Test-Find-Test Data Example

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

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.