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

8.4.5.1.

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:
  1. Estimating the MTBF (or repair rate/failure rate)
  2. How to use the MTBF confidence interval factors
  3. Tables of MTBF confidence interval factors 
  4. Confidence interval equation and "zero fails" case
  5. Calculation of confidence intervals
  6. Example
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
Confidence Interval Factors to Multiply MTBF Estimate

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 
Confidence Interval Factors to Multiply MTBF Estimate

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.

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