while defining the new fitness function for pso i am getting errors

while defining a new fitness function for pso

function F = tracklsqnewfit(pid) % Track the output of optsim to a signal of 1

         % Variables a1 and a2 are shared with RUNTRACKLSQ
         Kp = pid(1);
         Ki = pid(2);
         Kd = pid(3);

         sprintf('The value of interation Kp= %3.0f,Ki= %3.0f, Kd= %3.0f', pid(1),pid(2),pid(3)) 
         % Compute function value
         simopt = simset('solver','ode5','SrcWorkspace','Current','DstWorkspace','Current');  % Initialize sim options
         [tout,xout,yout] = sim('meenakshipid7',[0 10],simopt);
         S=stepinfo(yout,tout);
         tr=S.RiseTime;
         ts=S.SettlingTime;
         tm=S.Overshoot;
         sprintf('The value of interation tr=%3.0f,ts=%3.0f', tr,ts)
         e=yout-1 ;  % compute the error 
         sprintf('The value of interation e= %3.0f,tm=%3.0f', e,tm)

         Y=(1-exp(-.5))*(tm+e)+exp(-.5)*(ts-tr);
         F=1/Y;

      sprintf('The value of interation Y= %3.0f,F=%3.0f', Y,F)

end

this error exists ..... ??? In an assignment A(I) = B, the number of elements in B and I must be the same.

Error in ==> PSO at 47 current_fitness(i) = tracklsqnewfit(current_position(:,i));

i cannot understand why??

Answer 1

[Walter Roberson]

My speculation would be that the yout output from sim() is a vector. If so then e = yout - 1 would be a vector, and then since e appears in the definition of Y, Y would end up a vector. That would lead to F being a vector. And a vector of values cannot be stored into a single value.

Your function needs to return a scalar.

Answer 2

[meenakshi]

should i run a loop then to send individual valuee to function F......if theres ny ther solution please let me know sir