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.
 Figure 1 - Reliability
plot for design A and design B
 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:

Combining the integrals and factoring the integrand yields:

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

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

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

The calculation for P1 is:
 |
(5) |
The calculation for P2 is:
 |
(6) |
The sum of P1 and P2 is:

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

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

After simplifying, we are left with the following:

Rewriting in terms of the reliability functions, we obtain:

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.
|