Test-Fix-Find-Test Data Example

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This example appears in the Reliability Growth and Repairable System Analysis Reference.


Consider the data given in the first table below. There were 56 total failures and [math]\displaystyle{ T=400\,\! }[/math]. The effectiveness factors of the unique BD modes are given in the second table. Determine the following:

1) Calculate the demonstrated MTBF and failure intensity.

2) Calculate the projected MTBF and failure intensity.

3) What is the rate at which unique BD modes are being generated during this test?

4) If the test continues for an additional 50 hours, what is the minimum number of new unique BD modes expected to be generated?

Test-Fix-Find-Test Data
[math]\displaystyle{ i\,\! }[/math] [math]\displaystyle{ {{X}_{i}}\,\! }[/math] Mode [math]\displaystyle{ i\,\! }[/math] [math]\displaystyle{ {{X}_{i}}\,\! }[/math] Mode
1 0.7 BC17 29 192.7 BD11
2 3.7 BC17 30 213 A
3 13.2 BC17 31 244.8 A
4 15 BD1 32 249 BD12
5 17.6 BC18 33 250.8 A
6 25.3 BD2 34 260.1 BD1
7 47.5 BD3 35 263.5 BD8
8 54 BD4 36 273.1 A
9 54.5 BC19 37 274.7 BD6
10 56.4 BD5 38 282.8 BC27
11 63.6 A 39 285 BD13
12 72.2 BD5 40 304 BD9
13 99.2 BC20 41 315.4 BD4
14 99.6 BD6 42 317.1 A
15 100.3 BD7 43 320.6 A
16 102.5 A 44 324.5 BD12
17 112 BD8 45 324.9 BD10
18 112.2 BC21 46 342 BD5
19 120.9 BD2 47 350.2 BD3
20 121.9 BC22 48 355.2 BC28
21 125.5 BD9 49 364.6 BD10
22 133.4 BD10 50 364.9 A
23 151 BC23 51 366.3 BD2
24 163 BC24 52 373 BD8
25 164.7 BD9 53 379.4 BD14
26 174.5 BC25 54 389 BD15
27 177.4 BD10 55 394.9 A
28 191.6 BC26 56 395.2 BD16
Effectiveness Factors for the Unique BD Modes
BD Mode EF [math]\displaystyle{ {{d}_{i}}\,\! }[/math]
1 .67
2 .72
3 .77
4 .77
5 .87
6 .92
7 .50
8 .85
9 .89
10 .74
11 .70
12 .63
13 .64
14 .72
15 .69
16 .46


Solution

1) In order to obtain [math]\displaystyle{ {{\widehat{\lambda }}_{CA}}\,\! }[/math], use the traditional Crow-AMSAA model for test-fix-test to fit all 56 data points, regardless of the failure mode classification to get:

[math]\displaystyle{ \begin{align} \widehat{\beta }= & 0.91026 \\ \widehat{\lambda }= & 0.23969 \end{align}\,\! }[/math]

Thus the achieved or demonstrated failure intensity is estimated by:

[math]\displaystyle{ \begin{align} {{\widehat{\lambda }}_{CA}} = & \widehat{\lambda }\widehat{\beta }{{T}^{\widehat{\beta }-1}} \\ = & 0.23969\times 0.91026\times {{400}^{(0.91026-1)}} \\ = & 0.12744 \end{align}\,\! }[/math]

The achieved or demonstrated MTBF, [math]\displaystyle{ {{M}_{CA}}\,\! }[/math], is the system reliability attained at the end of test, [math]\displaystyle{ T=400\,\! }[/math], and is estimated by:

[math]\displaystyle{ {{\widehat{M}}_{CA}}={{[{{\widehat{\lambda }}_{CA}}]}^{-1}}=7.84708\,\! }[/math]

2) For this data set, [math]\displaystyle{ M=16\,\! }[/math] and [math]\displaystyle{ T=400\,\! }[/math].

[math]\displaystyle{ {{\widehat{\lambda }}_{BD}}=\frac{{{N}_{BD}}}{T}=\frac{32}{400}=0.08\,\! }[/math]
[math]\displaystyle{ \overline{d}=\underset{i=1}{\overset{M}{\mathop \sum }}\,{{d}_{i}}/M=0.72125\,\! }[/math]
[math]\displaystyle{ \underset{i=1}{\overset{16}{\mathop \sum }}\,(1-{{d}_{i}}){{N}_{i}}/T=0.01955\,\! }[/math]

Calculate maximum likelihood estimates, [math]\displaystyle{ \hat{\beta }\,\! }[/math] and [math]\displaystyle{ \hat{\lambda }\,\! }[/math], of the BD modes:

[math]\displaystyle{ \begin{align} {{{\hat{\beta }}}_{BD}}= & 0.74715 \\ {{{\hat{\lambda }}}_{BD}} = & 0.18197 \end{align}\,\! }[/math]

Then:

[math]\displaystyle{ \overline{d}\widehat{h}(T|BD)=0.0215\,\! }[/math]

Therefore:

[math]\displaystyle{ \begin{align} {{\widehat{\lambda }}_{EM}} = & {{\widehat{\lambda }}_{CA}}-{{\widehat{\lambda }}_{BD}}+\underset{i=1}{\overset{K}{\mathop \sum }}\,(1-{{d}_{i}})\frac{{{N}_{i}}}{T}+\overline{d}\widehat{h}(T|BD) \\ = & 0.12744-0.08+0.0196+0.0215 \\ = & 0.08854 \end{align}\,\! }[/math]

The Crow Extended model projected MTBF is:

[math]\displaystyle{ \begin{align} {{\widehat{M}}_{EM}} = & {{[{{\widehat{\lambda }}_{EM}}]}^{-1}} \\ = & 11.29418 \end{align}\,\! }[/math]

Consequently, based on the Crow Extended model and the data shown in the tables above, the MTBF grew to 7.85 as a result of the corrective actions for the BC failure modes during the test. The MTBF then jumped to 11.29 after the test as a result of the delayed corrective actions for the BD failure modes. The management strategy can be summarized by the Failure Mode Strategy plot, as shown next.

Failure Mode Strategy plot.

This pie chart shows that 9.48% of the system's failure intensity has been left in (A modes), 31.81% of the failure intensity due to the BC modes has not been seen yet and 13.40% was removed during the test (BC modes - seen). In addition, 33.23% of the failure intensity due to the BD modes has not been seen yet, 3.37% will remain in the system since the corrective actions will not be completely effective at eliminating the identified failure modes, and 8.72% will be removed after the delayed corrective actions.

3) The rate at which unique BD modes are being generated is equal to [math]\displaystyle{ h{{(T|BD)}^{-1}}\,\! }[/math], where:

[math]\displaystyle{ \begin{align} h{{(T|BD)}^{-1}} = & \frac{1}{{{\widehat{\lambda }}_{BD}}{{\widehat{\beta }}_{BD}}{{T}^{{{\widehat{\beta }}_{BD}}-1}}} \\ = & \frac{T}{M{{\widehat{\beta }}_{BD}}} \\ = & 33.4605 \end{align}\,\! }[/math]

4) Unique BD modes are being generated every 33.4605 hours. If the test continues for another 50 hours, then at least one new unique BD mode would be expected to be seen from this additional testing.

As shown in the next figure, the MTBF of each individual failure mode can be plotted, and the failure modes with the lowest MTBF can be identified. These are the failure modes that cause the majority of the system failures.

Individual Mode MTBF chart.