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

## Constant repair rate (HPP/exponential) model

This section covers estimating MTBF's and calculating upper and lower confidence bounds The HPP or exponential model is widely used for two reasons:
• Most systems spend most of their useful lifetimes operating in the flat constant repair rate portion of the bathtub curve
• It is easy to plan tests, estimate the MTBF and calculate confidence intervals when assuming the exponential model.
This section covers the following: Estimating the MTBF (or repair rate/failure rate)

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 non-repairable 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
1. Estimate the MTBF by the standard estimate (total unit test hours divided by total failures)
2. Pick a confidence level (i.e., pick 100(1-$$\alpha$$)). For 95 %, $$\alpha$$ = 0.05; for 90 %, $$\alpha$$ = 0.1; for 80 %, $$\alpha$$ = 0.2 and for 60 %, $$\alpha$$ = 0.4
3. Read off a lower and an upper factor from the confidence interval tables for the given confidence level and number of failures $$r$$
4. Multiply the MTBF estimate by the lower and upper factors to obtain MTBFlower and MTBFupper
5. When $$r$$ (the number of failures) = 0, multiply the total unit test hours by the "0 row" lower factor to obtain a 100(1-$$\alpha$$/2) % one-sided lower bound for the MTBF. There is no upper bound when $$r$$ = 0.
6. Use (MTBFlower, MTBFupper) as a 100(1-$$\alpha$$) % confidence interval for the MTBF $$\lambda$$ ($$r$$ > 0)
7. Use MTBFlower as a (one-sided) lower 100(1-$$\alpha$$/2) % limit for the MTBF
8. Use MTBFupper as a (one-sided) upper 100(1-$$\alpha$$/2) % limit for the MTBF
9. Use (1/MTBFupper, 1/MTBFlower) as a 100(1-$$\alpha$$) % confidence interval for $$\lambda$$
10. Use 1/MTBFupper as a (one-sided) lower 100(1-$$\alpha$$/2) % limit for $$\lambda$$
11. Use 1/MTBFlower as a (one-sided) upper 100(1-$$\alpha$$/2) % limit for$$\lambda$$
Tables of MTBF Confidence Interval Factors
Confidence bound factor tables for 60, 80, 90 and 95 % confidence
 60% 80% Num Fails r Lower for MTBF Upper for MTBF Lower for MTBF Upper for MTBF 0 0.6213 - 0.4343 - 1 0.3340 4.4814 0.2571 9.4912 2 0.4674 2.4260 0.3758 3.7607 3 0.5440 1.9543 0.4490 2.7222 4 0.5952 1.7416 0.5004 2.2926 5 0.6324 1.6184 0.5391 2.0554 6 0.6611 1.5370 0.5697 1.9036 7 0.6841 1.4788 0.5947 1.7974 8 0.7030 1.4347 0.6156 1.7182 9 0.7189 1.4000 0.6335 1.6567 10 0.7326 1.3719 0.6491 1.6074 11 0.7444 1.3485 0.6627 1.5668 12 0.7548 1.3288 0.6749 1.5327 13 0.7641 1.3118 0.6857 1.5036 14 0.7724 1.2970 0.6955 1.4784 15 0.7799 1.2840 0.7045 1.4564 20 0.8088 1.2367 0.7395 1.3769 25 0.8288 1.2063 0.7643 1.3267 30 0.8436 1.1848 0.7830 1.2915 35 0.8552 1.1687 0.7978 1.2652 40 0.8645 1.1560 0.8099 1.2446 45 0.8722 1.1456 0.8200 1.2280 50 0.8788 1.1371 0.8286 1.2142 75 0.9012 1.1090 0.8585 1.1694 100 0.9145 1.0929 0.8766 1.1439 500 0.9614 1.0401 0.9436 1.0603
 90% 95% Num Fails Lower for MTBF Upper for MTBF Lower for MTBF Upper for MTBF 0 0.3338 - 0.2711 - 1 0.2108 19.4958 0.1795 39.4978 2 0.3177 5.6281 0.2768 8.2573 3 0.3869 3.6689 0.3422 4.8491 4 0.4370 2.9276 0.3906 3.6702 5 0.4756 2.5379 0.4285 3.0798 6 0.5067 2.2962 0.4594 2.7249 7 0.5324 2.1307 0.4853 2.4872 8 0.5542 2.0096 0.5075 2.3163 9 0.5731 1.9168 0.5268 2.1869 10 0.5895 1.8432 0.5438 2.0853 11 0.6041 1.7831 0.5589 2.0032 12 0.6172 1.7330 0.5725 1.9353 13 0.6290 1.6906 0.5848 1.8781 14 0.6397 1.6541 0.5960 1.8291 15 0.6494 1.6223 0.6063 1.7867 20 0.6882 1.5089 0.6475 1.6371 25 0.7160 1.4383 0.6774 1.5452 30 0.7373 1.3893 0.7005 1.4822 35 0.7542 1.3529 0.7190 1.4357 40 0.7682 1.3247 0.7344 1.3997 45 0.7800 1.3020 0.7473 1.3710 50 0.7901 1.2832 0.7585 1.3473 75 0.8252 1.2226 0.7978 1.2714 100 0.8469 1.1885 0.8222 1.2290 500 0.9287 1.0781 0.9161 1.0938

Confidence Interval Equation and "Zero Fails" Case

Formulas for confidence bound factors - even for "zero fails" case Confidence bounds for the typical Type I censoring situation are obtained from chi-square 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 chi-square statistic with 2$$r$$ degrees of freedom is less than with probability $$\alpha$$/2. In other words, the left-hand 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 (one-sided) 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 MTBFlower, 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 MTBFlower.

Calculation of confidence limits A one-sided, 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 one-sided, 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-$$\alpha$$) % two-sided confidence interval for the true MTBF.

Please use caution when using CDF and inverse CDF functions in commercial software because some functions require left-tail probabilities and others require right-tail probabilities. In the left-tail 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 right-tail 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 left-tail probabilities.

Example

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 %, two-sided 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 one-sided 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.