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==Parameter Estimation==
#REDIRECT [[RGA Models for Repairable Systems Analysis]]
<br>
Suppose that the number of systems under study is  <math>K</math>  and the  <math>{{q}^{th}}</math>  system is observed continuously from time  <math>{{S}_{q}}</math>  to time  <math>{{T}_{q}}</math> ,  <math>q=1,2,\ldots ,K</math> . During the period  <math>[{{S}_{q}},{{T}_{q}}]</math> , let  <math>{{N}_{q}}</math>  be the number of failures experienced by the  <math>{{q}^{th}}</math>  system and let  <math>{{X}_{i,q}}</math>  be the age of this system at the  <math>{{i}^{th}}</math>  occurrence of failure,  <math>i=1,2,\ldots ,{{N}_{q}}</math> . It is also possible that the times  <math>{{S}_{q}}</math>  and  <math>{{T}_{q}}</math>  may be observed failure times for the  <math>{{q}^{th}}</math>  system. If  <math>{{X}_{{{N}_{q}},q}}={{T}_{q}}</math>  then the data on the  <math>{{q}^{th}}</math>  system is said to be failure terminated and  <math>{{T}_{q}}</math>  is a random variable with  <math>{{N}_{q}}</math>  fixed. If  <math>{{X}_{{{N}_{q}},q}}<{{T}_{q}}</math>  then the data on the  <math>{{q}^{th}}</math>  system is said to be time terminated with  <math>{{N}_{q}}</math>  a random variable. The maximum likelihood estimates of  <math>\lambda </math>  and  <math>\beta </math>  are values satisfying the Eqns. (lambdaPowerLaw) and (BetaPowerLaw).
 
 
::<math>\begin{align}
  & \widehat{\lambda }= & \frac{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{N}_{q}}}{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,\left( T_{q}^{\widehat{\beta }}-S_{q}^{\widehat{\beta }} \right)} \\
& \widehat{\beta }= & \frac{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{N}_{q}}}{\widehat{\lambda }\underset{q=1}{\overset{K}{\mathop{\sum }}}\,\left[ T_{q}^{\widehat{\beta }}\ln ({{T}_{q}})-S_{q}^{\widehat{\beta }}\ln ({{S}_{q}}) \right]-\underset{q=1}{\overset{K}{\mathop{\sum }}}\,\underset{i=1}{\overset{{{N}_{q}}}{\mathop{\sum }}}\,\ln ({{X}_{i,q}})} 
\end{align}</math>
 
 
where  <math>0\ln 0</math>  is defined to be 0. In general, these equations cannot be solved explicitly for  <math>\widehat{\lambda }</math>  and  <math>\widehat{\beta },</math>  but must be solved by iterative procedures. Once  <math>\widehat{\lambda }</math>  and  <math>\widehat{\beta }</math>  have been estimated, the maximum likelihood estimate of the intensity function is given by:
 
::<math>\widehat{u}(t)=\widehat{\lambda }\widehat{\beta }{{t}^{\widehat{\beta }-1}}</math>
 
If  <math>{{S}_{1}}={{S}_{2}}=\ldots ={{S}_{q}}=0</math>  and  <math>{{T}_{1}}={{T}_{2}}=\ldots ={{T}_{q}}</math>  <math>\,(q=1,2,\ldots ,K)</math>  then the maximum likelihood estimates  <math>\widehat{\lambda }</math>  and  <math>\widehat{\beta }</math>  are in closed form.
 
::<math>\begin{align}
  & \widehat{\lambda }= & \frac{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{N}_{q}}}{K{{T}^{\beta }}} \\
& \widehat{\beta }= & \frac{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{N}_{q}}}{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,\underset{i=1}{\overset{{{N}_{q}}}{\mathop{\sum }}}\,\ln (\tfrac{T}{{{X}_{iq}}})} 
\end{align}</math>
 
 
The following examples illustrate these estimation procedures.
<br>
<br>
=====Example 1=====
<br>
For the data in Table 13.1, the starting time for each system is equal to  <math>0</math>  and the ending time for each system is 2000 hours. Calculate the maximum likelihood estimates  <math>\widehat{\lambda }</math>  and  <math>\widehat{\beta }</math> .
 
<br>
{|system= align="center" border="1"
|-
|colspan="3" style="text-align:center"|Table 13.1 - Repairable system failure data
|-
!System 1 ( <math>{{X}_{i1}}</math> )
!System 2 ( <math>{{X}_{i2}}</math> )
!System 3 ( <math>{{X}_{i3}}</math> )
|-
|1.2|| 1.4|| 0.3
|-
|55.6|| 35.0|| 32.6
|-
|72.7|| 46.8|| 33.4
|-
|111.9|| 65.9|| 241.7
|-
|121.9|| 181.1|| 396.2
|-
|303.6|| 712.6|| 444.4
|-
|326.9|| 1005.7|| 480.8
|-
|1568.4|| 1029.9 ||588.9
|-
|1913.5|| 1675.7|| 1043.9
|-
| ||1787.5|| 1136.1
|-
| ||1867.0|| 1288.1
|-
| || ||1408.1
|-
| || ||1439.4
|-
| || ||1604.8
|-
|<math>{{N}_{1}}=9</math> || <math>{{N}_{2}}=11</math> ||<math>{{N}_{3}}=14</math>
|}
 
<br>
'''Solution'''
<br>
Since the starting time for each system is equal to zero and each system has an equivalent ending time, the general Eqns. (lambdaPowerLaw) and (BetaPowerLaw) reduce to the closed form Eqns. (sample1) and (sample2). The maximum likelihood estimates of  <math>\hat{\beta }</math>  and  <math>\hat{\lambda }</math>  are then calculated as follows:
 
::<math>\begin{align}
  & \widehat{\beta }= & \frac{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{N}_{q}}}{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,\underset{i=1}{\overset{{{N}_{q}}}{\mathop{\sum }}}\,\ln (\tfrac{T}{{{X}_{iq}}})} \\
& = & 0.45300 
\end{align}</math>
 
 
::<math>\begin{align}
  & \widehat{\lambda }= & \frac{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{N}_{q}}}{K{{T}^{\beta }}} \\
& = & 0.36224 
\end{align}</math>
 
 
[[Image:rga13.2.png|thumb|center|300px|Instantaneous Failure Intensity vs. Time plot.]]
 
<br>
The system failure intensity function is then estimated by:
 
::<math>\widehat{u}(t)=\widehat{\lambda }\widehat{\beta }{{t}^{\widehat{\beta }-1}},\text{ }t>0</math>
 
Figure wpp intensity is a plot of  <math>\widehat{u}(t)</math>  over the period (0, 3000). Clearly, the estimated failure intensity function is most representative over the range of the data and any extrapolation should be viewed with the usual caution.
 
===Goodness-of-Fit Tests for Repairable System Analysis===
<br>
It is generally desirable to test the compatibility of a model and data by a statistical goodness-of-fit test. A parametric Cramér-von Mises goodness-of-fit test is used for the multiple system and repairable system Power Law model, as proposed by Crow in [17]. This goodness-of-fit test is appropriate whenever the start time for each system is 0 and the failure data is complete over the continuous interval  <math>[0,{{T}_{q}}]</math>  with no gaps in the data. The Chi-Squared test is a goodness-of-fit test that can be applied under more general circumstances. In addition, the Common Beta Hypothesis test also can be used to compare the intensity functions of the individual systems by comparing the  <math>{{\beta }_{q}}</math>  values of each system. Lastly, the Laplace Trend test checks for trends within the data. Due to their general applicatoin, the Common Beta Hypothesis test and the Laplace Trend test are both presented in Appendix B. The Cramér-von Mises and Chi-Squared goodness-of-fit tests are illustrated next.
<br>
<br>
====Cramér-von Mises Test====
<br>
To illustrate the application of the Cramér-von Mises statistic for multiple system data, suppose that  <math>K</math>  like systems are under study and you wish to test the hypothesis  <math>{{H}_{1}}</math>  that their failure times follow a non-homogeneous Poisson process. Suppose information is available for the  <math>{{q}^{th}}</math>  system over the interval  <math>[0,{{T}_{q}}]</math>  , with successive failure times    ,  <math>(q=1,2,\ldots ,\,K)</math> . The Cramér-von Mises test can be performed with the following steps:
<br>
<br>
Step 1: If  <math>{{x}_{{{N}_{q}}q}}={{T}_{q}}</math>  (failure terminated) let  <math>{{M}_{q}}={{N}_{q}}-1</math> , and if  <math>{{x}_{{{N}_{q}}q}}<T</math>  (time terminated) let  <math>{{M}_{q}}={{N}_{q}}</math> . Then:
 
::<math>M=\underset{q=1}{\overset{K}{\mathop \sum }}\,{{M}_{q}}</math>
 
Step 2: For each system divide each successive failure time by the corresponding end time  <math>{{T}_{q}}</math> , <math>\,i=1,2,...,{{M}_{q}}.</math>  Calculate the  <math>M</math>  values:
 
::<math>{{Y}_{iq}}=\frac{{{X}_{iq}}}{{{T}_{q}}},i=1,2,\ldots ,{{M}_{q}},\text{ }q=1,2,\ldots ,K</math>
 
 
Step 3: Next calculate  <math>\overline{\beta }</math> , the unbiased estimate of  <math>\beta </math> , from:
 
::<math>\overline{\beta }=\frac{M-1}{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,\underset{i=1}{\overset{Mq}{\mathop{\sum }}}\,\ln \left( \tfrac{{{T}_{q}}}{{{X}_{i}}{{}_{q}}} \right)}</math>
 
 
Step 4: Treat the  <math>{{Y}_{iq}}</math>  values as one group and order them from smallest to largest. Name these ordered values  <math>{{z}_{1}},\,{{z}_{2}},\ldots ,{{z}_{M}}</math> , such that  <math>{{z}_{1}}<\ \ {{z}_{2}}<\ldots <{{z}_{M}}</math> .
<br>
<br>
Step 5: Calculate the parametric Cramér-von Mises statistic.
 
::<math>C_{M}^{2}=\frac{1}{12M}+\underset{j=1}{\overset{M}{\mathop \sum }}\,{{(Z_{j}^{\overline{\beta }}-\frac{2j-1}{2M})}^{2}}</math>
 
 
Critical values for the Cramér-von Mises test are presented in Table B.2 of Appendix B.
<br>
<br>
Step 6: If the calculated  <math>C_{M}^{2}</math>  is less than the critical value then accept the hypothesis that the failure times for the  <math>K</math>  systems follow the non-homogeneous Poisson process with intensity function  <math>u(t)=\lambda \beta {{t}^{\beta -1}}</math> .
<br>
<br>
=====Example 2=====
<br>
For the data from Example 1, use the Cramér-von Mises test to examine the compatibility of the model at a significance level  <math>\alpha =0.10</math>
<br>
<br>
''Solution''
<br>
Step 1:
 
::<math>\begin{align}
  & {{X}_{9,1}}= & 1913.5<2000,\,\ {{M}_{1}}=9 \\
& {{X}_{11,2}}= & 1867<2000,\,\ {{M}_{2}}=11 \\
& {{X}_{14,3}}= & 1604.8<2000,\,\ {{M}_{3}}=14 \\
& M= & \underset{q=1}{\overset{3}{\mathop \sum }}\,{{M}_{q}}=34 
\end{align}</math>
 
 
Step 2: Calculate  <math>{{Y}_{iq}},</math>  treat the  <math>{{Y}_{iq}}</math>  values as one group and order them from smallest to largest. Name these ordered values  <math>{{z}_{1}},\,{{z}_{2}},\ldots ,{{z}_{M}}</math> .
<br>
<br>
Step 3: Calculate  <math>\overline{\beta }=\tfrac{M-1}{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,\underset{i=1}{\overset{Mq}{\mathop{\sum }}}\,\ln \left( \tfrac{{{T}_{q}}}{{{X}_{i}}{{}_{q}}} \right)}=0.4397</math>
<br>
<br>
Step 4: Calculate  <math>C_{M}^{2}=\tfrac{1}{12M}+\underset{j=1}{\overset{M}{\mathop{\sum }}}\,{{(Z_{j}^{\overline{\beta }}-\tfrac{2j-1}{2M})}^{2}}=0.0611</math>
<br>
<br>
Step 5: Find the critical value (CV) from Table B.2 for  <math>M=34</math>  at a significance level  <math>\alpha =0.10</math> .  <math>CV=0.172</math> .
<br>
<br>
Step 6: Since  <math>C_{M}^{2}<CV</math> , accept the hypothesis that the failure times for the  <math>K=3</math>  repairable systems follow the non-homogeneous Poisson process with intensity function  <math>u(t)=\lambda \beta {{t}^{\beta -1}}</math> .
<br>
<br>
 
====Chi-Squared Test====
<br>
The parametric Cramér-von Mises test described above requires that the starting time,  <math>{{S}_{q}}</math> , be equal to 0 for each of the  <math>K</math>  systems. Although not as powerful as the Cramér-von Mises test, the Chi-Squared test can be applied regardless of the starting times. The expected number of failures for a system over its age  <math>(a,b)</math>  for the Chi-Squared test is estimated by  <math>\widehat{\lambda }{{b}^{\widehat{\beta }}}-\widehat{\lambda }{{a}^{\widehat{\beta }}}=\widehat{\theta }</math> , where  <math>\widehat{\lambda }</math>  and  <math>\widehat{\beta }</math>  are the maximum likelihood estimates.
The computed  <math>{{\chi }^{2}}</math>  statistic is:
 
::<math>{{\chi }^{2}}=\underset{j=1}{\overset{d}{\mathop \sum }}\,{{\frac{\left[ N(j)-\theta (j) \right]}{\widehat{\theta }(j)}}^{2}}</math>
 
where  <math>d</math>  is the total number of intervals. The random variable  <math>{{\chi }^{2}}</math>  is approximately Chi-Square distributed with  <math>df=d-2</math>  degrees of freedom. There must be at least three intervals and the length of the intervals do not have to be equal. It is common practice to require that the expected number of failures for each interval,  <math>\theta (j)</math> , be at least five. If  <math>\chi _{0}^{2}>\chi _{\alpha /2,d-2}^{2}</math>  or if  <math>\chi _{0}^{2}<\chi _{1-(\alpha /2),d-2}^{2}</math> , reject the null hypothesis.
 
===Confidence Bounds for Repairable Systems Analysis===
====Bounds on  <math>\beta </math>====
=====Fisher Matrix Bounds=====
The parameter  <math>\beta </math>  must be positive, thus  <math>\ln \beta </math>  is approximately treated as being normally distributed.
 
 
::<math>\frac{\ln (\widehat{\beta })-\ln (\beta )}{\sqrt{Var\left[ \ln (\widehat{\beta }) \right]}}\ \tilde{\ }\ N(0,1)</math>
 
 
::<math>C{{B}_{\beta }}=\widehat{\beta }{{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\widehat{\beta })}/\widehat{\beta }}}</math>
 
 
::<math>\widehat{\beta }=\frac{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{N}_{q}}}{\widehat{\lambda }\underset{q=1}{\overset{K}{\mathop{\sum }}}\,\left[ (T_{q}^{\widehat{\beta }}\ln ({{T}_{q}})-S_{q}^{\widehat{\beta }}\ln ({{S}_{q}}) \right]-\underset{q=1}{\overset{K}{\mathop{\sum }}}\,\underset{i=1}{\overset{{{N}_{q}}}{\mathop{\sum }}}\,\ln ({{X}_{i}}{{}_{q}})}</math>
 
 
All variance can be calculated using the Fisher Information Matrix.
<br>
<math>\Lambda </math>  is the natural log-likelihood function.
 
 
::<math>\Lambda =\underset{q=1}{\overset{K}{\mathop \sum }}\,\left[ {{N}_{q}}(\ln (\lambda )+\ln (\beta ))-\lambda (T_{q}^{\beta }-S_{q}^{\beta })+(\beta -1)\underset{i=1}{\overset{{{N}_{q}}}{\mathop \sum }}\,\ln ({{x}_{iq}}) \right]</math>
 
 
::<math>\frac{{{\partial }^{2}}\Lambda }{\partial {{\lambda }^{2}}}=-\frac{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{N}_{q}}}{{{\lambda }^{2}}}</math>
 
 
::<math>\frac{{{\partial }^{2}}\Lambda }{\partial \lambda \partial \beta }=-\underset{q=1}{\overset{K}{\mathop \sum }}\,\left[ T_{q}^{\beta }\ln ({{T}_{q}})-S_{q}^{\beta }\ln ({{S}_{q}}) \right]</math>
 
 
::<math>\frac{{{\partial }^{2}}\Lambda }{\partial {{\beta }^{2}}}=-\frac{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{N}_{q}}}{{{\beta }^{2}}}-\lambda \underset{q=1}{\overset{K}{\mathop \sum }}\,\left[ T_{q}^{\beta }{{(\ln ({{T}_{q}}))}^{2}}-S_{q}^{\beta }{{(\ln ({{S}_{q}}))}^{2}} \right]</math>
 
=====Crow Bounds=====
Calculate the conditional maximum likelihood estimate of  <math>\tilde{\beta }</math> :
 
 
::<math>\tilde{\beta }=\frac{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{M}_{q}}}{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,\underset{i=1}{\overset{M}{\mathop{\sum }}}\,\ln \left( \tfrac{{{T}_{q}}}{{{X}_{iq}}} \right)}</math>
 
 
The Crow 2-sided  <math>(1-a)</math> 100-percent confidence bounds on  <math>\beta </math>  are:
 
::<math>\begin{align}
  & {{\beta }_{L}}= & \tilde{\beta }\frac{\chi _{\tfrac{\alpha }{2},2M}^{2}}{2M} \\
& {{\beta }_{U}}= & \tilde{\beta }\frac{\chi _{1-\tfrac{\alpha }{2},2M}^{2}}{2M} 
\end{align}</math>
 
 
====Bounds on  <math>\lambda </math>====
=====Fisher Matrix Bounds=====
The parameter  <math>\lambda </math>  must be positive, thus  <math>\ln \lambda </math>  is approximately treated as being normally distributed. These bounds are based on:
 
 
::<math>\frac{\ln (\widehat{\lambda })-\ln (\lambda )}{\sqrt{Var\left[ \ln (\widehat{\lambda }) \right]}}\ \tilde{\ }\ N(0,1)</math>
 
<br>
The approximate confidence bounds on  <math>\lambda </math>  are given as:
 
 
::<math>C{{B}_{\lambda }}=\widehat{\lambda }{{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\widehat{\lambda })}/\widehat{\lambda }}}</math>
 
 
where  <math>\widehat{\lambda }=\tfrac{n}{T_{K}^{{\hat{\beta }}}}</math> .
The variance calculation is the same as Eqns. (var1), (var2) and (var3).
<br>
<br>
=====Crow Bounds=====
''Time Terminated''
<br>
The confidence bounds on  <math>\lambda </math>  for time terminated data are calculated using:
 
 
::<math>\begin{align}
  & {{\lambda }_{L}}= & \frac{\chi _{\tfrac{\alpha }{2},2N}^{2}}{2\cdot \underset{q=1}{\overset{K}{\mathop{\sum }}}\,T_{q}^{^{\beta }}} \\
& {{\lambda }_{u}}= & \frac{\chi _{1-\tfrac{\alpha }{2},2N+2}^{2}}{2\cdot \underset{q=1}{\overset{K}{\mathop{\sum }}}\,T_{q}^{^{\beta }}} 
\end{align}</math>
 
 
 
''Failure Terminated''
<br>
The confidence bounds on  <math>\lambda </math>  for failure terminated data are calculated using:
 
 
::<math>\begin{align}
  & {{\lambda }_{L}}= & \frac{\chi _{\tfrac{\alpha }{2},2N}^{2}}{2\cdot \underset{q=1}{\overset{K}{\mathop{\sum }}}\,T_{q}^{^{\beta }}} \\
& {{\lambda }_{u}}= & \frac{\chi _{1-\tfrac{\alpha }{2},2N}^{2}}{2\cdot \underset{q=1}{\overset{K}{\mathop{\sum }}}\,T_{q}^{^{\beta }}} 
\end{align}</math>
 
 
====Bounds on Growth Rate====
Since the growth rate is equal to  <math>1-\beta </math> , the confidence bounds are:
 
::<math>\begin{align}
  & Gr.\text{ }Rat{{e}_{L}}= & 1-{{\beta }_{U}} \\
& Gr.\text{ }Rat{{e}_{U}}= & 1-{{\beta }_{L}} 
\end{align}</math>
 
If Fisher Matrix confidence bounds are used then  <math>{{\beta }_{L}}</math>  and  <math>{{\beta }_{U}}</math>  are obtained from Eqn. (betafc). If Crow bounds are used then  <math>{{\beta }_{L}}</math>  and  <math>{{\beta }_{U}}</math>  are obtained from Eqn. (betacc).
<br>
<br>
 
====Bounds on Cumulative MTBF====
=====Fisher Matrix Bounds=====
The cumulative MTBF,  <math>{{m}_{c}}(t)</math> , must be positive, thus  <math>\ln {{m}_{c}}(t)</math>  is approximately treated as being normally distributed.
 
::<math>\frac{\ln ({{\widehat{m}}_{c}}(t))-\ln ({{m}_{c}}(t))}{\sqrt{Var\left[ \ln ({{\widehat{m}}_{c}}(t)) \right]}}\ \tilde{\ }\ N(0,1)</math>
 
The approximate confidence bounds on the cumulative MTBF are then estimated from:
 
 
::<math>CB={{\widehat{m}}_{c}}(t){{e}^{\pm {{z}_{\alpha }}\sqrt{Var({{\widehat{m}}_{c}}(t))}/{{\widehat{m}}_{c}}(t)}}</math>
 
:where:
 
::<math>{{\widehat{m}}_{c}}(t)=\frac{1}{\widehat{\lambda }}{{t}^{1-\widehat{\beta }}}</math>
 
 
::<math>\begin{align}
  & Var({{\widehat{m}}_{c}}(t))= & {{\left( \frac{\partial {{m}_{c}}(t)}{\partial \beta } \right)}^{2}}Var(\widehat{\beta })+{{\left( \frac{\partial {{m}_{c}}(t)}{\partial \lambda } \right)}^{2}}Var(\widehat{\lambda }) \\
&  & +2\left( \frac{\partial {{m}_{c}}(t)}{\partial \beta } \right)\left( \frac{\partial {{m}_{c}}(t)}{\partial \lambda } \right)cov(\widehat{\beta },\widehat{\lambda })\, 
\end{align}</math>
 
The variance calculation is the same as Eqns. (var1), (var2) and (var3).
 
::<math>\begin{align}
  & \frac{\partial {{m}_{c}}(t)}{\partial \beta }= & -\frac{1}{\widehat{\lambda }}{{t}^{1-\widehat{\beta }}}\ln (t) \\
& \frac{\partial {{m}_{c}}(t)}{\partial \lambda }= & -\frac{1}{{{\widehat{\lambda }}^{2}}}{{t}^{1-\widehat{\beta }}} 
\end{align}</math>
 
 
=====Crow Bounds=====
To calculate the Crow confidence bounds on cumulative MTBF, first calculate the Crow cumulative failure intensity confidence bounds:
 
::<math>C{{(t)}_{L}}=\frac{\chi _{\tfrac{\alpha }{2},2N}^{2}}{2\cdot t}</math>
 
 
::<math>C{{(t)}_{u}}=\frac{\chi _{1-\tfrac{\alpha }{2},2N+2}^{2}}{2\cdot t}</math>
 
:Then
 
::<math>\begin{align}
  & {{[MTB{{F}_{c}}]}_{L}}= & \frac{1}{C{{(t)}_{U}}} \\
& {{[MTB{{F}_{c}}]}_{U}}= & \frac{1}{C{{(t)}_{L}}} 
\end{align}</math>
 
 
====Bounds on Instantaneous MTBF====
=====Fisher Matrix Bounds=====
The instantaneous MTBF,  <math>{{m}_{i}}(t)</math> , must be positive, thus  <math>\ln {{m}_{i}}(t)</math>  is approximately treated as being normally distributed.
 
::<math>\frac{\ln ({{\widehat{m}}_{i}}(t))-\ln ({{m}_{i}}(t))}{\sqrt{Var\left[ \ln ({{\widehat{m}}_{i}}(t)) \right]}}\ \tilde{\ }\ N(0,1)</math>
 
 
The approximate confidence bounds on the instantaneous MTBF are then estimated from:
 
::<math>CB={{\widehat{m}}_{i}}(t){{e}^{\pm {{z}_{\alpha }}\sqrt{Var({{\widehat{m}}_{i}}(t))}/{{\widehat{m}}_{i}}(t)}}</math>
 
:where:
 
::<math>{{\widehat{m}}_{i}}(t)=\frac{1}{\lambda \beta {{t}^{\beta -1}}}</math>
 
::<math>\begin{align}
  & Var({{\widehat{m}}_{i}}(t))= & {{\left( \frac{\partial {{m}_{i}}(t)}{\partial \beta } \right)}^{2}}Var(\widehat{\beta })+{{\left( \frac{\partial {{m}_{i}}(t)}{\partial \lambda } \right)}^{2}}Var(\widehat{\lambda }) \\
&  & +2\left( \frac{\partial {{m}_{i}}(t)}{\partial \beta } \right)\left( \frac{\partial {{m}_{i}}(t)}{\partial \lambda } \right)cov(\widehat{\beta },\widehat{\lambda }) 
\end{align}</math>
 
 
The variance calculation is the same as (var1), (var2) and (var3).
 
::<math>\begin{align}
  & \frac{\partial {{m}_{i}}(t)}{\partial \beta }= & -\frac{1}{\widehat{\lambda }{{\widehat{\beta }}^{2}}}{{t}^{1-\widehat{\beta }}}-\frac{1}{\widehat{\lambda }\widehat{\beta }}{{t}^{1-\widehat{\beta }}}\ln (t) \\
& \frac{\partial {{m}_{i}}(t)}{\partial \lambda }= & -\frac{1}{{{\widehat{\lambda }}^{2}}\widehat{\beta }}{{t}^{1-\widehat{\beta }}} 
\end{align}</math>
 
 
=====Crow Bounds=====
''Failure Terminated Data''
<br>
To calculate the bounds for failure terminated data, consider the following equation:
 
::<math>G(\mu |n)=\mathop{}_{0}^{\infty }\frac{{{e}^{-x}}{{x}^{n-2}}}{(n-2)!}\underset{i=0}{\overset{n-1}{\mathop \sum }}\,\frac{1}{i!}{{\left( \frac{\mu }{x} \right)}^{i}}\exp (-\frac{\mu }{x})\,dx</math>
 
Find the values  <math>{{p}_{1}}</math>  and  <math>{{p}_{2}}</math>  by finding the solution  <math>c</math>  to  <math>G({{n}^{2}}/c|n)=\xi </math>  for  <math>\xi =\tfrac{\alpha }{2}</math>  and  <math>\xi =1-\tfrac{\alpha }{2}</math> , respectively. If using the biased parameters,  <math>\hat{\beta }</math>  and  <math>\hat{\lambda }</math> , then the upper and lower confidence bounds are:
 
::<math>\begin{align}
  & {{[MTB{{F}_{i}}]}_{L}}= & MTB{{F}_{i}}\cdot {{p}_{1}} \\
& {{[MTB{{F}_{i}}]}_{U}}= & MTB{{F}_{i}}\cdot {{p}_{2}} 
\end{align}</math>
 
where  <math>MTB{{F}_{i}}=\tfrac{1}{\hat{\lambda }\hat{\beta }{{t}^{\hat{\beta }-1}}}</math> . If using the unbiased parameters,  <math>\bar{\beta }</math>  and  <math>\bar{\lambda }</math> , then the upper and lower confidence bounds are:
 
::<math>\begin{align}
  & {{[MTB{{F}_{i}}]}_{L}}= & MTB{{F}_{i}}\cdot \left( \frac{N-2}{N} \right)\cdot {{p}_{1}} \\
& {{[MTB{{F}_{i}}]}_{U}}= & MTB{{F}_{i}}\cdot \left( \frac{N-2}{N} \right)\cdot {{p}_{2}} 
\end{align}</math>
 
where  <math>MTB{{F}_{i}}=\tfrac{1}{\hat{\lambda }\hat{\beta }{{t}^{\hat{\beta }-1}}}</math> .
<br>
<br>
''Time Terminated Data''
<br>
To calculate the bounds for time terminated data, consider the following equation where  <math>{{I}_{1}}(.)</math>  is the modified Bessel function of order one:
 
::<math>H(x|k)=\underset{j=1}{\overset{k}{\mathop \sum }}\,\frac{{{x}^{2j-1}}}{{{2}^{2j-1}}(j-1)!j!{{I}_{1}}(x)}</math>
 
Find the values  <math>{{\Pi }_{1}}</math>  and  <math>{{\Pi }_{2}}</math>  by finding the solution  <math>x</math>  to  <math>H(x|k)=\tfrac{\alpha }{2}</math>  and  <math>H(x|k)=1-\tfrac{\alpha }{2}</math>  in the cases corresponding to the lower and upper bounds, respectively. <br>
Calculate  <math>\Pi =\tfrac{{{n}^{2}}}{4{{x}^{2}}}</math>  for each case. If using the biased parameters,  <math>\hat{\beta }</math>  and  <math>\hat{\lambda }</math> , then the upper and lower confidence bounds are:
 
::<math>\begin{align}
  & {{[MTB{{F}_{i}}]}_{L}}= & MTB{{F}_{i}}\cdot {{\Pi }_{1}} \\
& {{[MTB{{F}_{i}}]}_{U}}= & MTB{{F}_{i}}\cdot {{\Pi }_{2}} 
\end{align}</math>
 
where  <math>MTB{{F}_{i}}=\tfrac{1}{\hat{\lambda }\hat{\beta }{{t}^{\hat{\beta }-1}}}</math> . If using the unbiased parameters,  <math>\bar{\beta }</math>  and  <math>\bar{\lambda }</math> , then the upper and lower confidence bounds are:
 
::<math>\begin{align}
  & {{[MTB{{F}_{i}}]}_{L}}= & MTB{{F}_{i}}\cdot \left( \frac{N-1}{N} \right)\cdot {{\Pi }_{1}} \\
& {{[MTB{{F}_{i}}]}_{U}}= & MTB{{F}_{i}}\cdot \left( \frac{N-1}{N} \right)\cdot {{\Pi }_{2}} 
\end{align}</math>
 
where  <math>MTB{{F}_{i}}=\tfrac{1}{\hat{\lambda }\hat{\beta }{{t}^{\hat{\beta }-1}}}</math> .
<br>
<br>
 
====Bounds on Cumulative Failure Intensity====
=====Fisher Matrix Bounds=====
The cumulative failure intensity,  <math>{{\lambda }_{c}}(t)</math>  must be positive, thus  <math>\ln {{\lambda }_{c}}(t)</math>  is approximately treated as being normally distributed.
 
::<math>\frac{\ln ({{\widehat{\lambda }}_{c}}(t))-\ln ({{\lambda }_{c}}(t))}{\sqrt{Var\left[ \ln ({{\widehat{\lambda }}_{c}}(t)) \right]}}\ \tilde{\ }\ N(0,1)</math>
 
The approximate confidence bounds on the cumulative failure intensity are then estimated using:
 
::<math>CB={{\widehat{\lambda }}_{c}}(t){{e}^{\pm {{z}_{\alpha }}\sqrt{Var({{\widehat{\lambda }}_{c}}(t))}/{{\widehat{\lambda }}_{c}}(t)}}</math>
 
:where:
 
::<math>{{\widehat{\lambda }}_{c}}(t)=\widehat{\lambda }{{t}^{\widehat{\beta }-1}}</math>
 
:and:
 
::<math>\begin{align}
  & Var({{\widehat{\lambda }}_{c}}(t))= & {{\left( \frac{\partial {{\lambda }_{c}}(t)}{\partial \beta } \right)}^{2}}Var(\widehat{\beta })+{{\left( \frac{\partial {{\lambda }_{c}}(t)}{\partial \lambda } \right)}^{2}}Var(\widehat{\lambda }) \\
&  & +2\left( \frac{\partial {{\lambda }_{c}}(t)}{\partial \beta } \right)\left( \frac{\partial {{\lambda }_{c}}(t)}{\partial \lambda } \right)cov(\widehat{\beta },\widehat{\lambda }) 
\end{align}</math>
 
 
The variance calculation is the same as Eqns. (var1), (var2) and (var3):
 
::<math>\begin{align}
  & \frac{\partial {{\lambda }_{c}}(t)}{\partial \beta }= & \widehat{\lambda }{{t}^{\widehat{\beta }-1}}\ln (t) \\
& \frac{\partial {{\lambda }_{c}}(t)}{\partial \lambda }= & {{t}^{\widehat{\beta }-1}} 
\end{align}</math>
 
<br>
=====Crow Bounds=====
The Crow cumulative failure intensity confidence bounds are given by:
 
::<math>C{{(t)}_{L}}=\frac{\chi _{\tfrac{\alpha }{2},2N}^{2}}{2\cdot t}</math>
 
 
::<math>C{{(t)}_{u}}=\frac{\chi _{1-\tfrac{\alpha }{2},2N+2}^{2}}{2\cdot t}</math>
 
 
====Bounds on Instantaneous Failure Intensity====
=====Fisher Matrix Bounds=====
The instantaneous failure intensity,  <math>{{\lambda }_{i}}(t)</math> , must be positive, thus  <math>\ln {{\lambda }_{i}}(t)</math>  is approximately treated as being normally distributed.
 
::<math>\frac{\ln ({{\widehat{\lambda }}_{i}}(t))-\ln ({{\lambda }_{i}}(t))}{\sqrt{Var\left[ \ln ({{\widehat{\lambda }}_{i}}(t)) \right]}}\sim N(0,1)</math>
 
<br>
The approximate confidence bounds on the instantaneous failure intensity are then estimated from:
 
::<math>CB={{\widehat{\lambda }}_{i}}(t){{e}^{\pm {{z}_{\alpha }}\sqrt{Var({{\widehat{\lambda }}_{i}}(t))}/{{\widehat{\lambda }}_{i}}(t)}}</math>
 
 
where  <math>{{\lambda }_{i}}(t)=\lambda \beta {{t}^{\beta -1}}</math>  and:
 
::<math>\begin{align}
  & Var({{\widehat{\lambda }}_{i}}(t))= & {{\left( \frac{\partial {{\lambda }_{i}}(t)}{\partial \beta } \right)}^{2}}Var(\widehat{\beta })+{{\left( \frac{\partial {{\lambda }_{i}}(t)}{\partial \lambda } \right)}^{2}}Var(\widehat{\lambda }) \\
&  & +2\left( \frac{\partial {{\lambda }_{i}}(t)}{\partial \beta } \right)\left( \frac{\partial {{\lambda }_{i}}(t)}{\partial \lambda } \right)cov(\widehat{\beta },\widehat{\lambda }) 
\end{align}</math>
 
<br>
The variance calculation is the same as Eqns. (var1), (var2) and (var3):
 
::<math>\begin{align}
  & \frac{\partial {{\lambda }_{i}}(t)}{\partial \beta }= & \hat{\lambda }{{t}^{\widehat{\beta }-1}}+\hat{\lambda }\hat{\beta }{{t}^{\widehat{\beta }-1}}\ln (t) \\
& \frac{\partial {{\lambda }_{i}}(t)}{\partial \lambda }= & \widehat{\beta }{{t}^{\widehat{\beta }-1}} 
\end{align}</math>
 
 
=====Crow Bounds=====
The Crow instantaneous failure intensity confidence bounds are given as:
 
::<math>\begin{align}
  & {{[{{\lambda }_{i}}(t)]}_{L}}= & \frac{1}{{{[MTB{{F}_{i}}]}_{U}}} \\
& {{[{{\lambda }_{i}}(t)]}_{U}}= & \frac{1}{{{[MTB{{F}_{i}}]}_{L}}} 
\end{align}</math>
 
 
====Bounds on Time Given Cumulative MTBF====
=====Fisher Matrix Bounds=====
The time,  <math>T</math> , must be positive, thus  <math>\ln T</math>  is approximately treated as being normally distributed.
 
::<math>\frac{\ln (\widehat{T})-\ln (T)}{\sqrt{Var\left[ \ln (\widehat{T}) \right]}}\ \tilde{\ }\ N(0,1)</math>
 
The confidence bounds on the time are given by:
 
::<math>CB=\widehat{T}{{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\widehat{T})}/\widehat{T}}}</math>
 
:where:
 
::<math>Var(\widehat{T})={{\left( \frac{\partial T}{\partial \beta } \right)}^{2}}Var(\widehat{\beta })+{{\left( \frac{\partial T}{\partial \lambda } \right)}^{2}}Var(\widehat{\lambda })+2\left( \frac{\partial T}{\partial \beta } \right)\left( \frac{\partial T}{\partial \lambda } \right)cov(\widehat{\beta },\widehat{\lambda })</math>
 
The variance calculation is the same as Eqns. (var1), (var2) and (var3).
 
::<math>\widehat{T}={{(\lambda \cdot {{m}_{c}})}^{1/(1-\beta )}}</math>
 
 
::<math>\begin{align}
  & \frac{\partial T}{\partial \beta }= & \frac{{{(\lambda \cdot {{m}_{c}})}^{1/(1-\beta )}}\ln (\lambda \cdot {{m}_{c}})}{{{(1-\beta )}^{2}}} \\
& \frac{\partial T}{\partial \lambda }= & \frac{{{(\lambda \cdot {{m}_{c}})}^{1/(1-\beta )}}}{\lambda (1-\beta )} 
\end{align}</math>
 
 
=====Crow Bounds=====
Step 1: Calculate:
 
::<math>\hat{T}={{\left( \frac{{{\lambda }_{c}}(T)}{{\hat{\lambda }}} \right)}^{\tfrac{1}{\beta -1}}}</math>
 
Step 2: Estimate the number of failures:
 
::<math>N(\hat{T})=\hat{\lambda }{{\hat{T}}^{{\hat{\beta }}}}</math>
 
Step 3: Obtain the confidence bounds on time given the cumulative failure intensity by solving for  <math>{{t}_{l}}</math>  and  <math>{{t}_{u}}</math>  in the following equations:
 
::<math>\begin{align}
  & {{t}_{l}}= & \frac{\chi _{\tfrac{\alpha }{2},2N}^{2}}{2\cdot {{\lambda }_{c}}(T)} \\
& {{t}_{u}}= & \frac{\chi _{1-\tfrac{\alpha }{2},2N+2}^{2}}{2\cdot {{\lambda }_{c}}(T)} 
\end{align}</math>
 
<br>
 
====Bounds on Time Given Instantaneous MTBF====
=====Fisher Matrix Bounds=====
The time,  <math>T</math> , must be positive, thus  <math>\ln T</math>  is approximately treated as being normally distributed.
 
::<math>\frac{\ln (\widehat{T})-\ln (T)}{\sqrt{Var\left[ \ln (\widehat{T}) \right]}}\ \tilde{\ }\ N(0,1)</math>
 
The confidence bounds on the time are given by:
 
::<math>CB=\widehat{T}{{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\widehat{T})}/\widehat{T}}}</math>
 
:where:
 
::<math>Var(\widehat{T})={{\left( \frac{\partial T}{\partial \beta } \right)}^{2}}Var(\widehat{\beta })+{{\left( \frac{\partial T}{\partial \lambda } \right)}^{2}}Var(\widehat{\lambda })+2\left( \frac{\partial T}{\partial \beta } \right)\left( \frac{\partial T}{\partial \lambda } \right)cov(\widehat{\beta },\widehat{\lambda })</math>
 
The variance calculation is the same as Eqns. (var1), (var2) and (var3).
 
 
::<math>\widehat{T}={{(\lambda \beta \cdot MTB{{F}_{i}})}^{1/(1-\beta )}}</math>
 
 
::<math>\begin{align}
  & \frac{\partial T}{\partial \beta }= & {{\left( \lambda \beta \cdot MTB{{F}_{i}} \right)}^{1/(1-\beta )}}[\frac{1}{{{(1-\beta )}^{2}}}\ln (\lambda \beta \cdot MTB{{F}_{i}})+\frac{1}{\beta (1-\beta )}] \\
& \frac{\partial T}{\partial \lambda }= & \frac{{{(\lambda \beta \cdot MTB{{F}_{i}})}^{1/(1-\beta )}}}{\lambda (1-\beta )} 
\end{align}</math>
 
<br>
=====Crow Bounds=====
Step 1: Calculate the confidence bounds on the instantaneous MTBF as presented in Section 5.5.2.
<br>
Step 2: Calculate the bounds on time as follows.
<br>
<br>
''Failure Terminated Data''
 
::<math>\hat{T}={{(\frac{\lambda \beta \cdot MTB{{F}_{i}}}{c})}^{1/(1-\beta )}}</math>
 
 
So the lower an upper bounds on time are:
 
 
::<math>{{\hat{T}}_{L}}={{(\frac{\lambda \beta \cdot MTB{{F}_{i}}}{{{c}_{1}}})}^{1/(1-\beta )}}</math>
 
 
::<math>{{\hat{T}}_{U}}={{(\frac{\lambda \beta \cdot MTB{{F}_{i}}}{{{c}_{2}}})}^{1/(1-\beta )}}</math>
 
 
''Time Terminated Data''
 
::<math>\hat{T}={{(\frac{\lambda \beta \cdot MTB{{F}_{i}}}{\Pi })}^{1/(1-\beta )}}</math>
 
 
So the lower and upper bounds on time are:
 
 
::<math>{{\hat{T}}_{L}}={{(\frac{\lambda \beta \cdot MTB{{F}_{i}}}{{{\Pi }_{1}}})}^{1/(1-\beta )}}</math>
 
 
::<math>{{\hat{T}}_{U}}={{(\frac{\lambda \beta \cdot MTB{{F}_{i}}}{{{\Pi }_{2}}})}^{1/(1-\beta )}}</math>
 
 
====Bounds on Time Given Cumulative Failure Intensity====
=====Fisher Matrix Bounds=====
The time,  <math>T</math> , must be positive, thus  <math>\ln T</math>  is approximately treated as being normally distributed.
 
::<math>\frac{\ln (\widehat{T})-\ln (T)}{\sqrt{Var\left[ \ln \widehat{T} \right]}}\ \tilde{\ }\ N(0,1)</math>
 
The confidence bounds on the time are given by:
 
::<math>CB=\widehat{T}{{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\widehat{T})}/\widehat{T}}}</math>
 
:where:
 
::<math>Var(\widehat{T})={{\left( \frac{\partial T}{\partial \beta } \right)}^{2}}Var(\widehat{\beta })+{{\left( \frac{\partial T}{\partial \lambda } \right)}^{2}}Var(\widehat{\lambda })+2\left( \frac{\partial T}{\partial \beta } \right)\left( \frac{\partial T}{\partial \lambda } \right)cov(\widehat{\beta },\widehat{\lambda })</math>
 
The variance calculation is the same as Eqns. (var1), (var2) and (var3):
 
::<math>\widehat{T}={{\left( \frac{{{\lambda }_{c}}(T)}{\lambda } \right)}^{1/(\beta -1)}}</math>
 
::<math>\begin{align}
  & \frac{\partial T}{\partial \beta }= & \frac{-{{\left( \tfrac{{{\lambda }_{c}}(T)}{\lambda } \right)}^{1/(\beta -1)}}\ln \left( \tfrac{{{\lambda }_{c}}(T)}{\lambda } \right)}{{{(1-\beta )}^{2}}} \\
& \frac{\partial T}{\partial \lambda }= & {{\left( \frac{{{\lambda }_{c}}(T)}{\lambda } \right)}^{1/(\beta -1)}}\frac{1}{\lambda (1-\beta )} 
\end{align}</math>
 
 
=====Crow Bounds=====
Step 1: Calculate:
 
 
::<math>\hat{T}={{\left( \frac{{{\lambda }_{c}}(T)}{{\hat{\lambda }}} \right)}^{\tfrac{1}{\beta -1}}}</math>
 
 
Step 2: Estimate the number of failures:
 
 
::<math>N(\hat{T})=\hat{\lambda }{{\hat{T}}^{{\hat{\beta }}}}</math>
 
 
Step 3: Obtain the confidence bounds on time given the cumulative failure intensity by solving for  <math>{{t}_{l}}</math>  and  <math>{{t}_{u}}</math>  in the following equations:
 
::<math>\begin{align}
  & {{t}_{l}}= & \frac{\chi _{\tfrac{\alpha }{2},2N}^{2}}{2\cdot {{\lambda }_{c}}(T)} \\
& {{t}_{u}}= & \frac{\chi _{1-\tfrac{\alpha }{2},2N+2}^{2}}{2\cdot {{\lambda }_{c}}(T)} 
\end{align}</math>
 
 
====Bounds on Time Given Instantaneous Failure Intensity====
=====Fisher Matrix Bounds=====
These bounds are based on:
 
::<math>\frac{\ln (\widehat{T})-\ln (T)}{\sqrt{Var\left[ \ln (\widehat{T}) \right]}}\sim N(0,1)</math>
 
 
The confidence bounds on the time are given by:
 
 
::<math>CB=\widehat{T}{{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\widehat{T})}/\widehat{T}}}</math>
 
:where:
 
::<math>\begin{align}
  & Var(\widehat{T})= & {{\left( \frac{\partial T}{\partial \beta } \right)}^{2}}Var(\widehat{\beta })+{{\left( \frac{\partial T}{\partial \lambda } \right)}^{2}}Var(\widehat{\lambda }) \\
&  & +2\left( \frac{\partial T}{\partial \beta } \right)\left( \frac{\partial T}{\partial \lambda } \right)cov(\widehat{\beta },\widehat{\lambda }) 
\end{align}</math>
 
The variance calculation is the same as Eqns. (var1), (var2) and (var3).
 
::<math>\widehat{T}={{\left( \frac{{{\lambda }_{i}}(T)}{\lambda \cdot \beta } \right)}^{1/(\beta -1)}}</math>
 
 
::<math>\begin{align}
  & \frac{\partial T}{\partial \beta }= & {{\left( \frac{{{\lambda }_{i}}(T)}{\lambda \cdot \beta } \right)}^{1/(\beta -1)}}[-\frac{\ln (\tfrac{{{\lambda }_{i}}(T)}{\lambda \cdot \beta })}{{{(\beta -1)}^{2}}}+\frac{1}{\beta (1-\beta )}] \\
& \frac{\partial T}{\partial \lambda }= & {{\left( \frac{{{\lambda }_{i}}(T)}{\lambda \cdot \beta } \right)}^{1/(\beta -1)}}\frac{1}{\lambda (1-\beta )} 
\end{align}</math>
 
 
=====Crow Bounds=====
Step 1: Calculate  <math>{{\lambda }_{i}}(T)=\tfrac{1}{MTB{{F}_{i}}}</math> .
<br>
Step 2: Use the equations from 13.1.7.9 to calculate the bounds on time given the instantaneous failure intensity.
<br>
<br>
====Bounds on Reliability====
=====Fisher Matrix Bounds=====
These bounds are based on:
 
::<math>\log it(\widehat{R}(t))\sim N(0,1)</math>
 
 
::<math>\log it(\widehat{R}(t))=\ln \left\{ \frac{\widehat{R}(t)}{1-\widehat{R}(t)} \right\}</math>
 
 
The confidence bounds on reliability are given by:
 
::<math>CB=\frac{\widehat{R}(t)}{\widehat{R}(t)+(1-\widehat{R}(t)){{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\widehat{R}(t))}/\left[ \widehat{R}(t)(1-\widehat{R}(t)) \right]}}}</math>
 
 
::<math>Var(\widehat{R}(t))={{\left( \frac{\partial R}{\partial \beta } \right)}^{2}}Var(\widehat{\beta })+{{\left( \frac{\partial R}{\partial \lambda } \right)}^{2}}Var(\widehat{\lambda })+2\left( \frac{\partial R}{\partial \beta } \right)\left( \frac{\partial R}{\partial \lambda } \right)cov(\widehat{\beta },\widehat{\lambda })</math>
 
 
The variance calculation is the same as Eqns. (var1), (var2) and (var3).
 
::<math>\begin{align}
  & \frac{\partial R}{\partial \beta }= & {{e}^{-[\widehat{\lambda }{{(t+d)}^{\widehat{\beta }}}-\widehat{\lambda }{{t}^{\widehat{\beta }}}]}}[\lambda {{t}^{\widehat{\beta }}}\ln (t)-\lambda {{(t+d)}^{\widehat{\beta }}}\ln (t+d)] \\
& \frac{\partial R}{\partial \lambda }= & {{e}^{-[\widehat{\lambda }{{(t+d)}^{\widehat{\beta }}}-\widehat{\lambda }{{t}^{\widehat{\beta }}}]}}[{{t}^{\widehat{\beta }}}-{{(t+d)}^{\widehat{\beta }}}] 
\end{align}</math>
 
 
=====Crow Bounds=====
''Failure Terminated Data''
<br>
With failure terminated data, the 100( <math>1-\alpha </math> )% confidence interval for the current reliability at time  <math>t</math>  in a specified mission time  <math>d</math>  is:
 
::<math>({{[\widehat{R}(d)]}^{\tfrac{1}{{{p}_{1}}}}},{{[\hat{R}(d)]}^{\tfrac{1}{{{p}_{2}}}}})</math>
 
:where
 
::<math>\widehat{R}(\tau )={{e}^{-[\widehat{\lambda }{{(t+\tau )}^{\widehat{\beta }}}-\widehat{\lambda }{{t}^{\widehat{\beta }}}]}}</math>
 
<math>{{p}_{1}}</math> and  <math>{{p}_{2}}</math>  can be obtained from Eqn. (ft).
<br>
<br>
''Time Terminated Data''
<br>
With time terminated data, the 100( <math>1-\alpha </math> )% confidence interval for the current reliability at time  <math>t</math>  in a specified mission time  <math>\tau </math>  is:
 
::<math>({{[\widehat{R}(d)]}^{\tfrac{1}{{{p}_{1}}}}},{{[\hat{R}(d)]}^{\tfrac{1}{{{p}_{2}}}}})</math>
 
:where:
 
::<math>\widehat{R}(d)={{e}^{-[\widehat{\lambda }{{(t+d)}^{\widehat{\beta }}}-\widehat{\lambda }{{t}^{\widehat{\beta }}}]}}</math>
 
<math>{{p}_{1}}</math>  and  <math>{{p}_{2}}</math>  can be obtained from Eqn. (tt).
 
====Bounds on Time Given Reliability and Mission Time====
=====Fisher Matrix Bounds=====
The time,  <math>t</math> , must be positive, thus  <math>\ln t</math>  is approximately treated as being normally distributed.
 
::<math>\frac{\ln (\hat{t})-\ln (t)}{\sqrt{Var\left[ \ln (\hat{t}) \right]}}\sim N(0,1)</math>
 
The confidence bounds on time are calculated by using:
 
::<math>CB=\hat{t}{{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\hat{t})}/\hat{t}}}</math>
 
:where:
 
::<math>Var(\hat{t})={{\left( \frac{\partial t}{\partial \beta } \right)}^{2}}Var(\widehat{\beta })+{{\left( \frac{\partial t}{\partial \lambda } \right)}^{2}}Var(\widehat{\lambda })+2\left( \frac{\partial t}{\partial \beta } \right)\left( \frac{\partial t}{\partial \lambda } \right)cov(\widehat{\beta },\widehat{\lambda })</math>
 
::<math>\hat{t}</math>  is calculated numerically from:
 
::<math>\widehat{R}(d)={{e}^{-[\widehat{\lambda }{{(\hat{t}+d)}^{\widehat{\beta }}}-\widehat{\lambda }{{{\hat{t}}}^{\widehat{\beta }}}]}}\text{ };\text{ }d\text{ = mission time}</math>
 
The variance calculations are done by:
 
::<math>\begin{align}
  & \frac{\partial t}{\partial \beta }= & \frac{{{{\hat{t}}}^{{\hat{\beta }}}}\ln (\hat{t})-{{(\hat{t}+d)}^{{\hat{\beta }}}}\ln (\hat{t}+d)}{\hat{\beta }{{(\hat{t}+d)}^{\hat{\beta }-1}}-\hat{\beta }{{{\hat{t}}}^{\hat{\beta }-1}}} \\
& \frac{\partial t}{\partial \lambda }= & \frac{{{{\hat{t}}}^{{\hat{\beta }}}}-{{(\hat{t}+d)}^{{\hat{\beta }}}}}{\hat{\lambda }\hat{\beta }{{(\hat{t}+d)}^{\hat{\beta }-1}}-\hat{\lambda }\hat{\beta }{{{\hat{t}}}^{\hat{\beta }-1}}} 
\end{align}</math>
 
=====Crow Bounds=====
''Failure Terminated Data''
<br>
Step 1: Calculate  <math>({{\hat{R}}_{lower}},{{\hat{R}}_{upper}})=({{R}^{\tfrac{1}{{{p}_{1}}}}},{{R}^{\tfrac{1}{{{p}_{2}}}}})</math> .
<br>
Step 2: Let  <math>R={{\hat{R}}_{lower}}</math>  and solve for  <math>{{t}_{1}}</math>  numerically using  <math>R={{e}^{-[\widehat{\lambda }{{({{{\hat{t}}}_{1}}+d)}^{\widehat{\beta }}}-\widehat{\lambda }\hat{t}_{1}^{\widehat{\beta }}]}}</math> .
<br>
Step 3: Let  <math>R={{\hat{R}}_{upper}}</math>  and solve for  <math>{{t}_{2}}</math>  numerically using  <math>R={{e}^{-[\widehat{\lambda }{{({{{\hat{t}}}_{2}}+d)}^{\widehat{\beta }}}-\widehat{\lambda }\hat{t}_{2}^{\widehat{\beta }}]}}</math> .
<br>
Step 4: If  <math>{{t}_{1}}<{{t}_{2}}</math> , then  <math>{{t}_{lower}}={{t}_{1}}</math>  and  <math>{{t}_{upper}}={{t}_{2}}</math> . If  <math>{{t}_{1}}>{{t}_{2}}</math> , then  <math>{{t}_{lower}}={{t}_{2}}</math>  and  <math>{{t}_{upper}}={{t}_{1}}</math> .
<br>
<br>
''Time Terminated Data''
<br>
Step 1: Calculate  <math>({{\hat{R}}_{lower}},{{\hat{R}}_{upper}})=({{R}^{\tfrac{1}{{{\Pi }_{1}}}}},{{R}^{\tfrac{1}{{{\Pi }_{2}}}}})</math> .
<br>
Step 2: Let  <math>R={{\hat{R}}_{lower}}</math>  and solve for  <math>{{t}_{1}}</math>  numerically using  <math>R={{e}^{-[\widehat{\lambda }{{({{{\hat{t}}}_{1}}+d)}^{\widehat{\beta }}}-\widehat{\lambda }\hat{t}_{1}^{\widehat{\beta }}]}}</math> .
<br>
Step 3: Let  <math>R={{\hat{R}}_{upper}}</math>  and solve for  <math>{{t}_{2}}</math>  numerically using  <math>R={{e}^{-[\widehat{\lambda }{{({{{\hat{t}}}_{2}}+d)}^{\widehat{\beta }}}-\widehat{\lambda }\hat{t}_{2}^{\widehat{\beta }}]}}</math> .
<br>
Step 4: If  <math>{{t}_{1}}<{{t}_{2}}</math> , then  <math>{{t}_{lower}}={{t}_{1}}</math>  and  <math>{{t}_{upper}}={{t}_{2}}</math> . If  <math>{{t}_{1}}>{{t}_{2}}</math> , then  <math>{{t}_{lower}}={{t}_{2}}</math>  and  <math>{{t}_{upper}}={{t}_{1}}</math> .
<br>
<br>
====Bounds on Mission Time Given Reliability and Time====
=====Fisher Matrix Bounds=====
The mission time,  <math>d</math> , must be positive, thus  <math>\ln \left( d \right)</math>  is approximately treated as being normally distributed.
 
::<math>\frac{\ln (\hat{d})-\ln (d)}{\sqrt{Var\left[ \ln (\hat{d}) \right]}}\sim N(0,1)</math>
 
 
The confidence bounds on mission time are given by using:
 
 
::<math>CB=\hat{d}{{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\hat{d})}/\hat{d}}}</math>
 
 
:where:
 
 
::<math>Var(\hat{d})={{\left( \frac{\partial d}{\partial \beta } \right)}^{2}}Var(\widehat{\beta })+{{\left( \frac{\partial d}{\partial \lambda } \right)}^{2}}Var(\widehat{\lambda })+2\left( \frac{\partial td}{\partial \beta } \right)\left( \frac{\partial d}{\partial \lambda } \right)cov(\widehat{\beta },\widehat{\lambda })</math>
 
 
Calculate  <math>\hat{d}</math>  from:
 
 
::<math>\hat{d}={{\left[ {{t}^{{\hat{\beta }}}}-\frac{\ln (R)}{{\hat{\lambda }}} \right]}^{\tfrac{1}{{\hat{\beta }}}}}-t</math>
 
 
The variance calculations are done by:
 
 
::<math>\begin{align}
  & \frac{\partial d}{\partial \beta }= & \left[ \frac{{{t}^{{\hat{\beta }}}}\ln (t)}{{{(t+\hat{d})}^{{\hat{\beta }}}}}-\ln (t+\hat{d}) \right]\cdot \frac{t+\hat{d}}{{\hat{\beta }}} \\
& \frac{\partial d}{\partial \lambda }= & \frac{{{t}^{{\hat{\beta }}}}-{{(t+\hat{d})}^{{\hat{\beta }}}}}{\hat{\lambda }\hat{\beta }{{(t+\hat{d})}^{\hat{\beta }-1}}} 
\end{align}</math>
 
 
=====Crow Bounds=====
''Failure Terminated Data''
<br>
Step 1: Calculate  <math>({{\hat{R}}_{lower}},{{\hat{R}}_{upper}})=({{R}^{\tfrac{1}{{{p}_{1}}}}},{{R}^{\tfrac{1}{{{p}_{2}}}}})</math> .
<br>
Step 2: Let  <math>R={{\hat{R}}_{lower}}</math>  and solve for  <math>{{d}_{1}}</math>  such that:
 
 
::<math>{{d}_{1}}={{\left( {{t}^{{\hat{\beta }}}}-\frac{\ln ({{R}_{lower}})}{{\hat{\lambda }}} \right)}^{\tfrac{1}{{\hat{\beta }}}}}-t</math>
 
 
Step 3: Let  <math>R={{\hat{R}}_{upper}}</math>  and solve for  <math>{{d}_{2}}</math>  such that:
 
 
::<math>{{d}_{2}}={{\left( {{t}^{{\hat{\beta }}}}-\frac{\ln ({{R}_{upper}})}{{\hat{\lambda }}} \right)}^{\tfrac{1}{{\hat{\beta }}}}}-t</math>
 
 
Step 4: If  <math>{{d}_{1}}<{{d}_{2}}</math> , then  <math>{{d}_{lower}}={{d}_{1}}</math>  and  <math>{{d}_{upper}}={{d}_{2}}</math> . If  <math>{{d}_{1}}>{{d}_{2}}</math> , then  <math>{{d}_{lower}}={{d}_{2}}</math>  and  <math>{{d}_{upper}}={{d}_{1}}</math> .
<br>
<br>
''Time Terminated Data''
<br>
Step 1: Calculate  <math>({{\hat{R}}_{lower}},{{\hat{R}}_{upper}})=({{R}^{\tfrac{1}{{{\Pi }_{1}}}}},{{R}^{\tfrac{1}{{{\Pi }_{2}}}}})</math> .
<br>
Step 2: Let  <math>R={{\hat{R}}_{lower}}</math>  and solve for  <math>{{d}_{1}}</math>  using Eqn. (CBR1).
<br>
Step 3: Let  <math>R={{\hat{R}}_{upper}}</math>  and solve for  <math>{{d}_{2}}</math>  using Eqn. (CBR2).
<br>
Step 4: If  <math>{{d}_{1}}<{{d}_{2}}</math> , then  <math>{{d}_{lower}}={{d}_{1}}</math>  and  <math>{{d}_{upper}}={{d}_{2}}</math> . If  <math>{{d}_{1}}>{{d}_{2}}</math> , then  <math>{{d}_{lower}}={{d}_{2}}</math>  and  <math>{{d}_{upper}}={{d}_{1}}</math> .
<br>
<br>
====Bounds on Cumulative Number of Failures====
=====Fisher Matrix Bounds=====
The cumulative number of failures,  <math>N(t)</math> , must be positive, thus  <math>\ln \left( N(t) \right)</math>  is approximately treated as being normally distributed.
 
::<math>\frac{\ln (\widehat{N}(t))-\ln (N(t))}{\sqrt{Var\left[ \ln \widehat{N}(t) \right]}}\sim N(0,1)</math>
 
 
::<math>N(t)=\widehat{N}(t){{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\widehat{N}(t))}/\widehat{N}(t)}}</math>
 
 
:where:
 
::<math>\widehat{N}(t)=\widehat{\lambda }{{t}^{\widehat{\beta }}}</math>
 
<br>
::<math>\begin{align}
  & Var(\widehat{N}(t))= & {{\left( \frac{\partial N(t)}{\partial \beta } \right)}^{2}}Var(\widehat{\beta })+{{\left( \frac{\partial N(t)}{\partial \lambda } \right)}^{2}}Var(\widehat{\lambda }) \\
&  & +2\left( \frac{\partial N(t)}{\partial \beta } \right)\left( \frac{\partial N(t)}{\partial \lambda } \right)cov(\widehat{\beta },\widehat{\lambda }) 
\end{align}</math>
 
 
The variance calculation is the same as Eqns. (var1), (var2) and (var3).
 
<br>
::<math>\begin{align}
  & \frac{\partial N(t)}{\partial \beta }= & \hat{\lambda }{{t}^{\widehat{\beta }}}\ln (t) \\
& \frac{\partial N(t)}{\partial \lambda }= & t\widehat{\beta } 
\end{align}</math>
 
<br>
=====Crow Bounds=====
::<math>\begin{array}{*{35}{l}}
  {{N}_{L}}(T)=\tfrac{T}{\widehat{\beta }}{{\lambda }_{i}}{{(T)}_{L}}  \\
  {{N}_{U}}(T)=\tfrac{T}{\widehat{\beta }}{{\lambda }_{i}}{{(T)}_{U}}  \\
\end{array}</math>
 
where  <math>{{\lambda }_{i}}{{(T)}_{L}}</math>  and  <math>{{\lambda }_{i}}{{(T)}_{U}}</math>  can be obtained from Eqn. (inr).
<br>
<br>
=====Example 3=====
Using the data from Example 1, calculate the mission reliability at  <math>t=2000</math>  hours and mission time  <math>d=40</math>  hours  along with the confidence bounds at the 90% confidence level.
<br>
''Solution''
<br>
The maximum likelihood estimates of  <math>\widehat{\lambda }</math>  and  <math>\widehat{\beta }</math>  from Example 1 are:
 
 
::<math>\begin{align}
  & \widehat{\beta }= & 0.45300 \\
& \widehat{\lambda }= & 0.36224 
\end{align}</math>
 
 
From Eq. (reliability), the mission reliability at  <math>t=2000</math>  for mission time  <math>d=40</math>  is:
 
::<math>\begin{align}
  & \widehat{R}(t)= & {{e}^{-\left[ \lambda {{\left( t+d \right)}^{\beta }}-\lambda {{t}^{\beta }} \right]}} \\
& = & 0.90292 
\end{align}</math>
 
 
At the 90% confidence level and  <math>T=2000</math>  hours, the Fisher Matrix confidence bounds for the mission reliability for mission time  <math>d=40</math>  are given by:
 
::<math>CB=\frac{\widehat{R}(t)}{\widehat{R}(t)+(1-\widehat{R}(t)){{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\widehat{R}(t))}/\left[ \widehat{R}(t)(1-\widehat{R}(t)) \right]}}}</math>
 
 
::<math>\begin{align}
  & {{[\widehat{R}(t)]}_{L}}= & 0.83711 \\
& {{[\widehat{R}(t)]}_{U}}= & 0.94392 
\end{align}</math>
 
 
The Crow confidence bounds for the mission reliability are:
 
::<math>\begin{align}
  & {{[\widehat{R}(t)]}_{L}}= & {{[\widehat{R}(\tau )]}^{\tfrac{1}{{{\Pi }_{1}}}}} \\
& = & {{[0.90292]}^{\tfrac{1}{0.71440}}} \\
& = & 0.86680 \\
& {{[\widehat{R}(t)]}_{U}}= & {{[\widehat{R}(\tau )]}^{\tfrac{1}{{{\Pi }_{2}}}}} \\
& = & {{[0.90292]}^{\tfrac{1}{1.6051}}} \\
& = & 0.93836 
\end{align}</math>
 
 
Figures ConfReliFish and ConfRelCrow show the Fisher Matrix and Crow confidence bounds on mission reliability for mission time  <math>d=40</math> .
 
[[Image:rga13.3.png|thumb|center|300px|Conditional Reliability vs. Time plot with Fisher Matrix confidence bounds.]]
<br>
<br>
[[Image:rga13.4.png|thumb|center|300px|Conditional Reliability vs. Time plot with Crow confidence bounds.]]
 
<br>
 
===Economical Life Model===
<br>
One consideration in reducing the cost to maintain repairable systems is to establish an overhaul policy that will minimize the total life cost of the system. However, an overhaul policy makes sense only if  <math>\beta >1</math> . It does not make sense to implement an overhaul policy if  <math>\beta <1</math>  since wearout is not present. If you assume that there is a point at which it is cheaper to overhaul a system than to continue repairs, what is the overhaul time that will minimize the total life cycle cost while considering repair cost and the cost of overhaul?
<br>
Denote  <math>{{C}_{1}}</math>  as the average repair cost (unscheduled),  <math>{{C}_{2}}</math>  as the replacement or overhaul cost and  <math>{{C}_{3}}</math>  as the average cost of scheduled maintenance. Scheduled maintenance is performed for every  <math>S</math>  miles or time interval. In addition, let  <math>{{N}_{1}}</math>  be the number of failures in  <math>[0,t]</math>  and let  <math>{{N}_{2}}</math>  be the number of replacements in  <math>[0,t]</math> . Suppose that replacement or overhaul occurs at times  <math>T</math> ,  <math>2T</math> ,  <math>3T</math> . The problem is to select the optimum overhaul time  <math>T={{T}_{0}}</math>  so as to minimize the long term average system cost (unscheduled maintenance, replacement cost and scheduled maintenance). Since  <math>\beta >1</math> , the average system cost is minimized when the system is overhauled (or replaced) at time  <math>{{T}_{0}}</math>  such that the instantaneous maintenance cost equals the average system cost.
The total system cost between overhaul or replacement is:
 
::<math>TSC(T)={{C}_{1}}E(N(T))+{{C}_{2}}+{{C}_{3}}\frac{T}{S}</math>
 
So the average system cost is:
 
::<math>C(T)=\frac{{{C}_{1}}E(N(T))+{{C}_{2}}+{{C}_{3}}\tfrac{T}{S}}{T}</math>
 
 
The instantaneous maintenance cost at time  <math>T</math>  is equal to:
 
::<math>IMC(T)={{C}_{1}}\lambda \beta {{T}^{\beta -1}}+\frac{{{C}_{3}}}{S}</math>
 
 
The following equation holds at optimum overhaul time  <math>{{T}_{0}}</math> :
 
 
::<math>\begin{align}
  & {{C}_{1}}\lambda \beta T_{0}^{\beta -1}+\frac{{{C}_{3}}}{S}= & \frac{{{C}_{1}}E(N(T))+{{C}_{2}}+{{C}_{3}}\tfrac{T}{S}}{T} \\
& = & \frac{{{C}_{1}}\lambda T_{0}^{\beta }+{{C}_{2}}+{{C}_{3}}\tfrac{{{T}_{0}}}{S}}{{{T}_{0}}} 
\end{align}</math>
 
 
:Therefore:
 
::<math>{{T}_{0}}={{\left[ \frac{{{C}_{2}}}{\lambda (\beta -1){{C}_{1}}} \right]}^{1/\beta }}</math>
 
 
When there is no scheduled maintenance, Eqn. (ecolm) becomes:
 
::<math>{{C}_{1}}\lambda \beta T_{0}^{\beta -1}=\frac{{{C}_{1}}\lambda T_{0}^{\beta }+{{C}_{2}}}{{{T}_{0}}}</math>
 
 
The optimum overhaul time,  <math>{{T}_{0}}</math> , is the same as Eqn. (optimt), so for periodic maintenance scheduled every  <math>S</math>  miles, the replacement or overhaul time is the same as for the unscheduled and replacement or overhaul cost model.

Latest revision as of 06:14, 23 August 2012

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