Parameter Estimation for the Lloyd-Lipow Model

When analyzing reliability data in RGA, you have the option to enter the reliability values in percent or in decimal format. However, $\hat{R}_{\infty }$ will always be returned in decimal format and not in percent. The estimated parameters in RGA are unitless.

Maximum Likelihood Estimators

For the kth stage:

And assuming that the results are independent between stages: MATH

Then taking the natural log gives: MATH

Differentiating with respect to $R_{\infty }$ and α yields: MATH

(2)

(5)Eqns. (4) and (5) can be solved simultaneously for $\widehat{\alpha }$ and $\hat{R}_{\infty }$ . It should be noted that a closed form solution does not exist for either of the parameters; thus they must be estimated numerically.

Least Squares Estimators

To obtain least squares estimators for $R_{\infty }$ and α, the sum of squares, Q, of the deviations of the observed success-ratio, Sk/nk, is minimized from its expected value,

, with respect to the parameters $R_{\infty }$ and α. Therefore, Q is expressed as: MATH

Taking the derivatives with respect to $R_{\infty }$ and α and setting equal to zero yields:

Solving Eqns. (6) and (7) simultaneously, the least squares estimates of $R_{\infty }$ and α are: MATH

or:

(8)and:

or:

(9)

Lloyd-Lipow Example 1

After a 20-stage reliability development test program, 20 groups of success/failure data were obtained and are given in Table 6.1. Do the following:

Table 6.1 - The test results and reliabilities of each stage calculated from raw data and the predicted reliability

Test Stage Number k	Number of Tests in Stage (nk)	Number of Successful Tests (Sk)	Raw Data Reliability	Lloyd-Lipow Reliability
1	9	6	0.667	0.7002
2	9	5	0.556	0.7369
3	8	7	0.875	0.7552
4	10	6	0.600	0.7662
5	9	7	0.778	0.7736

6	10	8	0.800	0.7788
7	10	7	0.700	0.7827
8	10	6	0.600	0.7858
9	11	7	0.636	0.7882
10	11	9	0.818	0.7902

11	9	9	1.000	0.7919
12	12	10	0.833	0.7933
13	12	9	0.750	0.7945
14	11	8	0.727	0.7956
15	10	7	0.700	0.7965

16	10	8	0.800	0.7973
17	11	10	0.909	0.7980
18	10	9	0.900	0.7987
19	9	8	0.889	0.7992
20	8	7	0.875	0.7998

Solution to Lloyd-Lipow Example 1

From Table 6.1, the least squares estimates are: MATH

and:

Substituting into Eqns. (8) and (9) yields: MATH

Therefore, the Lloyd-Lipow reliability growth model is as follows, where k is the test stage.

(10)

The reliabilities from the raw data and the reliabilities predicted from Eqn. 10 are given in the last two columns of Table 6.1. Figure 6.1 shows the plot. Based on the given data, the model cannot do much more than to basically fit a line through the middle of the points.