Derating a component to 50 percent of its operating value will generally decrease its failure rate by a factor of:

Question

Options

• AGreater than 50 percent.
• BGreater than 30 percent.
• CGreater than 10 percent but less than 30 percent.
• DLess than 10 percent.

• A
  7% (2)
• B
  78% (21)
• C
  4% (1)
• D
  11% (3)

Explanation

Derating a component to 50% of its rated operating value exploits the nonlinear relationship between stress and failure rate: failure rate doesn't drop proportionally with stress - it drops more steeply in the mid-stress range, which is why a 50% stress reduction typically yields a failure rate reduction greater than 30% (roughly in the 30–50% range), making B the best answer.

Why the distractors are wrong:

• A (>50%) overstates the general benefit - while certain specific components under voltage stress can achieve >50% reduction (due to high power-law exponents), as a general rule for a mixed component population, the reduction typically falls short of 50%
• C (10–30%) understates the effect - reliability engineering data (e.g., MIL-HDBK-217) consistently shows derating produces more than a modest 10–30% improvement, since the stress-failure curve is concave
• D (<10%) would make derating practically pointless - in reality it is one of the most cost-effective reliability techniques available

Memory tip: Use the "half-stress, third-less-failure" rule of thumb - derating to 50% buys you more than 30% (roughly a third) fewer failures. The improvement exceeds the intuitive "it should help by the same proportion" expectation because failure rate curves are nonlinear.

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