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b) Situation 2 (Lines 10–18 in Algorithm 2). Here is a simple example: suppose the labels of three parent individuals are (1,1,2), (2,3,4), and (3,5,4), respectively. As shown in Table 2, according to Rp, there are four “1”s. Thus the label of the offspring individual will be (1,2,3,4), which is no longer a triplet (I,J,K). Then, according to Dp, the response strategies corresponding to labels 1, 2, 3, 4 will have a negative redundancy (-1/3). Therefore, we will randomly select three labels from 1, 2, 3, and 4 to obtain the label of the offspring individual; that is, the offspring individual in this situation will be labeled as (1,2,3) or (1,2,4) or (1,3,4) or (2,3,4).

Table 2:

Determining the label of an offspring after crossover in Situation 2.

RDIMDILPSFPSPPS
ML 
NL 
p 2/9 2/9 2/9 2/9 1/9 
Ap (Rp2/3 (1) 2/3 (1) 2/3 (1) 2/3 (1) 1/3 (0) 
Dp -1/3 -1/3 -1/3 -1/3 1/3 
RDIMDILPSFPSPPS
ML 
NL 
p 2/9 2/9 2/9 2/9 1/9 
Ap (Rp2/3 (1) 2/3 (1) 2/3 (1) 2/3 (1) 1/3 (0) 
Dp -1/3 -1/3 -1/3 -1/3 1/3 

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