Problem in GA algorithm (objective function is not drectly related to the variables)
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Hello all,
I am writing an optimization code, in which three variables are considered. The variables are used to find eigenpairs of the problem, and lately, eigenpairs are developing objective function. The objective fuction part of my code is as below.
function [y] = obj_func(x)
%Input.........................................................
%..............................................................
%number of nodes
num_nod=6;
% nodes coordinates
nod_coor=[720 0;720 360;360 0;360 360;0 0;0 360];
% connectivity table
ele_nod=[6 4;4 2;5 3;3 1;3 4;1 2;6 3;5 4;4 1;3 2;5 6];
%number of elements
num_ele=size(ele_nod,1);
% elements degree of freedom (DOF)
ele_dof=[11 12 7 8;7 8 3 4;9 10 5 6;...
5 6 1 2;5 6 7 8;1 2 3 4;11 12 5 6;...
9 10 7 8;7 8 1 2;5 6 3 4;9 10 11 12];
% A, E, L are cross sectional area, Young's modulus, length of elements,respectively.
A(1)=14;
A(2)=1;
A(3)=11;
A(4)=7;
A(5)=1;
A(6)=1;
A(7)=6;
A(8)=3;
A(9)=14;
A(10)=1;
A(11)=1;
E(1)=30000;
E(2)=30000;
E(3)=30000;
E(4)=30000;
E(5)=30000;
E(6)=30000;
E(7)=30000;
E(8)=30000;
E(9)=30000;
E(10)=30000;
E(11)=30000;
rho = 9.8759999999999994e-3;
stiffness_damage=zeros(2*num_nod);
mass=zeros(2*num_nod);
%applied loads at DOFs
force(2)=-10;
force(6)=-10;
%Boundary conditions
BC = [9;10;11;12];
%Variables from Phase I
global f_uD;
global f_DM;
global vec_D
global val_D
V_0_col = 2*num_nod - size(BC,1);
%..............................................................
%..............................................................
x1 = x(1);
x2 = x(2);
x3 = x(3);
damage_ele_size = 3;
damage_ele_1 = @(x)[1,x1;5,x2;10,x3];
%Compute stiffness matrix for damaged structure
for e=1:num_ele
L(e)=sqrt((nod_coor(ele_nod(e,2),1)-nod_coor(ele_nod(e,1),1))^2+...
(nod_coor(ele_nod(e,2),2)-nod_coor(ele_nod(e,1),2))^2);
C(e)=(nod_coor(ele_nod(e,2),1)-nod_coor(ele_nod(e,1),1))/L(e);
S(e)=(nod_coor(ele_nod(e,2),2)-nod_coor(ele_nod(e,1),2))/L(e);
k_1=(A(e)*E(e)/L(e)*[C(e)*C(e) C(e)*S(e) -C(e)*C(e) -C(e)*S(e);C(e)*S(e) ...
S(e)*S(e) -C(e)*S(e) -S(e)*S(e);-C(e)*C(e) -C(e)*S(e) C(e)*C(e) ...
C(e)*S(e); -C(e)*S(e) -S(e)*S(e) C(e)*S(e) S(e)*S(e)]);
for kk = 1:damage_ele_size
if e == damage_ele_1(kk,1)
k_1 = k_1*(1-damage_ele_1(kk,2));
end
end
% Develope mass matrix
m(e) = rho*L(e)*A(e);
m_local = [m(e)/2 0 0 0;0 m(e)/2 0 0;0 0 m(e)/2 0;0 0 0 m(e)/2];
% extract the rows of ele_dof (for each element e)
ele_dof_vec=ele_dof(e,:);
for i=1:4
for j=1:4
stiffness_damage(ele_dof_vec(1,i),ele_dof_vec(1,j))=...
stiffness_damage(ele_dof_vec(1,i),ele_dof_vec(1,j))+k_1(i,j);
mass(ele_dof_vec(1,i),ele_dof_vec(1,j))=...
mass(ele_dof_vec(1,i),ele_dof_vec(1,j))+m_local(i,j);
end
end
end
%Damaged Stiffness matrix after applying B.C.
for i = 1:size(BC,1)
stiffness_damage(BC,:)=0;
stiffness_damage(:,BC)=0;
end
for j = 1:size(BC)
stiffness_damage(BC(j),BC(j))=1;
end
%Calculate Eigenpairs
[vec_D_m,val_D_m] = eig(stiffness_damage, mass, 'vector');
f_DA = val_D_m*(1/(2*pi));
%Define change in frequencies
delta_f = (f_uD - f_DM)/norm(f_uD);
round_f = (f_uD - f_DA)/norm(f_uD);
MDLAC = (abs(delta_f'*round_f)^2)/((delta_f'*delta_f)*(round_f'*round_f));
% Define Objective function
sum = zeros(2*num_nod,1);
for ki=1:V_0_col
sum = sum + (vec_D(:,ki)-vec_D_m(:,ki)/norm(vec_D(:,ki)));
end
y = 1 - MDLAC + sum + 10e4*(MDLAC+abs(MDLAC));
end
The optimization line I use for solving the problem is as follow:
{[x,fval] = ga(@obj_func,3);}
The code is not working and gives me this error:
Error using obj_func>@(x)[1,x1;5,x2;10,x3]
Too many input arguments.
Error in obj_func (line 85) if e == damage_ele_1(kk,1)
1 Commento
TIACHACHT SAMIR
il 27 Feb 2020
please how to calculate f_uD
Risposta accettata
Più risposte (1)
You have called the function damage_ele_1 with two arguments kk and 1 in the line
if e == damage_ele_1(kk,1)
However, you have defined damage_ele_1 as a function of only one argument, x,
damage_ele_1 = @(x)[1,x1;5,x2;10,x3];
It is not at all clear what damage_ele_1 is really supposed to compute. It returns the same 1x6 vector for every input x that you provide. The if statement condition above will never evaluate true.
1 Commento
Maryam
il 18 Ott 2018
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