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RELIABILITY Dr. Ron Lembke SCM 352. Reliability Ability to perform its intended function under a prescribed set of conditions Probability product will.

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Presentation on theme: "RELIABILITY Dr. Ron Lembke SCM 352. Reliability Ability to perform its intended function under a prescribed set of conditions Probability product will."— Presentation transcript:

1 RELIABILITY Dr. Ron Lembke SCM 352

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 Measuring Probability Depends on whether components are in series or in parallel Series – one fails, everything fails

4 Measuring Probability Parallel: one fails, everything else keeps going

5 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 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 = % chance system works

7 Parallel 90% 80%75%

8 Parallel 0.9 prob. first bulb works 0.1 * 0.8 First fails & 2 operates 0.1 * 0.2 * &2 fail, 3 operates = = % chance system works 90% 80%75%

9 Parallel – Different Order 80% 75%90%

10 Parallel 0.8 prob. first bulb works 0.2 * 0.75 First fails & 2 operates 0.2 * 0.25 * &2 fail, 3 operates = = % chance system works Same thing! 80% 75%90%

11 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 = = % chance system works

12 Practice

13 1: 0.95 * : Simplify 0.8 * 0.75 = 0.6 and 0.9 * 0.95 * 0.9 = Then * = = Solutions

14 Practice.9.95

15 Solution 2 Simplify: 0.9 * 0.95 = * 0.95 = Then * = = =

16 Practice

17 Simplify 0.9 * 0.95 = * 0.75 = * 0.95 * 0.9 =

18 Simplify * = = * = =

19 These 3 are in parallel * *0.4*.7695 = =

20 Lifetime Failure Rate 3 Distinct phases: Infant Mortality StabilityWear-out Failure rate 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 =

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 = Probability surviving 250 hrs = =8.21%

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


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