MFE634. Reliability Analysis. Group 5. Xiangyu Luo Ameya Palekar Krutin Parashuram Patil Chao Sa

Transcription

1 MFE634 Reliability Analysis Group 5 Xiangyu Luo Ameya Palekar Krutin Parashuram Patil Chao Sa

2 Contents. Generate Data and Verify.... Calculate and Analyze the Data.... Complete Sample % Confidence Interval for the Mean Time to Failure (MTTF) % Confidence Interval for the Failure Rate (FR) % the Confidence BOUNDS for MTTF and FR (Hint: Develop 80% CI)..... With Truncated at the 7th Failure % Confidence Interval for the Mean Time to Failure (MTTF) % Confidence Interval for the Failure Rate (FR) % Confidence BOUNDS for MTTF and FR (Hint: Develop 80% CI) With Truncated at time T= 0.5*MTTF % Confidence Interval for the Mean Time to Failure (MTTF) % Confidence Interval for the Failure Rate (FR) % Confidence BOUNDS for MTTF and FR (Hint: Develop 80% CI) Summary Implement A Fault Tree Analysis for Flood Barrier A Failure Mode and Effects Analysis for Flood Barrier... 8 Appendix A:... 9

3 Frequency. Generate Data and Verify Generate 5 exponentially distributed times to failure, with MTTF as defined by group number. Expon. MTTF =.5K + GNo*00 =.5K = 3000 Data Display Failure Sum of Failure Histogram of Failure 9 Sum of Failure = Mean of Failure 6 5 Mean of Failure = Standard Deviation of Failure Standard deviation of Failure = Failure Distribution ID Plot for Failure Goodness of Fit Test Distribution AD P Normal.668 <0.005 Exponential Weibull >0.50 Gamma Through the highest P-value and lowest AD value, we pick the Exponential distribution as the best fit for our data.

4 . Calculate and Analyze the Data. Complete Sample... 95% Confidence Interval for the Mean Time to Failure (MTTF) In this case, we have 5 samples and the confidence interval is 95%, and by adding all the test times of failures, we get total time is T = n = 5 α = T i i= = 7089 With these inputs, we can use the Chi-Squared distribution table to calculate the confidence interval in the experiment with the degree of freedom = n: Hence, the corresponding for the mean life is: ( T, χ α n, χ α n, = χ 50,0.975 = 7.40 χ α n, = χ 50,0.05 = T ) = ( , 478 ) = (990.79, ) χ n, α.. 95% Confidence Interval for the Failure Rate (FR) Failure Rate = Mean Life = ( , ) = ( , ) % the Confidence BOUNDS for MTTF and FR (Hint: Develop 80% CI) Often, we just need a lower or upper bound on reliability. Assume we are interested in a 90% reliability lower bound for this case and we know mission time = We re-

5 estimate the failure rate bound, for the error α = 0., for only one side. With Degree of Freedom = n, the new Chi-square value: χ n, α = χ 50,0.9 = Hence, the corresponding for the mean life and failure rate are: Mean = FR = T = = χ n, α Mean Life = = Which, in turn, allow us to calculate 90% Lower Bound for the reliability R of the device. R lower(t) = P{x T} = EXP { T θ } = EXP{ pt} = EXP{ } = With Truncated at the 7th Failure. Data Display.. 95% Confidence Interval for the Mean Time to Failure (MTTF) In this case, the experiment is truncated at the 7 th failure = We have 7 failures with known failure time. The confidence interval is 95%. By adding all the known test times of failures and assumed failure time for the last 8 samples, we get total time is 658. T = 7 T i i= k = 7 α = 0.05 The 7 th Failure (Sort smallest to largest) (5 7) = 658 3

6 With these inputs, we can use the Chi-Squared distribution table to calculate the confidence interval in the experiment with the degree of freedom = k: Hence, the corresponding for the mean life is: ( T, χ α k, χ α k, = χ 4,0.975 = 6.9 χ α k, = χ 4,0.05 = 5.69 T ) = ( , 3364 ) = (69.74, ) 5.69 χ k, α.. 95% Confidence Interval for the Failure Rate (FR) Failure Rate = Mean Life = ( , ) = ( , ) % Confidence BOUNDS for MTTF and FR (Hint: Develop 80% CI) In a 90% reliability lower bound with mission time = For the error α = 0., for only one side, with Degree of Freedom = k, the new Chi-square value: χ k, α = χ 4,0.9 =.064 Hence, the corresponding for the mean life and failure rate are: Mean = FR = T = = χ k, α Mean Life = = Which, in turn, allow us to calculate 90% Lower Bound for the reliability R of the device. R lower(t) = P{x T} = EXP { T θ } = EXP{ pt} = EXP{ } = 0.6 4

7 .3. With Truncated at time T= 0.5*MTTF. Data Display.3. 95% Confidence Interval for the Mean Time to Failure (MTTF) In this case, the experiment runs in a specific time = 0.5*3000 = 750, so we have 6 failures with known failure time. The confidence interval is 95%. We get total time is T = k = 6 α = 0.05 T = = 8750 With these inputs, we can use the Chi-Squared distribution table to calculate the confidence interval in the experiment with the degree of freedom = n+: Hence, the corresponding for the mean life is: ( T, χ α n+, χ α k+, = χ 4,0.975 = 6.9 χ α k+, = χ 4,0.05 = 5.69 T ) = ( , ) = ( , ) 5.69 χ n+, α.3. 95% Confidence Interval for the Failure Rate (FR) Failure Rate = 5 Mean Life Time=750 Failure (Sort smallest to largest) = ( , ) = ( , ) % Confidence BOUNDS for MTTF and FR (Hint: Develop 80% CI) In a 90% reliability lower bound with mission time = For the error α = 0., for only one side, with Degree of Freedom = k+, the new Chi-square value:

8 χ k+, α = χ 4,0.9 =.064 Hence, the corresponding for the mean life and failure rate are: Mean = FR = T = = χ k+, α Mean Life = = Which, in turn, allow us to calculate 90% Lower Bound for the reliability R of the device. R lower(t) = P{x T} = EXP { T θ } = EXP{ pt} = EXP{ } = Summary Collect the three sets of results in a table and compare. Complete Truncated at 7 th Truncated at 0.5 MTTF 95% CI for the MTTF (990.79, 4394) (69.74, ) ( ,666.53) 95% CI for the FR ( , ) ( , ) ( , ) 90% Lower Confidence Bounds for the MTTF & FR MTTF: FR: MTTF: FR: MTTF: FR: As we can see, the complete data has the most accurate 95% confidence interval and failure rate. The failure and time censored (truncated) data, which are both incomplete, have larger confidence interval and larger failure rate. 6

9 4. Implement 4. A Fault Tree Analysis for Flood Barrier Fail to Use Portable Flood Barriers Connective Parts Cracked Barrier Leaky Anchorage Corroded Bearing Heavier than Permitted Covering Face Broken Lack of Rustproof Paint Fail of Construction Fail of Design Fail of Construction Fail of Material 7

10 4. A Failure Mode and Effects Analysis for Flood Barrier Part / Design Parameter Portable Flood Barrier After-sale service Potential Failure Mode Connective part crack Barriers leaky Missing part Do not reach.5m height Service hotline is always busy Potential Failure Effects Barrier destroy Barrier can't be used Barrier can't be used Barrier can't prevent flood Customers disappointed SEV Potential Causes Part Corroded Damaged during delivery Fail to check Installed improperly Hotlines are few OCC Current Controls Protective paint Strength the package Offer backup parts Offer detailed Instruction Set more hotlines DET RPN Actions Recommended Redesign the structure of connective part Add protections during delivery Set up double-check mechanism 3 30 Training the QA department 9 70 Invest more money on After-sale apartment Severity (SEV) If SEV >= 8 Otherwise If SEV <= 3 Risk Priority Number (RPN) If RPN >= 300 Otherwise If RPN <= 00 8

11 Appendix A: MFE634 Group Take Home Final Exam Q-: S08 Group No: Date: Reliability Analysis Question: use the following Group Exp. MTTF: Group Expon. MTTF Group Expon. MTTF One,500 hours All others.5k+gno*00 Generate 5 exponentially distributed times to failure, with MTTF as defined above. Using as a Model the START sheets on Exponential Reliability and Censored Data Analyze such above defined complete sample, in the following manner: a) Obtain a 95% Confidence Interval for the Mean Time to Failure (MTTF) b) Obtain a 95% Confidence Interval for the Failure Rate (FR) c) Obtain 90% the Confidence BOUNDS for MTTF and FR (Hint: Develop 80% CI) Now, assume your sample has been Truncated at the 7 th failure. This means the test was suspended at that Failure Time. Thence, you will never know when the remaining units will fail. With such Failure Censored Data: a) Obtain a 95% Confidence Interval for the Mean Time to Failure (MTTF) b) Obtain a 95% Confidence Interval for the Failure Rate (FR) c) Obtain 90% Confidence BOUNDS for MTTF and FR (Hint: Develop 80% CI) Finally, assume your sample has been Truncated at time T= 0.5*MTTF. This means the test was suspended after that Truncation Time. Thence, you will never know when units still working after such time T, will fail. With such Time Censored Data: a) Obtain a 95% Confidence Interval for the Mean Time to Failure (MTTF) b) Obtain a 95% Confidence Interval for the Failure Rate (FR) c) Obtain 90% Confidence BOUNDS for MTTF and FR (Hint: Develop 80% CI) Compare the three sets of results in a table. Take Conclusions. Now, implement () a Fault Tree Analysis (FTA) and () a Failure Mode and Effects Analysis (FMEA) to an example of your Group Project Topic problem. Make it specific to your Topic problem issues. Use the Blueprints material in BB as a guide. This Take Home Part of the Final Exam is worth 0 Points (Companion, also 0 points). Each Group will deliver a printout/file of this Take Home at Entrance of the Final Exam. 9