i have this coding for quadratic assignment problem (assign facilities to location using flow and distant matrix)
i want to display all the answer that i get from 10 iteration
%%read data
clc clear %flow
f=[ 0 90 10 23 43 0 0 0 0 0 0 0 90 0 0 0 0 88 0 0 0 0 0 0 10 0 0 0 0 0 26 16 0 0 0 0 23 0 0 0 0 0 0 0 0 0 0 0 43 0 0 0 0 0 0 0 0 0 0 0 0 88 0 0 0 0 0 0 1 0 0 0 0 0 26 0 0 0 0 0 0 0 0 0 0 0 16 0 0 0 0 0 0 96 0 0 0 0 0 0 0 1 0 0 0 0 29 0 0 0 0 0 0 0 0 96 0 0 0 37 0 0 0 0 0 0 0 0 29 0 0 0 0 0 0 0 0 0 0 0 0 37 0 0]; %distance
d=[ 0 36 54 26 59 72 9 34 79 17 46 95 36 0 73 35 90 58 30 78 35 44 79 36 54 73 0 21 10 97 58 66 69 61 54 63 26 35 21 0 93 12 46 40 37 48 68 85 59 90 10 93 0 64 5 29 76 16 5 76 72 58 97 12 64 0 96 55 38 54 0 34 9 30 58 46 5 96 0 83 35 11 56 37 34 78 66 40 29 55 83 0 44 12 15 80 79 35 69 37 76 38 35 44 0 64 39 33 17 44 61 48 16 54 11 12 64 0 70 86 46 79 54 68 5 0 56 15 39 70 0 18 95 36 63 85 76 34 37 80 33 86 18 0]; [r,c]=size(f); max_i= r; max_j =r; max_k= r; max_q= r; MaxIt=10; for no=1:MaxIt ID=randperm(r); %PERMUTATION r NOMBOR ..r = 3, 1 2 3, 1 3 2...
x=zeros(r,r); n=length(ID); x1=ID; for i=1:n for j=1:r x(j,x1(j))=1; B(i).mat=x; % used to keep the memory when 0,1 permutation in (i),for BINARY RELATION
end x=zeros(r,r); end z=0;xa=B(no).mat; for i=1:max_i for j=1:max_j for k=1:max_k for q=1:max_q z= z+ f(i,k).*d(j,q).*xa(i,j).*xa(k,q); %find the min value of the assignment
end end end end no; F(no,:)=[no z]; %STOR ALL VALUE Z 1...UNTIL MaxIt IN MATRIX F
end %end
zmin=min(F(:,2)); ii=find(F(:,2)==zmin) ; x_initial = B(ii).mat; % xbin= B(ii).mat % xbin = initial value for facility location for minimun z
z_minimum = zmin Iter_zvalue = F % matrix F has no of iteration..col 1 dan z
Iter_MaxIt = ID % disp(ID)
i want to make it display the iteration as example below, but i didnt know how to do it
exp:
Iter_MaxIt 1=
11 4 7 5 12 3 9 1 2 8 6 10
F 1 =
31868
Iter_MaxIt 2=
11 4 7 5 12 3 9 1 2 8 6 10
F 2 =
31868
Best Answer