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

8.4.2.1.

Graphical estimation

This section will discuss the following: 

1. How to fit an Arrhenius model with graphical estimation
2. Graphical estimation: an Arrhenius model example
3. Fitting more complicated models

Estimate acceleration model parameters by estimating cell T₅₀'s (or 's) and then using regression to fit the model across the cells

How to fit an Arrhenius Model with Graphical Estimation

Graphical methods work best (and are easiest to describe) for a simple one-stress model like the widely used Arrhenius model

with T denoting temperature measured in degrees Kelvin (273.16 + degrees Celsius) and k is Boltzmann's constant (8.617 x 10^-5 in eV/°K). 

When applying an acceleration model to a distribution of failure times, we interpret the deterministic model equation to apply at any distribution percentile we want. This is equivalent to setting the life distribution scale parameter equal to the model equation (T₅₀ for the lognormal,  for the Weibull and the MTBF or 1/ for the exponential). For the lognormal, for example, we have

So, if we run several stress cells and compute T₅₀'s for each cell, a plot of the natural log of these T₅₀'s versus the corresponding 1/kT values should be roughly linear with a slope of H and an intercept of ln A. In practice, a computer fit of a line through these points is typically used to obtain the Arrhenius model estimates. There are even commercial Arrhenius graph papers that have a temperature scale in 1/kT units and a T₅₀ scale in log units, but it is easy enough to make the transformations and then use linear or log-linear papers. Remember that T is in Kelvin in the above equations. For temperature in Celsius, use the following for 1/kT: 11605/(TCELSIUS + 273.16) 

An example will illustrate the procedure. 

Graphical Estimation: An Arrhenius Model Example:

Arrhenius model example

Component life tests were run at 3 temperatures: 85°C, 105°C and 125°C. The lowest temperature cell was populated with 100 components; the 105° cell had 50 components and the highest stress cell had 25 components. All tests were run until either all the units in the cell had failed or 1000 hours was reached. Acceleration was assumed to follow an Arrhenius model and the life distribution model for the failure mode was believed to be lognormal. The normal operating temperature for the components is 25°C and it is desired to project the use CDF at 100,000 hours. 

Test results: 

Cell 1 (85°C): 5 failures at 401, 428, 695, 725 and 738 hours. 95 units were censored at 1000 hours running time. 

Cell 2 (105°C): 35 failures at 171, 187, 189, 266, 275, 285, 301, 302, 305, 316, 317, 324, 349, 350, 386, 405, 480, 493, 530, 534, 536, 567, 589, 598, 599, 614, 620, 650, 668, 685, 718, 795, 854, 917, and 926 hours. 15 units were censored at 1000 hours running time. 

Cell 3 (125°C): 24 failures at 24, 42, 92, 93, 141, 142, 143, 159, 181, 188, 194, 199, 207, 213, 243, 256, 259, 290, 294, 305, 392, 454, 502 and 696. 1 unit was censored at 1000 hours running time. 

Failure analysis confirmed that all failures were due to the same failure mechanism (if any failures due to another mechanism had occurred, they would have been considered censored run times in the Arrhenius analysis). 

Steps to Fitting the Distribution Model and the Arrhenius Model:

• Do graphical plots for each cell and estimate T₅₀'s and sigma's as previously discussed.
• Put all the plots on the same sheet of graph paper and check whether the lines are roughly parallel (a necessary consequence of true acceleration).
• If satisfied from the plots that both the lognormal model and the constant sigma from cell to cell are consistent with the data, plot the cell ln T₅₀'s versus the 11605/(TCELSIUS + 273.16) cell values, check for linearity and fit a straight line through the points. Since the points have different degrees of precision, because different numbers of failures went into their calculation, it is recommended that the number of failures in each cell be used as weights in a regression program, when fitting a line through the points.
• Use the slope of the line as the H estimate and calculate the Arrhenius A constant from the intercept using A = e^intercept .
• Estimate the common sigma across all the cells by the weighted average of the individual cell sigma estimates. Use the number of failures in a cell divided by the total number of failures in all cells as that cells weight. This will allow cells with more failures to play a bigger role in the estimation process. 

Dataplot solution for Arrhenius model example 

Dataplot Analysis of Multicell Arrhenius Model Data:

After creating text files DAT1.TXT, DAT2.TXT and DAT3.TXT of the failure times for the 3 stress cells, enter Dataplot and execute the following sequence of commands (individual cell plots have been skipped): 

READ DAT1.TXT CELL1
READ DAT2.TXT CELL2
READ DAT3.TXT CELL3
LET Y1 = LOG(CELL1)
LET Y2 = LOG(CELL2)
LET Y3 = LOG(CELL3)
LET POS1 = SEQUENCE 1 1 5
LET POS2 = SEQUENCE 1 1 35
LET POS3 = SEQUENCE 1 1 24
LET POS1 = (POS1 -.3)/100.4
LET POS2 = (POS2 -.3)/50.4
LET POS3 = (POS3 -.3)/25.4
LET X1 = NORPPF(POS1)
LET X2 = NORPPF(POS2)
LET X3 = NORPPF(POS3)
TITLE PROBABILITY PLOTS OF THREE TEMPERATURE CELLS 
PLOT Y1 X1 AND
PLOT Y2 X2 AND
PLOT Y3 X3

This will produce the following probability plot of all three stress cells on the same graph.

Note that the lines are somewhat straight (a check on the lognormal model) and the slopes are approximately parallel (a check on the acceleration assumption). 

The cell ln T₅₀ and sigma estimates are obtained from the FIT function as follows: 

FIT Y1 X1
FIT Y2 X2
FIT Y3 X3

Each FIT will yield a cell A_o, the ln T₅₀ estimate, and A₁, the cell sigma estimate. These are summarized in the table below. 

Summary of Least Squares Estimation of Cell Lognormal Parameters

Cell Number	ln T50	Sigma
1   (T = 85)	8.168	.908
2  (T = 105)	6.415	.663
3  (T = 125)	5.319	.805

The three cells have 11605/(T + 273.16) values of 32.40, 30.69 and 29.15 respectively, in cell number order. The Dataplot commands to generate the Arrhenius plot are: 

LET YARRH = DATA 8.168 6.415 5.319
LET XARRH = DATA 32.4 30.69 29.15
TITLE = ARRHENIUS PLOT OF CELL T50'S

With only three cells, it is unlikely a straight line through the points will present obvious visual lack of fit. However, in this case, the points appear to line up very well. 

Finally, the model coefficients are computed from 

LET SS = DATA 5 35 24
WEIGHT = SS
FIT YARRH XARRH

This will yield a ln A estimate of -18.312 (A = e^-18.312 = .1115x10^-7) and a H estimate of .808. With this value of H, the acceleration between the lowest stress cell of 85°C and the highest of 125°C is 

which is almost 14× acceleration. Acceleration from 125 to the use condition of 25°C is 3708× . The use T₅₀ is e^-18.312 x e^{.808x11605x1/298.16}= e^13.137 = 507383. 

A single sigma estimate for all stress conditions can be calculated as a weighted average of the 3 sigma estimates obtained from the experimental cells. The weighted average is (5/64) × .908 + (35/64) × .663 + (24/64) × .805 = .74. 

Fitting More Complicated models

Models involving several stresses can be fit using multiple regression

Two stress models, such as the temperature /voltage model given by 

need at least 4 or five carefully chosen stress cells to estimate all the parameters. The Backwards L design previously described is an example of a design for this model. The bottom row of the "backward L" could be used for a plot testing the Arrhenius temperature dependence, similar to the above Arrhenius example. The right hand column could be plotted using y = ln T₅₀ and x = ln V, to check the voltage term in the model. The overall model estimates should be obtained from fitting the multiple regression model 

The Dataplot command for fitting this model, after setting up the Y, X1 = X₁, X2 = X₂  data vectors, is simply 

FIT Y X1 X2

and the output gives the estimates for b₀, b₁ and b₂. 

Three stress models, and even Eyring models with interaction terms, can be fit by a direct extension of these methods. Graphical plots to test the model, however, are less likely to be meaningful as the model becomes more complex.