ITPEC IP April 2019 Question 67
A company wants to perform an acceptance inspection on an outsourced part with a non-defective probability of 0.9 and defective probability of 0.1. The table below shows four (4) proposed acceptance inspections with their respective cost per non-defective and defective part. Which of the following is the plan with the lowest expected cost?
Explanation
Plan C — its expected cost per part, 122, is the lowest of the four.
Expected cost = 0.9 × cost per non-defective part + 0.1 × cost per defective part:
- •A: 0.9 × 0 + 0.1 × 1500 = 0 + 150 = 150
- •B: 0.9 × 40 + 0.1 × 1000 = 36 + 100 = 136
- •C: 0.9 × 80 + 0.1 × 500 = 72 + 50 = 122
- •D: 0.9 × 120 + 0.1 × 200 = 108 + 20 = 128
Answer (c)
Why not others:
- •(a) A — 150: it spends nothing on good parts, but 1,500 on each defective part is the largest weighted term of any plan
- •(b) B — 136, a middle ground that is still 14 above plan C
- •(d) D — 128: it has the lowest cost per defective part, but paying 120 on each of the 90% good parts outweighs that saving
Key rule: Weight each outcome's cost by its probability before comparing; judging by one column alone (cheapest per good part gives A, cheapest per defective part gives D) ignores how often each case happens.
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