Zero Acceptance Does Not Mean Zero Risk
A contact-retention feature on incoming connectors is inspected on receipt. Lots are 20,000 pieces and the current plan is a single sample of n = 50 with an acceptance number of zero. The supplier reports a process average of 0.5% nonconforming. A retention failure escaped to the customer last quarter, and the customer is demanding 95% confidence of detecting a lot containing 1% defective. Purchasing objects that sampling cost is already too high and that a zero-acceptance plan should be enough because it never accepts a bad lot. The feature is critical to function but not safety-critical; a full automated retention test costs $0.04 per piece.
1. For n = 50, c = 0, compute the probability of accepting a lot at 5%, 2% and 0.5% nonconforming. What protection does the plan actually give?
Read the worked answer
For a c = 0 plan, P(accept) = (1 − p)^n.
- p = 0.05: 0.95^50 = e^(50 × ln 0.95) = e^(−2.565) = 0.077 (7.7%)
- p = 0.02: 0.98^50 = e^(−1.010) = 0.364 (36.4%)
- p = 0.005: 0.995^50 = e^(−0.251) = 0.778 (77.8%)
Interpretation. The plan rejects 92% of lots at 5% defective, but it still accepts more than a third of lots at 2% defective. Purchasing's claim that it 'never accepts a bad lot' is arithmetically false - a c = 0 plan bounds the sample, not the lot. The plan is also harsh on a good supplier: roughly one in five lots at the 0.5% process average will be rejected, generating investigations, sorting costs and supplier disputes.
At the 0.5% process average and 20,000 pieces per lot, the supplier is sending about 100 defective connectors per lot regardless of the inspection outcome, so any escape rate has to be read against a detection probability near 22% per lot - most defective lots are being passed.
2. How large must n be, with c = 0, to give 95% confidence of rejecting a lot at 1% defective? What producer risk does that create at the supplier's 0.5% average?
Read the worked answer
Solve (1 − 0.01)^n = 0.05:
n = ln(0.05) / ln(0.99) = −2.996 / −0.01005 = 298.1 → 299 pieces
At that sample size, at the supplier's honest 0.5% process average:
P(accept) = 0.995^299 = e^(299 × −0.0050125) = e^(−1.499) = 0.223
So 77.7% of conforming lots would be rejected. That is economically impossible and would destroy the supplier relationship within a month.
The structural lesson: the discrimination required here is between 0.5% and 1.0% - only a factor of two. No zero-acceptance plan can give 95% detection at 1% while protecting a 0.5% process. Widening n to satisfy the customer simply shifts the damage onto the supplier. A plan with an acceptance number greater than zero, or a change in the control strategy, is required.
3. Recommend a sampling and control strategy, and show the economics that justify it.
Read the worked answer
Option A - 100% automated test. Cost is $0.04 × 20,000 = $800 per lot. That figure has to be compared with the cost of one escape (containment, sort, freight, customer chargeback, typically thousands of dollars), plus the internal cost of a 299-piece sample, the disputes and the sorting of rejected good lots. If a single escape costs more than the annualized testing cost, 100% test wins outright, and it removes the sampling argument entirely. For a critical-to-function attribute, a 100% automated test or a validated error-proofing device is the appropriate control; attribute sampling is a detection control with a known and nonzero escape rate.
Option B - if sampling must continue.
- Use an AQL-indexed plan with an acceptance number above zero (single normal inspection with switching rules to tightened and reduced), so that both parties' risks are quantified and written into the purchase order.
- Acknowledge the sampling risk in writing: state the AQL, the LTPD and the consumer risk at the LTPD, rather than promising detection the plan cannot deliver.
Option C - remove the need to inspect. Require the supplier to run a capability study on retention force and submit ongoing control-chart data; move to skip-lot or reduced inspection only after demonstrated capability and a clean audit trail. This is the only route that actually reduces the 0.5% average, which is the root of the problem.
Recommended: 100% automated test in the near term, with Option C running in parallel and used to justify removing the test only when the supplier's demonstrated defect rate and capability support it.