Some variables not being shown in workspace

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Rakesh Jain 2017-8-18

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https://ww2.mathworks.cn/matlabcentral/answers/353164-some-variables-not-being-shown-in-workspace

评论： Rakesh Jain 2017-8-18

I am implementing an optimisation heuristic algorithm in matlab. When I run it, some of the variables are not coming up in workspace. When I run that variable (say xyz) from Command window, it says undefined variable xyz. Though I can see that variable xyz is defined in the program. Program was running well until day before yesterday but suddenly why are such problems coming. Why is this happening? The program is attached

clc;
clear;
close all;

INITIAL PARAMETERE VALUE

N=5;TS=100;LB=[1 0.1 0.1 0.1 0.1];UB=[50 30 25 0.99 0.99];ITER=2000;
Costfunction= @pidisefrac2;

INITIAL POPULATION

Individual.Position=zeros(1,N);
Individual.Cost=zeros(1,1);
Pop=repmat(Individual,TS,1);
for i=1:TS
    for j=1:5
    Pop(i).Position(j)=unifrnd(LB(j),UB(j),1,1);
    end
    Pop(i).Cost=Costfunction(Pop(i).Position);
end

EVALOTUNARY SECTION

for i=1:ITER
    for j=1:N
        sum1=0;
        sum2=0;
        for k=1:TS
            sum1=sum1+Pop(k).Position(j)*(1/Pop(k).Cost);
            sum2=sum2+1/Pop(k).Cost;
        end
        Center(j)=sum1/sum2;
    end
      for j = 1:5
      c=(0.5*(UB-LB))/(1+ITER/10);
      end
      for k=1:TS-1
          for j= 1:5
      Pop(k).Position(j)=Center(j)+c(j)*normrnd(0,1,1,1);
          end
          for j=1:N
              if Pop(k).Position(j)<LB
                  Pop(k).Position(j)=LB;
              end
              if Pop(k).Position(j)>UB
                  Pop(k).Position(j)=UB;
              end      
          end
      Pop(k).Cost=Costfunction(Pop(k).Position);
      end
      Pop(end).Position=Center;
      Pop(end).Cost=Costfunction(Center);
      Costs=[Pop.Cost];
      [BestCost(i) index] = min(Costs)
      BestmPosn (i,:) = Pop(index).Position;
end

RESULT

Costs=[Pop.Cost];  
BestCost = min(Costs)

pidisefrac2.m

function ise = pidisefrac2(x)
sys = fotf([0.08 1.88 0],[2 1 0],[1.8],[0]);
pidi = fotf([1 0],[x(4) 1],[x(3) x(1) x(2)],[x(4)+x(5) x(4) 0]);
sysnew = sys*pidi;
final = feedback(sysnew,1);
%finalWithoutPID = tf([0.01125],[0.0005 0.01175 0.01125]);
[y t]=step(final,0:0.001:10);
ac =stepinfo(y,t,1);
 % step(final)
for i = 1:101
error(i) =1 - y(i);
end
error1=error*error';
ise=abs(sum(error1));

7 个评论
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Stephen23 2017-8-18

编辑：Stephen23 2017-8-18

在 MATLAB Online 中打开

I do not have the FOTF toolbox which you seem to be using, but when I replace the cost function with:

Costfunction = @(varargin)1;

then the script runs without error, and returns:

BestCost =
     1
>> BestmPosn
BestmPosn =
       25.587       16.222       13.446      0.50843      0.51146
       25.423        16.34       13.621      0.50816      0.51123
       25.631       16.285        13.56      0.50519      0.51467
        25.42       16.233       13.448      0.50802      0.51374
       25.414       16.294       13.555      0.50112      0.51198
       25.436       16.287       13.507      0.50512      0.51118
       etc.

Rakesh Jain 2017-8-18

The mistake was mine. I had taken time to be 0:0.001:10. Thus the processing time increased and thus it was taking longer time. I was using ctrl+C earlier to break the program. Another question : if I uncomment the sixth last line step(final), the graph of the response which shows up is for 0.1 second. What if i want to check the response till 1 second or say 5 seconds

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