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| {{Probability Plotting}}
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| One method of calculating the parameter of the exponential distribution is by using probability plotting. To better illustrate this procedure, consider the following example.
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| <br>
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| {{Example:Probability Plotting:Exponential[1]}}
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| ====Example 1====
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| <br>
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| Let's assume six identical units are reliability tested at the same application and operation
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| stress levels. All of these units fail during the test after operating for the following times (in hours), <math>{{T}_{i}}</math> : 96, 257, 498, 763, 1051 and 1744.
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| <br>
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| The steps for determining the parameters of the exponential <math>pdf</math> representing the
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| data, using probability plotting, are as follows:
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| <br>
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| ::• Rank the times-to-failure in ascending order as shown next.
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| <br>
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| ::<math>\begin{matrix}
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| \text{Time-to-} & \text{Failure Order Number} \\
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| \text{failure, hr} & \text{out of a Sample Size of 6} \\
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| \text{96} & \text{1} \\
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| \text{257} & \text{2} \\
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| \text{498} & \text{3} \\
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| \text{763} & \text{4} \\
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| \text{1,051} & \text{5} \\
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| \text{1,744} & \text{6} \\
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| \end{matrix}</math>
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| <br>
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| ::• Obtain their median rank plotting positions.
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| <br>
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| Median rank positions are used instead of other ranking methods because median ranks are at a
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| specific confidence level (50%).
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| <br>
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| ::• The times-to-failure, with their corresponding median ranks, are shown next:
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| <br>
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| ::<math>\begin{matrix}
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| \text{Time-to-} & \text{Median} \\
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| \text{failure, hr} & \text{Rank, }% \\
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| \text{96} & \text{10}\text{.91} \\
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| \text{257} & \text{26}\text{.44} \\
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| \text{498} & \text{42}\text{.14} \\
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| \text{763} & \text{57}\text{.86} \\
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| \text{1,051} & \text{73}\text{.56} \\
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| \text{1,744} & \text{89}\text{.10} \\
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| \end{matrix}</math>
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| <br>
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| ::• On an exponential probability paper, plot the times on the x-axis and their corresponding
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| rank value on the y-axis. Fig. 4 displays an example of an exponential probability paper. The
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| paper is simply a log-linear paper. (The solution is given in Fig. 2.)
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| <br>
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| [[File:ALTA4.1.gif|center]]
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| <br>
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| <center>Fig. 4: Sample exponential probability paper.</center>
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| <br>
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| ::• Draw the best possible straight line that goes through the <math>t=0</math> and <math>
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| (t)=100%</math> point and through the plotted points (as shown in Fig. 5).
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| <br>
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| ::• At the <math>Q(t)=63.2%</math> or <math>R(t)=36.8%</math> ordinate point, draw a
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| straight horizontal line until this line intersects the fitted straight line. Draw a vertical line through this intersection until it crosses the abscissa. The value at the intersection of the abscissa is the estimate of the mean. For this case, <math>\widehat{\mu }=833</math> hr which means that <math>\lambda =\tfrac{1}{\mu }=0.0012</math> . (This is always at 63.2% since <math>(T)=1-{{e}^{-\tfrac{\mu }{\mu }}}=1-{{e}^{-1}}=0.632=63.2%).</math>
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| <br>
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| [[File:ALTA4.2.gif|center]] | |
| <br>
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| <center>Fig. 5: Probability plot for Example 1.</center>
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| <br>
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| <br>
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| Now any reliability value for any mission time <math>t</math> can be obtained. For example, the
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| reliability for a mission of 15 hr, or any other time, can now be obtained either from the plot or analytically (i.e. using the equations given in Section <math>5.1.1</math> ).
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| <br>
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| To obtain the value from the plot, draw a vertical line from the abscissa, at <math>t=15</math>
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| hr, to the fitted line. Draw a horizontal line from this intersection to the ordinate and read
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| <math>R(t)</math> . In this case, <math>R(t=15)=98.15%</math> . This can also be obtained
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| analytically, from the exponential reliability function.
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| <br>
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| ====MLE Parameter Estimation====
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| <br>
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| The parameter of the exponential distribution can also be estimated using the maximum likelihood estimation (MLE) method. This log-likelihood function is:
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| <br>
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| ::<math>\ln (L)=\Lambda =\underset{i=1}{\overset{{{F}_{e}}}{\mathop \sum }}\,{{N}_{i}}\ln \left[ \lambda {{e}^{-\lambda {{T}_{i}}}} \right]-\underset{i=1}{\overset{S}{\mathop \sum }}\,N_{i}^{\prime }\lambda T_{i}^{\prime }+\overset{FI}{\mathop{\underset{i=1}{\mathop{\underset{}{\overset{}{\mathop \sum }}\,}}\,}}\,N_{i}^{\prime \prime }\ln [R_{Li}^{\prime \prime }-R_{Ri}^{\prime \prime }]</math>
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| <br>
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| where:
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| <br>
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| ::<math>R_{Li}^{\prime \prime }={{e}^{-\lambda T_{Li}^{\prime \prime }}}</math>
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| <br>
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| ::<math>R_{Ri}^{\prime \prime }={{e}^{-\lambda T_{Ri}^{\prime \prime }}}</math>
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| <br>
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| and
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| <br>
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| ::• <math>{{F}_{e}}</math> is the number of groups of times-to-failure data points.
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| ::• <math>{{N}_{i}}</math> is the number of times-to-failure in the <math>{{i}^{th}}</math> time-to-failure data group.
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| ::• <math>\lambda </math> is the failure rate parameter (unknown a priori, the only parameter to be found).
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| ::• <math>{{T}_{i}}</math> is the time of the <math>{{i}^{th}}</math> group of time-to-failure data.
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| ::• <math>S</math> is the number of groups of suspension data points.
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| ::• <math>N_{i}^{\prime }</math> is the number of suspensions in the <math>{{i}^{th}}</math> group of suspension data points.
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| ::• <math>T_{i}^{\prime }</math> is the time of the <math>{{i}^{th}}</math> suspension data group.
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| ::• <math>FI</math> is the number of interval data groups.
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| ::• <math>N_{i}^{\prime \prime }</math> is the number of intervals in the i <math>^{th}</math> group of data intervals.
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| ::• <math>T_{Li}^{\prime \prime }</math> is the beginning of the i <math>^{th}</math> interval.
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| ::• <math>T_{Ri}^{\prime \prime }</math> is the ending of the i <math>^{th}</math> interval.
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| <br>
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| The solution will be found by solving for a parameter <math>\widehat{\lambda }</math> so that <math>\tfrac{\partial \Lambda }{\partial \lambda }=0</math> where:
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| <br>
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| ::<math>\frac{\partial \Lambda }{\partial \lambda }=\underset{i=1}{\overset{{{F}_{e}}}{\mathop \sum }}\,{{N}_{i}}\left( \frac{1}{\lambda }-{{T}_{i}} \right)-\underset{i=1}{\overset{S}{\mathop \sum }}\,N_{i}^{\prime }T_{i}^{\prime }-\overset{FI}{\mathop{\underset{i=1}{\mathop{\underset{}{\overset{}{\mathop \sum }}\,}}\,}}\,N_{i}^{\prime \prime }\frac{T_{Li}^{\prime \prime }R_{Li}^{\prime \prime }-T_{Ri}^{\prime \prime }R_{Ri}^{\prime \prime }}{R_{Li}^{\prime \prime }-R_{Ri}^{\prime \prime }}</math>
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| <br>
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| ====Example 2====
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| <br>
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| Using the same data as in the probability plotting example (Example 1), and assuming an exponential distribution, estimate the parameter using the MLE method.
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| <br>
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| ''Solution''
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| <br>
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| In this example we have non-grouped data without suspensions. Thus Eqn. (exp-mle) becomes:
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| <br>
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| ::<math>\frac{\partial \Lambda }{\partial \lambda }=\underset{i=1}{\overset{{{F}_{e}}}{\mathop \sum }}\,\left[ \frac{1}{\lambda }-\left( {{T}_{i}} \right) \right]=\underset{i=1}{\overset{14}{\mathop \sum }}\,\left[ \frac{1}{\lambda }-\left( {{T}_{i}} \right) \right]=0</math>
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| <br>
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| Substituting the values for <math>T</math> we get:
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| <br>
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| ::<math>\begin{align}
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| \frac{6}{\lambda }= & 4409,\text{ or:} \\
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| \lambda = & 0.00136\text{ failure/hr}
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| \end{align}</math>
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