Template:ReliaSoft's Alternate Ranking Method: Difference between revisions

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::<math>{{\widehat{\beta }}_{RRX}}=1.82890,\text{ }{{\widehat{\eta }}_{RRX}}=41.69774</math>
::<math>{{\widehat{\beta }}_{RRX}}=1.82890,\text{ }{{\widehat{\eta }}_{RRX}}=41.69774</math>

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Chapter 4D: ReliaSoft's Alternate Ranking Method


Weibullbox.png

Chapter 4D  
ReliaSoft's Alternate Ranking Method  

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Available Software:
Weibull++

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More Resources:
Weibull++ Examples Collection


ReliaSoft's Alternate Ranking Method (RRM) Step-by-Step Example

This section illustrates the ReliaSoft ranking method (RRM), which is an iterative improvement on the standard ranking method (SRM). This method is illustrated in this section using an example for the two-parameter Weibull distribution. This method can also be easily generalized for other models.

Consider the following test data, as shown in the following Table B.1.


Table B.1- The test data
Number of Items Type Last Inspection Time
1 Exact Failure 10
1 Right Censored 20
2 Left Censored 0 30
2 Exact Failure 40
1 Exact Failure 50
1 Right Censored 60
1 Left Censored 0 70
2 Interval Failure 20 80
1 Interval Failure 10 85
1 Left Censored 0 100


Initial parameter estimation

As a preliminary step, we need to provide a crude estimate of the Weibull parameters for this data. To begin, we will extract the exact times-to-failure: 10, 40, and 50 and append them to the midpoints of the interval failures: 50 (for the interval of 20 to 80) and 47.5 (for the interval of 10 to 85). Now, our extracted list consists of the data in Table B.2.

Using the traditional rank regression, we obtain the first initial estimates:

[math]\displaystyle{ \begin{align} & {{\widehat{\beta }}_{0}}= & 1.91367089 \\ & {{\widehat{\eta }}_{0}}= & 43.91657736 \end{align} }[/math]


Table B.2- The Union of Exact times-to-failure with the "midpoint" of the interval failures
Number of Items Type Last Inspection Time
1 Exact Failure 10
2 Exact Failure 40
1 Exact Failure 47.5
3 Exact Failure 50


Step 1

For all intervals, we obtain a weighted ``midpoint using:

[math]\displaystyle{ \begin{align} {{{\hat{t}}}_{m}}\left( \hat{\beta },\hat{\eta } \right)= & \frac{\int_{LI}^{TF}t\text{ }f(t;\hat{\beta },\hat{\eta })dt}{\int_{LI}^{TF}f(t;\hat{\beta },\hat{\eta })dt}, \\ = & \frac{\int_{LI}^{TF}t\tfrac{{\hat{\beta }}}{{\hat{\eta }}}{{\left( \tfrac{t}{{\hat{\eta }}} \right)}^{\hat{\beta }-1}}{{e}^{-{{\left( \tfrac{t}{{\hat{\eta }}} \right)}^{{\hat{\beta }}}}}}dt}{\int_{LI}^{TF}\tfrac{{\hat{\beta }}}{{\hat{\eta }}}{{\left( \tfrac{t}{{\hat{\eta }}} \right)}^{\hat{\beta }-1}}{{e}^{-{{\left( \tfrac{t}{{\hat{\eta }}} \right)}^{{\hat{\beta }}}}}}dt} \end{align} }[/math]


This transforms our data into the format in Table B.3.


Table B.3- The Union of Exact times-to-failure with the "midpoint" of the interval failures, based upon the parameters [math]\displaystyle{ \beta }[/math] and [math]\displaystyle{ \eta }[/math].
Number of Items Type Last Inspection Time Weighted "Midpoint"
1 Exact Failure 10
2 Exact Failure 40
1 Exact Failure 50
2 Interval Failure 20 80 42.837
1 Interval Failure 10 85 39.169

Step 2

Now we arrange the data as in Table B.4.


Table B.4- The Union of Exact times-to-failure with the "midpoint" of the interval failures, in ascending order.
Number of Items Time
1 10
1 39.169
2 40
2 42.837
1 50

Step 3

We now consider the left and right censored data, as in Table B.5.


Table B.5- Computation of increments, in a matrix format, for computing a revised Mean Order Number.
Number of items Time of Failure 2 Left Censored t = 30 1 Left Censored t = 70 1 Left Censored t = 100 1 Right Censored t = 20 1 Right Censored t = 60
1 10 [math]\displaystyle{ 2 \frac{\int_0^{10} f_0(t)dt}{F_0 (30)-F_0 (0)} }[/math] [math]\displaystyle{ \frac{\int_0^{10} f_0 (t)dt}{F_0(70)-F_1(0)} }[/math] [math]\displaystyle{ \frac{\int_0^{10} f_0(t)dt}{F_0(100)-F_0(0)} }[/math] 0 0
1 39.169 [math]\displaystyle{ 2 \frac{\int_{10}^{30} f_0(t)dt}{F_0(30)-F_0(0)} }[/math] [math]\displaystyle{ \frac{\int_{10}^{39.169} f_0(t)dt}{F_0(70)-F_0(0)} }[/math] [math]\displaystyle{ \frac{\int_{10}^{39.169} f_0(t)dt}{F_0(100)-F_0(0)} }[/math] [math]\displaystyle{ \frac{\int_{20}^{39.169} f_0(t)dt}{F_0(\infty)-F_0(20)} }[/math] 0
2 40 0 [math]\displaystyle{ \frac{\int_{39.169}^{40} f_0(t)dt}{F_0(70)-F_0(0)} }[/math] [math]\displaystyle{ \frac{\int_{39.169}^{40} f_0(t)dt}{F_0(100)-F_0(0)} }[/math] [math]\displaystyle{ \frac{\int_{39.169}^{40} f_0(t)dt}{F_0(\infty)-F_0(20)} }[/math] 0
2 42.837 0 [math]\displaystyle{ \frac{\int_{40}^{42.837} f_0(t)dt}{F_0(70)-F_0(0)} }[/math] [math]\displaystyle{ \frac{\int_{40}^{42.837} f_0(t)dt}{F_0(100)-F_0(0)} }[/math] [math]\displaystyle{ \frac{\int_{40}^{42.837} f_0(t)dt}{F_0(\infty)-F_0(0)} }[/math] 0
1 50 0 [math]\displaystyle{ \frac{\int_{42.837}^{50} f_0(t)dt}{F_0(70)-F_0(0)} }[/math] [math]\displaystyle{ \frac{\int_{42.837}^{50} f_0(t)dt}{F_0(100)-F_0(0)} }[/math] [math]\displaystyle{ \frac{\int_{42.837}^{50} f_0(t)dt}{F_0(\infty)-F_0(0)} }[/math] 0


In general, for left censored data:

• The increment term for [math]\displaystyle{ n }[/math] left censored items at time [math]\displaystyle{ ={{t}_{0}}, }[/math] with a time-to-failure of .. when [math]\displaystyle{ {{t}_{0}}\le {{t}_{i-1}} }[/math] is zero.

• When [math]\displaystyle{ {{t}_{0}}\gt {{t}_{i-1}}, }[/math] the contribution is:

[math]\displaystyle{ \frac{n}{{{F}_{0}}({{t}_{0}})-{{F}_{0}}(0)}\underset{{{t}_{i-1}}}{\overset{MIN({{t}_{i}},{{t}_{0}})}{\mathop \int }}\,{{f}_{0}}\left( t \right)dt }[/math]

or:

[math]\displaystyle{ n\frac{{{F}_{0}}(MIN({{t}_{i}},{{t}_{0}}))-{{F}_{0}}({{t}_{i-1}})}{{{F}_{0}}({{t}_{0}})-{{F}_{0}}(0)} }[/math]

where [math]\displaystyle{ {{t}_{i-1}} }[/math] is the time-to-failure previous to the [math]\displaystyle{ {{t}_{i}} }[/math] time-to-failure and [math]\displaystyle{ n }[/math] is the number of units associated with that time-to-failure (or units in the group).

In general, for right censored data:

• The increment term for [math]\displaystyle{ n }[/math] right censored at time [math]\displaystyle{ ={{t}_{0}}, }[/math] with a time-to-failure of [math]\displaystyle{ {{t}_{i}} }[/math], when [math]\displaystyle{ {{t}_{0}}\ge {{t}_{i}} }[/math] is zero.

• When [math]\displaystyle{ {{t}_{0}}\lt {{t}_{i}}, }[/math] the contribution is:

[math]\displaystyle{ \frac{n}{{{F}_{0}}(\infty )-{{F}_{0}}({{t}_{0}})}\underset{MAX({{t}_{0}},{{t}_{i-1}})}{\overset{{{t}_{i}}}{\mathop \int }}\,{{f}_{0}}\left( t \right)dt }[/math]

or:

[math]\displaystyle{ n\frac{{{F}_{0}}({{t}_{i}})-{{F}_{0}}(MAX({{t}_{0}},{{t}_{i-1}}))}{{{F}_{0}}(\infty )-{{F}_{0}}({{t}_{0}})} }[/math]

where [math]\displaystyle{ {{t}_{i-1}} }[/math] is the time-to-failure previous to the [math]\displaystyle{ {{t}_{i}} }[/math] time-to-failure and [math]\displaystyle{ n }[/math] is the number of units associated with that time-to-failure (or units in the group).

Step 4

Sum up the increments (horizontally in rows), as in Table B.6.


Table B.6- Increments solved numberically, along with the sum of each row.
Number of items Time of Failure 2 Left Censored t=30 1 Left Censored t=70 1 Left Censored t=100 1 Right Censored t=20 1 Right Censored t=60 Sum of row(increment)
1 10 0.299065 0.062673 0.057673 0 0 0.419411
1 39.169 1.700935 0.542213 0.498959 0.440887 0 3.182994
2 40 0 0.015892 0.014625 0.018113 0 0.048630
2 42.831 0 0.052486 0.048299 0.059821 0 0.160606
1 50 0 0.118151 0.108726 0.134663 0 0.361540

Step 5

Compute new mean order numbers (MON), as shown Table B.7, utilizing the increments obtained in Table B.6, by adding the ``number of items plus the ``previous MON plus the current ``increment.


Table B.7- Mean Order Numbers (MON)
Number of items Time of Failure Sum of row(increment) Mean Order Number
1 10 0.419411 1.419411
1 39.169 3.182994 5.602405
2 40 0.048630 7.651035
2 42.837 0.160606 9.811641
1 50 0.361540 11.173181

Step 6

Compute the median ranks based on these new MONs as shown in Table B.8.


Table B.8- Mean Order Numbers with their ranks for a sample size of 13 units.
Time MON Ranks
10 1.419411 0.0825889
39.169 5.602405 0.3952894
40 7.651035 0.5487781
42.837 9.811641 0.7106217
50 11.173181 0.8124983

Step 7

Compute new [math]\displaystyle{ \beta }[/math] and [math]\displaystyle{ \eta , }[/math] using standard rank regression and based upon the data as shown in Table B.9.

Time Ranks
10 0.0826889
39.169 0.3952894
40 0.5487781
42.837 0.7106217
50 0.8124983

Step 8

Return and repeat the process from Step 1 until an acceptable convergence is reached on the parameters (i.e. the parameter values stabilize).

Results

The results of the first five iterations are shown in Table B.10. Using Weibull++ with rank regression on X yields:



Table B.10-The parameters after the first five iterations
Iteration [math]\displaystyle{ \beta }[/math] [math]\displaystyle{ \eta }[/math]
1 1.845638 42.576422
2 1.830621 42.039743
3 1.828010 41.830615
4 1.828030 41.749708
5 1.828383 41.717990


[math]\displaystyle{ {{\widehat{\beta }}_{RRX}}=1.82890,\text{ }{{\widehat{\eta }}_{RRX}}=41.69774 }[/math]


The direct MLE solution yields:


[math]\displaystyle{ {{\widehat{\beta }}_{MLE}}=2.10432,\text{ }{{\widehat{\eta }}_{MLE}}=42.31535 }[/math]