A manufacturer provides a warranty against failure of a carbon steel product within the first 30 days after sale. Out of 1000 sold, 10 were found to have failed by corrosion during the warranty period. Total cost of replacement for each failed product is approximately $100,000, including the cost of environ-mental clean-up, loss of product, downtime, repair, and replacement. (A) Calculate the risk of failure by corrosion, in dollars. (B) If a corrosion-resistant alloy would prevent failure by corrosion, is an incremental cost of $100 to manufacture the product using such an alloy justified? What would be the maximum incremental cost that would be justified in using an alloy that would prevent failures by corrosion?

Question

Laci Holmes

Physics

10 August, 02:50

A manufacturer provides a warranty against failure of a carbon steel product within the first 30 days after sale. Out of 1000 sold, 10 were found to have failed by corrosion during the warranty period. Total cost of replacement for each failed product is approximately $100,000, including the cost of environ-mental clean-up, loss of product, downtime, repair, and replacement. (A) Calculate the risk of failure by corrosion, in dollars. (B) If a corrosion-resistant alloy would prevent failure by corrosion, is an incremental cost of $100 to manufacture the product using such an alloy justified? What would be the maximum incremental cost that would be justified in using an alloy that would prevent failures by corrosion?

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Answers (1)

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Kiana Blevins · Answer 1

Answer:A) Risk (R) = $1000

B) There is justification for spending an additional cost of $100 to prevent a corrosion whose consequence in monetary terms is $1000

Explanation:R = Risk,

P=Probability of failure

C = Consequence of failure

Mathematically, R=P * C

10 out of 1000 carbon-steal products failed

Probability of failure = 10/1000 = 0.01

The consequence of failure by corrosion given in monetary term = $100,000

Risk of failure = 0.01 * $100,000

R=$1000