Accepting a Pump Module on a Zero-Failure Demonstration Test
Your company must accept a new pump module from a supplier who warrants an MTBF of 2,000 hours, and the contract calls for demonstrating that value at 80% one-sided confidence under a constant failure rate assumption. The agreed plan is time-terminated: run three units, permit zero failures, and stop when accumulated unit-hours reach the precomputed requirement. The test engineer reports 2,900 accumulated hours - two units ran 1,100 hours each, and the third was pulled at 700 hours when its test stand lost cooling water - with no failures. The program manager wants to sign acceptance because each unit essentially reached 1,100 hours, and because calendar time already exceeds the schedule. Manufacturing is holding the first production lot for the signature, and the supplier has offered a 4% price reduction if acceptance is signed this week. The contract is silent on how censored units and test-stand downtime are to be accounted for, and the supplier's quality engineer has already emailed that zero failures means the requirement is met.
1. Compute the test time the agreed plan actually requires and the one-sided lower confidence bound demonstrated by the data. Should the module be accepted?
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The demonstrated time is short of plan. For a time-terminated test with zero allowed failures, the one-sided lower confidence limit on the exponential MTBF is theta_L = T / ln(1/alpha), where alpha = 1 - C.
- Required: T = 2,000 x ln(1/0.20) = 2,000 x 1.609 = 3,219 unit-hours.
- Achieved: T = 1,100 + 1,100 + 700 = 2,900 unit-hours. The pulled unit contributes only the 700 hours it actually ran; censoring reduces information, it does not preserve it.
- theta_L = 2,900 / 1.609 = 1,802 hours, below the 2,000-hour requirement.
Equivalently, the confidence demonstrated at 2,000 hours is 1 - exp(-2,900/2,000) = 0.765, or 76.5%, short of the contractual 80%. Do not sign acceptance on this data.
The test is close rather than hopeless: 319 additional failure-free unit-hours (about 107 hours on each of the three units, or 319 hours on one) would close it. Note also what the 80% plan means. If the true MTBF were exactly 2,000 hours, a correct plan still rejects 20% of the time by design; that is the consumer risk the contract accepted, and it is not evidence of a bad supplier.
2. The program manager argues that the 700-hour unit should be credited with a full 1,100 hours and that test-stand downtime should count toward accumulated test time because the schedule is already long. Evaluate both arguments and state the legitimate ways to close the demonstration.
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Both arguments fail, for different reasons.
- Crediting the pulled unit with 1,100 hours would add 400 hours of exposure that never occurred. The confidence bound is a function of failure opportunity; inflating T without exposure produces a bound that is not statistically valid and would not survive a supplier audit or a field failure. A censored unit contributes its actual on-test time, no more.
- Test-stand downtime from the cooling water loss is not unit operating time. The unit is not powered, not loaded, and not accumulating failure risk, so it adds no information about MTBF. Calendar time matters for schedule, not for the demonstration.
Legitimate ways to close it:
- Continue the test to 3,219 accumulated unit-hours with zero failures - the cheapest path given how close the test already is.
- Accept at the 76.5% confidence actually demonstrated, with a written deviation approved by the customer and reflected in warranty terms or price. That is a commercial decision, and it should be recorded as one rather than disguised as a statistical result.
- Renegotiate the plan itself - a lower MTBF target at 80% confidence, or a failure-terminated plan - before more data are taken, not after.
The 4% price reduction is not evidence of reliability and must not enter the technical decision. If it does, the plant buys a number it cannot defend.
3. A single failure occurs at 1,500 cumulative test hours before the test is complete. How does the evaluation change, and what should be done before deciding whether to continue?
Read the worked answer
One failure changes the plan materially. With r = 1 the time-terminated lower limit becomes theta_L = 2T / chi-square(1-alpha; 2r+2).
- chi-square(0.80; 4) = 5.989, so theta_L = 2 x 2,900 / 5.989 = 968 hours.
- To demonstrate 2,000 hours with one failure: T = theta_L x chi-square / 2 = 2,000 x 5.989 / 2 = 5,989 unit-hours, more than double the zero-failure plan.
This is the exam-relevant point: at low failure counts the demonstrated MTBF is extremely sensitive to a single event, so a zero-failure plan buys its efficiency by carrying high risk. The sensitivity, not just the arithmetic, should be understood before signing that kind of plan.
Before deciding, classify the failure.
- If it is an assignable cause - design defect, supplier lot, test fixture problem - suspend the test, correct and validate the fix, then restart the clock or switch to a reliability growth model. Continuing to accumulate hours on an uncorrected design measures the wrong configuration.
- If it is a chance failure consistent with a 2,000-hour MTBF, the honest conclusion is that the demonstration has failed. For reference, the probability of at least one failure in 2,900 hours when theta = 2,000 is about 76%, so a single failure is unremarkable and tells you very little about whether the supplier is bad - but it does tell you the contract requirement is not met.