# Difference between revisions of "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.

### 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:

\begin{align} & {{\widehat{\beta }}_{0}}= & 1.91367089 \\ & {{\widehat{\eta }}_{0}}= & 43.91657736 \end{align}

Step 1 For all intervals, we obtain a weighted midpoint using:

\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}

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

Step 2 Now we arrange the data as in Table B.4.

Step 3 We now consider the left and right censored data, as in Table B.5.

In general, for left censored data:

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

• When ${{t}_{0}}\gt {{t}_{i-1}},$ the contribution is:

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

or:

$n\frac{{{F}_{0}}(MIN({{t}_{i}},{{t}_{0}}))-{{F}_{0}}({{t}_{i-1}})}{{{F}_{0}}({{t}_{0}})-{{F}_{0}}(0)}$

where ${{t}_{i-1}}$ is the time-to-failure previous to the ${{t}_{i}}$ time-to-failure and $n$ 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 $n$ right censored at time $={{t}_{0}},$ with a time-to-failure of ${{t}_{i}}$, when ${{t}_{0}}\ge {{t}_{i}}$ is zero.

• When ${{t}_{0}}\lt {{t}_{i}},$ the contribution is:

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

or:

$n\frac{{{F}_{0}}({{t}_{i}})-{{F}_{0}}(MAX({{t}_{0}},{{t}_{i-1}}))}{{{F}_{0}}(\infty )-{{F}_{0}}({{t}_{0}})}$

where ${{t}_{i-1}}$ is the time-to-failure previous to the ${{t}_{i}}$ time-to-failure and $n$ 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 $\beta$ and $\eta ,$ 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:

${{\widehat{\beta }}_{RRX}}=1.82890,\text{ }{{\widehat{\eta }}_{RRX}}=41.69774$

The direct MLE solution yields:

${{\widehat{\beta }}_{MLE}}=2.10432,\text{ }{{\widehat{\eta }}_{MLE}}=42.31535$