f   func_b: The number of functions where the DFPSO algorithm obtains significantly better results。

g   func_w: The number of functions where the DFPSO algorithm obtains significantly worse results。

Table 5

Statistical results obtained by the Wilcoxon signed ranks test for pairwise comparisons based on 25 benchmark functions。

Comparison Rþa R— b p value Comparison R— p value

DFPSO versus PSO 275 50 0。0016 DFPSO versus PSOw 280。5 44。5 0。0011

DFPSO versus ALC-PSO 201。5 123。5 P 0:2 DFPSO versus SPSO2007 226。5 98。5 0。0951

DFPSO versus GPSO 269。5 55。5 0。0029 DFPSO versus SPSO2011 247 78 0。0219

DFPSO versus CLPSO 227。5 97。5 0。0826

a    Rþ: The sum of ranks for the functions in which the first algorithm outperforms the second one。

b    R— : The  sum of ranks for the  functions in which the  second algorithm outperforms the  first    one。

5。Odor source localization

In this section, we will deal with the problem of odor source localization by using the proposed continuous-time FPSO algorithm, which consists of a decision algorithm pi ðkÞ and a finite-time cooperative control algorithm ui ðtÞ given by (17)。 In terms of the characteristics of the odor source localization problem, the decision algorithm has been developed in our pre- vious  studies  [23,24]。  In  order  to  let  the  reader  easily  understand  the  proposed  continuous-time  FPSO  algorithm,  we   will

briefly restate the decision  algorithm。

5。1。The  decision algorithm

The probable position of the odor source estimated by the ith robot based on wind information and concentration infor- mation can be calculated  by

hi ðkÞ¼ c1^xi ðkÞþ c2  a    ð  Þþ a ð  Þ

where cj   ðj ¼ 1; 2Þ are weighted parameters to be selected to satisfy c1  þ c2  ¼ 1;  ^xi ðkÞ is the position of the odor source esti-

mated by the ith robot based on wind information while robot based on concentration information。

a1 þa2 is the position of the odor source estimated by the   ith

In terms of wind information [23,24], the dynamics model of the position of the odor source is directly given by

。 xs ðkÞ¼ xs ðk — 1Þ

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