if it were a series chain.
The series formula R = R1·R2·…·Rn is the first one learned and gets over-applied; candidates do not pause to classify each block as series, active-parallel, or standby before doing arithmetic, and the block diagram is mentally reduced to a single chain.
What to do instead
Before any arithmetic, redraw the diagram and label every block S (series) or P (parallel). Compute each parallel group first with 1 − Π(1 − Ri), collapse it to one equivalent block, then multiply the resulting series chain. Sanity-check: adding redundancy must raise reliability, adding series elements must lower it.
The candidate computes MTBF as the arithmetic mean of a set of observed failure times when the data are time-terminated (test stopped at a fixed time) or the failed units are a subset of units on test, producing a biased estimate.
Mean-of-failure-times is the intuitive calculation and matches textbook examples that quietly assume all units failed; candidates miss that censored (suspended) units contribute their operating time to the denominator but no failure to the numerator.
What to do instead
Write down total time on test and total failures separately before dividing. Use total operating time (including suspended units' run time) divided by number of failures. For time-terminated tests, confirm whether the exponential MTTF estimate is appropriate and check the assumption of constant failure rate before reporting a single number.
The candidate confuses the bathtub curve regions and reports a decreasing failure rate (infant mortality) as the steady-state assumption, or applies the exponential distribution/constant-hazard formulas to wear-out data.
The bathtub curve is memorized as a picture with three regions but the boundaries and the hazard behaviour in each region are not tied to which distribution and which reliability formula apply; candidates also assume 'exponential is always used' because it dominates early coursework.
What to do instead
For each problem, state the life phase explicitly (infant mortality = decreasing hazard, useful life = constant hazard/exponential, wear-out = increasing hazard, typically Weibull with β>1 or normal). Match the model to the phase, and if a Weibull shape parameter is given, use β<1, =1, >1 to justify the choice rather than defaulting to exponential.
In an FMEA the candidate assigns severity, occurrence, and detection scores by gut feel, then ranks items by RPN alone and recommends action on the highest RPN without considering severity or criticality.
RPN is a single memorable number and candidates treat it as a strict priority ranking; they also score severity relative to how bad the failure looks rather than against the effect on the customer/end user, and they inflate detection scores because they cannot picture the current controls.
What to do instead
Score severity against the defined effect categories (with the customer as reference), occurrence against demonstrated failure rate or field data, and detection against the specific current control and its proven effectiveness. Then prioritize by severity first (high-severity items always get action regardless of RPN) and use criticality (severity × occurrence) or an explicit RPN threshold rule. State the scoring scale used.
The candidate treats maintainability and availability as interchangeable, computing availability as MTBF/(MTBF+MTTR) but then using MTTR where maintainability (a probability of repair within time t) is asked for, or omitting logistics/downtime from operational availability.
The three terms share the same time quantities and the inherent-availability formula is over-learned; candidates do not distinguish inherent availability (MTBF, MTTR only) from operational availability (includes administrative and logistic delay) and do not recognize maintainability as a distribution question.
What to do instead
Classify the question before computing: availability is a ratio of uptime to total time (state whether inherent or operational and which downtimes are included); maintainability is the probability a repair is completed within a specified time, usually from a lognormal or exponential repair-time distribution. Write the formula and define each term (MTBF, MTTR, MDT, logistics delay) before substituting numbers.
When asked to allocate a system reliability target to subsystems, the candidate divides the target equally among subsystems or assigns each subsystem the full system target, instead of allocating by complexity, criticality, or demonstrated capability.
Equal allocation is the simplest arithmetic and candidates do not recall that allocation methods (equal, ARINC/apportionment by failure rate, AGREE by complexity and importance) exist and are chosen deliberately; they also forget that series subsystems must each exceed the system target.
What to do instead
State the allocation method and its inputs (number of parts, complexity factors, importance, current demonstrated reliability). Remember that in a series system each subsystem's allocated reliability must be higher than the system goal, and verify by recombining the allocations to confirm the product meets the target.
The candidate reads a confidence interval or demonstration test result as a point estimate and declares the requirement met (or failed) without applying the confidence level, or uses the wrong side of a one-sided test.
MTBF and reliability are reported as single numbers in most practice problems, so the distinction between the estimate and the lower confidence bound is blurred; candidates also mix up consumer's risk (β, producer's risk of rejection) and producer's risk (α) in demonstration tests.
What to do instead
Identify the decision rule explicitly: for a reliability demonstration, the lower confidence bound must exceed the requirement. Write the test as H0/H1, identify α (producer's risk) and β (consumer's risk), and confirm whether the test is fixed-duration or failure-terminated. Report the estimate and its confidence interval separately and state which one the decision uses.
The candidate builds a fault tree with the wrong gate logic-using an AND gate where events are redundant (either alone causes the top event is OR) or an OR gate for independent failures that must both occur-and fails to cut the tree into minimal cut sets.
The visual convention of AND/OR is easy to invert under time pressure, and candidates conflate 'both failures needed' with 'either failure suffices'; minimal cut sets are often skipped because the tree looks small enough to evaluate directly.
What to do instead
State the top event and, for each gate, ask 'does one input alone cause the output?' If yes it is OR; if all inputs are required it is AND. Convert the tree to minimal cut sets (single-element cut sets flag single points of failure) and compute top-event probability from the cut sets, checking that OR gates combine by 1−Π(1−Pi) and AND gates by ΠPi.
The candidate assumes a constant failure rate (exponential) for a component in a redundant or standby configuration and computes standby redundancy reliability with the active-parallel formula, ignoring the switching and dormancy effects.
The exponential model is the default and standby redundancy looks like ordinary parallel redundancy on a diagram; candidates do not account for the standby unit being unpowered (different failure rate, possible failure to switch on) or for the fact that a standby system's reliability is not simply 1−(1−R)².
What to do instead
Distinguish active (both units powered) from standby (one unit idle until switchover). For standby, include the probability the switch operates and use the standby reliability expression (e.g., Poisson-based for exponential units) rather than the active-parallel formula. State the assumptions about dormancy and switching before computing.
The candidate answers a reliability-growth or test-analysis question by treating successive fixes as independent and reporting the final demonstrated MTBF without accounting for the growth model (Duane/Crow-AMSAA) or for the fact that fixes change the population being measured.
Growth curves are less practiced than static reliability formulas, and candidates apply a single MTBF to data collected across several test phases with different configurations; they also confuse instantaneous and cumulative MTBF.
What to do instead
Plot cumulative failures against cumulative test time on log-log axes and check for a straight line (Duane) before quoting a growth rate. Distinguish cumulative MTBF from instantaneous MTBF at the end of the test, note which configuration each data point belongs to, and state whether the reported figure is demonstrated (lower confidence bound) or projected.