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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 = 0.729 72.9% 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 * 0.75 1&2 fail, 3 operates =0.9 + 0.08 + 0.015 = 0.995 99.5% 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 * 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 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 Practice.95.9.75.80.95.9.95

13 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 Solutions

14 Practice.9.95

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 Practice.90.95.9.75.80.9.95

17 Simplify 0.9 * 0.95 = 0.855 0.8 * 0.75 = 0.6 0.9 * 0.95 * 0.9 = 0.7695.855.60.7695

18 Simplify 0.6 + 0.4 * 0.7695 = 0.6 + 0.3078 =0.9078.8550.9078 0.855 + 0.145 * 0.9078 = 0.855+0.13631 =0.986631

19 These 3 are in parallel 0.855 + 0.145 * 0.6 + 0.145*0.4*.7695 =0.855 + 0.087 + 0.044631 = 0.986631.855.60.7695

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 = 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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