8.
Assessing Product Reliability
8.4. Reliability Data Analysis 8.4.5. How do you fit system repair rate models?


This section covers estimating MTBF's and calculating upper and lower confidence bounds 
The HPP
or
exponential model
is widely used for two reasons:
For the HPP system model, as well as for the non repairable exponential population model, there is only one unknown parameter \(\lambda\) (or equivalently, the MTBF = 1/\(\lambda\). The method used for estimation is the same for the HPP model and for the exponential population model. 

The best estimate of the MTBF is just "Total Time" divided by "Total Failures"  The estimate of the MTBF is
$$ \widehat{\mbox{MTBF}} = \frac{\mbox{Total System(s) operation time}}{\mbox{Total number of failures}} $$
$$ \hat{\lambda} = \frac{1}{\widehat{\mbox{MTBF}}}
= \frac{\mbox{Total number of failures}}{\mbox{Total System(s) (or units) operation time}} $$
This estimate is the maximum likelihood estimate whether the data are censored or complete, or from a repairable system or a nonrepairable population. 

Confidence Interval Factors multiply the estimated MTBF to obtain lower and upper bounds on the true MTBF 
How To Use the MTBF Confidence Interval Factors


Confidence bound factor tables for 60, 80, 90 and 95 % confidence 


Formulas for confidence bound factors  even for "zero fails" case  Confidence bounds for the typical Type I censoring
situation are obtained from chisquare distribution tables or programs.
The formula for calculating confidence intervals is:
$$
P \left[
\frac{\mbox{MTBF } \cdot 2r}{\chi^2_{1\alpha/2, \, 2(r+1)}} \le \mbox { True MTBF }
\le \frac{\mbox{MTBF } \cdot 2r}{\chi^2_{\alpha/2, \, 2r}} \right] \ge 1 \alpha \, .$$
In this formula, \(\chi^2_{\alpha/2, \, 2r}\) is a value that the chisquare statistic with 2\(r\) degrees of freedom is less than with probability \(\alpha\)/2. In other words, the lefthand tail of the distribution has probability \(\alpha\)/2. An even simpler version of this formula can be written using \(T\) = the total unit test time: $$ P \left[ \frac{2T}{\chi^2_{1\alpha/2, \, 2(r+1)}} \le \mbox{ True MTBF } \le \frac{2T}{\chi^2_{\alpha/2, \, 2r}} \right] \ge 1 \alpha \, . $$ These bounds are exact for the case of one or more repairable systems on test for a fixed time. They are also exact when non repairable units are on test for a fixed time and failures are replaced with new units during the course of the test. For other situations, they are approximate. When there are zero failures during the test or operation time, only a (onesided) MTBF lower bound exists, and this is given by $$ \mbox{MTBF}_{\mbox{lower}} = \frac{T}{\mbox{ln } \alpha} \, . $$ The interpretation of this bound is the following: if the true MTBF were any lower than MTBF_{lower}, we would have seen at least one failure during \(T\) hours of test with probability at least 1\(\alpha\). Therefore, we are 100(1\(\alpha\)) % confident that the true MTBF is not lower than MTBF_{lower}. 

Calculation of confidence limits 
A onesided, lower 100(1\(\alpha\)/2) %
confidence bound for the MTBF is given by
$$ \mbox{LOWER } = \frac{2T}{G^{1}(1\alpha/2, \, 2(r+1))} \, , $$
where \(T\) is the total unit or system test time, \(r\)
is the total
number of failures, and \(G(q, \, \nu)\)
is the \(\chi^2\)
distribution function with shape parameter \(\nu\).
A onesided, upper 100(1\(\alpha\)/2) % confidence bound for the MTBF is given by $$ \mbox{UPPER } = \frac{2T}{G^{1}(\alpha/2, \, 2r)} \, . $$ The two intervals together, (LOWER, UPPER), are a 100(1\(\alpha\)) % twosided confidence interval for the true MTBF. Please use caution when using CDF and inverse CDF functions in commercial software because some functions require lefttail probabilities and others require righttail probabilities. In the lefttail case, \(\alpha\)/2 is used for the upper bound because 2\(T\) is being divided by the smaller percentile, and 1\(\alpha\)/2 is used for the lower bound because 2\(T\) is divided by the larger percentile. For the righttail case, 1\(\alpha\)/2 is used to compute the upper bound and \(\alpha\)/2 is used to compute the lower bound. Our formulas for \(G^{1}(q, \, \nu)\) assume the inverse CDF function requires lefttail probabilities. 

Example showing how to calculate confidence limits  A system was observed for two calendar months
of operation, during which time it was in operation for 800 hours and had
2 failures.
The MTBF estimate is 800/2 = 400 hours. A 90 %, twosided confidence interval is given by (400×0.3177, 400×5.6281) = (127, 2251). The same interval could have been obtained using $$ \begin{eqnarray} \mbox{LOWER } & = & \frac{1600}{G^{1}(0.95, \, 6)} \\ & & \\ \mbox{UPPER } & = & \frac{1600}{G^{1}(0.05, \, 4)} \, . \end{eqnarray} $$ Note that 127 is a 95 % lower limit for the true MTBF. The customer is usually only concerned with the lower limit and onesided lower limits are often used for statements of contractual requirements. 

Zero fails confidence limit calculation 
What could we have said if the system had no failures? For a 95 % lower
confidence limit on the true MTBF, we either use the 0 failures factor from
the 90 % confidence interval table and calculate 800 × 0.3338 = 267, or
we use \(T / \mbox{ ln } \alpha\) = 800/(ln 0.05) = 267.
The analyses in this section can can be implemented using both Dataplot code and R code. 