# Cumulative Damage Model for Step Stress Profiles

 Cumulative Damage Model for Step Stress Profiles

This example validates the results for a cumulative damage model for a step stress test in an ALTA standard folio.

Reference Case

The data set is from Table 2.1 on page 496 in the book Accelerated Testing: Statistical Models, Test Plans, and Data Analysis by Dr. Nelson, John Wiley & Sons, 1990.

Data

A step-stress test is conducted for one type of cable insulation to estimate the insulation life at a constant design stress of 400 volts/mil. The test cables are of different thickness, in units of mils (0.001 inch). The stress is the applied step voltage divided by the thickness.

A total of 6 step-stress profiles are used. These stress profiles are calculated based on the applied voltage (Kilovolts) and the insulation thickness given on page 495 and Table 2.1. For all the stress profiles, the holding time for the first 4 steps is 10 mins. From step 5 onwards, a different a holding time is applied at each step for each of the stress profiles. These profiles are given in the following tables.

Profile 1 (named "G1"): The holding time after step 4 is 15 mins and the thickness is 27 mils.

Segment Start Time Segment End Time Stress (Kvolts/mils)
0 10 0.185185185
10 20 0.37037037
20 30 0.555555556
30 40 0.740740741
40 55 0.962962963
55 70 1.055555556
70 85 1.148148148
85 100 1.233333333
100 115 1.333333333
115 130 1.425925926
130 145 1.518518519

Profile 2 (named "G2"): The holding time after step 4 is 60 mins and the thickness is 29.5 mils.

Segment Start Time Segment End Time Stress (Kvolts/mils)
0 10 0.169491525
10 20 0.338983051
20 30 0.508474576
30 40 0.677966102
40 100 0.881355932
100 160 0.966101695
160 220 1.050847458
220 280 1.128813559
280 340 1.220338983
340 400 1.305084746
400 460 1.389830508

Profile 3 (named "G3"): The holding time after step 4 is 60 mins and the thickness is 28 mils.

Segment Start Time Segment End Time Stress (Kvolts/mils)
0 10 0.178571429
10 20 0.357142857
20 30 0.535714286
30 40 0.714285714
40 100 0.928571429
100 160 1.017857143
160 220 1.107142857
220 280 1.189285714
280 340 1.285714286
340 400 1.375
400 460 1.464285714

Profile 4 (named "G4"): The holding time after step 4 is 240 mins and the thickness is 29 mils.

Segment Start Time Segment End Time Stress (Kvolts/mils)
0 10 0.172413793
10 20 0.344827586
20 30 0.517241379
30 40 0.689655172
40 280 0.896551724
280 520 0.982758621
520 760 1.068965517
760 1000 1.148275862
1000 1240 1.24137931
1240 1480 1.327586207
1480 1720 1.413793103

Profile 5 (named "G5"): The holding time after step 4 is 240 mins and the thickness is 30 mils.

Segment Start Time Segment End Time Stress (Kvolts/mils)
0 10 0.166666667
10 20 0.333333333
20 30 0.5
30 40 0.666666667
40 280 0.866666667
280 520 0.95
520 760 1.033333333
760 1000 1.11
1000 1240 1.2
1240 1480 1.283333333
1480 1720 1.366666667

Profile 6 (named "G6"): The holding time after step 4 is 960 mins and the thickness is 30 mils.

Segment Start Time Segment End Time Stress (Kvolts/mils)
0 10 0.166666667
10 20 0.333333333
20 30 0.5
30 40 0.666666667
40 1000 0.866666667
1000 1960 0.95
1960 2920 1.033333333
2920 3880 1.11
3880 4840 1.2
4840 5800 1.283333333
5800 6760 1.366666667

The following table shows the test results.

Status F/S Time to F/S Test Profile Subset ID
F 102 G1 1
F 113 G1 2
F 113 G1 3
S 370 G2 4
F 345 G2 5
S 345 G3 6
F 1249 G4 7
F 1333 G4 8
S 1333 G4 9
F 1096.6 G4 10
F 1250.8 G5 11
F 1097.9 G4 12
F 2460.9 G6 13
S 2460.9 G6 14
F 2700.4 G6 15
F 2923.9 G6 16
F 1160 G6 17
F 1962.9 G6 18
S 363.9 G6 19
F 898.4 G6 20
F 4142.1 G6 21

Result

The power law life stress relationship and the Weibull distribution are used to analyze the data. At a constant stress V, the $\eta\,\!$ is:

$\eta(V) = \left(\frac{V_{0}}{V} \right)^p\,\!$

where $V_{0}\,\!$ and $p\,\!$ are the model parameters used in the book. The above equation can be rewritten as:

$\eta(V) = e^{\alpha_{0}+\alpha_{1}ln(V)}\,\!$

where $\alpha_{0} = p \cdot ln(V_{0})\,\!$  and  $\alpha_{1} = -p\,\!$

The reliability function at time t and stress V is:

$R(t,V) = e^{-\left(\frac{t}{\eta(V)} \right)^\beta}\,\!$

When stress is varying with time, the reliability at time t is given as:

$R(t,V) = e^{-\left(\int_{0}^{t}\frac{1}{\eta(x)} dx\right)^{\beta}}\,\!$

In the book, the following results are provided:

• ML solution for the parameters are $\beta\,\!$ = 0.75597, $V_{0}\,\!$ = 1616.4 (1.6164 Kvolts), and $p\,\!$ = 19.937.
• The maximum log likelihood is -103.53.
• The 1% percentile point (B1 life) at 0.4 Kvolts/mil is 2.81 x 109.
• The normal distribution approximation two-sided 95% confidence intervals are $\beta\,\!$ = [0.18, 1.33], $V_{0}\,\!$ = [1291, 1941.8], $p\,\!$ = [6.2, 33.7], and the B1 life is [2.65 x 104, 2.98 x 1014].

Results in ALTA

First, we create each stress profile in ALTA. For example, the following picture shows the data for Profile G1.

The following picture shows the plot for this stress profile.

The next step is to enter the failure data into an ALTA standard folio and use the stress profiles to define the stress values, as shown next.

The log transformation is used for the stress. To compare the parameter values used in the book with the results obtained by ALTA, we need do convert the values in the book in terms of the parameters used in ALTA. Therefore the parameters in the book are: $\beta\,\!$ = 0.75597,   $\alpha_{0}\,\!$ = 9.573776,   $\alpha_{1}\,\!$ = -19.937. Comparing these values to the value shown in the picture above, we see that the results obtained in ALTA are close to the results shown in the book.

The picture above also shows that the maximum log likelihood (LK Value) is -114.932651. This value is smaller than the value given in the book. To validate what the LK value would be if the results in the book were used in ALTA, we use the Alter Parameters tool, as shown next.

The resulting LK Value is -115.502692, as shown next.

Altered Parameters

This value is different from the value of -13.53 given in the book. Later, we will validate the calculations for this result.

Using the parameters values originally calculated in ALTA:

• The 1% percentile point (B1 life) at 0.4 Kvolts/mil is 4.8342 X 107 hours, as shown next. The value given in the book is 2.81 X 109. The results are different because of the differences in the values of the parameters of the model.

• The 2-sided 95% confidence intervals for the model parameters are shown next. These results are also different from the results in the book due to the estimated parameter values.

Validate the Likelihood Calculation in ALTA

For the ML solution given in the book, the parameters are: $\beta\,\!$ = 0.75597,  $\alpha_{0}\,\!$ = 9.573776, $\alpha_{1}\,\!$ = -19.937, and the log likelihood value is -103.53. Entering these parameters in ALTA, we get a log likelihood value of -115.502692. We will validate the calculation for the log likelihood value in ALTA to make sure it is correct.

Define the stress value at step i as $S_{i}\,\!$. If the test unit stayed in this stress level for a time period of $\Delta t_{i}\,\!$, then the cumulative damage at this step will be $\frac{1}{\eta(S_{i})}\times \Delta t_{i}\,\!$ and

$A_{i} = \frac{1}{\eta(S_{i})} \times \Delta t_{i} = e^{-(\alpha_{0}+\alpha_{1}ln(S_{i}))} \times \Delta t_{i}\,\!$

For example, for the first failure, the failure occurred at 102, which is at the 9th step of profile G1. The test unit accumulates damage from each step stress it experienced before it fails. The first step stress value is 0.185185, which is the voltage divided by the thickness (5 Kvolts/27 mils). The length of the first step is 10 mins. Therefore, the damage at the 1st step segment is:

$A_{1} = e^{-(\alpha_{0}+\alpha_{1}ln(S_{1}))} \times \Delta t_{i} = e^{-(8.757876+16.095189 \times 1.6864)}\times 10 = 2.56 \times 10^{-15}\,\!$

We can use this method to calculate the damage at each stress segment. The total damage will be the sum of the damage at each segment. For the first failure at time 102 under stress profile G1, the damage at each stress segment is calculated, as given below.

Step Profile G1 Segment End Voltage Thickness Stress ln(S) Cum Damage
1 0 10 5 27 0.185185 -1.68639895 2.50025E-15
2 10 20 10 27 0.37037 -0.99325177 1.76011E-10
3 20 30 15 27 0.555556 -0.58778666 1.20553E-07
4 30 40 20 27 0.740741 -0.30010459 1.23907E-05
5 40 55 26 27 0.962963 -0.03774033 0.0012707
6 55 70 28.5 27 1.055556 0.054067221 0.005573181
7 70 85 31 27 1.148148 0.138150338 0.021584512
8 85 100 33.4 27 1.233333 0.209720531 0.071720247
9 100 102 36 27 1.333333 0.287682072 0.031977014
A = 0.132138165

The total cumulative damage at a failure time of 102 is A in the above table.

Since the reliability function is:

$R(t) = e^{-\left[\int^{t}_{0}e^{-(\alpha_{0}+\alpha_{1}ln(S(x)))} dx \right]^{\beta}} = e^{-A^{\beta}}\,\!$

The probability density function (pdf) at time t is:

\begin{align} f(t) =& R(t) \cdot \beta \cdot \left[\int^{t}_{0}e^{-(\alpha_{0}+\alpha_{1}ln(S(x)))} dx \right]^{\beta-1}\cdot e^{-(\alpha_{0}+\alpha_{1}ln(S(t)))} \\ =& e^{-A^{\beta}}\cdot \beta \cdot A^{\beta-1}\cdot e^{-(\alpha_{0}+\alpha_{1}ln(S(t)))} \\ \end{align}\,\!

If an observation is a failure, then the pdf will be used to calculate its log likelihood value. If an observation is a suspension, then the reliability will be used to calculate its log likelihood value.

Since the 1st observation is a failure, the pdf is used. Using the model parameters and the cumulative damage A, the value of the pdf is:

\begin{align} f(t) =& e^{-A^{\beta}} \cdot \beta \cdot A^{\beta - 1} \cdot e^{-(\alpha_{0}+\alpha_{1}ln(S(t)))} \\ =& e^{-0.13214^{0.7906}} \times 0.7906 \times 0.13214^{0.7906-1} \times e^{-(8.7685 - 16.103 \times 0.2877)} \\ =& 0.016 \end{align}\,\!

Take the logarithm of the above value and we get the log likelihood value for the first observation as $log(0.016) = -4.1489\,\!$.

Repeating the above procedure, we can get the log likelihood value for each observation, as given in the following table.

Status F/S Time To F/S Profile Log likelihood Value
F 102 G1 -4.148929577
F 113 G1 -4.518387096
F 113 G1 -4.518387096
S 370 G2 -0.72559186
F 345 G2 -4.983506849
S 345 G3 -0.910644661
F 1249 G4 -6.199936981
F 1333 G4 -7.129057207
S 1333 G4 -2.451259934
F 1096.6 G4 -6.518744419
F 1250.8 G5 -6.075214111
F 1097.9 G4 -6.523803805
F 2460.9 G6 -8.444820079
S 2460.9 G6 -0.29378846
F 2700.4 G6 -8.566484581
F 2923.9 G6 -7.469922325
F 1160 G6 -9.119207934
F 1962.9 G6 -8.086491838
S 363.9 G6 -0.015237126
F 898.4 G6 -10.4367511
F 4142.1 G6 -7.897236924
Total -115.033404

The log likelihood value calculated by hand in Microsoft Excel is -115.0334, as given in the above table. It matches the value in ALTA. The slight difference is caused by the precision error. Therefore, the log likelihood value calculated in ALTA is correct.

In fact, the optimization tool in Excel can also be used to find the ML solutions. Using the “Solver Add-in” program in Excel and the “GRG” nonlinear optimization, we get the same results in Excel as in ALTA.

The following reliability plot shows that the predicted values (the line) by the model can match the observed values (the points) very well.