Template:Non-parametric LDA Examples: Difference between revisions

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====Non-Parametric LDA Example (Kaplan-Meier)====
====Non-Parametric LDA Example (Kaplan-Meier)====
====Problem Statement(Kaplan-Meier)====
A group of 20 units are put on a life test with the following results.
A group of 20 units are put on a life test with the following results.


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Use the Kaplan-Meier estimator to determine the reliability estimates for each failure time.
Use the Kaplan-Meier estimator to determine the reliability estimates for each failure time.


=====Solution to Non-Parametric LDA Example (Kaplan-Meier)=====
=====Solution(Kaplan-Meier)=====
Using the data and Eqn. (kapmeier), the following table can be constructed:  
Using the data and Eqn. (kapmeier), the following table can be constructed:  


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\end{matrix}</math></center>
\end{matrix}</math></center>


====Example 10====
====Non-Parametric LDA Example (Actuarial Method)====
====Problem Statement(Actuarial Method)====
 
A group of 55 units are put on a life test during which the units are evaluated every 50 hours, with the following results:
A group of 55 units are put on a life test during which the units are evaluated every 50 hours, with the following results:


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====Solution (Actuarial Method)====
=====Solution to Example 10=====
The reliability estimates for the simple actuarial method can be obtained by expanding the data table to include terms used in calculation of the reliability estimates for Eqn. (simpact):  
The reliability estimates for the simple actuarial method can be obtained by expanding the data table to include terms used in calculation of the reliability estimates for Eqn. (simpact):  


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   650 & 8.2%  \\
   650 & 8.2%  \\
\end{matrix}</math></center>
\end{matrix}</math></center>
====Example 11====
Find reliability estimates for the data in Example 10 using the standard actuarial method.


=====Solution to Example 11=====
====Non-Parametric LDA Example (Standard Actuarial Method)====
====Problem Statement(Standard Actuarial Method)====
 
Find reliability estimates for the data in Example for the Simple Actuarial Method using the standard actuarial method.
 
====Solution(Standard Actuarial Method)====
The solution to this example is similar to that of Example 10, with the exception of the inclusion of the  <math>n_{i}^{\prime }</math>  term, which is used in Eqn. (standact). Applying this equation to the data, we can generate the following table:  
The solution to this example is similar to that of Example 10, with the exception of the inclusion of the  <math>n_{i}^{\prime }</math>  term, which is used in Eqn. (standact). Applying this equation to the data, we can generate the following table:  



Revision as of 18:22, 6 February 2012

Non-Parametric LDA Example (Kaplan-Meier)

Problem Statement(Kaplan-Meier)

A group of 20 units are put on a life test with the following results.

[math]\displaystyle{ \begin{matrix} Number & State & State \\ in State & (F or S) & End Time \\ 3 & F & 9 \\ 1 & S & 9 \\ 1 & F & 11 \\ 1 & S & 12 \\ 1 & F & 13 \\ 1 & S & 13 \\ 1 & S & 15 \\ 1 & F & 17 \\ 1 & F & 21 \\ 1 & S & 22 \\ 1 & S & 24 \\ 1 & S & 26 \\ 1 & F & 28 \\ 1 & F & 30 \\ 1 & S & 32 \\ 2 & S & 35 \\ 1 & S & 39 \\ 1 & S & 41 \\ \end{matrix} }[/math]

Use the Kaplan-Meier estimator to determine the reliability estimates for each failure time.

Solution(Kaplan-Meier)

Using the data and Eqn. (kapmeier), the following table can be constructed:

[math]\displaystyle{ \begin{matrix} State & Number of & Number of & Available & {} & {} \\ End Time & Failures, {{r}_{i}} & Suspensions, {{s}_{i}} & Units, {{n}_{i}} & \tfrac{{{n}_{i}}-{{r}_{i}}}{{{n}_{i}}} & \mathop{}_{}^{}\tfrac{{{n}_{i}}-{{r}_{i}}}{{{n}_{i}}} \\ 9 & 3 & 1 & 20 & 0.850 & 0.850 \\ 11 & 1 & 0 & 16 & 0.938 & 0.797 \\ 12 & 0 & 1 & 15 & 1.000 & 0.797 \\ 13 & 1 & 1 & 14 & 0.929 & 0.740 \\ 15 & 0 & 1 & 12 & 1.000 & 0.740 \\ 17 & 1 & 0 & 11 & 0.909 & 0.673 \\ 21 & 1 & 0 & 10 & 0.900 & 0.605 \\ 22 & 0 & 1 & 9 & 1.000 & 0.605 \\ 24 & 0 & 1 & 8 & 1.000 & 0.605 \\ 26 & 0 & 1 & 7 & 1.000 & 0.605 \\ 28 & 1 & 0 & 6 & 0.833 & 0.505 \\ 30 & 1 & 0 & 5 & 0.800 & 0.404 \\ 32 & 0 & 1 & 4 & 1.000 & 0.404 \\ 35 & 0 & 1 & 3 & 1.000 & 0.404 \\ 39 & 0 & 1 & 2 & 1.000 & 0.404 \\ 41 & 0 & 1 & 1 & 1.000 & 0.404 \\ \end{matrix} }[/math]

As can be determined from the preceding table, the reliability estimates for the failure times are:

[math]\displaystyle{ \begin{matrix} Failure Time & Reliability Est. \\ 9 & 85.0% \\ 11 & 79.7% \\ 13 & 74.0% \\ 17 & 67.3% \\ 21 & 60.5% \\ 28 & 50.5% \\ 30 & 40.4% \\ \end{matrix} }[/math]

Non-Parametric LDA Example (Actuarial Method)

Problem Statement(Actuarial Method)

A group of 55 units are put on a life test during which the units are evaluated every 50 hours, with the following results:

[math]\displaystyle{ \begin{matrix} Start & End & Number of & Number of \\ Time & Time & Failures, {{r}_{i}} & Suspensions, {{s}_{i}} \\ 0 & 50 & 2 & 4 \\ 50 & 100 & 0 & 5 \\ 100 & 150 & 2 & 2 \\ 150 & 200 & 3 & 5 \\ 200 & 250 & 2 & 1 \\ 250 & 300 & 1 & 2 \\ 300 & 350 & 2 & 1 \\ 350 & 400 & 3 & 3 \\ 400 & 450 & 3 & 4 \\ 450 & 500 & 1 & 2 \\ 500 & 550 & 2 & 1 \\ 550 & 600 & 1 & 0 \\ 600 & 650 & 2 & 1 \\ \end{matrix} }[/math]


Solution (Actuarial Method)

The reliability estimates for the simple actuarial method can be obtained by expanding the data table to include terms used in calculation of the reliability estimates for Eqn. (simpact):

[math]\displaystyle{ \begin{matrix} Start & End & Number of & Number of & Available & {} & {} \\ Time & Time & Failures, {{r}_{i}} & Suspensions, {{s}_{i}} & Units, {{n}_{i}} & 1-\tfrac{{{r}_{j}}}{{{n}_{j}}} & \mathop{}_{}^{}1-\tfrac{{{r}_{j}}}{{{n}_{j}}} \\ 0 & 50 & 2 & 4 & 55 & 0.964 & 0.964 \\ 50 & 100 & 0 & 5 & 49 & 1.000 & 0.964 \\ 100 & 150 & 2 & 2 & 44 & 0.955 & 0.920 \\ 150 & 200 & 3 & 5 & 40 & 0.925 & 0.851 \\ 200 & 250 & 2 & 1 & 32 & 0.938 & 0.798 \\ 250 & 300 & 1 & 2 & 29 & 0.966 & 0.770 \\ 300 & 350 & 2 & 1 & 26 & 0.923 & 0.711 \\ 350 & 400 & 3 & 3 & 23 & 0.870 & 0.618 \\ 400 & 450 & 3 & 4 & 17 & 0.824 & 0.509 \\ 450 & 500 & 1 & 2 & 10 & 0.900 & 0.458 \\ 500 & 550 & 2 & 1 & 7 & 0.714 & 0.327 \\ 550 & 600 & 1 & 0 & 4 & 0.750 & 0.245 \\ 600 & 650 & 2 & 1 & 3 & 0.333 & 0.082 \\ \end{matrix} }[/math]

As can be determined from the preceding table, the reliability estimates for the failure times are:

[math]\displaystyle{ \begin{matrix} Failure Period & Reliability \\ End Time & Estimate \\ 50 & 96.4% \\ 150 & 92.0% \\ 200 & 85.1% \\ 250 & 79.8% \\ 300 & 77.0% \\ 350 & 71.1% \\ 400 & 61.8% \\ 450 & 50.9% \\ 500 & 45.8% \\ 550 & 32.7% \\ 600 & 24.5% \\ 650 & 8.2% \\ \end{matrix} }[/math]

Non-Parametric LDA Example (Standard Actuarial Method)

Problem Statement(Standard Actuarial Method)

Find reliability estimates for the data in Example for the Simple Actuarial Method using the standard actuarial method.

Solution(Standard Actuarial Method)

The solution to this example is similar to that of Example 10, with the exception of the inclusion of the [math]\displaystyle{ n_{i}^{\prime } }[/math] term, which is used in Eqn. (standact). Applying this equation to the data, we can generate the following table:

[math]\displaystyle{ \begin{matrix} Start & End & Number of & Number of & Adjusted & {} & {} \\ Time & Time & Failures, {{r}_{i}} & Suspensions, {{s}_{i}} & Units, n_{i}^{\prime } & 1-\tfrac{{{r}_{j}}}{n_{j}^{\prime }} & \mathop{}_{}^{}1-\tfrac{{{r}_{j}}}{n_{j}^{\prime }} \\ 0 & 50 & 2 & 4 & 53 & 0.962 & 0.962 \\ 50 & 100 & 0 & 5 & 46.5 & 1.000 & 0.962 \\ 100 & 150 & 2 & 2 & 43 & 0.953 & 0.918 \\ 150 & 200 & 3 & 5 & 37.5 & 0.920 & 0.844 \\ 200 & 250 & 2 & 1 & 31.5 & 0.937 & 0.791 \\ 250 & 300 & 1 & 2 & 28 & 0.964 & 0.762 \\ 300 & 350 & 2 & 1 & 25.5 & 0.922 & 0.702 \\ 350 & 400 & 3 & 3 & 21.5 & 0.860 & 0.604 \\ 400 & 450 & 3 & 4 & 15 & 0.800 & 0.484 \\ 450 & 500 & 1 & 2 & 9 & 0.889 & 0.430 \\ 500 & 550 & 2 & 1 & 6.5 & 0.692 & 0.298 \\ 550 & 600 & 1 & 0 & 4 & 0.750 & 0.223 \\ 600 & 650 & 2 & 1 & 2.5 & 0.200 & 0.045 \\ \end{matrix} }[/math]


As can be determined from the preceding table, the reliability estimates for the failure times are: