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Application of an Adaptive Differential Evolution Algorithm ... - Koszalin

SLOWIK: APPLICATION OF ADAPTIVE DIFFERENTIAL EVOLUTION ALGORITHM WITH MULTIPLE TRIAL VECTORS 3165Fig. 4.Memory consumption in bytes for the DE+ **an**d LM algorithms.during the next experiment, the value NT equal to 3 wasassumed. In the second experiment, the DE-ANNT+ algorithmwas executed tenfold, **an**d the average values **of** the resultsobtained for ϕ =0.90 **an**d ϕ =0.99 are presented in Table II(ϕ =0.90) **an**d in Table III (ϕ =0.99). The comparative resultsin both tables are taken from [15] (the structures **of** trainedANN are presented in Fig. 3). The ϕ values were chosenexperimentally according to the author’s previous experience.From Tables II **an**d III, it c**an** be seen that, for the threshold**of** training correctness values ϕ =0.90 **an**d ϕ =0.99, theapplication **of** the proposed DE-ANNT+ method caused **an**increase in the correct classification **of** data as compared to theDE-ANNT method. In six out **of** the eight possible cases, betterresults were obtained using the proposed method th**an** by usingthe DE-ANNT method. Also, the results obtained using theDE-ANNT+ algorithm are better th**an** the results obtained byusing the EBP **an**d EA algorithms. The results obtained by theDE-ANNT+ algorithm is better (having a higher percentage **of**correctly classified data) or comparable in four out **of** the eightpossible cases, as compared to the results obtained by using theLM algorithm. Also, it c**an** be seen from Tables II **an**d III thatthe number **of** iterations (NI) **of** EBP increases as the maximaltime increases, but the NI **of** LM does not. This is caused by thefact that, for the EBP method, the ANN training error value wasnot lower th**an** ɛ =0.0001 after the maximal time. Therefore, inall cases, the EBP method was stopped after the same number**of** iterations, but with the LM algorithm, the computations were**of**ten stopped before the maximal time was reached.In the LM algorithm, the memory consumption c**an** be estimatedas N 2 , where N is the number **of** weights in the ANN[29]. In the DE+ algorithm, the memory consumption c**an**be estimated as PopSize· N. If we assume that the floatingpointnumber is represented by 4 bytes, then the estimatedvalues for memory consumption must be multiplied by 4. InFig. 4, the memory consumption as the function **of** the paritypproblem (for p ∈ [3; 12]) for the LM **an**d DE+ algorithms ispresented.VII. CONCLUSIONBased on the results shown in Tables II **an**d III, it c**an** beseen that training **of** ANNs by using the DE-ANNT+ algorithmincreases the efficiency **of** the data classification in the sametime period when compared with the EA, EBP, or DE-ANNTalgorithms. In comparison with the LM algorithm, the resultsobtained when using the DE-ANNT+ algorithm are comparable.However, in the case **of** the DE+ algorithm, memoryconsumption grows more slowly th**an** with the LM algorithmas regard to increasing the values **of** p in parity-p problems. It isnecessary to point out that the algorithm presented c**an** also beused easily to train multioutput ANNs, ANNs with nonst**an**dardarchitectures (for example, the tower architecture [16]), **an**dnetworks with a nondifferentiable neuron activation functionfor which applications **of** the EBP or LM algorithms are notpossible. Additionally, the introduction **of** adaptive ch**an**ges **of**F **an**d CR parameter values, together with the introduction**of** multiple trial vectors in the presented DE-ANNT+ algorithm,allows one to increase the effectiveness **of** the proposedalgorithm in relation to the previously elaborated DE-ANNTalgorithm [15]. Also, it is worth saying that the proposedDE-ANNT+ algorithm c**an** be used in m**an**y industrial electronicsapplications in which the use **of** ANN is needed. Asexamples, based on the literature review, the DE-ANNT+algorithm c**an** be used for the following: state-**of**-charge estimationin battery string systems [32], shape recognition systems[33], estimation **of** nonlinear load harmonic currents [34],sensorless control **of** single switch-based switched reluct**an**cemotor drives [35], **an**d modeling **of** embedded fuel-cell powergenerators [36].APPENDIXEXAMPLE OF THE DE-ANNT+ METHOD IN OPERATIONIt c**an** be assumed that the ANN from Fig. 2 is trained forthe classification **of** a parity-2 problem. Therefore, the ANNhas nine values **of** weights. Additionally, it is assumed thatPopSize =5, ɛ =0.0001, **an**d NT =2.In the first step, the population **of** x i individuals containingPopSize =5vectors (individual) is r**an**domly createdx 1 = {0.8; 0.7; 0.6; −0.3; 0.4; −0.5; 0.1; 0.2; −0.3}x 2 = {−0.4; 0.3; 0.8; −0.9; −0.2; −0.1; 0.1; 0.2; 0.4}x 3 = {−0.2; 0.2; 0.6; 0.7; 0.9; −0.3; 0.4; 0.5; 0.2}x 4 = {0.7; −0.8; −0.9; 0.4; 0.5; −0.6; 0.2; −0.1; 0.1}x 5 = {0.5; −0.5; −0.2; 0.4; −0.7; 0.8; −0.1; 0.3; 0.6}.In the second step, the NT =2 mutated vectors V i,m arecreated for each individual x i , **an**d next, only one vector v i(having the lowest value **of** the objective function ERR(.)) isdetermined from trial vectors V i,m for each individual x i .At the beginning, the best individual in the population isdetermined. Therefore, the value **of** the objective function

3166 IEEE TRANSACTIONS ON INDUSTRIAL ELECTRONICS, VOL. 58, NO. 8, AUGUST 2011ERR(.) is computed for each individual x i according to (3)ERR(x 1 )=2.013ERR(x 2 )=2.017ERR(x 3 )=2.105ERR(x 4 )=2.039ERR(x 5 )=2.087.It c**an** be seen that the best individual is individual x 1 ;therefore r1 =1.Thetwo(NT =2)trial vectors V i,1 **an**d V i,2for particular individuals x i obtained using (1) are as follows(in parentheses, the value **of** coefficient r1 is presented, **an**dthe r**an**domly chosen values **of** coefficients r2, r3, **an**d F forparticular individuals x i are also shown):Individual x 1 :(r1 =1;r2 =4;r3 =5;F =1.5)V 1,1 = {1.1; 0.25; −0.45; −0.3; 2.2; −2.6; 0.55; −0.4; −1.05}(r1 =1;r2 =2;r3 =3;F =0.8)V 1,2 ={0.64; 0.78; 0.76; −1.58; 0.48; −0.34; −0.14; −0.04; −0.14}ERR(V 1,1 )=2.552; ERR(V 1,2 )=1.999Individual x 2 :(r1 =1;r2 =3;r3 =5;F =0.7)V 2,1 = {0.31; 1.19; 1.16; −0.09; 1.52; −1.27; 0.45; 0.34; −0.58}(r1 =1;r2 =4;r3 =5;F =0.2)V 2,2 = {0.84; 0.64; 0.46; −0.3; 0.64; −0.78; 0.16; 0.12; −0.4}ERR(V 2,1 )=2.186; ERR(V 2,2 )=2.059Individual x 3 :(r1 =1;r2 =2;r3 =4;F =0.6)V 3,1 = {0.14; 1.36; 1.62; −1.08; −0.02; −0.2; 0.04; 0.38; −0.12}(r1 =1;r2 =5;r3 =4;F =1.2)V 3,2 = {0.56; 1.06; 1.44; −0.3; −1.04; 1.18; −0.26; 0.68; 0.3}ERR(V 3,1 )=2.013; ERR(V 3,2 )=1.999Individual x 4 :(r1 =1;r2 =2;r3 =3;F =0.3)V 4,1 = {0.74; 0.73; 0.66; −0.78; 0.07; −0.44; −0.01; 0.11; −0.24}(r1 =1;r2 =3;r3 =5;F =1.6)v 4,2 = {−0.32; 1.82; 1.88; 0.18; 2.96; −2.26; 0.9; 0.52; −0.94}ERR(V 4,1 )=2.000; ERR(V 4,2 )=2.533Individual x 5 :(r1 =1;r2 =4;r3 =2;F =0.9)V 5,1 ={1.79; −0.29; −0.93; 0.87; 1.03; −0.95; 0.19; −0.07; −0.57}(r1 =1;r2 =3;r3 =4;F =1.7)V 5,2 = {−0.73; 2.4; 3.15; 0.21; 1.08; 0.01; 0.44; 1.22 − 0.13}ERR(V 5,1 )=1.927; ERR(V 5,2 )=2.583.Next, from each group **of** two (NT =2)vectors, namely:V i,1 **an**d V i,2 connected to particular individuals x i , we choosethe best vector having a lower value **of** the objective functionERR(.). After this selection, the following vectors v i areassigned to vectors x i corresponding to them:Individual x 1 :v 1 = V 1,2 = {0.64; 0.78; 0.76; −1.58; −0.48;− 0.34; −0.14; −0.04; −0.14}Individual x 2 :v 2 = V 2,2 = {0.84; 0.64; 0.46; −0.3; 0.64;− 0.78; 0.16; 0.12; −0.4}Individual x 3 :v 3 = V 3,2 = {0.56; 1.06; 1.44; −0.3;− 1.04; 1.18; −0.26; 0.68; 0.3}Individual x 4 :v 4 = V 4,1 = {0.74; 0.73; 0.66; −0.78; 0.07;− 0.44; −0.01; 0.11; −0.24}Individual x 5 :v 5 = V 5,1 = {1.79; −0.29; −0.93; 0.87;1.03; −0.95; 0.19; −0.07; −0.57}.In the third step, the individuals x i are crossed over with theircorresponding individuals v i (see Section IV, step three). As aresult **of** the crossover operation, individuals u i are created.For individual x i , if we assume that the r**an**domly chosenvalue CR is equal to 0.4, **an**d that the set r**an**d **of** r**an**domlychosen numbers is equal to: {0.5; 0.6; 0.9; 0.1; 0.3; 0.3; 0.7;0.8; 0.6}, then the values **of** vectors u i are (in practice thevalue **of** the CR coefficient, **an**d the values in the set r**an**dare r**an**domly chosen for each individual separately; in thisexample, these values are the same for all the individuals, forexample simplification)u 1 = {0.8; 0.7; 0.6; −1.58; −0.48; −0.34; 0.1; 0.2; −0.3}ERR(u 1 )=2.027u 2 = {−0.4; 0.3; 0.8; −0.3; 0.64; −0.78; 0.1; 0.2; 0.4}ERR(u 2 )=1.997u 3 = {−0.2; 0.2; 0.6; −0.3; −1.04; 1.18; 0.4; 0.5; 0.2}ERR(u 3 )=2.067u 4 = {0.7; −0.8; −0.9; −0.78; 0.07; −0.44; 0.2; −0.1; 0.1}ERR(u 4 )=2.024u 5 = {0.5; −0.5; −0.2; 0.87; 1.03; −0.95; −0.1; 0.3; 0.6}ERR(u 5 )=2.174.In the fourth step, a selection **of** individuals for the newgeneration is performed (see Section IV, step four). The followingindividuals {x 1 ; u 2 ; u 3 ; u 4 ; x 5 } are selected for the new

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