TecnoParts: supplier defects, imperfect testing and scrapping policy
TecnoParts buys identical microchips from two suppliers. Supplier A provides 60% of the chips and has a defective rate of 2%; supplier B provides 40% and has a defective rate of 5%. A new automatic tester flags a chip as suspicious. The tester detects a defective chip with probability 90%, but it also flags a good chip with probability 8%. The quality manager must decide what to do with flagged chips. Scrapping a flagged chip costs €2. Letting a defective chip reach production causes an average failure cost of €20; a good chip causes no extra cost. The tester itself costs €0.05 per chip. The manager proposes scrapping every flagged chip, arguing that because the tester is not perfect, the safest policy is to discard all of them.
1. Calculate the overall defective rate and the probability that a defective chip came from supplier B. Explain why this conditional probability is larger than B's share of supply.
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Let D be defective and B be supplier B. By the law of total probability, P(D) = P(D|A)P(A) + P(D|B)P(B) = 0.02(0.60) + 0.05(0.40) = 0.012 + 0.020 = 0.032. So 3.2% of all chips are defective. By Bayes' rule, P(B|D) = P(D|B)P(B)/P(D) = 0.05(0.40)/0.032 = 0.020/0.032 = 0.625. Thus 62.5% of defective chips come from B, even though B supplies only 40% of the chips. The reason is that B's defect rate, 5%, is more than twice A's 2%, so B is over-represented among defective chips.
2. A chip is flagged by the tester. Calculate the probability that it is actually defective. Interpret the result for the quality manager.
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Let F be 'flagged'. We are given P(F|D) = 0.90 and P(F|good) = 0.08. Since P(good) = 1 - 0.032 = 0.968, P(F) = 0.90(0.032) + 0.08(0.968) = 0.0288 + 0.07744 = 0.10624. Then P(D|F) = P(F|D)P(D)/P(F) = 0.0288/0.10624 ≈ 0.2711. Only about 27.1% of flagged chips are truly defective. The tester catches most defectives, but because good chips are far more numerous, most flags are false positives. A flag is therefore evidence of defectiveness, but it is not proof.
3. Using expected costs per chip, evaluate the manager's proposal to scrap every flagged chip. Include the tester cost and compare with the policy of using all chips without scrapping.
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If every flagged chip is scrapped, the expected scrap cost per chip is P(F) × €2 = 0.10624 × 2 = €0.21248. Adding the tester cost of €0.05 gives €0.26248 per chip. If no chips are scrapped, the expected failure cost per chip is P(D) × €20 = 0.032 × 20 = €0.64. Scrapping flagged chips therefore saves about €0.64 - €0.2625 = €0.3775 per chip on these figures, so the policy is economically justified. However, about 72.9% of scrapped chips are good, so a second test or a rework option could reduce waste. If the €2 scrap cost already includes losing the purchase value of the chip, the comparison above is appropriate; if not, the scrap cost should be increased and the conclusion rechecked.