why i get this error Size vector must be a row vector with real elements.

Dear,

i have test this code to solve mixed integer problem but it give error her in this step

CoefMatZX = [repmat(-CoefZ,p,1),-U(1:p,:)*A; zeros(q,k+1),-V(1:q,:)*A];

it says error Size vector must be a row vector with real elements, that's because k is complex number and when i take

zeros(q,real(k)+1)

the loop doesn't stop, i don't know why.

%%Bender’s decompostion
% Application benders Decomposition algorithm to solve mixed- integer programming problem
% [OptX,OptY,OptValue]=BendersDecomposition(C,D,A,B,b)
% : To solve programming problems of the form :
%         min  C*x+D*y
%         s.t. A*x+B*y>=b; x,0-1??variable,y>=0
% Iterative stop determination threshold value:LB/UB More than epsilon Can be flexibly adjusted according     %to the complexity of the problem
function [OptX,OptY,OptValue]=BendersDecomposition(C,D,A,B,b)
C=[-1,-4];D=[-2,-3];A=[1 -3;-1 -3];B=[1 -2;-1 -1];b=[-2;-3];
epsilon = 0.99;   %Stop iteration. 
[nRowA,nColA] = size(A);
[nRowB,nColB] = size(B);
%step 1: Initialization
x0 = zeros(nColA,1);%The initial value
LB = -1e10;
UB = inf;
p = 0;
q = 0;
U = zeros(100,nRowA);
V = zeros(100,nRowA);
options = optimset('LargeScale', 'off', 'Simplex', 'on');
OptionsBint = optimset('MaxRLPIter',100000,'NodeSearchStrategy','bn',...
    'MaxTime',50000);
while LB/UB<epsilon
    %step 2: Subproblems
    [UorV,fval,exitflag] = linprog((A*x0-b),B',D',[],[],zeros(nRowB,1),...
        inf(nRowB,1),[],options)
    if exitflag == 1   % Get extreme point
        p = p+1;
        U(p,:) = UorV';  % Extreme point
        UB = C*x0 + U(p,:)*(b-A*x0);
    elseif exitflag == -3   % Unbounded
        q = q+1
        V(q,:) = UorV'/1e16 % Pole direction
          V(q,V(q,:)<0.0001) = 0;
      else
          msgbox('No feasible solution!') 
          return;
      end
      %step 3:  Solving the main problem
      if isinf(UB)
          k = 17
      else
          k = ceil(log2(UB+1))-1  %k is complex number??According to 2^(k+1)-1 > UB Determines the minimum 
                                                                                     %value of k                                  
      end                                   
      CoefZ =2.^(0:k)                %z0,, The coefficient of vector zk
       f = [CoefZ';zeros(nColA,1)]   % Main objective function coefficient vector of  
      CoefMatZX = [repmat(-CoefZ,p,1),-U(1:p,:)*A; zeros(q,k+1),-V(1:q,:)*A];
      bZX = [-(repmat(C*x0,p,1) + U(1:p,:)*b);-V(1:q,:)*b];
      [OptZX,minZ,ExitflagBint] = bintprog(f,CoefMatZX,bZX,[],[],[],...
          OptionsBint);
      if ExitflagBint ==1
          LB = minZ;
          x0 = OptZX(k+2:end);
      else
          break;
      end
      LB,UB %#ok<NOPRT>
  end
  OptX = x0;
  options = optimset('LargeScale', 'off', 'Simplex', 'on');
  [OptY,OptValue] = linprog(D',-B,A*OptX-b,[],[],zeros(nColB,1),inf(nColB,1),...
      [],options);
  OptValue = C*OptX+D*OptY;

    if true
      % code
    end
    if true
      % code
    end

I need to run this code because there is no other one in matlab and it is useful to my work. thanks for any help.

regards,

Imola

[Alka Nair]

Hi, real(k) does not guarantee that it is an integer. Try explicitly making the dimension an integer value. Thank, Alka

[Walter Roberson]

In order for k to become complex, UB would have to be less than -1. Between -1 (inclusive) and 0 (inclusive) for UB, k would become negative and that would give problems. I do not understand the linear algebra well enough to see any reason why k < -1 should not be possible.