Crow Extended Model Examples

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These examples appear in the Reliability Growth and Repairable System Analysis Reference book.

Parameter Estimation

Test-Find-Test Data

Consider the data in the first table below. A system was tested for [math]\displaystyle{ T=400\,\! }[/math] hours. There were a total of [math]\displaystyle{ N=42\,\! }[/math] failures and all corrective actions will be delayed until after the end of the 400 hour test. Each failure has been designated as either an A failure mode (the cause will not receive a corrective action) or a BD mode (the cause will receive a corrective action). There are [math]\displaystyle{ {{N}_{A}}=10\,\! }[/math] A mode failures and [math]\displaystyle{ {{N}_{BD}}=32\,\! }[/math] BD mode failures. In addition, there are [math]\displaystyle{ M=16\,\! }[/math] distinct BD failure modes, which means 16 distinct corrective actions will be incorporated into the system at the end of test. The total number of failures for the [math]\displaystyle{ {{j}^{th}}\,\! }[/math] observed distinct BD mode is denoted by [math]\displaystyle{ {{N}_{j}}\,\! }[/math], and the total number of BD failures during the test is [math]\displaystyle{ {{N}_{BD}}=\underset{j=1}{\overset{M}{\mathop{\sum }}}\,{{N}_{j}}\,\! }[/math]. These values and effectiveness factors are given in the second table

Do the following:

  1. Determine the projected MTBF and failure intensity.
  2. Determine the growth potential MTBF and failure intensity.
  3. Determine the demonstrated MTBF and failure intensity.
Test-Find-Test Data
[math]\displaystyle{ i\,\! }[/math] [math]\displaystyle{ {{X}_{i}}\,\! }[/math] Mode [math]\displaystyle{ i\,\! }[/math] [math]\displaystyle{ {{X}_{i}}\,\! }[/math] Mode
1 15 BD1 22 260.1 BD1
2 25.3 BD2 23 263.5 BD8
3 47.5 BD3 24 273.1 A
4 54 BD4 25 274.7 BD6
5 56.4 BD5 26 285 BD13
6 63.6 A 27 304 BD9
7 72.2 BD5 28 315.4 BD4
8 99.6 BD6 29 317.1 A
9 100.3 BD7 30 320.6 A
10 102.5 A 31 324.5 BD12
11 112 BD8 32 324.9 BD10
12 120.9 BD2 33 342 BD5
13 125.5 BD9 34 350.2 BD3
14 133.4 BD10 35 364.6 BD10
15 164.7 BD9 36 364.9 A
16 177.4 BD10 37 366.3 BD2
17 192.7 BD11 38 373 BD8
18 213 A 39 379.4 BD14
19 244.8 A 40 389 BD15
20 249 BD12 41 394.9 A
21 250.8 A 42 395.2 BD16

 

Effectiveness Factors for the Unique BD Modes
BD Mode Number [math]\displaystyle{ {{N}_{j}}\,\! }[/math] First Occurrence EF [math]\displaystyle{ {{d}_{i}}\,\! }[/math]
1 2 15.0 .67
2 3 25.3 .72
3 2 47.5 .77
4 2 54.0 .77
5 3 54.0 .87
6 2 99.6 .92
7 1 100.3 .50
8 3 112.0 .85
9 3 125.5 .89
10 4 133.4 .74
11 1 192.7 .70
12 2 249.0 .63
13 1 285.0 .64
14 1 379.4 .72
15 1 389.0 .69
16 1 395.2 .46

Solution

  1. The maximum likelihood estimates of [math]\displaystyle{ {{\beta }_{BD}}\,\! }[/math] and [math]\displaystyle{ {{\lambda }_{BD}}\,\! }[/math] are determined to be:
    [math]\displaystyle{ \begin{align} {{{\hat{\beta }}}_{BD}} = & \frac{M}{\underset{i=1}{\overset{M}{\mathop{\sum }}}\,\ln (\tfrac{T}{{{X}_{i}}})} \\ = & 0.7970 \\ {{{\hat{\lambda }}}_{BD}} = & 0.1350 \end{align}\,\! }[/math]
    The unbiased estimate of [math]\displaystyle{ \beta \,\! }[/math] is:
    [math]\displaystyle{ \begin{align} {{\overline{\beta }}_{BD}} = & \frac{M-1}{M}{{{\hat{\beta }}}_{BD}} \\ = & 0.7472 \end{align}\,\! }[/math]
    Based on the test data, [math]\displaystyle{ \overline{d}=\tfrac{1}{M}\underset{i=1}{\overset{M}{\mathop{\sum }}}\,{{d}_{i}}= 0.72125\,\! }[/math]. Therefore, [math]\displaystyle{ B(T)=\overline{d}\tfrac{M{{\overline{\beta }}_{BD}}}{T}=0.0215\,\! }[/math]. The projected failure intensity due to incorporating the 16 corrective actions is:
    [math]\displaystyle{ \begin{align} r(T) = & \left( \frac{{{N}_{A}}}{T}+\underset{i=1}{\overset{M}{\mathop \sum }}\,(1-{{d}_{i}})\frac{{{N}_{i}}}{T} \right)+\overline{d}\left( \frac{M}{T}{{\overline{\beta }}_{BD}} \right) \\ = & 0.0661 \end{align}\,\! }[/math]
    The projected MTBF is:
    [math]\displaystyle{ M\widehat{T}B{{F}_{P}}={{[r(T)]}^{-1}}=15.127\,\! }[/math]
  2. To estimate the maximum reliability that can be attained with this management strategy, use the following calculations.
    [math]\displaystyle{ \begin{align} {{N}_{A}}/T=0.0250 \end{align}\,\! }[/math]
    [math]\displaystyle{ \frac{1}{T}\underset{i=1}{\overset{16}{\mathop \sum }}\,(1-{{d}_{i}}){{N}_{i}}=0.0196\,\! }[/math]
    The growth potential failure intensity is estimated by:
    [math]\displaystyle{ \begin{align} {{\widehat{r}}_{GP}}(T) = & \left( \frac{{{N}_{A}}}{T}+\underset{i=1}{\overset{M}{\mathop \sum }}\,(1-{{d}_{i}})\frac{{{N}_{i}}}{T} \right) \\ = & 0.0250+0.0196 \\ = & 0.0446 \end{align}\,\! }[/math]
    The growth potential MTBF is:
    [math]\displaystyle{ M\widehat{T}B{{F}_{GP}}={{[{{\widehat{r}}_{GP}}]}^{-1}}=22.4467\,\! }[/math]
  3. The demonstrated failure intensity and MTBF are estimated by:
    [math]\displaystyle{ \begin{align} {{\widehat{\lambda }}_{D}}(T) = & \frac{{{N}_{A}}+{{N}_{BD}}}{T} \\ = & \frac{42}{400} \\ = & 0.1050 \end{align}\,\! }[/math]
    [math]\displaystyle{ \begin{align} M\widehat{T}B{{F}_{D}} = & {{[{{\widehat{\lambda }}_{D}}(T)]}^{-1}} \\ = & 9.5238 \end{align}\,\! }[/math]
    The first chart below shows the demonstrated, projected and growth potential MTBF. The second shows the demonstrated, projected and growth potential failure intensity.
    Rga9.3.png
    Rga9.4.png


Confidence Bounds

Calculate the 2-sided 90% confidence bounds on the demonstrated, projected and growth potential failure intensity for the Test-Find-Test data given above.

Solution

The estimated demonstrated failure intensity is [math]\displaystyle{ {{\widehat{\lambda }}_{D}}(T)=\tfrac{{{N}_{A}}+{{N}_{B}}}{T}=0.1050\,\! }[/math]. Based on this value, the Fisher Matrix confidence bounds for the demonstrated failure intensity at the 90% confidence level are:

[math]\displaystyle{ \begin{align} {{[{{\lambda }_{D}}(T)]}_{L}} = & {{{\hat{\lambda }}}_{D}}(T)+\frac{{{C}^{2}}}{2}-\sqrt{{{{\hat{\lambda }}}_{D}}(T){{C}^{2}}+\frac{{{C}^{4}}}{4}} \\ = & 0.08152 \end{align}\,\! }[/math]
[math]\displaystyle{ \begin{align} {{[{{\lambda }_{D}}(T)]}_{U}} = & {{{\hat{\lambda }}}_{D}}(T)+\frac{{{C}^{2}}}{2}+\sqrt{{{{\hat{\lambda }}}_{D}}(T){{C}^{2}}+\frac{{{C}^{4}}}{4}} \\ = & 0.13525 \end{align}\,\! }[/math]

The Crow confidence bounds for the demonstrated failure intensity at the 90% confidence level are:

[math]\displaystyle{ \begin{align} {{[{{\lambda }_{D}}(T)]}_{L}} = & {{\widehat{\lambda }}_{D}}(T)\frac{\chi _{(2N,1-\alpha /2)}^{2}}{2N} \\ = & 0.07985 \\ {{[{{\lambda }_{D}}(T)]}_{U}} = & {{\widehat{\lambda }}_{D}}(T)\frac{\chi _{(2N,\alpha /2)}^{2}}{2N} \\ = & 0.13299 \end{align}\,\! }[/math]

The projected failure intensity is:

[math]\displaystyle{ \begin{align} \hat{\lambda_{p}} &= \frac{N_{i}}{T}+\sum_{i=1}^{M}(1-d_{i})\frac{N}{T}+\overline{d}\left(\frac{M}{T}\overline{\beta} \right )\\ &= 0.06611 \end{align} }[/math]

Based on this value, the Fisher Matrix confidence bounds at the 90% confidence level for the projected failure intensity are:

[math]\displaystyle{ \begin{align} {{[{{{\hat{\lambda }}}_{P}}(T)]}_{L}} = & {{{\hat{\lambda }}}_{P}}(T){{e}^{{{z}_{\alpha }}\sqrt{Var({{{\hat{\lambda }}}_{P}}(T))}/{{{\hat{\lambda }}}_{P}}(T)}} \\ = & 0.04902 \end{align}\,\! }[/math]
[math]\displaystyle{ \begin{align} {{[{{{\hat{\lambda }}}_{P}}(T)]}_{U}} = & {{{\hat{\lambda }}}_{P}}(T){{e}^{-{{z}_{\alpha }}\sqrt{Var({{{\hat{\lambda }}}_{P}}(T))}/{{{\hat{\lambda }}}_{P}}(T)}} \\ = & 0.08915 \end{align}\,\! }[/math]

The Crow confidence bounds for the projected failure intensity are:

[math]\displaystyle{ \begin{align} {{[{{\lambda }_{P}}(T)]}_{L}} = & {{{\hat{\lambda }}}_{P}}(T)+\frac{{{C}^{2}}}{2}-\sqrt{{{{\hat{\lambda }}}_{P}}(T)\cdot {{C}^{2}}+\frac{{{C}^{4}}}{4}} \\ = & 0.04807 \\ {{[{{\lambda }_{P}}(T)]}_{U}} = & {{{\hat{\lambda }}}_{P}}(T)+\frac{{{C}^{2}}}{2}+\sqrt{{{{\hat{\lambda }}}_{P}}(T)\cdot \ \,{{C}^{2}}+\frac{{{C}^{4}}}{4}} \\ = & 0.09090 \end{align}\,\! }[/math]

The growth potential failure intensity is:

[math]\displaystyle{ \widehat{r}_{GP} (T) = \left (\frac{N_A}{T} + \sum_{i=1}^M (1-d_i) \tfrac{N_i}{T} \right ) = 0.04455 \,\! }[/math].

Based on this value, the Fisher Matrix and Crow confidence bounds at the 90% confidence level for the growth potential failure intensity are:

[math]\displaystyle{ \begin{align} {{r}_{L}} = & {{{\hat{r}}}_{GP}}+\frac{{{C}^{2}}}{2}-\sqrt{{{{\hat{r}}}_{GP}}{{C}^{2}}+\frac{{{C}^{4}}}{4}} \\ = & 0.03020 \\ {{r}_{U}} = & {{{\hat{r}}}_{GP}}+\frac{{{C}^{2}}}{2}+\sqrt{{{{\hat{r}}}_{GP}}{{C}^{2}}+\frac{{{C}^{4}}}{4}} \\ = & 0.0656 \end{align}\,\! }[/math]

The figure below shows the Fisher Matrix confidence bounds at the 90% confidence level for the demonstrated, projected and growth potential failure intensity.

Rga9.8.png

The following figure shows these bounds based on the Crow method.

Rga9.9.png


Test-Fix-Find-Test Data

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.
    Rga9.6.png

    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.
    Rga9.7.png


Confidence Bounds

Calculate the 2-sided confidence bounds at the 90% confidence level on the demonstrated, projected and growth potential MTBF for the Test-Fix-Find-Test data given above.

Solution
For this example, there are A, BC and BD failure modes, so the estimated demonstrated failure intensity, [math]\displaystyle{ {{\hat{\lambda }}_{D}}(T)\,\! }[/math], is simply the Crow-AMSAA model applied to all A, BC, and BD data.

[math]\displaystyle{ {{\hat{\lambda }}_{D}}(T)={{\widehat{\lambda }}_{CA}}=\widehat{\lambda }\widehat{\beta }{{T}^{\widehat{\beta }-1}}=0.12744\,\! }[/math]

Therefore, the demonstrated MTBF is:

[math]\displaystyle{ MTB{{F}_{D}}={{[{{\hat{\lambda }}_{D}}(T)]}^{-1}}=7.84708\,\! }[/math]

Based on this value, the Fisher Matrix confidence bounds for the demonstrated failure intensity at the 90% confidence level are:

[math]\displaystyle{ \begin{align} {{[{{\lambda }_{D}}(T)]}_{L}} = & {{{\hat{\lambda }}}_{CA}}(T){{e}^{{{z}_{\alpha }}\sqrt{Var({{{\hat{\lambda }}}_{CA}}(T))}/{{{\hat{\lambda }}}_{CA}}(T)}} \\ = & 0.09339 \end{align}\,\! }[/math]
[math]\displaystyle{ \begin{align} {{[{{\lambda }_{D}}(T)]}_{U}} = & {{{\hat{\lambda }}}_{CA}}(T){{e}^{-{{z}_{\alpha }}\sqrt{Var({{{\hat{\lambda }}}_{CA}}(T))}/{{{\hat{\lambda }}}_{CA}}(T)}} \\ = & 0.17390 \end{align}\,\! }[/math]

The Fisher Matrix confidence bounds for the demonstrated MTBF at the 90% confidence level are:

[math]\displaystyle{ \begin{align} MTB{{F}_{{{D}_{L}}}} = & \frac{1}{{{[{{\lambda }_{D}}(T)]}_{U}}} \\ = & 5.75054 \\ MTB{{F}_{{{D}_{U}}}} = & \frac{1}{{{[{{\lambda }_{D}}(T)]}_{L}}} \\ = & 10.70799 \end{align}\,\! }[/math]

The Crow confidence bounds for the demonstrated MTBF at the 90% confidence level are:

[math]\displaystyle{ \begin{align} MTB{{F}_{{{D}_{L}}}} = & \frac{1}{{{[{{\lambda }_{D}}(T)]}_{U}}} \\ = & \frac{1}{{{\widehat{\lambda }}_{D}}(T)\tfrac{{{\chi }^{2}}(2N,\alpha /2)}{2N}} \\ = & 5.6325 \\ MTB{{F}_{{{D}_{U}}}} = & \frac{1}{{{[{{\lambda }_{D}}(T)]}_{L}}} \\ = & \frac{1}{{{\widehat{\lambda }}_{D}}(T)\tfrac{{{\chi }^{2}}(2N,1-\alpha /2)}{2N}} \\ = & 10.8779 \end{align}\,\! }[/math]

The projected failure intensity is:

[math]\displaystyle{ \begin{align} \hat{\lambda}_P (T) &= \widehat{\lambda}_{CA} - \widehat{\lambda}_{BD} + \sum_{i=1}^M (1-d_i) \tfrac{N_i}{T} + \bar{d}\widehat{h}(T|BD) \\ &= 0.0885 \,\! \end{align} }[/math]

Based on this value, the Fisher Matrix confidence bounds at the 90% confidence level for the projected failure intensity are:

[math]\displaystyle{ \begin{align} {{[{{\lambda }_{P}}(T)]}_{L}} = & {{{\hat{\lambda }}}_{P}}(T){{e}^{{{z}_{\alpha }}\sqrt{Var({{{\hat{\lambda }}}_{P}}(T))}/{{{\hat{\lambda }}}_{P}}(T)}} \\ = & 0.0681 \end{align}\,\! }[/math]
[math]\displaystyle{ \begin{align} {{[{{\lambda }_{P}}(T)]}_{U}} = & {{{\hat{\lambda }}}_{P}}(T){{e}^{-{{z}_{\alpha }}\sqrt{Var({{{\hat{\lambda }}}_{P}}(T))}/{{{\hat{\lambda }}}_{P}}(T)}} \\ = & 0.1152 \end{align}\,\! }[/math]

The Fisher Matrix confidence bounds for the projected MTBF at the 90% confidence level are:

[math]\displaystyle{ \begin{align} MTB{{F}_{{{P}_{L}}}} = & \frac{1}{{{[{{\lambda }_{P}}(T)]}_{U}}} \\ = & 8.6818 \\ MTB{{F}_{{{P}_{U}}}} = & \frac{1}{{{[{{\lambda }_{P}}(T)]}_{L}}} \\ = & 14.6926 \end{align}\,\! }[/math]

The Crow confidence bounds for the projected failure intensity are:

[math]\displaystyle{ \begin{align} {{[{{\lambda }_{P}}(T)]}_{L}} = & {{{\hat{\lambda }}}_{P}}(T)+\frac{{{C}^{2}}}{2}-\sqrt{{{{\hat{\lambda }}}_{P}}(T)\cdot \ \,{{C}^{2}}+\frac{{{C}^{4}}}{4}} \\ = & 0.0672 \\ {{[{{\lambda }_{P}}(T)]}_{U}} = & {{{\hat{\lambda }}}_{P}}(T)+\frac{{{C}^{2}}}{2}+\sqrt{{{{\hat{\lambda }}}_{P}}(T)\cdot {{C}^{2}}+\frac{{{C}^{4}}}{4}} \\ = & 0.1166 \end{align}\,\! }[/math]

The Crow confidence bounds for the projected MTBF at the 90% confidence level are:

[math]\displaystyle{ \begin{align} MTB{{F}_{{{P}_{L}}}} = & \frac{1}{{{[{{\widehat{\lambda }}_{P}}(T)]}_{U}}} \\ = & 8.5743 \\ MTB{{F}_{{{P}_{U}}}} = & \frac{1}{{{[{{\widehat{\lambda }}_{P}}(T)]}_{L}}} \\ = & 14.8769 \end{align}\,\! }[/math]

The growth potential failure intensity is:

[math]\displaystyle{ \widehat{\lambda}_{GP} = \widehat{\lambda}_{CA} - \widehat{\lambda}_{BD} + \sum_{i=1}^M (1-d_i) \tfrac{N_i}{T} = 0.0670 \,\! }[/math]

Based on this value, the Fisher Matrix and Crow confidence bounds at the 90% confidence level for the growth potential failure intensity are:

[math]\displaystyle{ \begin{align} {{r}_{L}} = & {{{\hat{r}}}_{GP}}+\frac{{{C}^{2}}}{2}-\sqrt{{{{\hat{r}}}_{GP}}{{C}^{2}}+\frac{{{C}^{4}}}{4}} \\ = & 0.0488 \\ {{r}_{U}} = & {{{\hat{r}}}_{GP}}+\frac{{{C}^{2}}}{2}+\sqrt{{{{\hat{r}}}_{GP}}{{C}^{2}}+\frac{{{C}^{4}}}{4}} \\ = & 0.0919 \end{align}\,\! }[/math]

The Fisher Matrix and Crow confidence bounds for the growth potential MTBF at the 90% confidence level are:

[math]\displaystyle{ \begin{align} MTB{{F}_{G{{P}_{L}}}} = & \frac{1}{{{r}_{U}}} \\ = & 10.8790 \\ MTB{{F}_{G{{P}_{U}}}} = & \frac{1}{{{r}_{L}}} \\ = & 20.4855 \end{align}\,\! }[/math]

The figure below shows the Fisher Matrix confidence bounds at the 90% confidence level for the demonstrated, projected and growth potential MTBF.

Rga9.10.png

The next figure shows these bounds based on the Crow method.

Rga9.11.png


Grouped Failure Times Data Example

A reliability growth test was conducted for 200 hours. Some of the corrective actions were applied during the test while others were delayed until after the test was completed. The tables below give the data set and the effectiveness factors for the BD modes. Do the following:

  1. Estimate the parameters of the Crow Extended model.
  2. Determine the average effectiveness factor of the BC modes using the Function Wizard.
  3. What percent of the failure intensity will be left in the system due to the BD modes after implementing the delayed fixes?
Grouped Failure Times Data
Number at Event Time to Event Classification Mode
3 25 BC 1
1 25 BD 9
1 25 BC 2
1 50 BD 10
1 50 BD 11
1 75 BD 12
1 75 BC 3
2 75 BD 13
1 75 A
1 100 BC 4
1 100 BD 14
1 125 BD 15
1 125 A
1 125 A
1 125 BC 5
1 125 BD 10
1 125 BC 6
1 150 A
1 150 BD 16
1 175 BC 4
1 175 BC 8
1 175 A
1 175 BC 7
1 200 BD 16
1 200 BC 3
1 200 BD 17
Effectiveness Factors
BD Mode Effectiveness Factor
9 0.75
10 0.5
11 0.9
12 0.6
13 0.8
14 0.8
15 0.25
16 0.75
17 0.8

Solution

  1. The next figure shows the estimated parameters of the Crow Extended model.
    Rga9.33.png
  2. Insert a general spreadsheet into the folio, and then access the Function Wizard. In the Function Wizard, select Average Effectiveness Factor from the list of available functions and under Avg. Eff. Factor select BC modes, as shown next. Click Insert and the result will be inserted into the general spreadsheet. The average effectiveness factor for the BC modes is 0.6983.
    Rga9.34.png
  3. The failure intensity left in the system due to the BD modes can be determined using the Failure Mode Strategy plot, as shown next. Therefore, the failure intensity left in the system due to the BD modes is 1.79%.   
    Rga9.35.png


Discrete Mixed Data Example

A one-shot system underwent reliability growth testing for a total of 20 trials. The test was performed as a combination of groups of units with the same configuration and individual trials. The following table shows the data set. The Failures in Interval column specifies the number of failures that occurred in each interval and the Cumulative Trials column specifies the cumulative number of trials at the end of that interval. In other words, the first three rows contain the data from the first trial, in which 8 units with the same configuration were tested and 3 failures (with different failure modes) were observed. The next row contains data from the second trial, in which 2 units with the same configuration were tested and no failures occurred. And so on.

Mixed Data
Failures in Interval Cumulative Trials Classification Mode
1 8 BD 1
1 8 BD 2
1 8 BD 3
0 10
0 11
0 12
1 13 BD 2
0 14
0 15
1 16 BD 4
0 17
0 18
0 19
1 20 BD 5


The table also gives the classifications of the failure modes. There are 5 BD modes. The average effectiveness factor for the BD modes is 0.7. Do the following:

  1. Calculate the demonstrated reliability at the end of the test.
  2. Calculate the growth potential reliability.

Solution

  1. Based on the equations presented in Crow-AMSAA (NHPP), the parameters of the Crow-AMSAA (NHPP) model are estimated as follows:
    [math]\displaystyle{ \widehat{\beta }=0.8572\,\! }[/math]
    and:
    [math]\displaystyle{ \widehat{\lambda }=0.4602\,\! }[/math]
    However, because there are only A or BD modes for mixed data, there is no growth during the test. In other words, the hypothesis for the [math]\displaystyle{ \hat{\beta }\,\! }[/math] parameter is that [math]\displaystyle{ \hat{\beta }=1.\,\! }[/math] From the Crow-AMSAA (NHPP) model, we know that:
    [math]\displaystyle{ \hat{\lambda }=\frac{n}{T_{n}^{\hat{\beta }}}\,\! }[/math]
    or, if [math]\displaystyle{ \hat{\beta }=1\,\! }[/math], this becomes:
    [math]\displaystyle{ \begin{align} \hat{\lambda }= & \frac{n}{{{T}_{n}}} \\ = & \frac{6}{20} \\ = & 0.3 \end{align}\,\! }[/math]
    As we have seen, the Crow-AMSAA instantaneous failure intensity, [math]\displaystyle{ {{\lambda }_{i}}(T)\,\! }[/math], is defined as:
    [math]\displaystyle{ \begin{align} {{\lambda }_{i}}(T)=\lambda \beta {{T}^{\beta -1}},\text{with }T\gt 0,\text{ }\lambda \gt 0\text{ and }\beta \gt 0 \end{align}\,\! }[/math]
    Using trials instead of time, and accommodating for [math]\displaystyle{ \hat{\beta }=1\,\! }[/math], we can calculate the instantaneous failure probability at the end of the test, or [math]\displaystyle{ T=20\,\! }[/math] :
    [math]\displaystyle{ {{Q}_{i}}(20)=\widehat{\lambda }=0.3\,\! }[/math]
    So the instantaneous reliability at the end of the test, or demonstrated reliability, is:
    [math]\displaystyle{ \begin{align} {{R}_{i}}(20)= & 1-{{Q}_{i}}(20) \\ = & 1-0.3 \\ = & 0.7 \end{align}\,\! }[/math]
    The next figure shows the data sheet as calculated in the RGA software.
    Rga9.12.png
  2. The growth potential unreliability is:
    [math]\displaystyle{ \begin{align} {{\widehat{Q}}_{GP}}(T)= & \left( \frac{{{N}_{A}}}{T}+\underset{i=1}{\overset{M}{\mathop \sum }}\,(1-{{d}_{i}})\frac{{{N}_{i}}}{T} \right) \\ = & \underset{i=1}{\overset{M}{\mathop \sum }}\,(1-0.7)\frac{{{N}_{i}}}{T} \\ = & 0.3*(\frac{1+1+1+1+1+1)}{20} \\ = & 0.09 \end{align}\,\! }[/math]
    So the growth potential reliability is:
    [math]\displaystyle{ \begin{align} {{\widehat{R}}_{GP}}(T) = & 1-{{\widehat{Q}}_{GP}}(T) \\ = & 1-0.09 \\ = & 0.91 \end{align}\,\! }[/math]
    The figures below show the calculation of the growth potential reliability for the mixed data using the RGA software's QCP, followed by the growth potential plot.
    Rga9.13.png


    Rga9.14.png


Multiple Systems with Event Codes

Consider the data set shown in the following figure.

Rga9.15.png

Solution

The first step to creating the equivalent system is to isolate each failure mode and its implemented fixes independently from each other. The numbered steps follow the five rules and are presented in the same numbering sequence. 

  1. The next figure illustrates the application of rule #1 for mode BD1. The mode in the equivalent single system is calculated as [math]\displaystyle{ (75+75+75)=225\,\! }[/math] or [math]\displaystyle{ 75*3=225.\,\! }[/math]
    Rga9.16.png

    The next figure illustrates the application of rule #1 for mode A110. The mode in the equivalent single system is calculated as [math]\displaystyle{ (280+280+280)=840\,\! }[/math] or [math]\displaystyle{ 280*3=840.\,\! }[/math]

    Rga9.17.png
  2. The next figure illustrates the application of rule #2 for the first occurrence of the mode BC10 in system 1. The mode in the equivalent single system is calculated as [math]\displaystyle{ (150+150+150)=450\,\! }[/math] or [math]\displaystyle{ 150*3=450.\,\! }[/math]
    Rga9.18.png
  3. The next figure illustrates the application of rule #3 for implemented fixes (I events) of the mode BC10. In the graph, the I events are symbolized by having the letter "I" before the naming of the mode. In this case, IBC10 is for the implemented fix of mode BC10. The IBC10 in the equivalent single system is calculated as [math]\displaystyle{ (200+175+200)=575\,\! }[/math].
    Rga9.19.png
  4. The next figure illustrates the application of rule #4 for the mode BC20 in system 1, which occurs after a fix for the same mode was implemented in system 2. The mode in the equivalent single system is calculated as [math]\displaystyle{ (350+300+350)=1000.\,\! }[/math]
    Rga9.20.png
  5. The next figure illustrates the application of rule #5 for the second occurrence of the mode BC10 in system 1, which occurs after an implemented fix (I event) has occurred for that mode in the same system. The mode in the equivalent single system is calculated as [math]\displaystyle{ 575+(175+200+175)=1125.\,\! }[/math]
    Rga9.21.png

    After having transferred the data set to the Crow Extended equivalent system, the data set is analyzed using the Crow Extended model. The last figure shows the growth potential MTBF plot.

    Rga9.22.png


Multiple Systems - Concurrent Operating Times

Three systems were subjected to a reliability growth test to evaluate the prototype of a new product. Based on a failure analysis on the results of the test, the proposed management strategy is to delay corrective actions until after the test. The tables shown next display the data set and the associated effectiveness factors for the unique BD modes.

Multiple Systems (Concurrent Operating Times) Data
System 1 System 2 System 3
Start Time 0 0 0
End Time 541 454 436
Times-to-Failure 83 BD37 26 BD25 23 BD30
83 BD43 26 BD43 46 BD49
83 BD46 57 BD37 127 BD47
169 A45 64 BD19 166 A2
213 A18 169 A45 169 BD23
299 A42 213 A32 213 BD7
375 A1 231 BD8 213 BD29
431 BD16 231 BD25 255 BD26
231 BD27 369 A33
231 A28 374 BD29
304 BD24 380 BD22
383 BD40 415 BD7
Effectiveness factors
BD Mode Effectiveness Factor
30 0.75
43 0.5
25 0.5
49 0.75
37 0.9
19 0.75
46 0.75
47 0.25
23 0.5
7 0.25
29 0.25
8 0.5
27 0.5
26 0.75
24 0.5
22 0.5
40 0.75
16 0.75

The prototype is required to meet a projected MTBF goal of 55 hours. Do the following:

  1. Estimate the parameters of the Crow Extended model.
  2. Based on the current management strategy what is the projected MTBF?
  3. If the projected MTBF goal is not met, alter the current management strategy to meet this requirement with as little adjustment as possible and without changing the EFs of the existing BD modes. Assume an EF = 0.7 for any newly assigned BD modes.

Solution

  1. The next figure shows the estimated Crow Extended parameters.
    Rga9.29.png
  2. There are a couple of ways to calculate the projected MTBF, but the easiest is via the Quick Calculation Pad (QCP). The following result shows that the projected MTBF is estimated to be 53.9390 hours, which is below the goal of 55 hours.
    Rga9.30.png
  3. To reach our goal, or to see if we can even get there, the management strategy must be changed. The effectiveness factors for the existing BD modes cannot be changed; however, it is possible to change an A mode to a BD mode, but which A mode(s) should be changed? To answer this question, create an Individual Mode Failure Intensity plot with just the A modes displayed, as shown next. As you can see from the plot, failure mode A45 has the highest failure intensity. Therefore, among the A modes, this particular failure mode has the greatest negative effect on the system MTBF.
    Rga9.31.png

    Change A45 to BD45. Be sure to change all instances of A45 to a BD mode. Enter an effectiveness factor of 0.7 for BD45, and then recalculate the parameters of the Crow Extended model. Now go back to the QCP to calculate the projected MTBF, as shown below. The projected MTBF is now estimated to be 55.5903 hours. Based on the change in the management strategy, the projected MTBF goal is now expected to be met.

    Rga9.32.png


Multiple Systems - Known Operating Times

Two prototypes of a new design are tested simultaneously. Whenever a failure is observed for one unit, the current operating time of the other unit is also recorded. The test is terminated after 300 hours. All of the design changes for the prototypes were delayed until after completing the test. The data set is given in the table below. Assume a fixed effectiveness factor equal to 0.7. The MTBF goal for the new design is 30 hours. Do the following:

  1. Estimate the parameters of the Crow Extended model.
  2. What is the projected MTBF and growth potential?
  3. Under the current management strategy, is it even possible to reach the MTBF goal of 30 hours?
Multiple Systems (Known Operating Times) Data
Failed Unit ID Time Unit 1 Time Unit 2 Classification Mode
1 16.5 0 BD seal leak
1 16.5 0 BD valve
1 17 0 A
2 20.5 0.9 A
2 25.3 3.8 BD hose
2 28.7 4.6 BD operator error
1 41.8 14.7 BD bearing
1 45.5 17.6 A
2 48.6 22 A
2 49.6 23.4 BD seal leak
1 51.4 26.3 A
1 58.2 35.7 BD seal leak
2 59 36.5 A
2 60.6 37.6 BD hose
1 61.9 39.1 BD seal leak
1 76.6 55.4 BD bearing
2 81.1 61.1 A
1 84.1 63.6 A
1 84.7 64.3 A
1 94.6 72.6 A
2 100 78.1 BD valve
1 104 81.4 BD bearing
2 104.8 85.9 BD spring
2 105.9 87.1 BD operator error
1 108.8 89.9 BD hose
2 132.4 119.5 BD spring
2 132.4 150.1 BD operator error
2 132.4 153.7 A


Solution

  1. The next figure shows the estimated Crow Extended parameters.
    Rga9.36.png
  2. One possible method for calculating the projected MTBF and growth potential is to use the Quick Calculation Pad, but you can also view these two values at the same time by viewing the Growth Potential MTBF plot, as shown next. From the plot, the projected MTBF is equal to 16.87 hours and the growth potential is equal to 18.63 hours.
    Rga9.37.png
  3. The current projected MTBF and growth potential MTBF are both well below the required goal of 30 hours. To check if this goal can even be reached, you can set the effectiveness factor equal to 1. In other words, if all of the corrective actions were to remove the failure modes completely, what would be the projected and growth potential MTBF? Change the fixed effectiveness factor to 1, then recalculate the parameters and refresh the Growth Potential plot, as shown next. Even if you assume an effectiveness factor equal to 1, the growth potential is still only 27.27 hours. Based on the current design process, it will not be possible to reach the MTBF goal of 30 hours. Therefore, you have two options: start a new design stage or reduce the required MTBF goal.
    Rga9.38.png
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