Using Genetic Algorithm with bounds and non linear constraints

I have a problem that I am trying to solve with a genetic algorithm. I am able to run the code with just the bounds, but when I add the non linear constraints, MATLAB gives the following errors:

"The following error occurred converting from function_handle to double: Error using double Conversion to double from function_handle is not possible.

Error in constrValidate (line 55) Iterate.cineq(:) = cineq;

Error in gacommon (line 133) [LinearConstr, Iterate,nineqcstr,neqcstr,ncstr] = constrValidate(NonconFcn, ...

Error in ga (line 327) [x,fval,exitFlag,output,population,scores,FitnessFcn,nvars,Aineq,bineq,Aeq,beq,lb,ub, ...

Error in GenericzTurbo_only_ga (line 120) [xopt,fval,eflag,output,solutions] = ga(problem);"

I have my code defined as follows

     x0 = [5000  0.5  .2  70  .2 70 .1   .177  32.19  .002  .95  .002  .95   1.38];  % defaults of variables 
     lb = [1000  0    .1  50  .1 50 .001 .138  27.24  .002  .40  .002  .40   0.6];  % lower boundaries of variables
     ub = [10000 1.0  .3  80  .3 80 .29  .209  39.81  .002  1.50 .002  1.50   2.5];  % upper boundaries of variables 

     ObjectiveFunction = @zTurbo;
     nvars = length(x0);    % Number of variables
     ConstraintFunction = @simple_constraint; % the non linear constraints

     tic
     [xopt,fval] = ga(ObjectiveFunction,nvars,[],[],[],[],lb,ub, ConstraintFunction);
     toc

The constraint function itself is defined as follows

function [ c, ceq ] = simple_constraint( x )
c  = @(x) [x(:,8) - (0.0001*x(:,9)^2  -0.0123*x(:,9) + 0.4690); ...
           x(:,7) - (0.0003*x(:,9)^2  -0.0258*x(:,9) + 0.7458)];

ceq = @(x) [];

end

I have no idea what the errors actually mean, and why the genetic algorithm works fine without the non linear inequality constraints. Could someone please help me with this?

Answer 1

[Alan Weiss]

Your simple_constraint function should not return function handles for c and ceq, but instead should return double values:

function [ c, ceq ] = simple_constraint( x )
c  = [x(:,8) - (0.0001*x(:,9)^2  -0.0123*x(:,9) + 0.4690); ...
           x(:,7) - (0.0003*x(:,9)^2  -0.0258*x(:,9) + 0.7458)];

ceq = [];

end

I am also not sure that you have written it correctly in a vectorized form. Perhaps you should not have things like x(:,9). You certainly did not include dot-exponentiation, such as x(:,9).^2. I suggest that you do not, at first, use vectorized functions, and just assume that x is a row vector, so you would have things like x(9) instead of x(:,9).

Alan Weiss

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