Crow-AMSAA Confidence Bounds Example: Difference between revisions

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The partial derivatives for the Fisher Matrix confidence bounds are:
The partial derivatives for the Fisher Matrix confidence bounds are:


:<math>\begin{align}
:<math>\begin{align}
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   \frac{{{\partial }^{2}}\Lambda }{\partial \lambda \partial \beta }  = & -{{620}^{0.6142}}\ln 620=-333.64   
   \frac{{{\partial }^{2}}\Lambda }{\partial \lambda \partial \beta }  = & -{{620}^{0.6142}}\ln 620=-333.64   
\end{align}\,\!</math>
\end{align}\,\!</math>


The Fisher Matrix then becomes:  
The Fisher Matrix then becomes:  


:<math>\begin{align}
:<math>\begin{align}
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& = \begin{bmatrix} 0.13519969 & -0.046614609\\ -0.046614609 & 0.017105343 \end{bmatrix}
& = \begin{bmatrix} 0.13519969 & -0.046614609\\ -0.046614609 & 0.017105343 \end{bmatrix}
\end{align}\,\!</math>
\end{align}\,\!</math>


For <math>T=620\,\!</math> hours, the partial derivatives of the cumulative and instantaneous failure intensities are:  
For <math>T=620\,\!</math> hours, the partial derivatives of the cumulative and instantaneous failure intensities are:  


:<math>\begin{align}
:<math>\begin{align}
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   =  & 0.083694185   
   =  & 0.083694185   
\end{align}\,\!</math>
\end{align}\,\!</math>


:<math>\begin{align}
:<math>\begin{align}
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   =  & 0.17558519   
   =  & 0.17558519   
\end{align}\,\!</math>
\end{align}\,\!</math>


:<math>\begin{align}
:<math>\begin{align}
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   =  & 0.051404969   
   =  & 0.051404969   
\end{align}\,\!</math>
\end{align}\,\!</math>


Therefore, the variances become:  
Therefore, the variances become:  


:<math>\begin{align}
:<math>\begin{align}
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&= 0.0000431393
&= 0.0000431393
\end{align}\,\!</math>
\end{align}\,\!</math>


The cumulative and instantaneous failure intensities at <math>T=620\,\!</math> hours are:  
The cumulative and instantaneous failure intensities at <math>T=620\,\!</math> hours are:  


:<math>\begin{align}
:<math>\begin{align}
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  {{\lambda }_{i}}(T)= & 0.02179   
  {{\lambda }_{i}}(T)= & 0.02179   
\end{align}\,\!</math>
\end{align}\,\!</math>


So, at the 90% confidence level and for <math>T=620\,\!</math> hours, the Fisher Matrix confidence bounds for the cumulative failure intensity are:
So, at the 90% confidence level and for <math>T=620\,\!</math> hours, the Fisher Matrix confidence bounds for the cumulative failure intensity are:


:<math>\begin{align}
:<math>\begin{align}
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   {{[{{\lambda }_{c}}(T)]}_{U}}= & 0.05039   
   {{[{{\lambda }_{c}}(T)]}_{U}}= & 0.05039   
\end{align}\,\!</math>
\end{align}\,\!</math>


The confidence bounds for the instantaneous failure intensity are:
The confidence bounds for the instantaneous failure intensity are:


:<math>\begin{align}
:<math>\begin{align}
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   {{[{{\lambda }_{i}}(T)]}_{U}}= & 0.03579   
   {{[{{\lambda }_{i}}(T)]}_{U}}= & 0.03579   
\end{align}\,\!</math>
\end{align}\,\!</math>


The following figures display plots of the Fisher Matrix confidence bounds for the cumulative and instantaneous failure intensity, respectively.
The following figures display plots of the Fisher Matrix confidence bounds for the cumulative and instantaneous failure intensity, respectively.


[[Image:rga5.2.png|center|500px|Cumulative failure intensity with 2-sided 90% Fisher Matrix confidence bounds.]]
[[Image:rga5.2.png|center|500px|Cumulative failure intensity with 2-sided 90% Fisher Matrix confidence bounds.]]
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[[Image:rga5.3.png|center|500px|Instantaneous failure intensity with 2-sided 90% Fisher Matrix confidence bounds.]]
[[Image:rga5.3.png|center|500px|Instantaneous failure intensity with 2-sided 90% Fisher Matrix confidence bounds.]]


'''Crow Bounds'''
'''Crow Bounds'''


Given that the data is failure terminated, the Crow confidence bounds for the cumulative failure intensity at the 90% confidence level and for <math>T=620\,\!</math> hours are:
Given that the data is failure terminated, the Crow confidence bounds for the cumulative failure intensity at the 90% confidence level and for <math>T=620\,\!</math> hours are:


:<math>\begin{align}
:<math>\begin{align}
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     = & 0.048775   
     = & 0.048775   
\end{align}\,\!</math>
\end{align}\,\!</math>


The Crow confidence bounds for the instantaneous failure intensity at the 90% confidence level and for <math>T=620\,\!</math> hours are calculated by first estimating the bounds on the instantaneous MTBF. Once these are calculated, take the inverse as shown below. Details on the confidence bounds for instantaneous MTBF are presented [[Crow-AMSAA Confidence Bounds#Crow_Bounds_4|here]].
The Crow confidence bounds for the instantaneous failure intensity at the 90% confidence level and for <math>T=620\,\!</math> hours are calculated by first estimating the bounds on the instantaneous MTBF. Once these are calculated, take the inverse as shown below. Details on the confidence bounds for instantaneous MTBF are presented [[Crow-AMSAA Confidence Bounds#Crow_Bounds_4|here]].


:<math>\begin{align}
:<math>\begin{align}
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  = & 0.01179   
  = & 0.01179   
\end{align}\,\!</math>
\end{align}\,\!</math>


:<math>\begin{align}
:<math>\begin{align}
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   =  & 0.03253   
   =  & 0.03253   
\end{align}\,\!</math>
\end{align}\,\!</math>


The following figures display plots of the Crow confidence bounds for the cumulative and instantaneous failure intensity, respectively.
The following figures display plots of the Crow confidence bounds for the cumulative and instantaneous failure intensity, respectively.


[[Image:rga5.4.png|center|500px|Cumulative failure intensity with 2-sided 90% Crow confidence bounds.]]
[[Image:rga5.4.png|center|500px|Cumulative failure intensity with 2-sided 90% Crow confidence bounds.]]
Line 118: Line 142:


[[Image:rga5.5.png|center|500px|Instantaneous failure intensity with 2-sided 90% Crow confidence bounds.]]
[[Image:rga5.5.png|center|500px|Instantaneous failure intensity with 2-sided 90% Crow confidence bounds.]]


====Failure Times - Example 3====
====Failure Times - Example 3====
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[[Image:rga5.7.png|center|500px|Instantaneous MTBF with 2-sided Fisher Matrix confidence bounds.]]
[[Image:rga5.7.png|center|500px|Instantaneous MTBF with 2-sided Fisher Matrix confidence bounds.]]


'''Crow Bounds'''
'''Crow Bounds'''

Revision as of 22:39, 29 January 2014

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


Using the values of [math]\displaystyle{ \hat{\beta }\,\! }[/math] and [math]\displaystyle{ \hat{\lambda }\,\! }[/math] estimated in the Failure Times - Example 1, calculate the 90% 2-sided confidence bounds on the cumulative and instantaneous failure intensity.

Solution

Fisher Matrix Bounds

The partial derivatives for the Fisher Matrix confidence bounds are:


[math]\displaystyle{ \begin{align} \frac{{{\partial }^{2}}\Lambda }{\partial {{\lambda }^{2}}} = & -\frac{22}{{{0.4239}^{2}}}=-122.43 \\ \frac{{{\partial }^{2}}\Lambda }{\partial {{\beta }^{2}}} = & -\frac{22}{{{0.6142}^{2}}}-0.4239\cdot {{620}^{0.6142}}{{(\ln 620)}^{2}}=-967.68 \\ \frac{{{\partial }^{2}}\Lambda }{\partial \lambda \partial \beta } = & -{{620}^{0.6142}}\ln 620=-333.64 \end{align}\,\! }[/math]


The Fisher Matrix then becomes:


[math]\displaystyle{ \begin{align} \begin{bmatrix}122.43 & 333.64\\ 333.64 & 967.68\end{bmatrix}^{-1} & = \begin{bmatrix}Var(\hat{\lambda}) & Cov(\hat{\beta},\hat{\lambda})\\ Cov(\hat{\beta},\hat{\lambda}) & Var(\hat{\beta})\end{bmatrix} \\ & = \begin{bmatrix} 0.13519969 & -0.046614609\\ -0.046614609 & 0.017105343 \end{bmatrix} \end{align}\,\! }[/math]


For [math]\displaystyle{ T=620\,\! }[/math] hours, the partial derivatives of the cumulative and instantaneous failure intensities are:


[math]\displaystyle{ \begin{align} \frac{\partial {{\lambda }_{c}}(T)}{\partial \beta }= & \hat{\lambda }{{T}^{\hat{\beta }-1}}\ln (T) \\ = & 0.4239\cdot {{620}^{-0.3858}}\ln 620 \\ = & 0.22811336 \\ \frac{\partial {{\lambda }_{c}}(T)}{\partial \lambda }= & {{T}^{\hat{\beta }-1}} \\ = & {{620}^{-0.3858}} \\ = & 0.083694185 \end{align}\,\! }[/math]


[math]\displaystyle{ \begin{align} \frac{\partial {{\lambda }_{i}}(T)}{\partial \beta }= & \hat{\lambda }{{T}^{\hat{\beta }-1}}+\hat{\lambda }\hat{\beta }{{T}^{\hat{\beta }-1}}\ln T \\ = & 0.4239\cdot {{620}^{-0.3858}}+0.4239\cdot 0.6142\cdot {{620}^{-0.3858}}\ln 620 \\ = & 0.17558519 \end{align}\,\! }[/math]


[math]\displaystyle{ \begin{align} \frac{\partial {{\lambda }_{i}}(T)}{\partial \lambda }= & \hat{\beta }{{T}^{\hat{\beta }-1}} \\ = & 0.6142\cdot {{620}^{-0.3858}} \\ = & 0.051404969 \end{align}\,\! }[/math]


Therefore, the variances become:


[math]\displaystyle{ \begin{align} Var(\hat{\lambda_{c}}(T)) & = 0.22811336^{2}\cdot 0.017105343\ + 0.083694185^{2} \cdot 0.13519969\ -2\cdot 0.22811336\cdot 0.083694185\cdot 0.046614609 \\ & = 0.00005721408 \\ Var(\hat{\lambda_{i}}(T)) & = 0.17558519^{2}\cdot 0.01715343\ + 0.051404969^{2}\cdot 0.13519969\ -2\cdot 0.17558519\cdot 0.051404969\cdot 0.046614609 \\ &= 0.0000431393 \end{align}\,\! }[/math]


The cumulative and instantaneous failure intensities at [math]\displaystyle{ T=620\,\! }[/math] hours are:


[math]\displaystyle{ \begin{align} {{\lambda }_{c}}(T)= & 0.03548 \\ {{\lambda }_{i}}(T)= & 0.02179 \end{align}\,\! }[/math]


So, at the 90% confidence level and for [math]\displaystyle{ T=620\,\! }[/math] hours, the Fisher Matrix confidence bounds for the cumulative failure intensity are:


[math]\displaystyle{ \begin{align} {{[{{\lambda }_{c}}(T)]}_{L}}= & 0.02499 \\ {{[{{\lambda }_{c}}(T)]}_{U}}= & 0.05039 \end{align}\,\! }[/math]


The confidence bounds for the instantaneous failure intensity are:


[math]\displaystyle{ \begin{align} {{[{{\lambda }_{i}}(T)]}_{L}}= & 0.01327 \\ {{[{{\lambda }_{i}}(T)]}_{U}}= & 0.03579 \end{align}\,\! }[/math]


The following figures display plots of the Fisher Matrix confidence bounds for the cumulative and instantaneous failure intensity, respectively.


Cumulative failure intensity with 2-sided 90% Fisher Matrix confidence bounds.


Instantaneous failure intensity with 2-sided 90% Fisher Matrix confidence bounds.


Crow Bounds

Given that the data is failure terminated, the Crow confidence bounds for the cumulative failure intensity at the 90% confidence level and for [math]\displaystyle{ T=620\,\! }[/math] hours are:


[math]\displaystyle{ \begin{align} {{[{{\lambda }_{c}}(T)]}_{L}} = & \frac{\chi _{\tfrac{\alpha }{2},2N}^{2}}{2\cdot t} \\ = & \frac{29.787476}{2*620} \\ = & 0.02402 \\ {{[{{\lambda }_{c}}(T)]}_{U}} = & \frac{\chi _{1-\tfrac{\alpha }{2},2N}^{2}}{2\cdot t} \\ = & \frac{60.48089}{2*620} \\ = & 0.048775 \end{align}\,\! }[/math]


The Crow confidence bounds for the instantaneous failure intensity at the 90% confidence level and for [math]\displaystyle{ T=620\,\! }[/math] hours are calculated by first estimating the bounds on the instantaneous MTBF. Once these are calculated, take the inverse as shown below. Details on the confidence bounds for instantaneous MTBF are presented here.


[math]\displaystyle{ \begin{align} {{[{{\lambda }_{i}}(t)]}_{L}} = & \frac{1}{{{[MTB{{F}_{i}}]}_{U}}} \\ = & \frac{1}{MTB{{F}_{i}}\cdot U} \\ = & 0.01179 \end{align}\,\! }[/math]


[math]\displaystyle{ \begin{align} {{[{{\lambda }_{i}}(t)]}_{U}}= & \frac{1}{{{[MTB{{F}_{i}}]}_{L}}} \\ = & \frac{1}{MTB{{F}_{i}}\cdot L} \\ = & 0.03253 \end{align}\,\! }[/math]


The following figures display plots of the Crow confidence bounds for the cumulative and instantaneous failure intensity, respectively.


Cumulative failure intensity with 2-sided 90% Crow confidence bounds.


Instantaneous failure intensity with 2-sided 90% Crow confidence bounds.


Failure Times - Example 3

Calculate the confidence bounds on the cumulative and instantaneous MTBF for the data from the example given above.

Solution

Fisher Matrix Bounds

From the previous example:


[math]\displaystyle{ \begin{align} Var(\hat{\lambda }) = & 0.13519969 \\ Var(\hat{\beta }) = & 0.017105343 \\ Cov(\hat{\beta },\hat{\lambda }) = & -0.046614609 \end{align}\,\! }[/math]


And for [math]\displaystyle{ T=620\,\! }[/math] hours, the partial derivatives of the cumulative and instantaneous MTBF are:


[math]\displaystyle{ \begin{align} \frac{\partial {{m}_{c}}(T)}{\partial \beta }= & -\frac{1}{\hat{\lambda }}{{T}^{1-\hat{\beta }}}\ln T \\ = & -\frac{1}{0.4239}{{620}^{0.3858}}\ln 620 \\ = & -181.23135 \\ \frac{\partial {{m}_{c}}(T)}{\partial \lambda } = & -\frac{1}{{{\hat{\lambda }}^{2}}}{{T}^{1-\hat{\beta }}} \\ = & -\frac{1}{{{0.4239}^{2}}}{{620}^{0.3858}} \\ = & -66.493299 \\ \frac{\partial {{m}_{i}}(T)}{\partial \beta } = & -\frac{1}{\hat{\lambda }{{\hat{\beta }}^{2}}}{{T}^{1-\beta }}-\frac{1}{\hat{\lambda }\hat{\beta }}{{T}^{1-\hat{\beta }}}\ln T \\ = & -\frac{1}{0.4239\cdot {{0.6142}^{2}}}{{620}^{0.3858}}-\frac{1}{0.4239\cdot 0.6142}{{620}^{0.3858}}\ln 620 \\ = & -369.78634 \\ \frac{\partial {{m}_{i}}(T)}{\partial \lambda } = & -\frac{1}{{{\hat{\lambda }}^{2}}\hat{\beta }}{{T}^{1-\hat{\beta }}} \\ = & -\frac{1}{{{0.4239}^{2}}\cdot 0.6142}\cdot {{620}^{0.3858}} \\ = & -108.26001 \end{align}\,\! }[/math]


Therefore, the variances become:


[math]\displaystyle{ \begin{align} Var({{\hat{m}}_{c}}(T)) = & {{\left( -181.23135 \right)}^{2}}\cdot 0.017105343+{{\left( -66.493299 \right)}^{2}}\cdot 0.13519969 \\ & -2\cdot \left( -181.23135 \right)\cdot \left( -66.493299 \right)\cdot 0.046614609 \\ = & 36.113376 \end{align}\,\! }[/math]


[math]\displaystyle{ \begin{align} Var({{\hat{m}}_{i}}(T)) = & {{\left( -369.78634 \right)}^{2}}\cdot 0.017105343+{{\left( -108.26001 \right)}^{2}}\cdot 0.13519969 \\ & -2\cdot \left( -369.78634 \right)\cdot \left( -108.26001 \right)\cdot 0.046614609 \\ = & 191.33709 \end{align}\,\! }[/math]


So, at 90% confidence level and [math]\displaystyle{ T=620\,\! }[/math] hours, the Fisher Matrix confidence bounds are:


[math]\displaystyle{ \begin{align} {{[{{m}_{c}}(T)]}_{L}} = & {{{\hat{m}}}_{c}}(t){{e}^{-{{z}_{\alpha }}\sqrt{Var({{{\hat{m}}}_{c}}(t))}/{{{\hat{m}}}_{c}}(t)}} \\ = & 19.84581 \\ {{[{{m}_{c}}(T)]}_{U}} = & {{{\hat{m}}}_{c}}(t){{e}^{{{z}_{\alpha }}\sqrt{Var({{{\hat{m}}}_{c}}(t))}/{{{\hat{m}}}_{c}}(t)}} \\ = & 40.01927 \end{align}\,\! }[/math]


[math]\displaystyle{ \begin{align} {{[{{m}_{i}}(T)]}_{L}} = & {{{\hat{m}}}_{i}}(t){{e}^{-{{z}_{\alpha }}\sqrt{Var({{{\hat{m}}}_{i}}(t))}/{{{\hat{m}}}_{i}}(t)}} \\ = & 27.94261 \\ {{[{{m}_{i}}(T)]}_{U}} = & {{{\hat{m}}}_{i}}(t){{e}^{{{z}_{\alpha }}\sqrt{Var({{{\hat{m}}}_{i}}(t))}/{{{\hat{m}}}_{i}}(t)}} \\ = & 75.34193 \end{align}\,\! }[/math]


The following two figures show plots of the Fisher Matrix confidence bounds for the cumulative and instantaneous MTBFs.


Cumulative MTBF with 2-sided 90% Fisher Matrix confidence bounds.


Instantaneous MTBF with 2-sided Fisher Matrix confidence bounds.


Crow Bounds

The Crow confidence bounds for the cumulative MTBF and the instantaneous MTBF at the 90% confidence level and for [math]\displaystyle{ T=620\,\! }[/math] hours are:


[math]\displaystyle{ \begin{align} {{[{{m}_{c}}(T)]}_{L}} = & \frac{1}{{{[{{\lambda }_{c}}(T)]}_{U}}} \\ = & 20.5023 \\ {{[{{m}_{c}}(T)]}_{U}} = & \frac{1}{{{[{{\lambda }_{c}}(T)]}_{L}}} \\ = & 41.6282 \end{align}\,\! }[/math]


[math]\displaystyle{ \begin{align} {{[MTB{{F}_{i}}]}_{L}} = & MTB{{F}_{i}}\cdot {{\Pi }_{1}} \\ = & 30.7445 \\ {{[MTB{{F}_{i}}]}_{U}} = & MTB{{F}_{i}}\cdot {{\Pi }_{2}} \\ = & 84.7972 \end{align}\,\! }[/math]


The figures below show plots of the Crow confidence bounds for the cumulative and instantaneous MTBF.

Cumulative MTBF with 2-sided 90% Crow confidence bounds.


Instantaneous MTBF with 2-sided 90% Crow confidence bounds.

Confidence bounds can also be obtained on the parameters [math]\displaystyle{ \hat{\beta }\,\! }[/math] and [math]\displaystyle{ \hat{\lambda }\,\! }[/math]. For Fisher Matrix confidence bounds:


[math]\displaystyle{ \begin{align} {{\beta }_{L}} = & \hat{\beta }{{e}^{{{z}_{\alpha }}\sqrt{Var(\hat{\beta })}/\hat{\beta }}} \\ = & 0.4325 \\ {{\beta }_{U}} = & \hat{\beta }{{e}^{-{{z}_{\alpha }}\sqrt{Var(\hat{\beta })}/\hat{\beta }}} \\ = & 0.8722 \end{align}\,\! }[/math]


and:


[math]\displaystyle{ \begin{align} {{\lambda }_{L}} = & \hat{\lambda }{{e}^{{{z}_{\alpha }}\sqrt{Var(\hat{\lambda })}/\hat{\lambda }}} \\ = & 0.1016 \\ {{\lambda }_{U}} = & \hat{\lambda }{{e}^{-{{z}_{\alpha }}\sqrt{Var(\hat{\lambda })}/\hat{\lambda }}} \\ = & 1.7691 \end{align}\,\! }[/math]


For Crow confidence bounds:


[math]\displaystyle{ \begin{align} {{\beta }_{L}}= & 0.4527 \\ {{\beta }_{U}}= & 0.9350 \end{align}\,\! }[/math]


and:


[math]\displaystyle{ \begin{align} {{\lambda }_{L}}= & 0.2870 \\ {{\lambda }_{U}}= & 0.5827 \end{align}\,\! }[/math]