8.
Assessing Product Reliability
8.4. Reliability Data Analysis 8.4.2. How do you fit an acceleration model?


If you can fit models using degradation data, you don't need actual test failures 
When failure can be related directly to a change over time in a measurable product
parameter, it opens up the possibility of measuring degradation over time and using
that data to extrapolate when failure will occur. That allows us to fit acceleration
models and life distribution models without actually waiting for failures
to occur.
This overview of degradation modeling assumes you have chosen a life distribution model and an acceleration model and offers an alternative to the accelerated testing methodology based on failure data, previously described. The following topics are covered. More details can be found in Nelson (1990, pages 521544) or Tobias and Trindade (1995, pages 197203). 

You need a measurable parameter that drifts (degrades) linearly to a critical failure value 
Two common assumptions typically made when degradation data are modeled are the following:


Plot of linear degradation trends for five units read out at four time points 
In many practical situations, \(D\) starts at 0 at time zero, and all the linear theoretical degradation lines start at the origin. This is the case when \(D\) is a "% change" parameter, or failure is defined as a change of a specified magnitude in a parameter, regardless of its starting value. Lines all starting at the origin simplify the analysis since we don't have to characterize the population starting value for \(D\), and the "distance" any unit "travels" to reach failure is always the constant \(DF\). For these situations, the degradation lines would look as follows. 

Often, the degradation lines go through the origin  as when % change is the measurable parameter increasing to a failure level 
It is also common to assume the effect of measurement error, when reading values of \(D\), has relatively little impact on the accuracy of model estimates. 

Modeling based on complete samples of measurement data, even with low stress cells, offers many advantages 


Degradation may not proceed in a smooth, linear fashion towards what the customer calls "failure" 


A simple approach is to extend each unit's degradation line until a projected "failure time" is obtained 
A More Accurate Regression Approach For the Case When \(D\) = 0 at time 0 and the "Distance To Fail" \(DF\) is the Same for All Units 

Models can be fit using all the degradation readings and linear regression 
Let the degradation measurement for the \(i\)th
unit at the \(j\)th
readout time in the \(k\)th
stress cell be given by \(D_{ijk}\),
and let the corresponding readout time be denoted by \(t_{jk}\).
That readout gives a degradation rate (or slope)
estimate of \(D_{ijk} / t_{jk}\).
This follows from the linear assumption or:
Based on that readout alone, an estimate of the natural logarithm of the time to fail for that unit is $$ y_{ijk} = \mbox{ln } DF  \left( \mbox{ln } D_{ijk}  \mbox{ln } t_{jk} \right) \, . $$ This follows from the basic formula connecting linear degradation with failure time by solving for (time of failure) and taking natural logarithms. For an Arrhenius model analysis, with $$ t_f = A \cdot \mbox{exp}\left( \frac{\Delta H}{KT} \right) \, , $$ $$ y_{ijk} = a + b x_k \, , $$ with the \(x_k\) values equal to \(1/KT\). Here \(T\) is the temperature of the \(k\)th cell, measured in Kelvin (273.16 + degrees Celsius) and \(K\) is Boltzmann's constant (8.617 × 10^{5} in eV/ unit Kelvin). Use a linear regression program to estimate \(a = \mbox{ln } A\) and \(b = \Delta H\). If we further assume \(t_f\) has a lognormal distribution, the mean square residual error from the regression fit is an estimate of \(\sigma^2\) (with \(\sigma\) the lognormal sigma). One way to think about this model is as follows: each unit has a random rate \(R\) of degradation. Since \(t_f = DF/R\), it follows from a characterization property of the normal distribution that if is lognormal, then \(R\) must also have a lognormal distribution (assuming \(DF\) and \(R\) are independent). After we take logarithms, \(\mbox{ln } R\) has a normal distribution with a mean determined by the acceleration model parameters. The randomness in \(R\) comes from the variability in physical characteristics from unit to unit, due to material and processing differences. Note: The estimate of sigma based on this simple graphical approach might tend to be too large because it includes an adder due to the measurement error that occurs when making the degradation readouts. This is generally assumed to have only a small impact. 

An example using the regression approach to fit an Arrhenius model 
A component has a critical parameter that studies
show degrades linearly over time at a rate that varies with operating temperature.
A component failure based on this parameter occurs when the parameter value
changes by 30 % or more. Fifteen components were tested under 3 different
temperature conditions (5 at 65 °C, 5 at 85 °C and the
last 5 at 105 °C). Degradation percent values were read out at
200, 500 and 1000 hours. The readings are given by unit in the following
three temperature cell tables.
65 °C
85 °C
105 °C
Note that one unit failed in the 85 °C cell and four units failed in the 105 °C cell. Because there were so few failures, it would be impossible to fit a life distribution model in any cell but the 105 °C cell, and therefore no acceleration model can be fit using failure data. We will fit an Arrhenius/lognormal model, using the degradation data. Solution: 

Fit the model to the degradation data 
From the above tables, first create a variable (\(DEG\)) with 45 degradation values
starting with the first row in the first table and proceeding to the last
row in the last table. Next, create a temperature variable (\(TEMP\)) that has
15 repetitions of 65, followed by 15 repetitions of 85 and then 15 repetitions of 105.
Finally, create a time variable (\(TIME\)) that corresponds to readout times.
Fit the Arrhenius/lognormal equation, \(y_{ijk} = a + bx_{ijk}\), where $$ y_{ijk} = \mbox{ln } 30  \left( \mbox{ln } DEG  \mbox{ln } TIME \right) $$ and $$ x_{ijk} = \frac{100000}{8.617(TEMP + 273.16)} \, . $$ The linear regression results are the following. Parameter Estimate Stan. Dev t Value     a 18.94337 1.83343 10.33 b 0.81877 0.05641 14.52 Residual standard deviation = 0.5611 Residual degrees of freedom = 45 The Arrhenius model parameter estimates are: \(\mbox{ln } A\) = 18.94; \(\Delta H\) = 0.82. An estimate of the lognormal sigma is \(\sigma\) = 0.56. The analyses in this section can can be implemented using both Dataplot code and R code. 