Example of Accelerated Life Test Plan

An engineer tests capacitors at accelerated temperatures of 85 degrees, 105 degrees, and 125 degrees to determine the 360-day reliability under normal-use conditions of 45 degrees. 100 capacitors are available to test. The engineer uses the following information for the test plans.
  • Previous studies used the Weibull distribution with a shape parameter value of 3.5 to model failure time data for similar capacitors.
  • From past experience, the engineer knows that the Arrhenius relationship models the relationship over the range of moderate to high temperatures for these capacitors.
  • The engineer anticipates that about .5% of the capacitors will fail in 90 days at 50 degrees C, and that about 70% will fail in 90 days at 125 degrees C.
  • The engineer has only 90 days to complete the test.
  1. Choose Stat > Reliability/Survival > Test Plans > Accelerated Life Testing.
  2. From Calculate, select Standard errors of estimates.
  3. From Relationship between life and stress, select Arrhenius.
  4. Select Specify test stresses.
  5. In Design stress, enter 45.
  6. From Type of censoring for each test stress, select Right censoring time.
  7. From Percent allocation of units for each test stress, select Search for best allocations only.
  8. In Number of best plans to output, enter 3.
  9. Below Test stress, select Add Row.
  10. In each row, enter a test stress and a censor time.
    Test stress Censor time
    85 90
    105 90
    125 90
  11. Click OK.
  12. From Distribution, select Weibull.
  13. In Planning shape parameter value, enter 3.5.
  14. From Specify planning values for regression coefficients, select Percentile points for two distinct stress levels.
  15. Enter the following values for the percentile points.
    Percentile value Percent Stress level
    90 .5 50
    90 70 125
  16. Under Parameter to estimate, select Reliability at time. In Time value, enter 360.
  17. In Sample sizes, enter 100.
  18. Click OK.

Interpret the results

Minitab evaluates the resulting test plans and displays the best plans with respect to minimizing the precision or the sample size. If the sample size is specified, the 1st Best "Optimum" Allocations Test Plan is the plan with the smallest standard error of the parameter of interest.

To estimate the 360-day reliability at the design stress of 45 degrees using the optimal plan, the engineer should test the following number of units at each accelerated temperature:
  • Test 63 units at 85 degrees
  • Test 4 units at 105 degrees
  • Test 33 units at 125 degrees

The settings for this test plan have time censoring, so the plan is that the test continues for 90 days.

Because the standard error for all three plans is very close, the engineer should also consider additional criteria when choosing a plan. Examples of criteria are which plan yields more failures and which plan is the least expensive to execute.

Planning Distribution

DistributionInterceptSlopeShape
Weibull-2.301140.2315113.5
Type I right-censored data (Single Censoring)
Arrhenius model
Estimated parameter: Reliability at time = 360
Calculated planning estimate = 0.666228
Design stress value = 45
Planning Values
Percentile values = 90, 90 for percents = 0.5, 70 at stresses = 50, 125
Selected test plans:  “Optimum” allocations test plans
Total available sample units = 100

1st Best “Optimum” Allocations Test Plan

Test StressCensor TimePercent
Failure
Percent
Alloc
Sample UnitsExpected
Failures
85908.250863.3333635
1059029.19563.958341
1259070.000032.70833323
Standard error of the parameter of interest = 0.338939

2nd Best “Optimum” Allocations Test Plan

Test StressCensor TimePercent
Failure
Percent
Alloc
Sample UnitsExpected
Failures
85908.250858.3333585
1059029.19568.958393
1259070.000032.70833323
Standard error of the parameter of interest = 0.339922

3rd Best “Optimum” Allocations Test Plan

Test StressCensor TimePercent
Failure
Percent
Alloc
Sample UnitsExpected
Failures
85908.250868.3333686
1059029.19563.958341
1259070.000027.70832820
Standard error of the parameter of interest = 0.341185