Order of Addition changes answer? Approx 0 or actually 0

I've been trying to figure out for half a day why my Solution was not 0 by an order of 1.0e-14 or e-15. When I added several matrices together I noticed that the order to which I add them changes the value. Sol_1=0 (3X3) Matrix but Sol_2=~0(3X3). Please help, I can't figure it out.

clear
E1=80.32
E2=5.590
v12=0.338
G12=3.590

Composite_Properties=[E1 E2 v12 G12 0 0.254;
    E1 E2 v12 G12 -45 0.254;
    E1 E2 v12 G12 45 0.254;
    E1 E2 v12 G12 45 0.254;
    E1 E2 v12 G12 -45 0.254;
    E1 E2 v12 G12 0 0.254]
[r,c]=size(Composite_Properties)
    A=cell(1,r)
for i = 1:r
    A{i}=Q_BAR(Composite_Properties(i,1),Composite_Properties(i,2),Composite_Properties(i,3),Composite_Properties(i,4),Composite_Properties(i,5))*Composite_Properties(i,6)
end

A_SUM=A{1};
for i=2:(c)
    A_SUM=A_SUM+A{i};
end
    A_SUM

   z_k_bar=zeros(1,r,'double')
   for j=1:r
       z_k_bar(j)=(((-r/2)+((2*j-1)/2))*Composite_Properties(j,6))

   end

clear i j 
%Making B Matices
B=cell(1,r)
for i = 1:r
B{i}=Q_BAR(Composite_Properties(i,1),Composite_Properties(i,2),Composite_Properties(i,3),Composite_Properties(i,4),Composite_Properties(i,5))*z_k_bar(1,i)*Composite_Properties(i,6)
end

%Summation of B Matrices
 B_SUM=B{1}
for j=2:c
B_SUM=B_SUM + B{j};
end
B_SUM

Sol_1=B{1}+B{6}+B{2}+B{5}+B{3}+B{4} %0 Matrix%
Sol_2=B{1}+B{2}+B{3}+B{4}+B{5}+B{6}  %not Zero why???%

[the cyclist]

Can't run it without definition of Q_BAR.

[aldburg]

Sorry!

function [Z] = Q_BAR( E1,E2,v12,G12,theta)
  Z=inv(T(theta))*Q(E1,E2,v12,G12)*T(theta).*[1 1 0.5; 1 1 0.5;1 1 0.5];
  %Q_BAR for Laminate Theory
  %   Detailed explanation goes here
    function Y=T(theta)
                Y=[(cos(C(theta)))^2  (sin(C(theta)))^2 2*cos(C(theta))*sin(C(theta));
  (sin(C(theta)))^2 (cos(C(theta)))^2 -2*cos(C(theta))*sin(C(theta));
   -cos(C(theta))*sin(C(theta)) cos(C(theta))*sin(C(theta)) (cos(C(theta)))^2-(sin(C(theta)))^2];
      function Z=C(theta)
      Z=theta*pi/180;
          end
  end
      function [ W ] = Q( E1,E2,v12,G12 )
      W=inv(S(E1,E2,v12,G12)).*[1 1 1; 1 1 1;1 1 2];
      %Q is stiffness matrix 
      %   Needs Compliance matrix S first
          function [X]=S(E1,E2,v12,G12)
              X=[S11(E1) S12(v12,E1) 0; S21(v12,E1) S22(E2) 0; 0 0 S66(G12)];
      %S-> compliance Matrix for v12
      %   epsilon_12=S*Sigma_12
              function [a]= S11(E1)
                  a=1/E1;
              end
              function [b]=S12(v12,E1)
                  b=-v12/E1;
              end
              function[c]=S21(v12,E1)
                  c=-v12/E1;
              end
              function[d]=S22(E2)
                  d=1/E2;
              end
              function[e]=S66(G12)
                  e=1/G12;
              end
          end
      end

    end

[Stephen23]

"Order of Addition changes answer?"

Yes, quite possibly. Floating-point operations accumulate floating-point error.

[aldburg]

Any way I can improve accuracy or do I have to implement an if statement so if its less than some number (epsilon) then the elements of the matrix is 0?

[John D'Errico]

You can essentially never improve the accuracy so much that floating point computations will be perfect when using floating point arithmetic. Use of higher precision simply pushes the errors down to smaller levels. Yes, you can use my HPF toolbox to do high precision computations, and with some effort, you can do better, at a cost of time.

Instead, you really need to understand how precision impacts your computations, learn what are acceptable errors in computations, and learn how to minimize any errors by the use of robust computational schemes.

Answer 1

[Walter Roberson]

You can make some small changes for higher accuracy if you have the Symbolic Toolbox. Modified version enclosed.

[aldburg]

Amazing! Just a simple sym() made all the difference.

[Walter Roberson]

There was also the change to how z_k_bar was initialized; you were always initializing it as double, and I change that to initialize to the same as whatever the table is (so if you commented out the sym of the table then the z_k_bar would quietly go back to using double.)

[aldburg]

I didn't notice that, thank you for pointing that out. You have helped me a lot the past few weeks. Started out not knowing what a double or a string was and now I am here :)