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Table 5.7
Set of fitness cases used in the “V” function problem.
f
(
a
)
a
-0.2639725157548
3.19498066265276
0.0578905532656938
1.99052001725998
0.334025290109634
8.39663703997286
-0.236334577564462
3.07088976972825
-0.855744382566804
5.87946763695703
-0.0194437136332785
-0.775326322328458
-0.192134388183304
2.83470225774408
0.529307910124627
12.2154726642137
-0.00788974118728459
-2.49803983418635
0.438969804950631
10.4071734858808
-0.107559292698039
2.09413635645908
-0.274556994377163
3.23927278010839
-0.0595333219604528
1.19701284767347
0.384492993958352
9.35580769189855
-0.874923020736333
6.00642453001302
-0.236546636250546
3.07189729043837
-0.167875941704557
2.67440053130986
0.950682181822091
22.4819639844149
0.946979159577362
22.3750161187355
0.639339910059591
14.5701285332337
For the breast cancer problem, we are going to use the same datasets of
section 4.2.1, Diagnosis of Breast Cancer. In this case, however, the fitness
function will be based on the number of hits and will be evaluated by equa-
tion (3.10). Thus, for this problem with 350 training samples,
f
max
= 350. This
experiment, with its three different approaches, is summarized in Table 5.9.
For the analog circuit design, we are going to find the transfer function
expressing the yield in terms of three parameter tolerances. A training set
consisting of
n
= 40 pairs of tolerances and their corresponding yields ob-
tained from
n
runs of Monte Carlo simulations, kindly provided by Lukasz
Zielinski (Zielinski and Rutkowski 2004), will be used. The mean squared
error, evaluated by equation (3.4a), will be again used as basis for the fitness
function, with the fitness being evaluated by equation (3.5) and, therefore,
f
max
= 1000. This experiment, with its three different approaches, is summa-
rized in Table 5.10.
