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RELIABILITY

RELIABILITY. Dr. Ron Tibben-Lembke SCM 494. Reliability. Ability to perform its intended function under a prescribed set of conditions Probability product will function when activated Probability will function for a given length of time. Grace-o-vac. Measuring Probability.

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RELIABILITY

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  1. RELIABILITY Dr. Ron Tibben-Lembke SCM 494

  2. Reliability • Ability to perform its intended function under a prescribed set of conditions • Probability product will function when activated • Probability will function for a given length of time

  3. Grace-o-vac Measuring Probability • Depends on whether components are in series or in parallel • Series – one fails, everything fails

  4. Grace-o-vac Measuring Probability • Parallel: one fails, everything else keeps going

  5. Grace-o-vac Reliability • Light bulbs have 90% chance of working for 2 days. • System operates if at least one bulb is working • What is the probability system works?

  6. Grace-o-vac Reliability • Light bulbs have 90% chance of working for 2 days. • System operates if at least one bulb is working • What is the probability system works? Pr = 0.9 * 0.9 * 0.9 = 0.729 72.9% chance system works

  7. Grace-o-vac Parallel 90% 80% 75%

  8. Grace-o-vac Parallel • 0.9 prob. first bulb works • 0.1 * 0.8 First fails & 2 operates • 0.1 * 0.2 * 0.75 1&2 fail, 3 operates • =0.9 + 0.08 + 0.015 = 0.995 99.5% chance system works 90% 80% 75%

  9. Grace-o-vac Parallel – Different Order 80% 75% 90%

  10. Grace-o-vac Parallel • 0.8 prob. first bulb works • 0.2 * 0.75 First fails & 2 operates • 0.2 * 0.25 * 0.90 1&2 fail, 3 operates • =0.8 + 0.15 + 0.045 = 0.995 99.5% chance system works Same thing! 80% 75% 90%

  11. Grace-o-vac Parallel – All 3 90% • 0.9 prob. first bulb works • 0.1 * 0.9 First fails & 2 operates • 0.1 * 0.1 * 0.9 1&2 fail, 3 operates • =0.9 + 0.09 + 0.009 = 0.999 99.9% chance system works

  12. .95 .9 .75 .80 .95 .9 .95 Practice

  13. Solutions • 1: 0.95 * 0.95 • 2: Simplify • 0.8 * 0.75 = 0.6 and • 0.9 * 0.95 * 0.9 = 0.7695 • Then 0.6 + 0.4 * 0.7695 • = 0.6 + 0.3078 = 0.9078

  14. .9 .95 .95 .95 Practice

  15. Solution 2 • Simplify: • 0.9 * 0.95 = 0.855 • 0.95 * 0.95 = 0.9025 • Then • 0.855 + 0.145 * 0.9025 = • = 0.855 + 0.130863 = 0.985863

  16. .90 .95 .9 .75 .80 .9 .95 Practice

  17. .855 .60 .7695 Simplify • 0.9 * 0.95 = 0.855 • 0.8 * 0.75 = 0.6 • 0.9 * 0.95 * 0.9 = 0.7695

  18. .855 0.9078 Simplify • 0.6 + 0.4 * 0.7695 = 0.6 + 0.3078 =0.9078 • 0.855 + 0.145 * 0.9078 = 0.855+0.13631 =0.986631

  19. .855 .60 .7695 These 3 are in parallel • 0.855 + 0.145 * 0.6 + 0.145*0.4*.7695 =0.855 + 0.087 + 0.044631 = 0.986631

  20. Lifetime Failure Rate • 3 Distinct phases: Failure rate Infant Mortality Stability Wear-out time, T

  21. Exponential Distribution • MTBF = mean time between failures • Probability no failure before time T • Probability does not survive until time T • = 1- f(T) • e = 2.718281828459045235360287471352662497757

  22. Example • Product fails, on average, after 100 hours. • What is the probability it survives at least 250 hours? • T/MTBF = 250 / 100 = 2.5 • e^-T/MTBF = 0.0821 • Probability surviving 250 hrs = 0.0821 =8.21%

  23. Normally Distributed Lifetimes • Product failure due to wear-out may follow Normal Distribution

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