Crow-AMSAA Confidence Bounds Example: Difference between revisions
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<noinclude>{{Banner RGA Examples}} | <noinclude>{{Banner RGA Examples}} | ||
''This example appears in the [ | ''This example appears in the [https://help.reliasoft.com/reference/reliability_growth_and_repairable_system_analysis Reliability growth reference]''. | ||
</noinclude> | </noinclude> | ||
Using the values of <math>\ | ====Example - Confidence Bounds on Failure Intensity==== | ||
Using the values of <math>\hat{\beta }\,\!</math> and <math>\hat{\lambda }\,\!</math> estimated in the <noinclude>[[Crow-AMSAA Parameter Estimation Example]]</noinclude><includeonly>example given above</includeonly>, calculate the 90% 2-sided confidence bounds on the cumulative and instantaneous failure intensity. | |||
'''Solution''' | '''Solution''' | ||
| Line 10: | Line 10: | ||
'''Fisher Matrix Bounds''' | '''Fisher Matrix Bounds''' | ||
The | The partial derivatives for the Fisher Matrix confidence bounds are: | ||
:<math>\begin{align} | |||
\frac{{{\partial }^{2}}\Lambda }{\partial {{\lambda }^{2}}} = & -\frac{22}{{{0.4239}^{2}}}=-122.43 \\ | \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 {{\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 | \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} | |||
\begin{bmatrix}122.43 & 333.64\\ 333.64 & 967.68\end{bmatrix}^{-1} & = \begin{bmatrix}Var(\hat{\lambda}) & Cov(\hat{\beta},\hat{\lambda})\\ | \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} \\ | Cov(\hat{\beta},\hat{\lambda}) & Var(\hat{\beta})\end{bmatrix} \\ | ||
& = \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} | |||
\frac{\partial {{\lambda }_{c}}(T)}{\partial \beta }= & \ | \frac{\partial {{\lambda }_{c}}(T)}{\partial \beta }= & \hat{\lambda }{{T}^{\hat{\beta }-1}}\ln (T) \\ | ||
= & 0.4239\cdot {{620}^{-0.3858}}\ln 620 \\ | = & 0.4239\cdot {{620}^{-0.3858}}\ln 620 \\ | ||
= & 0.22811336 \\ | = & 0.22811336 \\ | ||
\frac{\partial {{\lambda }_{c}}(T)}{\partial \lambda }= & {{T}^{\ | \frac{\partial {{\lambda }_{c}}(T)}{\partial \lambda }= & {{T}^{\hat{\beta }-1}} \\ | ||
= & {{620}^{-0.3858}} \\ | = & {{620}^{-0.3858}} \\ | ||
= & 0.083694185 | = & 0.083694185 | ||
\end{align}\,\!</math> | \end{align}\,\!</math> | ||
:<math>\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 \\ | |||
\frac{\partial {{\lambda }_{i}}(T)}{\partial \beta }= & \ | |||
= & 0.4239\cdot {{620}^{-0.3858}}+0.4239\cdot 0.6142\cdot {{620}^{-0.3858}}\ln 620 \\ | = & 0.4239\cdot {{620}^{-0.3858}}+0.4239\cdot 0.6142\cdot {{620}^{-0.3858}}\ln 620 \\ | ||
= & 0.17558519 | = & 0.17558519 | ||
\end{align}\,\!</math> | \end{align}\,\!</math> | ||
:<math>\begin{align} | |||
\frac{\partial {{\lambda }_{i}}(T)}{\partial \lambda }= & \hat{\beta }{{T}^{\hat{\beta }-1}} \\ | |||
\frac{\partial {{\lambda }_{i}}(T)}{\partial \lambda }= & \ | |||
= & 0.6142\cdot {{620}^{-0.3858}} \\ | = & 0.6142\cdot {{620}^{-0.3858}} \\ | ||
= & 0.051404969 | = & 0.051404969 | ||
\end{align}\,\!</math> | \end{align}\,\!</math> | ||
Therefore, the variances become: | Therefore, the variances become: | ||
:<math>\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 \\ | 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 \\ | & = 0.00005721408 \\ | ||
| Line 62: | Line 57: | ||
&= 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} | |||
{{\lambda }_{c}}(T)= & 0.03548 \\ | {{\lambda }_{c}}(T)= & 0.03548 \\ | ||
{{\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} | |||
{{[{{\lambda }_{c}}(T)]}_{L}}= & 0.02499 \\ | {{[{{\lambda }_{c}}(T)]}_{L}}= & 0.02499 \\ | ||
{{[{{\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} | |||
{{[{{\lambda }_{i}}(T)]}_{L}}= & 0.01327 \\ | {{[{{\lambda }_{i}}(T)]}_{L}}= & 0.01327 \\ | ||
{{[{{\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| | [[Image:rga5.2.png|center|450px]] | ||
[[Image:rga5.3.png|center|450px]] | |||
'''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: | |||
:<math>\begin{align} | |||
{{[{{\lambda }_{c}}(T)]}_{L}} = & \frac{\chi _{\tfrac{\alpha }{2},2N}^{2}}{2\cdot t} \\ | {{[{{\lambda }_{c}}(T)]}_{L}} = & \frac{\chi _{\tfrac{\alpha }{2},2N}^{2}}{2\cdot t} \\ | ||
= & \frac{29.787476}{2*620} \\ | = & \frac{29.787476}{2*620} \\ | ||
= & 0.02402 \\ | = & 0.02402 \\ | ||
{{[{{\lambda }_{c}}(T)]}_{U}} = & \frac{\chi _{1-\tfrac{\alpha }{2},2N | {{[{{\lambda }_{c}}(T)]}_{U}} = & \frac{\chi _{1-\tfrac{\alpha }{2},2N}^{2}}{2\cdot t} \\ | ||
= & \frac{ | = & \frac{60.48089}{2*620} \\ | ||
= & 0. | = & 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 | 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} | |||
{{[{{\lambda }_{i}}(t)]}_{L}} = & \frac{1}{{{[MTB{{F}_{i}}]}_{U}}} \\ | {{[{{\lambda }_{i}}(t)]}_{L}} = & \frac{1}{{{[MTB{{F}_{i}}]}_{U}}} \\ | ||
= & \frac{1}{MTB{{F}_{i}}\cdot U} \\ | = & \frac{1}{MTB{{F}_{i}}\cdot U} \\ | ||
| Line 117: | Line 106: | ||
\end{align}\,\!</math> | \end{align}\,\!</math> | ||
:<math>\begin{align} | |||
{{[{{\lambda }_{i}}(t)]}_{U}}= & \frac{1}{{{[MTB{{F}_{i}}]}_{L}}} \\ | {{[{{\lambda }_{i}}(t)]}_{U}}= & \frac{1}{{{[MTB{{F}_{i}}]}_{L}}} \\ | ||
= & \frac{1}{MTB{{F}_{i}}\cdot L} \\ | = & \frac{1}{MTB{{F}_{i}}\cdot L} \\ | ||
| Line 125: | Line 114: | ||
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| | [[Image:rga5.4.png|center|450px]] | ||
[[Image:rga5.5.png|center|450px]] | |||
==== | ====Example - Confidence Bounds on MTBF==== | ||
Calculate the confidence bounds on the cumulative and instantaneous MTBF for the data from the example given above. | Calculate the confidence bounds on the cumulative and instantaneous MTBF for the data from the example given above. | ||
'''Solution''' | '''Solution''' | ||
| Line 141: | Line 127: | ||
From the previous example: | From the previous example: | ||
:<math>\begin{align} | |||
Var(\ | Var(\hat{\lambda }) = & 0.13519969 \\ | ||
Var(\ | Var(\hat{\beta }) = & 0.017105343 \\ | ||
Cov(\ | Cov(\hat{\beta },\hat{\lambda }) = & -0.046614609 | ||
\end{align}\,\!</math> | \end{align}\,\!</math> | ||
And for <math>T=620\,\!</math> hours, the partial derivatives of the cumulative and instantaneous MTBF are: | And for <math>T=620\,\!</math> hours, the partial derivatives of the cumulative and instantaneous MTBF are: | ||
:<math>\begin{align} | |||
\frac{\partial {{m}_{c}}(T)}{\partial \beta }= & -\frac{1}{\ | \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 \\ | = & -\frac{1}{0.4239}{{620}^{0.3858}}\ln 620 \\ | ||
= & -181.23135 \\ | = & -181.23135 \\ | ||
\frac{\partial {{m}_{c}}(T)}{\partial \lambda } = & -\frac{1}{{{\ | \frac{\partial {{m}_{c}}(T)}{\partial \lambda } = & -\frac{1}{{{\hat{\lambda }}^{2}}}{{T}^{1-\hat{\beta }}} \\ | ||
= & -\frac{1}{{{0.4239}^{2}}}{{620}^{0.3858}} \\ | = & -\frac{1}{{{0.4239}^{2}}}{{620}^{0.3858}} \\ | ||
= & -66.493299 \\ | = & -66.493299 \\ | ||
\frac{\partial {{m}_{i}}(T)}{\partial \beta } = & -\frac{1}{\ | \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 \\ | = & -\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 \\ | = & -369.78634 \\ | ||
\frac{\partial {{m}_{i}}(T)}{\partial \lambda } = & -\frac{1}{{{\ | \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}} \\ | = & -\frac{1}{{{0.4239}^{2}}\cdot 0.6142}\cdot {{620}^{0.3858}} \\ | ||
= & -108.26001 | = & -108.26001 | ||
\end{align}\,\!</math> | \end{align}\,\!</math> | ||
Therefore, the variances become: | Therefore, the variances become: | ||
:<math>\begin{align} | |||
Var({{\ | 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 \\ | & -2\cdot \left( -181.23135 \right)\cdot \left( -66.493299 \right)\cdot 0.046614609 \\ | ||
= & 36.113376 | = & 36.113376 | ||
\end{align}\,\!</math> | \end{align}\,\!</math> | ||
:<math>\begin{align} | |||
Var({{\ | 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 \\ | & -2\cdot \left( -369.78634 \right)\cdot \left( -108.26001 \right)\cdot 0.046614609 \\ | ||
= & 191.33709 | = & 191.33709 | ||
\end{align}\,\!</math> | \end{align}\,\!</math> | ||
So, at 90% confidence level and <math>T=620\,\!</math> hours, the Fisher Matrix confidence bounds are: | So, at 90% confidence level and <math>T=620\,\!</math> hours, the Fisher Matrix confidence bounds are: | ||
:<math>\begin{align} | |||
{{[{{m}_{c}}(T)]}_{L}} = & {{{\hat{m}}}_{c}}(t){{e}^{-{{z}_{\alpha }}\sqrt{Var({{{\hat{m}}}_{c}}(t))}/{{{\hat{m}}}_{c}}(t)}} \\ | {{[{{m}_{c}}(T)]}_{L}} = & {{{\hat{m}}}_{c}}(t){{e}^{-{{z}_{\alpha }}\sqrt{Var({{{\hat{m}}}_{c}}(t))}/{{{\hat{m}}}_{c}}(t)}} \\ | ||
= & 19.84581 \\ | = & 19.84581 \\ | ||
| Line 190: | Line 173: | ||
\end{align}\,\!</math> | \end{align}\,\!</math> | ||
:<math>\begin{align} | |||
{{[{{m}_{i}}(T)]}_{L}} = & {{{\hat{m}}}_{i}}(t){{e}^{-{{z}_{\alpha }}\sqrt{Var({{{\hat{m}}}_{i}}(t))}/{{{\hat{m}}}_{i}}(t)}} \\ | {{[{{m}_{i}}(T)]}_{L}} = & {{{\hat{m}}}_{i}}(t){{e}^{-{{z}_{\alpha }}\sqrt{Var({{{\hat{m}}}_{i}}(t))}/{{{\hat{m}}}_{i}}(t)}} \\ | ||
= & 27.94261 \\ | = & 27.94261 \\ | ||
| Line 197: | Line 179: | ||
= & 75.34193 | = & 75.34193 | ||
\end{align}\,\!</math> | \end{align}\,\!</math> | ||
The following two figures show plots of the Fisher Matrix confidence bounds for the cumulative and instantaneous MTBFs. | The following two figures show plots of the Fisher Matrix confidence bounds for the cumulative and instantaneous MTBFs. | ||
[[Image:rga5.6.png|center| | [[Image:rga5.6.png|center|450px]] | ||
[[Image:rga5.7.png|center|450px]] | |||
[[Image:rga5.7.png|center| | |||
'''Crow Bounds''' | '''Crow Bounds''' | ||
| Line 211: | Line 190: | ||
The Crow confidence bounds for the cumulative MTBF and the instantaneous MTBF at the 90% confidence level and for <math>T=620\,\!</math> hours are: | The Crow confidence bounds for the cumulative MTBF and the instantaneous MTBF at the 90% confidence level and for <math>T=620\,\!</math> hours are: | ||
:<math>\begin{align} | |||
{{[{{m}_{c}}(T)]}_{L}} = & \frac{1}{{{[{{\lambda }_{c}}(T)]}_{U}}} \\ | {{[{{m}_{c}}(T)]}_{L}} = & \frac{1}{{{[{{\lambda }_{c}}(T)]}_{U}}} \\ | ||
= & 20.5023 \\ | = & 20.5023 \\ | ||
| Line 218: | Line 197: | ||
\end{align}\,\!</math> | \end{align}\,\!</math> | ||
:<math>\begin{align} | |||
{{[MTB{{F}_{i}}]}_{L}} = & MTB{{F}_{i}}\cdot {{\Pi }_{1}} \\ | {{[MTB{{F}_{i}}]}_{L}} = & MTB{{F}_{i}}\cdot {{\Pi }_{1}} \\ | ||
= & 30.7445 \\ | = & 30.7445 \\ | ||
| Line 225: | Line 203: | ||
= & 84.7972 | = & 84.7972 | ||
\end{align}\,\!</math> | \end{align}\,\!</math> | ||
The figures below show plots of the Crow confidence bounds for the cumulative and instantaneous MTBF. | The figures below show plots of the Crow confidence bounds for the cumulative and instantaneous MTBF. | ||
[[Image:rga5.8.png|center| | [[Image:rga5.8.png|center|450px]] | ||
[[Image:rga5.9.png|center|450px]] | |||
[[Image:rga5.9.png|center| | |||
Confidence bounds can also be obtained on the parameters <math>\ | Confidence bounds can also be obtained on the parameters <math>\hat{\beta }\,\!</math> and <math>\hat{\lambda }\,\!</math>. For Fisher Matrix confidence bounds: | ||
:<math>\begin{align} | |||
{{\beta }_{L}} = & \hat{\beta }{{e}^{{{z}_{\alpha }}\sqrt{Var(\hat{\beta })}/\hat{\beta }}} \\ | {{\beta }_{L}} = & \hat{\beta }{{e}^{{{z}_{\alpha }}\sqrt{Var(\hat{\beta })}/\hat{\beta }}} \\ | ||
= & 0.4325 \\ | = & 0.4325 \\ | ||
| Line 243: | Line 219: | ||
\end{align}\,\!</math> | \end{align}\,\!</math> | ||
and: | |||
:<math>\begin{align} | |||
{{\lambda }_{L}} = & \hat{\lambda }{{e}^{{{z}_{\alpha }}\sqrt{Var(\hat{\lambda })}/\hat{\lambda }}} \\ | {{\lambda }_{L}} = & \hat{\lambda }{{e}^{{{z}_{\alpha }}\sqrt{Var(\hat{\lambda })}/\hat{\lambda }}} \\ | ||
= & 0.1016 \\ | = & 0.1016 \\ | ||
| Line 251: | Line 227: | ||
= & 1.7691 | = & 1.7691 | ||
\end{align}\,\!</math> | \end{align}\,\!</math> | ||
For Crow confidence bounds: | For Crow confidence bounds: | ||
:<math>\begin{align} | |||
{{\beta }_{L}}= & 0.4527 \\ | {{\beta }_{L}}= & 0.4527 \\ | ||
{{\beta }_{U}}= & 0.9350 | {{\beta }_{U}}= & 0.9350 | ||
\end{align}\,\!</math> | \end{align}\,\!</math> | ||
and: | |||
:<math>\begin{align} | |||
{{\lambda }_{L}}= & 0.2870 \\ | {{\lambda }_{L}}= & 0.2870 \\ | ||
{{\lambda }_{U}}= & 0.5827 | {{\lambda }_{U}}= & 0.5827 | ||
\end{align}\,\!</math> | \end{align}\,\!</math> | ||
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This example appears in the Reliability growth reference.
Example - Confidence Bounds on Failure Intensity
Using the values of [math]\displaystyle{ \hat{\beta }\,\! }[/math] and [math]\displaystyle{ \hat{\lambda }\,\! }[/math] estimated in the Crow-AMSAA Parameter Estimation Example, 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.
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.
Example - Confidence Bounds on MTBF
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.
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.
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]