Predicting the Number of Failures for Repairable
Systems
[Editor's Note: This article has been updated
since its original publication to reflect a more recent
version of the software interface.]
Many complex systems, such as military vehicles and
mining equipment, are repairable. For repairable systems,
the number of failures at a given operation period is
one of the most important reliability metrics. Based on
the predicted number of failures, proper resources can
be allocated. ReliaSoft’s
RGA
(reliability growth analysis) software package provides
tools for repairable system modeling and prediction. In
this article, we will explain how to predict the number
of failures, with confidence bounds, for repairable systems.
Modeling
The model used in RGA is a power law
non-homogeneous Poisson process (NHPP) model. This model
assumes that the rate of occurrence of failure (ROCOF)
is a power function of time. For NHPP, the ROCOFs are
different at different time periods. The ROCOF for a
power law NHPP is:
 |
(1) |
where λ(t) is the ROCOF at time
t, and β and λ are the model
parameters.
From Eqn. (1), the expected number of failures from
time 0 to t is calculated by:
 |
(2) |
Therefore, the expected number of failures from time
t1 to t2 is:
 |
(3) |
where Δt = t2-t1.
For an NHPP, if a time interval is given, the number
of failures in this given interval follows a Poisson
process with parameter of (E(N(Δt)).
Defining m = (E(N(Δt)), the probability of
observing exactly i failures in this interval is:
 |
(4) |
Please note that Eqns. (2) and (3) calculate the
expected number of failures, while Eqn. (4) calculates the
probability of obtaining i failures for a Poisson
process with parameter of (E(N(Δt)). The
"expected number of
failures" and the "number of failures" are different--
the "number of failures" value is derived from a Poisson
process that uses the "expected number of failures"
value as a parameter.
Parameter Estimation
Once we have failure data, the model parameters and
their variance can be estimated using maximum likelihood
estimation (MLE). For details, please refer to the
Reliability Basics article in last month's issue of
HotWire. In this article, we will focus on calculating the
confidence bounds for the number of failures and for the
expected number of failures. An example is
used to illustrate the calculations.
Example
The following figure shows failure data for a repairable system.
Figure 1 - Failure data for a repairable system
The estimated β = 1.3009 and λ = 0.0052.
The variance and covariance matrix for these two
parameters can be obtained from a general spreadsheet,
as shown next.
Figure 2
- Variance/covariance matrix for model
parameters
Using the calculated results from RGA, we can
calculate the bounds for the expected number of failures.
Based on these calculated bounds, which are then
used as the parameter for the Poisson distribution, we
can further calculate the confidence bounds for the
number of failures. These calculations are not available
in RGA; all of the calculations are done in
Excel®.
Predicting the
Confidence Bounds for the Expected Number of Failures
Assume we want to calculate the expected number of
failures and its confidence bounds for the next 200
hours. The current time is 410 hours.
Applying Eqn. (3), the expected number of
failures is:
 |
(5) |
The variance of
 is:
 |
(6) |
Once we have obtained the value and variance of the
expected number of failures, we can calculate the bounds
using the following equation:
 |
(7) |
where Z1-α/2 is the 1-alpha/2 percentile of a standard
normal distribution. 1-alpha is the confidence level.
Here we use 90% two-sided confidence bounds.
For this example, the upper bound and the lower bound
for the expected number of failure are:
 |
(8) |
Many engineers mistakenly think that these two values
are the bounds for the number of failures in the next
200 hours. In fact, they are the bounds
for the expected number of failures. In other words,
they are the bounds for the mean value of the number of
failures. One may therefore ask how to obtain the bounds for the
number of failures. This is shown in the following
section.
Predicting the
Confidence Bounds for the Number of Failures
As we discussed in Eqn. (4), the number of failures is
a random number from a Poisson distribution with a parameter of
m(Δt). Using Eqn. (4) and
 = 8.8185 from Eqn. (5), we
can calculate the probability of observing i failures in
the interval time of Δt. The results are given in the next
table.
Table 1 - Predicting the number of
failures using the
estimated mean
Number of
Failures
= 8.8185 |
Probability of Failure |
Cumulative
Probability
of Failure |
| 0 |
0.00015 |
0.00015 |
| 1 |
0.0013 |
0.00145 |
| 2 |
0.00575 |
0.00721 |
|
3 |
0.01691 |
0.02412 |
| 4 |
0.03729 |
0.0614 |
| 5 |
0.06576 |
0.12717 |
| 6 |
0.09665 |
0.22382 |
| 7 |
0.12176 |
0.34558 |
| 8 |
0.13422 |
0.4798 |
| 9 |
0.13151 |
0.61131 |
| 10 |
0.11597 |
0.72728 |
| 11 |
0.09297 |
0.82026 |
| 12 |
0.06832 |
0.88858 |
| 13 |
0.04635 |
0.93493 |
|
14 |
0.02919 |
0.96412 |
| 15 |
0.01716 |
0.98128 |
| 16 |
0.00946 |
0.99074 |
| 17 |
0.00491 |
0.99565 |
| 18 |
0.0024 |
0.99805 |
| 19 |
0.00112 |
0.99917 |
| 20 |
0.00049 |
0.99966 |
| 21 |
0.00021 |
0.99987 |
| 22 |
0.00008 |
0.99995 |
| 23 |
0.00003 |
0.99998 |
| 24 |
0.00001 |
0.99999 |
The first column is the number of failures; the
second column is the probability of getting this number
of failures; the third column is the cumulative
probability of failure. For the 90% two-sided
confidence bounds for the number of failures, we can see
from the above table that the lower bound is 3
failures and the upper bound is 14 failures. Due to the
discrete nature of the Poisson random variable, the
lower bound is the largest integer whose cumulative
probability is less than 5%; the upper bound is the
smallest integer whose cumulative probability is greater
than 95%.
Table 1 uses
, the mean value of the expected number
of failures, for the calculation. However, as given in
Eqn. (6), there is uncertainty associated with the
estimation of the mean value. Eqn. (7) provides the upper
and lower bounds for the mean value of the expected
number of failures. To get more conservative results,
instead of using the mean value, we can use the upper
bound of the expected number of failures to calculate
the probability of failures. The calculation is similar
to the calculation in Table 1. In Table 2, we calculate
the probability of failures using
 .
Table 2 - Predicting the number of
failures using the
estimated upper bound
Number of Failures
= 18.61 |
Probability of Failure |
Cumulative Probability
of Failure |
|
0 |
0 |
0 |
|
1 |
0 |
0 |
|
2 |
0.000001 |
0.000002 |
|
3 |
0.000009 |
0.00001 |
|
4 |
0.000041 |
0.000052 |
|
5 |
0.000154 |
0.000206 |
|
6 |
0.000477 |
0.000682 |
|
7 |
0.001268 |
0.00195 |
|
8 |
0.00295 |
0.0049 |
|
9 |
0.0061 |
0.011001 |
|
10 |
0.011354 |
0.022354 |
|
11 |
0.01921 |
0.041564 |
|
12 |
0.029794 |
0.071359 |
|
13 |
0.042656 |
0.114014 |
|
14 |
0.056707 |
0.170721 |
|
15 |
0.070361 |
0.241082 |
|
16 |
0.081846 |
0.322928 |
|
17 |
0.089606 |
0.412533 |
|
18 |
0.092651 |
0.505184 |
|
19 |
0.090757 |
0.595942 |
|
20 |
0.084458 |
0.680399 |
|
21 |
0.074852 |
0.755251 |
|
22 |
0.063324 |
0.818576 |
|
23 |
0.051242 |
0.869818 |
|
24 |
0.039738 |
0.909556 |
|
25 |
0.029584 |
0.939139 |
|
26 |
0.021177 |
0.960316 |
|
27 |
0.014598 |
0.974914 |
|
28 |
0.009703 |
0.984617 |
|
29 |
0.006227 |
0.990844 |
|
30 |
0.003863 |
0.994708 |
Table 2 shows that the upper and lower bounds for
the number of failures are 26 and 11, respectively. The results in
Table 2 can be summarized as: the probability of getting
no more than 26 failures is 0.960316 with a confidence
level of 95% for the expected number of failures. This
is because
= 18.61 is the one-sided upper bound for the
expected number of failures with a confidence level of
95%.
Conclusion
This article explains the differences between the
confidence bounds for the expected number of failures
and the confidence bounds for the number of failures.
The number of failures is a Poisson random variable with
its parameters estimated from the data. Due to the
limited sample size, the estimated parameter, which is
the expected number of failures, has uncertainty
associated with it. To calculate the probability of
obtaining a certain number of failures, both the
uncertainty of the parameter and the uncertainty due to
the random Poisson process should be considered. Results
in Table 2 include both uncertainties.
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