Chapter 1: LDA Case Studies
This chapter is dedicated to present some examples and case studies of using R4All platform to perform LDA analysis.
1.1 Case Study 1: Truck Manufacturer
A truck manufacturer is developing a new vehicle to be released next year. The company manufactured 20 of this new truck to be tested in real condition in the field. The objective of this test is to determine the reliability of several important components of this new vehicle.
In the table below, the test data referring to one of these components are presented: items marked with an “F” represent components that failed and items with an “S” represent components still in operation at the given time for this sample fleet (20 vehicles).
Time in kilometers (Km)
To analyze this data set, the LDA module was used. A correct data type has been defined, including time to event and status (failure or suspensions).
The next image shows the Data tab of the LDA module with the test data of the analyzed component and its respective reliability results.
The annual production of this vehicle is planned to be 120,000 units, and the warranty guaranteed for the first 100,000 Km. The manpower cost for a warranty return is $20 and for the new part is $30.
Next, some reliability plots for the component are presented:
As the plots and parameters indicate, this failure has a premature pattern, also known as “wear-in”. To determine the reliability for the warranty period (100,000 Km), the user can use the Reliability Calculator:
If the bounds are used and the annual production of vehicles and the total cost of replacement for this item are considered, the cost of warranty for this item is expected to range from $3,393,000 to $4,815,000.
1.2 Case Study 2: Gearbox Bearing
A gearbox bearing operates on run-to-failure mode and the company’s managers want to implement a planned maintenance on this asset. The registered times to failure are listed below:
The next image shows the Data tab of the LDA module with the test data of the analyzed component and its respective reliability results.
The calculator can be used to calculate the mean time to failure.
The calculator can be also used to calculate other results, such as reliability for the mean time to failure (the MTTF calculated previously).
Analyzing these results, we can conclude that if we use the MTTF as reliability reference to determine the preventive maintenance interval, we are assuming a reliability of 45.59%, that is, 54.41% probability of failure.
A better way to determine the optimum preventive maintenance interval is by taking in account a cost benefit approach.
Let´s then combine the reliability distribution of this item with the costs involved in its maintenance. The cost to perform a preventive maintenance for this item is $ 1,000 and the cost to perform a corrective maintenance is $ 5,000 (run to fail). Using the Optimum Replacement Interval equation, the preventive maintenance interval can be determined.
On the R4All platform, this equation can be calculated through the LDA Calculator or the LDA Plot tab.
Optimum Replacement Interval in the LDA Calculator:
Optimum Replacement Interval on the LDA Plot tab:
We can also determine the probability of failure for this item (2583 hours of operation).
Note that the risk to fail reduced dramatically (22.99% probability of failure) and at the same time we got the best results in terms of maintenance costs.
1.3 Case Study 3: Comparing Vendors Reliability
A company has two vendors of alumina refractory fire brick used in a production line. Due to changes in the purchasing department policies, the company intended to work with just one vendor. Then, they decided to apply the life data analysis over failure history of the refractory fire brick. The reliability engineer separated the data of each vendor, called Vendor A and Vendor B. The table below presents the failure history identifying each vendor.
The company’s manager decided they would pick the supplier with the greater reliability for 1500 days of operation. Each oven group needs to be analyzed separately, so in the R4All platform, two items were added: Vendor A and Vendor B.
In the image bellow, the life data analysis for Vendor A is presented.
In the image below, the life data analysis for Vendor B is presented.
Plotting the probability density function, we can visualize the reliability difference between Vendor A and B.
Based on the analysis, Vendor A is the best choice, as it has the lowest probability of failure in 1500 days of operation, as the plots show. Knowing that each set of refractory fire brick costs around $6,200, and that this company has around 300 ovens, the expected cost for each vendor can be estimated for the replacement time determined by the management.
In the image below, Vendor A probability of failure for 1500 hours of operation is presented.
Vendor A estimated replacement cost = 300 * 6200 * 0.210 = $390,600.
In the image below, Vendor B probability of failure for 1500 hours of operation is presented.
Vendor B estimated replacement cost: 300 * 6200 * 0.350 = $651,000.
1.4 Case Study 4: Pumps in Line Production
An assembly line operates using pumps in its production, in addition to other types of equipment. After an FMEA analysis of this plant, it was concluded that the failure mode “bomb break” was one of the most critical in relation to its occurrences and effects, which would cause the entire operation to stop. It was also observed that the “pump failure” was caused by the wear of its components. There is already a maintenance strategy to replace this pump every 6 months. This table shows the failure events recorded, in days:
During the data collecting period, 10 pumps were replaced during the preventive maintenance (6 months), and they were all still in good conditions.
The Probability of Failure and Reliability plots of these pumps were obtained and are shown below:
See below the probability of failure for the preventive maintenance period is calculated, considering that the year has 360 days:
The probability of failure for the 6 months (preventive maintenance period) is 54%. The company decided that this interval and its respective risk is too high. So, the reliability engineer proposed to diminish the risk down to just 15%, and then calculated the new preventive maintenance period for this pump.
The new preventive maintenance period to achieve 15% probability of failure is 120.9498 days. Rounding down to 120 days, the new preventive maintenance period for replacing the pump will run every 3 months.
1.5 Case Study 5: Measuring Instruments
A company has pachymeters as measuring instruments for its products. Currently, they gauge these instruments every 15 days of use. This checking period is based on an internal standard, which persisted for years. Trying to know the best length of this period, the company’s lab named a sample of 50 pachymeters to be monitored for a period and the obtained data are presented in the table below:
The next image presents the Data tab for this instrument. Note the Data Type includes grouped data, time to failure and suspension, and interval data.
Taking the current gauging time, of 15 days, the probability of failure is calculated using the LDA Calculator:
The company then decided they would make the gauging when the risk of the pachymeter out of specification is 2% (2% of probability of failure considering the pachymeter loses its standardization). So, the time for 2% of probability of failure results between 26 to 41 days, with the median value pointed at 33 days with 90% of two-sided confidence bounds.
1.6 Case Study 6: Component Failure
This table below contains the life data of a component from a truck model:
These data numbers are in kilometers. The truck manufacturers wish to know which the best distribution model is for this data set as well as the reliability and probability of failure for the warranty period (100,000 Km).
The best fit to treat these data is the Mixed Weibull with 3 subpopulations:
For this type of mathematical modeling, the curves and confidence bounds show different behaviors along the graphs plotting:
Each portion of the curve will have its own parameter settings:
The probability of failure for the warranty period of 100,000 Km:
1.7 Case Study 7: AC01 Equipment
The vendor of the equipment AC01 is competing on a bidding to provide the AC01 to a large, multinational company. Required in the bidding rules is an item referring to the “reliability” of the equipment. The documents requirements include the following information:
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Mean Time Between Failure.
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The curves for reliability, probability density function and failure rate.
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Probability of failure for a 2-year warranty period.
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Mean Time to Repair.
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The maintainability curve and the probability of repairing this equipment in 3 hours or less.
The following table represents the life data of the AC01 made by this company:
Data displayed in days.
The following table represents the maintenance data of the AC01 made by this company:
Data displayed in hours.
The best life distribution fitted to the data set was a Normal one:
Mean time between failures:
Reliability, probability density function and failure rate plots are shown in this order:
Now, the repair data were made in the same analysis; but, for that, we need to add a new item in our analysis:
Mean time to repair:
Maintainability Curve:
Probability of repairing the AC01 in 3 hours or less is 22.82% in the median value:
1.8 Case Study 8: Using One Parameter Distribution for a Small Sample Dataset
A company wants to determine the reliability of a new product, so they run a lab life test with just three samples. The life test was executed until all three items had failed.
See the Time to Failure in hours for the current sample (ACME132) below.
The next picture presents the life data analysis applying the Weibull 2P distribution.
Using the Probability of Failure plot (see below), we can see the confidence bounds range (90% - Two Sided) are very wide. This happen because the sample is too small (just three items).
However, the company already had analyzed a similar product in the past. So, the reliability consultant decided to use this previous knowledge to get a better result for this new product. He got the historical beta value from the analysis done in the past and by doing that, he could apply the Weibull with 1 parameter.
He chose the Weibull 1P and informed the value of the historical beta (beta = 3.5).
Using Weibull 1P distribution in Life Data Analysis module, the platform opens a pop-up so user can input the beta value.
The next picture show the Weibull parameter results where the beta is a fixed value informed by user and the Eta will be calculated based on the dataset.
Now when we see the Probability of Failure plot (see below) the confidence bounds range (90% - Two Sided) are a lot less wide.
1.9 Case Study 9: Item Presenting no Failures in Analysis Period
On several occasions, an industrial installation will contain some items which did not fail for a considered time. On these cases, if the company has some knowledge about the shape parameter of the item, the approach presented on this case study will allow to determine its reliability distribution. In order to do that, the Weibull 1 parameter will be used.
The table below presents 9 items with their respective operation time. As we can see all 9 items did not fail yet (just censored in the dataset).
The Reliability4All platform will not allow other distributions than Weibull 1P when using only censored data.
For the item was considered a shape parameter (beta) equal to 1. When using the Weibull 1P with zero failures in the dataset, it will be mandatory to inform the Confidence Level.
Inputting the Beta and Confidence Level the life distribution can be determined for a dataset with zero failures.
Inputting the Beta and Confidence Level the life distribution can be determined for a dataset with zero failures.
1.10 Case Study 10: Maintenance Strategy for a Filter
A petrochemical company decided to review their maintenance strategy for a specific type of filter applying the life data analysis. The table below presents this historical data in terms of failure (filters had lost their function).
First step was to determine the filter of reliability distribution using the LDA module. Using the RRX as analysis method to estimate the parameters, the distribution better fitted to the dataset was the Lognormal. In a practical perspective the Weibull 2 could be also used, due its Rho value (0.979) is close to the Lognormal Rho value (0.981).
The next picture shows the life data analysis results applying Lognormal distribution.
Once the filter reliability distribution is determined, the maintenance strategy calculation can be applied to determine the most cost-effective time to replace the filter. To perform this calculation was necessary to gather the cost to replace the filter during a planned maintenance and the filter replacement when it fails.
Follows the costs information:
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Planned Maintenance Cost: $ 2,500
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Unplanned Maintenance Cost: 5,113
Choosing the Optimum Replacement Interval Plot at the Plot tab the R4All platform will determine the optimal inspection time. This calculation can also be done at the “LDA Calculator”.
Results:
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Optimum Replacement Interval: 2377.90 hours
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Minimum Cost: $1.28 (cost per hour)
The filter can also be inspected for imminent failures, so the company decided to perform the Optimum Inspection Interval calculation. For this calculation was necessary additional information, including:
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Inspection Task Cost: $150
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Failure Detection: 0.9 (the fraction of the failure time for detecting an oncoming failure is 90%)
Choosing the Optimum Inspection Interval Plot at the Plot tab the R4All platform will determine the optimal inspection time.
Results:
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Optimum Inspection Interval: 306.01 hours
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Minimum Cost: $1.40 (cost per hour)
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This calculation can also be done at the “LDA Calculator”
Comparing the Optimum Replacement Interval and Optimum Inspection Interval results, the company decided to implement the planned maintenance strategy, due the minimum cost is lower:
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Planned Maintenance every 2400 hours of operation (rounding 2377.90) => Minimum Cost: $1.28 (cost per hour)
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Inspection Task every 310 hours of operation (rounding 306.01) => Minimum Cost: $1.40 (cost per hour)