MRA Trend Analysis

8/19/2019 MRA Trend Analysis

1/50


2/50


3/50

The next 2 slides cover some of the terminology you will encounter in this topic.

Stochastic means involving or subject to probabilistic behaviour. That is, the

behaviour cannot be predicted exactly. A stochastic model is contrasted to a

deterministic model where the relationships between the variables affecting the

system behaviour are known and defined.

Component failures represented as a point process tell us when on a time line the

failure events occurred and for a single component the time between the failures

gives us the times to failure for the component. To model these times with a

probability distribution the failures must be independent and identically distributed.


4/50

System failures represented as a point process give us when on the time line the

failure events occurred but tell us nothing about the time to failure of the individual

components causing the system failure.

If a system with a large number of components is in a steady state it can be modelled

using a homogeneous Poisson Process. Homogeneous in this context means that the

expected number of failures in any interval is constant and the distribution of the

time between failures is exponential.


5/50

The close conjunction of failures might be a chance phenomena but the

independence of failures occurring in quick succession needs be ascertained to gain a

better appreciation of the system or component reliability.

Independent means the failures are not related.

One failure does not contribute to the other or they do not have a common cause.

I have given some examples here of situations where failures may not be

independent.

GG960 Peter Gordon UOW


6/50

Identically distributed means that all components drawn from the population have

the same probability of surviving any given period. This must hold for components

concurrently in operation or components placed in operation sequentially over a

period of time.

I have listed some situations in which components lives may not be identically

distributed.

Note that although it is not helpful for the analysis we are doing in most situations

we would like components to become increasingly reliable. In fact as a maintenance

engineer you probably spend a lot of your time trying to establish positive trends in

component and system reliability.


7/50


8/50

This slide shows a diagrammatic representation of a stochastic point process for a

system that comprises 3 components. The failure of any of the components causes a

system failure.

If the component failures are independent renewal processes the point process for

the system is termed a superimposed renewal process.

The average rate of occurrence of failure (ROCOF) for each component is the inverse

of that component’s mean life. (MTTF).

The average rate of occurrence of failure for the system is the sum of the average

ROCOF of the components.


9/50


10/50

The data from the plot was derived from simulation of a system with 1000

components in series. The components fail by wear out mechanisms with the

individual component lives normally distributed.

The system is not preventively maintained. Components are replaced on failure. The

system commences operation with all components in new condition.

The plot shows that the ROCOF increases from zero to reach a constant average rate

after about 150 days.

The apparent initial deteriorating trend is characteristic of a superimposed renewal

system operated from new and is not indicative of reliability problems with the

system components.


11/50

This plot magnifies the apparent initial deteriorating trend from the previous plot by

showing the first 300 failures only.

The regression line of best fit shown is a power function of the operating time.

It can be seen more clearly here that the average ROCOF in this time period is not

constant. (The line of best fit would have to be a straight line)

The function for the ROCOF is obtained by differentiating the power function for the

cumulative number of failures.

The failure arrivals form a non-homogeneous Poisson process and with this form of

ROCOF (intensity) function the process is often referred to as a power law process.

If there was no trend we would expect the average ROCOF to be constant. I.e. the

power index would be 1.

Clearly in fitting a power function to the plotted data we will never get an index of

exactly 1.

To test whether the non-homogeneous Poisson process is in fact the mostappropriate model we can use the AMSAA model.

The application of this trend test is described in Ebeling (Ch 16.5)


12/50


13/50

Note using the MLE for b gives a value of b = 3.29

This does not change the conclusion.


14/50


15/50

This is the similar to the diagram given in O’Connor .

It shows a point process for a single component.

The variable capital Xi gives the individual times to failure.

The variable lower case xi gives the cumulative time to each failure from the start of

observation.

In this example observation continues beyond the last observed failure. The

observation interval is designated x0.

The test compares the mean of the cumulative times to failure for the sample data

with the mean that would be obtained if the times to failure were drawn from the

same population. (identically distributed)



16/50

If there is no trend the average of the cumulative times to failure will equal half the

period of observation and the Std Dev will be given by the period of observation

divided by the square root of 1 upon 12n where n is the number of failures.

The statistic U is a standard normal variable.*

If U is positive it means that more failures were observed in the second half of the

period of observation and the times to failure are decreasing . This would be a bad

trend. Ie it could indicate the component is becoming less reliable.

If U is negative it means that more failures were observed in the first half of the

period of observation and the times to failure are increasing = good trend.

However the indicated trend may not be statistically significant.

Even in the situation where there is no trend the average of the cumulative times to

failure for any sample will show variation from the population mean.

We know from the normal distribution that there is a 15.87% chance that the variable

(in our case the average of the cumulative times to failure) will have a value that ismore than 1 standard deviation from the mean. (U>1) This means that if we

concluded from the fact that U = 1 that the data was trending we would be making an

error 15.87% of the time.

*We know from the Central Limit Theorem that if Sn is the sum of n mutually

independent random variables, then the distribution function of Sn is well-

approximated by the normal distribution



17/50

The probability we are making an error in rejecting the null hypothesis is termed the

level of significance.

In our case the null hypothesis is that the data is not trending.

15.87% would normally be considered too large for a level of significance.

A value of U of 1.645 or -1.645 gives a 5% level of significance. In most cases this

would be an acceptable level at which to reject the null hypothesis.



18/50

If the period of observation is stopped at a failure the statistic is calculated in the

manner shown.



19/50

The test was shown at the 5% significance level (U>1.645)

The probability of getting a value of the test statistic of 15.89 is however effectively

zero.


20/50

In a fleet situation we may record the times between failures for each individual

system but if our approach to managing and maintaining each system is the same we

will be interested in the overall picture of trends for the fleet.

In this situation we can use a Laplace test for the pooled data.


21/50


22/50

This method uses the data from all the samples to calculate the average number of

repairs that could be expected to have occurred by a given mileage. This plot is

described as the Nelson-Aalen plot in most references and the mean cumulative

function is also referred to as the cumulative intensity function. If the slope of the

plot is increasing with age the population rate of occurrence of failure is increasing.

(Sad trend)


23/50

O’Connor explains that “exploratory data analysis is a simple graphical technique for

searching for connections between time series data and explanatory factors. In the

reliability context, the failure data are plotted on a time line, along with other

information. For example, overhaul intervals, seasonal changes, or different operating

patterns can be shown on the chart.” This diagram from O’Connor depicts point

processes for 6 sub-systems and a superimposed point process for the overall system.

The system is overhauled every 1000 hours and from visual examination you can see

failures clustered after each overhaul indicating the overhaul is actually adversely

affecting reliability and also failures paired in a number of places for the sub-systems.

This could indicate that the failures are not independent of each other. A pattern like

this would warrant further investigation of the nature and cause of the failures.


24/50

This diagram is reproduced from O’Connor’s text. O’Connor shows the value of the

Laplace statistic for each of the parts comprising the system and also for the

superimposed system at the bottom.

You are invited to determine whether the data-sets are trending from visual

examination of the point process for each component and for the superimposed

process for the system. Compare this with O’Connor’s assessment based on the U

value.


25/50


26/50


27/50

This is a diagrammatic representation of a stochastic point process for a pump that is

replaced on failure. The pump has been replaced 5 times since first installation. The

crosses indicate the times the failures occur measured from the time of the original

installation. The times to failure are shown in the table in chronological sequence.

In this type of situation it could take many years to gather failure data for a

component


28/50


29/50


30/50


31/50

Although you have the answer I suggest you attempt this calculation yourself.

Sample avg of cumulative times to failure ∑n-1x / n-1 = 358Mean = x0/2 = 798/2 = 399

SD = x0 √(1/(12(n-1)) = 798 x √(1/(12x11) = 69.46

Cv = −

2

−1/ = 0.216

We would reject the null hypothesis and conclude the data was trending. The level of

significance is 0.29%. This means that if we conclude the data is trending we will be

wrong 0.29% of the time. This is well within the 5% level of significance normally

used to test such an hypothesis.

3


32/50


33/50


34/50


35/50

Another situation that you may encounter quite frequently is where a number of

identical components are in operation concurrently.

In this situation more renewal data will be available but we must be confident that

the loads on the components are similar.

3


36/50

A fleet situation is similar to the previous case of many components in one system.

In a fleet we are looking at one or more identical components in concurrent

operation in many systems

This type of situation could generate lots of renewal records.

3


37/50 3


38/50

When we test for trend we are testing to see whether the times to failure are

dependent on the order in which they occur.

We test for trend when the data is in chronological sequence.

If we have a point process for a single component the sequence defined by

installation date is the same as that defined by replacement date.

If however we have “identical” components sourced from the same supplier that are

operating concurrently we get a different chronological sequence defined by the

installation dates from that defined by the replacement dates.

These are shown on the next slide.

3


39/50

If we sequence the ages at renewal based on the installation date we are in effect

looking to see whether the times to failure are dependent on the date the

components were supplied or installed. There are a number of things that occur in

supply, repair or installation of a component that could affect the life that it achieves.

If these factors are random the mean life will not be affected. Variability in repair and

installation processes is reflected in the variance of the times to failure. If over a

period of time however, there are systemic changes to methods or standards related

to supply, repair or installation the mean life is likely to show an increasing or

decreasing trend.

It is not so obvious how the times to failure could be dependent on the date the

components were replaced as this is after the event. In the example above the times

to failure are reasonably tightly banded and consequently the sequences are very

nearly the same.

In grouping data in this manner for multiple single component renewal processes weare essentially analysing the data as a single sequence originating from a supplier or a

repairer. The linear regression trend test can be used to test for trend in the sequence

and also the Lewis-Robinson test can be applied to the combined sequence as if it

were a single process.

3


40/50

The P-Value is half the value shown in the excel table for a 2 tailed t-test. (0.37)

4


41/50 4


42/50

The LR test confirms that the apparent deteriorating trend is not statistically

significant. This result is very close to that from the linear regression trend test (P-

value = 18.5%)

Note the P-value is the lowest level of significance at which the observed value of the

test statistic is significant.

4


43/50 4


44/50

From the statistical test the evidence of trend is borderline with a significance level

in rejecting the null hypothesis of 4.99%

4


45/50


46/50 4


47/50


48/50


49/50 4


50/50

MRA Trend Analysis

Documents

Transcript of MRA Trend Analysis