Template:ReliaSoft's Alternate Ranking Method

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 β and η.

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



0
0

1

39.169




0

2

40
0



0

2

42.837
0



0

1

50
0



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.

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.

Step 6

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

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.

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:

[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]