2.5.3.2 Algorithm: Warranty Analysis

The data of the app are start and end times in the form $(tl_1, tr_1), (tl_1, tr_2),...,(tl_k, tr_k)$ and each interval $(tl_i, tr_i)$ contains $n_i$ failures (if $tr_i < \infty$ ) or $n_i$ suspensions (if $tri = \infty$ ), $i = 1, 2,..., k$ . This arbitrarily censored dataset is first fitted with a weibull distribution using either the MLE(maximum likelihood estimation) or LS (least squares) method to obtain the parameters $\beta$ and $\eta$ . The results shown below are then computed based on the fitted model.

1 Summary of the warranty claims
2 Predicted number of future failures
3 Predicted cost of future failures

Summary of the warranty claims

Total number of units: $\sum_{i=1}^k n_i$

Observed number of failures: $\sum_{i=1}^l n_i$

where

$k$ : total times.
$l$ : the number of distinct interval censored times.

Expected Number of Failures

Reliability Function:
$R(t) = 1 - F(t;\beta,\eta)$ where $F(t;\beta,\eta)$ is the CDF of weibull distribution.

Expected number of failures without known warranty length (L):
$ENF = \sum n_i\mathcal{L}_{i}( \beta, \eta \,;tl_i, tr_i )$

if $tl_i < tr_i$ , then $\mathcal{L}_{i}( \beta, \eta \,;tl_i, tr_i) = R(tl_i)-R(tr_i)$

if $tl_i = tr_i$ or $tr_i = \infty$ , then $\mathcal{L}_{i}( \beta, \eta \,;tl_i, tr_i) = 1-R(tl_i)$

Expected number of gailures with known warranty length (L):
$ENF = \sum n_i\mathcal{L}_{i}( \beta, \eta \,;tl_i, tr_i )$

if $tl_i < tr_i$ , then $\mathcal{L}_{i}( \beta, \eta \,;tl_i, tr_i) = R(tl_i)-R(min(L,tr_i))$

if $tl_i = tr_i$ or $tr_i = \infty$ , then $\mathcal{L}_{i}( \beta, \eta \,;tl_i, tr_i) = 1-R(min(L,tl_i))$

Confidence intervals for the expected number of failures

Two-sided $100(1-\alpha)\%$ confidence interval
$x_{L} = \frac{1}{2}\chi^2_{2s, 1-\alpha/2}$

$x_{U} = \frac{1}{2}\chi^2_{2(s+1), \alpha/2}$
One-sided $100(1-\alpha)\%$ lower confidence bound
$x_{L} = \frac{1}{2}\chi^2_{2s, 1-\alpha}$
One-sided $100(1-\alpha)\%$ upper confidence bound
$x_{U}= \frac{1}{2}\chi^{2}_{2(s+1),\alpha}$

where

$s$ is the predicted number of future failures $PNF$ .

Number of units at risk for future time periods

Without known warranty length (L): $\sum_{i=1}^m n_i$
With known warranty length (L): $\sum_{i=1}^{m, tl_i < L} n_i$
$m$ : the number of distinct right censored times.

Predicted number of future failures

Predicted Number of failures without known warranty length (L):

If production quantity for each future period $d_i$ is not provided:
$PNF(\Delta) = \sum_{i=1}^{m} n_i(1-\frac{R(tl_i + \Delta)}{R(tl_i)})$
If production quantity for each future period $d_i$ is provided:
$PNF(\Delta) = \sum_{i=1}^{m} n_i(1-\frac{R(tl_i + \Delta)}{R(tl_i)}) + \sum_{i=1}^{q}d_j(1-R(\Delta+1-i))$

Predicted Number of failures without known warranty length (L):

If production quantity for each future period $d_i$ is not provided:
$PNF(\Delta) = \sum_{i=1}^{m, tl_i<L} n_i(1-\frac{R(tl_i + \min(\Delta, L-tl_i)}{R(tl_i)})$
If production quantity for each future period $d_i$ is provided:
$PNF(\Delta) = \sum_{i=1}^{m, tl_i<L} n_i(1-\frac{R(tl_i + \min(\Delta, L-tl_i)}{R(tl_i)})+\sum_{i=1}^{q}d_j(1-R(\min(\Delta+1-i,L))$