Reliability HotWire

Reliability HotWire

Issue 127, September 2011

Hot Topics

Truncation of the Integral Used in the Comparison Wizard in Weibull++: Part II

In the July 2011 issue of HotWire, the Hot Topics article used simulations in RENO to illustrate the results of the Comparison Wizard in Weibull++ and study the effect of the upper limit in the calculation to the output of the comparison. In this article, we will provide an analytical explanation of the results.

Table 1 shows the test data of the two designs.

Table 1 - Failure times for designs A and B

Design A Design B
1059 4869 
2245 5348
2781 5951
3130 6257
3535 6561
4512 6763
5632 6796
6218 7774
6819 7992
6891 8133

The data of these two designs are fitted with the Weibull distribution. Figure 1 shows the reliability plots and Figure 2 shows the pdf (probability density function) plots.

Reliability Plot
Figure 1 - Reliability plot for design A and design B

pdf Plot
Figure 2 - pdf plot for design A and design B

From Figures 1 and 2, it can be seen that most of the time design B will last longer than design A. The probability of B outlasting A can be calculated using the following equation:

(1)

where fA(t) is the pdf for design A and RB(t) is the reliability function for design B.

The above equation is used in the Comparison Wizard in Weibull++. We will discuss two cases related to the value of the upper limit in the integral in Eqn. (1):

  • When the upper limit in Eqn. (1) is infinite, the Comparison Wizard calculates the probability that design B outlasts A when the warranty is a lifetime warranty, and the sum of the probability of B outlasting A and the probability of A outlasting B is 1.
  • When the upper limit in Eqn. (1) is a limited value, the Comparison Wizard calculates the probability that design B outlasts A within a limited warranty time period, and the sum of the probability of B outlasting A and the probability of A outlasting B is less than 1.

In the following sections, we will discuss these two cases and explain why the sum of the two probabilities can have different values depending on the value of the upper limit in the integral.

Case 1: When the Upper Limit in Eqn. (1) is Infinite

The probability that B outlasts A is:

(2)

The probability that B fails before A is:

(3)

where FB(x) is the distribution function of B.

The sum of Eqns. (2) and (3) is:

Equation

Combining the integrals and factoring the integrand yields:

Equation

Because the sum of reliability and unreliability is 1, the integrand becomes:

Equation

Using the definition of a pdf, we see that the right side of the equation reduces to 1. Thus, we have the following result:

Equation

This can be confirmed in Weibull++. Figure 3 shows the calculation settings.

Comparison Wizard
Figure 3 - The settings of the Comparison Wizard in Weibull++

Figure 4 shows that the probability that B will last longer than A is 80.7452%.

Comparison Wizard
Figure 4 - The probability that B will last longer than A

Figure 5 shows that the probability that A will last longer than B is 19.2547%.

Comparison Wizard
Figure 5 - The probability that A will last longer than B

The sum of these two probabilities is 0.99999. Theoretically, it should be 1. The difference is caused by the precision of the numerical integration.

Case 2: When the Upper Limit in Eqn. (1) is T, a Finite Number

If we set the upper limit to T, using the Comparison Wizard in Weibull++, we get the results shown in Table 2.

Table 2 - Comparison Wizard results for warranty times T from 1,000 to 10,000 hours

Warranty Time
(Hours)
A Outlasts B
(%)
B Outlasts A
(%)
1,000 0.0002 4.6180
2,000 0.0183 16.3207
3,000 0.2339 32.0842
4,000 1.2799 48.7213
5,000 4.2155 63.3380
6,000 9.5036 73.7164
7,000 15.3323 79.0172
8,000 18.5812 80.5640
9,000 19.2281 80.7408
10,000 19.2546 80.7452

From Table 2, we can see that the sum of these two probabilities is not even close to 1 when the warranty time T is small. For example, when T=5,000, the sum of 4.2155% and 63.3380% is 67.5535%. Since the sum is not 1, there must be something missing from the calculations in the table. To explain why the sum is not 1, we need to find out the space of the defined probabilities. The entire space for the probability should be:

(4)

The second part of Eqn. (4) can be divided into two exclusive events:

Equation

The calculation for P1 is:

(5)

The calculation for P2 is:

(6)

The sum of P1 and P2 is:

Equation

Combining the fractions and rewriting the integrands in terms of the unreliability functions we obtain:

Equation

The first and third terms can be integrated directly, and the last term can be integrated by parts to yield:

Equation

After simplifying, we are left with the following:

Equation

Rewriting in terms of the reliability functions, we obtain:

Equation

Simplifying and cancelling yields:

(7)

The entire probability space in Eqn. (4) can be written as:

(8)

The second term in Eqn. (8) is the value in the second column in Table 2 (the probability that A will last longer than B); the third term in Eqn. (8) is the value in the third column in Table 2 (the probability that B will last longer than A). From Eqn. (8), we know that the missing probability in Table 2 is RA(T) x RB(T) (the probability that both designs will last longer than the warranty time).

Conclusion:

In this article, we discussed two cases for comparing two designs. When the comparison is for the entire life span, the sum of the probability of B outlasting A and the probability of A outlasting B is 1; when the comparison is for a given warranty time, the sum of the probability of B outlasting A and the probability of A outlasting B is not 1. This is because the probability of both designs outlasting the warranty time T is not considered. When both A and B are greater than the warranty time, it is not possible to see which design is better in terms of the given warranty time period.