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=Duane=
#REDIRECT [[Duane_Model]]
<br>
==Model History and Development==
In 1962, J. T. Duane published a report in which he presented failure data of different systems during their development programs [8]. While analyzing the data, it was observed that the cumulative MTBF versus cumulative operating time followed a straight line when plotted on log-log paper (Figure oldpic71).
Based on that observation, Duane developed his model as follows. If  <math>N(T)</math>  is the number of failures by time  <math>T</math> , the observed mean (average) time between failures,  <math>MTB{{F}_{c}},</math>  at time  <math>T</math>  is: 
 
::<math>MTB{{F}_{c}}=\frac{T}{N(T)}</math>
 
The equation of the line can be expressed as:
 
::<math>y=mx+c</math>
 
:Setting:
 
::<math>\begin{align}
  & y= & \ln (MTB{{F}_{c}}) \\
& x= & \ln (T) \\
& m= & \alpha  \\
& c= & \ln b 
\end{align}</math>
 
:yields:
<br>
 
::<math>\ln (MTB{{F}_{c}})=\ln b+\alpha \ln (T)</math>
<br>
<br>
[[Image:rga4.1.png|thumb|left|400px|Cumulative MTBF vs. Cumulative Test Time postulated by Duane.]]
<br>
 
Then equating  <math>MTB{{F}_{c}}</math>  to its expected value, and assuming an exact linear relationship, gives:
 
::<math>E(MTB{{F}_{c}})=b{{T}^{\alpha }}</math>
<br>
 
:or:
<br>
 
::<math>MTB{{F}_{c}}=b{{T}^{\alpha }}</math>
 
And, if you assume a constant failure intensity, then the cumulative failure intensity,  <math>{{\lambda }_{c}}</math> , is:
 
<br>
::<math>E({{\lambda }_{c}})=\frac{1}{b}{{T}^{-\alpha }}</math>
<br>
<br>
:or:
 
<br>
::<math>{{\widehat{{\bar{\lambda }}}}_{c}}=\frac{1}{b}{{T}^{-\alpha }}</math>
 
<br>
Also, the expected number of failures up to time  <math>T</math>  is:
 
<br>
::<math>\begin{align}
  & E(N(T))= & {{\widehat{{\bar{\lambda }}}}_{c}}\cdot T \\
& = & \frac{1}{b}{{T}^{1-\alpha }} 
\end{align}</math>
<br>
<br>
:where:
 
<br>
::<math>\begin{align}
  & {{\widehat{{\bar{\lambda }}}}_{c}}= & \text{the average estimate of the cumulative failure intensity, failures/hr}\text{.} \\
& T= & \text{the total accumulated unit hours of test and/or development time}\text{.} \\
& 1/b= & \text{the cumulative failure intensity at }T=1\text{, or at the beginning of the test,} \\
&  & \text{or the earliest time at which the first }\widehat{{\bar{\lambda }}}\text{ is predicted, or the }\widehat{{\bar{\lambda }}}\text{ for the} \\
&  & \text{equipment at the start of the design and development process}\text{.} \\
& \alpha = & \text{the improvement rate in the }\widehat{{\bar{\lambda }}}\text{, }0\le \alpha \le 1. 
\end{align}</math>
 
 
The corresponding  <math>MTB{{F}_{c}}</math> , or  <math>{{\hat{m}}_{c}}</math> , is equal to:
<br>
<br>
::<math>{{\hat{m}}_{c}}=b{{T}^{\alpha }}</math>
<br>
<br>
where  <math>b=</math>  cumulative MTBF at  <math>T=1</math>  or at the beginning of the test, or the earliest time at which the first  <math>\hat{m}</math>  can be determined, or the  <math>\hat{m}</math>  predicted at the start of the design and development process ( <math>b>0</math> ).
The cumulative MTBF,  <math>{{\hat{m}}_{c}}</math> , and  <math>{{\widehat{{\bar{\lambda }}}}_{c}}</math>  tell whether  <math>m</math>  is increasing or  <math>\lambda </math>  is decreasing with time, utilizing all data up to that time. You may want to know, however, the instantaneous  <math>{{\hat{m}}_{i}}</math>  or  <math>{{\widehat{{\bar{\lambda }}}}_{i}}</math>  to see what you are doing at a specific instant or after a specific test and development time. The instantaneous failure intensity,  <math>{{\lambda }_{i}}</math> , is:
 
::<math>\begin{align}
  & {{\lambda }_{i}}= & \frac{d(E(N(T)))}{dT} \\
& = & \frac{1}{b}(1-\alpha ){{T}^{-\alpha }} \\
& = & (1-\alpha ){{\lambda }_{c}} 
\end{align}</math>
 
<br>
Similarly, using Eqn. (duanecnew), this procedure yields:
 
<br>
::<math>\begin{align}
  & {{m}_{i}}= & \frac{1}{1-\alpha }b{{T}^{\alpha }} \\
& = & \frac{1}{1-\alpha }{{{\hat{m}}}_{c}},:\ \ \alpha \not{=}1 
\end{align}</math>
<br>
<br>
where  <math>\alpha =1</math>  implies infinite MTBF growth.
It can be seen from Eqn. (duane0) that the instantaneous failure intensity improvement line is obtained by shifting the cumulative failure intensity line down, parallel to itself, by a distance of  <math>(1-\alpha )</math> . Similarly, it can be seen from Eqn. (eq76) that the current or instantaneous MTBF growth line is obtained by shifting the cumulative MTBF line up, parallel to itself, by a distance of  <math>\tfrac{1}{1-\alpha }</math> , as illustrated in Figure oldpic71.
<br>
 
{{parameter estimation duane}}
 
==Confidence Bounds==
<br>
Least squares confidence bounds can be computed for both the model parameters and metrics of interest for the Duane model.
<br>
===Parameter Bounds===
<br>
Apply least squares analysis on the Duane model:
 
<br>
::<math>\ln ({{\hat{m}}_{c}})=\ln (b)+\alpha \ln (t)</math>
 
<br>
The unbiased estimator of    can be obtained from:
 
<br>
::<math>{{\sigma }^{2}}=Var\left[ \ln {{m}_{c}}(t) \right]=\frac{SSE}{(n-2)}</math>
 
<br>
:where:
 
<br>
::<math>SSE=\underset{i=1}{\overset{n}{\mathop \sum }}\,{{\left[ \ln {{{\hat{m}}}_{c}}({{t}_{i}})-\ln {{m}_{c}}({{t}_{i}}) \right]}^{2}}</math>
 
<br>
Thus, the confidence bounds on  <math>\alpha </math>  and  <math>b</math>  are:
 
 
::<math>C{{B}_{\alpha }}=\hat{\alpha }\pm {{t}_{n-2,\alpha /2}}SE(\hat{\alpha })</math>
 
 
 
::<math>C{{B}_{b}}=\hat{b}{{e}^{\pm {{t}_{n-2,\alpha /2}}SE\left[ \ln (\hat{b}) \right]}}</math>
 
<br>
where  <math>{{t}_{n-2,\alpha /2}}</math>  denotes the percentage point of the  <math>t</math>  distribution with  <math>n-2</math>  degrees of freedom such that  <math>P\{{{t}_{n-2}}\ge {{t}_{\alpha /2,n-2}}\}=\alpha /2</math>  and:
 
<br>
::<math>SE(\hat{\alpha })=\frac{\sigma }{\sqrt{{{S}_{xx}}}}</math>
 
 
::<math>SE\left[ \ln (\hat{b}) \right]=\sigma \cdot \sqrt{\frac{\underset{i=1}{\overset{n}{\mathop{\sum }}}\,{{(\ln {{t}_{i}})}^{2}}}{n\cdot {{S}_{xx}}}}</math>
 
 
::<math>{{S}_{xx}}=\left[ \underset{i=1}{\overset{n}{\mathop \sum }}\,{{(\ln {{t}_{i}})}^{2}} \right]-\frac{1}{n}{{\left( \underset{i=1}{\overset{n}{\mathop \sum }}\,\ln ({{t}_{i}}) \right)}^{2}}</math>
 
<br>
 
===Other Bounds===
<br>
Confidence bounds also can be obtained on the cumulative MTBF and the cumulative failure intensity:
 
<br>
::<math>C{{B}_{{{m}_{c}}}}={{\hat{m}}_{c}}(t){{e}^{\pm {{z}_{\alpha }}\sqrt{Var\left[ \ln ({{{\hat{m}}}_{c}}) \right]}}}</math>
 
 
 
::<math>\begin{align}
  & {{[{{\lambda }_{c}}(t)]}_{L}}= & \frac{1}{{{[{{m}_{c}}(t)]}_{u}}} \\
& {{[{{\lambda }_{c}}(t)]}_{U}}= & \frac{1}{{{[{{m}_{c}}(t)]}_{l}}} 
\end{align}</math>
 
<br>
When  <math>n</math>  is large, the approximate  <math>100(1-\alpha )%</math>  confidence bounds for instantaneous MTBF are given by:
 
<br>
<br>
::<math>\begin{align}
  & {{m}_{i}}{{(t)}_{L}}= & \frac{{{[{{m}_{c}}(t)]}_{L}}}{{\hat{\beta }}} \\
& {{m}_{i}}{{(t)}_{U}}= & \frac{{{[{{m}_{c}}(t)]}_{U}}}{{\hat{\beta }}} 
\end{align}</math>
 
<br>
and therefore, the confidence bounds on the instantaneous failure intensity are:
 
<br>
<br>
::<math>\begin{align}
  & {{[{{\lambda }_{i}}(t)]}_{L}}= & \frac{1}{{{[{{m}_{i}}(t)]}_{U}}} \\
& {{[{{\lambda }_{c}}(t)]}_{U}}= & \frac{1}{{{[{{m}_{i}}(t)]}_{L}}} 
\end{align}</math>
 
 
'''Example 5<math>{{\lambda }_{i}}(t)=\tfrac{1}{{{m}_{i}}(t)}</math>'''
<br>
For the data given in Table 4.3, calculate the 90% confidence bounds for:
<br>
#The parameters <math>\alpha\text{and} b</math>.
#The cumulative and instantaneous failure intensity.
#The cumulative and instantaneous MTBF.
 
<br>
<br>
'''Solution'''
:1.  Using the values of  <math>\widehat{b}</math>  and  <math>\widehat{\alpha }</math>  estimated from the least squares analysis in Example 3:
<br>
<br>
::<math>\widehat{b}=1.9453</math>
::<math>\widehat{\alpha}=0.6133
 
<br>
Eqn. (duanec9) is:
 
<br>
::<math>\begin{align}
  & {{S}_{xx}}= & 1400.9084-1301.4545 \\
& = & 99.4539 
\end{align}</math>
 
 
Eqn. (duanec7) is:
 
<br>
::<math>\begin{align}
  & SE(\hat{\alpha })= & \frac{\sigma }{\sqrt{{{S}_{xx}}}} \\
& = & \frac{0.08428}{9.9727} \\
& = & 0.008452 
\end{align}</math>
 
<br>
Eqn. (duanec8) is:
 
<br>
::<math>\begin{align}
  & SE(\ln \hat{b})= & \sigma \cdot \sqrt{\frac{\underset{i=1}{\overset{n}{\mathop{\sum }}}\,{{(\ln {{T}_{i}})}^{2}}}{n\cdot {{S}_{xx}}}} \\
& = & 0.065960 
\end{align}</math>
 
Thus, 90% confidence bounds on parameter  <math>\alpha </math>  using Eqn. (duanec1) are:
 
<br>
::<math>\begin{align}
  & {{\alpha }_{L}}= & 0.602050 \\
& {{\alpha }_{U}}= & 0.624417 
\end{align}</math>
 
<br>
And 90% confidence bounds on parameter  <math>b</math>  using Eqn. (duanec2) are:
<br>
 
<br>
::<math>\begin{align}
  & {{b}_{L}}= & 1.7831 \\
& {{b}_{U}}= & 2.1231 
\end{align}</math>
 
<br>
:2.  The cumulative failure intensity is:
 
<br>
::<math>\begin{align}
  & {{\lambda }_{c}}= & \frac{1}{1.9453}\cdot {{22000}^{-0.6133}} \\
& = & 0.00111689 
\end{align}</math>
 
<br>
And the instantaneous failure intensity is equal to:
 
<br>
::<math>\begin{align}
  & {{\lambda }_{i}}= & \frac{1}{1.9453}\cdot (1-0.6133)\cdot {{22000}^{-0.6133}} \\
& = & 0.00043198 
\end{align}</math>
 
<br>
So, at the 90% confidence level and for  <math>T=22,000</math>  hr, the confidence bounds on cumulative failure intensity are:
 
<br>
::<math>\begin{align}
  & {{[{{\lambda }_{c}}(t)]}_{L}}= & 0.00100254 \\
& {{[{{\lambda }_{c}}(t)]}_{U}}= & 0.00124429 
\end{align}</math>
 
<br>
For the instantaneous failure intensity:
 
<br>
::<math>\begin{align}
  & {{[{{\lambda }_{i}}(t)]}_{L}}= & 0.00038775 \\
& {{[{{\lambda }_{c}}(t)]}_{U}}= & 0.00048125 
\end{align}</math>
 
<br>
Figures figure75 and figure76 show the graphs of the cumulative and instantaneous failure intensity. Both are plotted with confidence bounds.
<br>
<br>
[[Image:rga4.7.png|thumb|center|400px|Cumulative Failure Intensity plot with 2-sided 90% confidence bounds.]]
<br>
<br>
[[Image:rga4.8.png|thumb|center|400px|Instantaneous Failure Intensity plot with 2-sided 90% confidence bounds.]]
 
<br>
:3.  The cumulative MTBF is:
 
<br>
::<math>\begin{align}
  & {{m}_{c}}(T)= & 1.9453\cdot {{22000}^{0.6133}} \\
& = & 895.3395 
\end{align}</math>
 
<br>
And the instantaneous MTBF is:
 
<br>
::<math>\begin{align}
  & {{m}_{i}}(T)= & \frac{1.9453}{1-0.6133}\cdot {{22000}^{0.6133}} \\
& = & 2314.9369 
\end{align}</math>
 
<br>
So, at 90% confidence level and for  <math>T=22,000</math>  hr, the confidence bounds on the cumulative MTBF are:
 
<br>
::<math>\begin{align}
  & {{m}_{c}}{{(t)}_{l}}= & 803.6695 \\
& {{m}_{c}}{{(t)}_{u}}= & 997.4658 
\end{align}</math>
 
<br>
The confidence bounds for the instantaneous MTBF are:
 
<br>
::<math>\begin{align}
  & {{m}_{i}}{{(t)}_{l}}= & 2077.9204 \\
& {{m}_{i}}{{(t)}_{u}}= & 2578.9886 
\end{align}</math>
 
<br>
Figure CumMTBFCB displays the cumulative MTBF while Figure InstMTBFCB displays the instantaneous MTBF. Both are plotted with confidence bounds.
<br>
<br>
[[Image:rga4.9.png|thumb|center|400px|Cumulative MTBF plot with 2-sided 90% condfidence bounds.]]
<br>
<br>
[[Image:rga4.10.png|thumb|center|400px|Instantaneous MTBF plot with 2-sided 90% confidence bounds.]]
<br>
 
==General Examples==
<br>
===Example 6===
<br>
A prototype of a system was tested with design changes incorporated during the test. A total of 12 failures occurred. The data set is given in Table 4.5.
<br>
<br>
#Estimate the Duane parameters.
#Plot the cumulative and instantaneous MTBF curves.
#How many cumulative test and development hours are required to meet an instantaneous MTBF goal #of 500 hours?
#How many cumulative test and development hours are required to meet a cumulative MTBF goal of 500 hours?
 
<br>
<br>
<br>
{|style= align="center" border="1"
|+Table 4.5 - Developmental test data for Example 6
!Failure Number
!Cumulative Test Time(hr)
|-
|1 ||80
|-
|2 ||175
|-
|3 ||265
|-
|4 ||400
|-
|5 ||590
|-
|6|| 1100
|-
|7|| 1650
|-
|8|| 2010
|-
|9|| 2400
|-
|10|| 3380
|-
|11|| 5100
|-
|12|| 6400
|}
 
====Solution to Example 6====
 
#Figure figuaneex11 shows the data entered into RGA along with the estimated Duane parameters.
#Figure figuaneex12 shows the cumulative and instantaneous MTBF curves.
#Figure figuaneex14 shows the cumulative test and development hours needed for an instantaneous MTBF goal of 500 hours.
#Figure figuaneex15 shows the cumulative test and development hours needed for a cumulative MTBF goal of 500 hours.
<br>
<br>
[[Image:rga4.11.png|thumb|center|400px|Entered data and the estimated Duane parameters.]]
<br>
 
<br>
[[Image:rga4.12.png|thumb|center|400px|The cumulative and instantaneous MTBF curves.]]
<br>
<br>
<br>
[[Image:rga4.13.png|thumb|center|400px|Required test time for an instantaneous MTBF of 500 hours.]]
<br>
<br>
<br>
[[Image:rga4.14.png|thumb|center|400px|Required test time for a cumulative MTBF of 500 hours.]]
<br>
 
===Example 7===
 
Two identical systems were tested. Any design changes made to improve the reliability of these systems were incorporated into both systems when any system failed. A total of 29 failures occurred. The data set is given in Table 4.6. Do the following:
<br>
#Estimate the Duane parameters.
#Assume both units are tested for an additional 100 hrs each. How many failures do you expect in that period?
#If testing/development were halted at this point, what would the reliability equation for this system be
<br>
{|style= align="center" border="1"
|+Table 4.6 - Developmental test data
!Failure Number
!Failed Unit
!Test Time Unit 1(hr)
!Test Time Unit 2 (hr)
|-
|1|| 1|| 0.2|| 2.0
|-
|2|| 2|| 1.7|| 2.9
|-
|3|| 2|| 4.5|| 5.2
|-
|4|| 2|| 5.8|| 9.1
|-
|5|| 2|| 17.3|| 9.2
|-
|6|| 2|| 29.3|| 24.1
|-
|7|| 1|| 36.5|| 61.1
|-
|8|| 2|| 46.3|| 69.6
|-
|9|| 1|| 63.6|| 78.1
|-
|10|| 2|| 64.4|| 85.4
|-
|11|| 1|| 74.3|| 93.6
|-
|12|| 1|| 106.6|| 103
|-
|13|| 2|| 195.2 ||117
|-
|14|| 2|| 235.1|| 134.3
|-
|15|| 1|| 248.7 ||150.2
|-
|16|| 2|| 256.8 ||164.6
|-
|17|| 2|| 261.1|| 174.3
|-
|18|| 2|| 299.4 ||193.2
|-
|19|| 1|| 305.3|| 234.2
|-
|20|| 1|| 326.9|| 257.3
|-
|21|| 1|| 339.2|| 290.2
|-
|22|| 1|| 366.1|| 293.1
|-
|23|| 2|| 466.4|| 316.4
|-
|24|| 1|| 504|| 373.2
|-
|25|| 1|| 510|| 375.1
|-
|26|| 2|| 543.2|| 386.1
|-
|27|| 2|| 635.4 ||453.3
|-
|28|| 1|| 641.2|| 485.8
|-
|29|| 2|| 755.8|| 573.6
|}
<br>
====Solution to Example 7====
:1) Figure figuaneex21 shows the data entered into RGA along with the estimated Duane parameters.
<br>
<br>
[[Image:rga4.15.png|thumb|center|400px|Entered data and the estimated parameters.]]
<br>
:2) The current accumulated test time for both units is 1329.4 hr. If the process were to continue for an additional combined time of 200 hr, the expected cumulative number of failures at  <math>T=1529.4</math>  is 31.2695, as shown in Figure figuaneex22. At T = 1329.4, the expected number of failures is 29.2004. Therefore, the expected number of failures that would be observed over the additional 200 hr is  <math>31.2695-29.2004=2.0691\approx 2</math> .
<br>
:3) If testing/development were halted at this point, the system failure intensity would be equal to the instantaneous failure intensity at that time, or  <math>\lambda =0.0107</math>  failures/hr. See Figure figuaneex23. An exponential distribution can be assumed since the value of the failure intensity at that instant in time is known. Therefore:
<br>
::<math>\begin{align}
  & R(t)= & {{e}^{-\lambda t}} \\
& = & {{e}^{-(0.0107)t}} 
\end{align}</math>
<br>
Weibull++ can be utilized (from within RGA) to provide a Reliability vs. Time plot. This is shown in Figure figuaneex24.
<br>
<br>
 
===Example 8===
<br>
Given the sequential success/failure data in the Table 4.7, do the following:
<br>
:1) Estimate the Duane parameters.
:2) What is the instantaneous MTBF at the end of the test?
:3) How many additional test runs with a one-sided 90% confidence level are required to meet an instantaneous MTBF goal of 5 hours?
<br>
<br>
<br>
[[Image:rga4.16.png|thumb|center|400px|The expected cumulative number of failures at <math>T=1529.4</math>.]]
<br>
<br>
<br>
[[Image:rga4.17.png|thumb|center|400px|Calculate the instantaneous failure intensity at the end of the test.]]
<br>
<br>
<br>
[[Image:rga4.18.png|thumb|center|400px|Reliability vs. Time plot.]]
<br>
<br>
<br>
<br>
{|style= align="center" border="1"
|+Table 4.7 - Sequential data for Example 8
!Run Number
!Result
|-
|1|| F
|-
|2|| F
|-
|3|| S
|-
|4|| S
|-
|5|| S
|-
|6|| F
|-
|7|| S
|-
|8|| F
|-
|9|| F
|-
|10|| S
|-
|11|| S
|-
|12|| S
|-
|13|| F
|-
|14|| S
|-
|15|| S
|-
|16|| S
|-
|17|| S
|-
|18|| S
|-
|19|| S
|-
|20|| S
|}
 
 
====Solution to Example 8====
:1) Figure figuaneex31 shows the data set entered into RGA along with the estimated Duane parameters.
:2) The MTBF at the end of the test is equal to 4.5904 hours. Note that  this is the DMTBF that is shown in the Control Panel in Figure figuaneex31.
:3) Figure figuaneex34 shows the number of test runs with both one-sided confidence bounds at 90% confidence level to achieve an instantaneous MTBF of 5 hours. Therefore, the number of additional test runs required with a 90% confidence level is equal to  <math>42.2481-20=22.2481\approx 23</math>  test runs.
<br>
[[Image:rga4.19.png|thumb|center|400px|Entered data and the estimated parameters.]]
<br>
[[Image:rga4.20.png|thumb|center|400px|Number of test runs with a one-sided 90% confidence level required to meet an instantaneous MTBF goal of 5 hours.]]

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