General Examples Using the Lloyd-Lipow Model
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General Examples Using the Lloyd-Lipow Model |
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In this section, the methods used in RGA to
estimate the confidence bounds under the Lloyd-Lipow
model will be presented. One of the properties of maximum likelihood estimators
is that they are asymptotically normal. This indicates that they are normally
distributed for large samples [6][7]. Additionally, since the
parameter α must
be positive, lnα
is treated as being normally distributed as well. The parameter 
represents
the ultimate reliability that would be attained if 
. Rk
is the actual reliability during the kth
stage of testing. Therefore, 
and Rk
will be between 0 and 1. Consequently, the endpoints of the confidence
intervals of the parameters 
and Rk
also will be between 0 and 1. To obtain the confidence interval, it is
common practice to use the logit transformation.
The confidence bounds on the parameters α
and 
are given
by:
(11)
(12)
where zα/2 represents the percentage points of the N(0,1) distribution such that P{z ≥ zα/2} = α/2.
The confidence bounds on reliability are given by:
(13)
where: 
(14)
All the variances can be calculated using the Fisher Matrix: 
From Eqns. (4) and (5), taking the second partial derivatives yields:
(15)
(16)and:
(17)Now
the confidence bounds can be obtained after calculating Eqns. (14) through
(17) and substituting into the Fisher Matrix.
Plot the confidence bounds for the data in Table 6.1.
From Eqns. (15), (16) and (17): 
The variances can be calculated using the Fisher Matrix: 
The variance of Rk
is obtained from Eqn. 14 such that: 
(18)
Now Eqn. (13) can be calculated and the associated confidence bounds at the 90% confidence level are plotted in Figure 6.2 with the predicted reliability, Rk.
Figure 6.2: Predicted reliability with 90% confidence bounds |
Consider the success/failure data given in Table 6.2. Solve for the Lloyd-Lipow parameters using least squares analysis and plot the Lloyd-Lipow reliability with 2-sided confidence bounds at the 90% confidence level.
Table 6.2 - Success/failure
data for a variable number
of tests performed in each test stage
Test
Stage |
Result |
Number of |
Successful |
1 |
F |
1 |
0 |
2 |
F |
1 |
0 |
3 |
F |
1 |
0 |
4 |
S |
1 |
0.2500 |
5 |
F |
1 |
0.2000 |
6 |
F |
1 |
0.1667 |
7 |
S |
1 |
0.2857 |
8 |
S |
1 |
0.3750 |
9 |
S |
1 |
0.4444 |
10 |
S |
1 |
0.5000 |
11 |
S |
1 |
0.5455 |
12 |
S |
1 |
0.5833 |
13 |
S |
1 |
0.6154 |
14 |
S |
1 |
0.6429 |
15 |
S |
1 |
0.6667 |
16 |
S |
1 |
0.6875 |
17 |
F |
1 |
0.6471 |
18 |
S |
1 |
0.6667 |
19 |
F |
1 |
0.6316 |
20 |
S |
1 |
0.6500 |
21 |
S |
1 |
0.6667 |
22 |
S |
1 |
0.6818 |
Note that the data set contains three consecutive failures at the beginning of the test. These failures will be ignored throughout the analysis because it is considered that the test starts when the reliability is not equal to zero or one. The number of data points is now reduced to 19. Also note that the only time that the first three first failures are considered is to calculate the observed reliability in the test. For example, given this data set, the observed reliability at stage 4 is 1/4 = 0.25. This is considered to be the reliability at stage 1.
From Table 6.2, the least squares estimates can be calculated as follows:
and: 
Substituting into Eqns. (8) and (9)
yields: 
and:
Therefore,
the Lloyd-Lipow reliability growth model is as follows, where k is the number of the test stage.
From Eqns. (15), (16), (17) and the data given in Table 6.2: 
The variances can be calculated using
the Fisher Matrix: 
The variance of Rk
is obtained from Eqn. 14: 
Now Eqn. (13) can be calculated and the associated confidence bounds on reliability at the 90% confidence level are plotted in Figure 6.3 with the predicted reliability, Rk.
Figure 6.3: Reliability vs. Time plot with 90% confidence bounds |
