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maintenanceStatistics4

Optimal Maintenance Decisions Inc.

Asset Reliability

Maintenance Statistics 4

Part 1 Part 2 Part 3 Part 5


Slide 1

Suspended data is handled by assigning an average order number to each failure time. If we tested five items to failure, the ordered failure times would have integer order numbers of one through five.

Now suppose that four items are placed on test with results as given in the table shown on the slide. The table shows that the first failure was at 84 hours. Then at 91 hours an item was taken off of test for reasons other than failure. (that is, it was suspended). Two more failures occurred at 122 and 274 hours.

The slide shows that the average order number for the second failure is 2.33. We use these average position values to calculate the median rank from Benard's formula (Slide 3. For example, the median rank for the second failure time (F2) would be:

Obviously, finding all the sequences for a mixture of several suspensions and failures and then calculatiing the averge order numbers is a time-consuming process. Fortunately a formula is available for calculating the order numbers. The formula produces what is termed a new increment. The new Increment I is given by:



Slide 2

Let's illustrate the use of the equation for the new increment. Referring to the table in the slide, although 10 items were originally placed on test, only six failures resulted.

The first two failure times would have order numbers 1 and 2 respectively. For the third failure a new increment must be calculated to be 1.29, as illustrated in the slide.

Adding I to the previous order number, 2, gives the order number of 3.29 to the third failure.

We continue with the same increment until another suspended item is encountered. Thus the order number for the fourth failure is 3.29+2.39=4.58. Applying the formula for the fifth failure gives:

The position for the fifth failure is 4.58+1.60=6.18, and the position for the sixth failure is 6.18+1.60=7.78. The median rank may then be used to plot the failures on Weibull paper.



Slide 3

Suppose that 5 components fail at times 67, 120, 130, 220, and 290 hours. To plot on Weibull paper we need the corresponding values of the CDF. That is we must know the fraction of the population failing prior to each sample. We cannot simply say that the percentage failed at time 120 hours is 2/5 because that would imply that the cumulative probability of failure at 290 hours, a random variable, is 100%. Such a small sample size does not justify such a blanket, definitive statement.

Extending this idea to the absurd, suppose the sample size is 1. We would not expect the age of this failure to represent the age by which 100% of items in the underlying population would fail. It would be more realistic to regard the single age at failure as representing the age by which 50% of the underlying population would fail. Benard’s formula gives (1-0.3)/(1+0.4)=50% which is intuitively reasonable.

Therefore we need a better way of estimating the CDF for plotting. The most popular approach of estimating the Y-axis plotting positions is the median rank. We obtain, from the median ranks table or from Benard’s approximation of the Median Ranks, the respective cumulative probabilities of failure which are .13, .31, .5, .69, and .87.

When we use reliability analysis software we do not have to look up the median ranks in tables. Not only does the program automatically adjust the orders due to suspensions, but it also calculates and applies the median ranks to each observation.



Slide 4

When the Weibull plot curves concave downwards, that’s an indication that the component may have a guaranteed minimum life below which it is physically impossible for it to fail. That means that the 3 Parameter Weibull model applies to this population. That is there is an additional parameter to estimate..

The procedure is to subtract the time of the first observation from all the observation times and re-plot.



Slide 5

That gives us a curve upwards rather than downwards. Therefore the location parameter (guarantee life parameter) is somewhere between 0 and 550.

Try γ=375.



Slide 6

Our guess of 375 for the location parameter was dead on. A rather tedious procedure, but it does get the job done. Obviously, this trial and error process along with manual plotting of data on Weibull paper, although instructive, has long since been superseded by reliability analysis software.

The next slide shows what the PDF would look like.



Slide 7

The PDF was generated with the Relcode software, giving us a Gamma, guarantee life, or location parameter value of 357.68.


Slide 8

We have plotted the points with CDF values according to the median rank.

The median rank is an estimate of the CDF for a sample size of 12 (we need to include the suspensions). It has been calculated such that the 50% of the time the true percentage of failures above and below this value. (The median of the rank distribution.)..

But we can also plot the CDF according to a 95% rank which is an estimate such that 95% of the time the percentage of failures will be below this value. Likewise for the 5% rank line 95% of the time the actual percentage of failures will be above this value..

Therefore, by plotting all three lines, we obtain a confidence interval which says that the cumulative probability of failure will be in the range from a to b with 90% confidence (that is 90% of the time).



Slide 9

In performing 2 and 3 parameter Weibull analysis, we have been creating models of reliability. They are predictors of reliability as a function of working age. These models are known as "age-reliability relationships".

We have been assuming that the equation relating reliability to working age respects the Weibull form. We have been using Weibull paper to estimate the parameters of the Weibull relationship that best represents or fits the data.

Now we want to know how good the fit is, so that we can decide whether to accept or reject the model.

Acceptance or rejection is based on a criterion (or significance level) that we have adopted. And the test we use is a goodness-of-fit test.




Slide 10

By following the procedure of the previous slide we have found that the maximum d is 0.261.

We proceed to the K-S critical value table on the next slide.




Slide 11

In statistics, a null hypothesis is a hypothesis that we set up to be nullified or refuted in order to support an alternative hypothesis.

The null hypothesis in this case is that "the model fits the data".

The alternative hypothesis is that "the model does not fit the data".

The test directly quantifies an error (called a "Type I error") that it may make:

Type I Error - Rejecting the null hypothesis when it is true.
The result of the K-S test is that the null hypothesis is:

"Not rejected at the 20% significance level."

Thus the test indirectly supports the hypothesis of a good fit because:

  • The K-S test (with a significance level of 20%) will accept a good model 80% of the time (and reject a good model 20% of the time).
A significance level is applied by some authority or standard governing the situation. Either 0.01, 0.05, 0.10, 0.15, or 0.20.

We apply our K-S statistic, 0.261 to the row for a sample size of 5. Assuming we wish to conform to a significance level of 0.20 we note that 0.261 is not greater than 0.446.

Hence the hypothesis of a "good" model is not rejected based on a 20% significance level.




Slide 12




Slide 13

. Which, plotted in Excel gives:



Slide 14



Slide 15



Slide 16



Slide 17

From Slide 39

We have an infant mortality kind of failure, due to one or more of substandard material quality, inadequate storage, or improper installation. Each of these possibilities should be investigated, and corrected as appropriate, in order to improve reliability.



Slide 18



Slide 19




Slide 20

In our manual Weibull plots we have been using eyeball fitting. Computer solutions require a numerical method. These are the two most well known and used.

The MLE method needs an iterative solution for the Beta estimate. Statisticians prefer maximum likelihood estimates to all other methods because MLE has excellent statistical properties. They recommend MLE as the primary method..

In contrast most engineers recommend the method of least squares method.

In general both methods should be used since each has advantages and disadvantages in different situations. MLE is more precise. On the other hand for small samples, it will be more biased than rank regression estimates.



Slide 21

Software should remove the tedium of analysis so that we may direct the maximum portion of human energy to understanding failure behavior. Following understanding we may select or design tasks that will mitigate the consequences of failure to an acceptable level.

With the exception of REWOP, all reliability analysis software requires that we transpose life data from the CMMS (computerized maintenance management system) to the software. This, it turns out, is an onerous task and the largest obstacle to using the techniques of reliabiltiy analysis.

In break-through technology, the REWOP methodology and software overcomes this difficulty by generating an "Events" table directly from the CMMS / work order database. We subsequently conduct analyses at will, with minimal data cleansing required. REWOP achieves its performance by linking work orders to the RCM knowledge base. Each siginficant work order becomes, thereby, an instance of a knowledge record. Using REWOP as a fully CMMS (SAP, Maximo, Datastream, Ellipse, etc) integrated module we analyze those instances.



Slide 22



Slide 23



Slide 24



Slide 25



Slide 26



Maintenance Statistics
Part 1 Part 2 Part 3 Part 5

Index, OMDEC Home



Created by: admin. Last Modification: Saturday 24 of November, 2007 11:10:06 MST by admin.

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