By Bernard Moret, Henry D. Shapiro

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**Example text**

Assuming that it is likely that an average order of more than two is very rare, the parameter k can be approximated by a two-term weighted average as follows: 2 cx · x. 6) x=1 The reason for this assumption is explained below. , quite misleading). , k = 2) is relatively rare in practice. Determining the coeﬃcients is a very diﬃcult problem. They are also sensitive to network size and topology. 7) c2 = A · |V|B . Here, A and B are domain-dependent constants. 0. 8) Therefore, the average order may be calculated as follows: k = 1 · c1 + 2 · c2 = 1 + c2 = 1 + 10−2 · (1 − α)2 · |V|.

13) 2 2π √ From Eq. 4), z is found to be 2/( 2m (χk − 1)). Thus, a fairly general, practical population-sizing model can be written as follows: p= N =− =− χk ln(α) z −1 2 χk ln(α) 2 π +1 2 χk − 1 √ πm + 1 . , average order) becomes large, the probability of disrupting the BBs is increased; thus, the population size may be increased to reach a particular quality of solution. This is the reason why a higher probability of disrupting the BBs drives the probability of making the correct decision on a single trial p towards smaller values so that the population size N must be increased for achieving the same GA failure probability α.

The chosen node is removed from the topological information database to prevent the node from being selected twice, thereby avoiding loops in the path. This process continues until the destination node is reached. Note that an encoding is possible only if each step of a path passes through a physical link in the network. 2 Population Initialization Heuristic initialization may be beneﬁcial to the SP routing problem because the topological information for computing the SP is already collected before the algorithm starts.