Help with this code.

          for i = 1:5
               b1(i,:)= round(((1860-0)*(rand(1)))+0)
          end
          b1
          for i = 1:5
               b2(i,:)= round(((1908-0)*(rand(1)))+0);
          end
          b2
          for i = 1:5
               b3(i)= round(((1316-0)*(rand(1)))+0);
          end
          b3
          for i = 1:5
               b4(i)= round(((2940-0)*(rand(1)))+0);
          end
          b4
          for i=1:5
              b5(i,:)= round(((1860-0)*(rand(1)))+0);
          end
          b5
          for i=1:5
              b6(i,:)= round(((1860-0)*(rand(1)))+0);
          end
          b6

In this code I want a condition that if b5+b6=b1 then only it should execute the result otherwise search for solutions until this criterion matches. Please help me with this.

[Guillaume]

Each loop could be replaced with:

bn = randi([minval maxval], 5, 1);

which is a lot more readable.

Since b5+b6 == b1 has only one solution, why not generate b1 as the sum of the other two? You could be there forever if you generate all three randomly.

[Abhinav]

This is a supply chain problem. b1 are the set of products which are produced by manufacturer. b5 and b6 are the quantities of product sent to two different customers. Hence I have to generate b1 first.

[Guillaume]

Then generate b6 as b1-b5?

Given two of b1, b5 or b6 there is only one of the other that work, so why generate random invalid sets for that other, when you could just calculate the unique set that would work.

Answer 1

[Guillaume]

As per my comment, it seems very silly to generate random sets that have no way of working when you could just generate sets that always work, so this will get you there faster:

invalidset = true;
while invalidset
   b1 = randi([0 1860], 5, 1);
   %... same for b2, b3, ...
   b5 = randi([0 1860], 5, 1);
   b6 = b1 - b5;
   invalidset = any(b6 <0 | b6 > 1860);
end

But if you really want to waste time generating a random bunch of 3 sets that have no chance of working, it's:

invalidset = true;
while invalidset
   b1 = randi([0 1860], 5, 1);
   %... same for b2, b3, ...
   b5 = randi([0 1860], 5, 1);
   b6 = randi([0 1860], 5, 1);
   invalidset = ~isequal(b1, b5 + b6);
end

First option took 0.007044 seconds to generate a set on my machine. Second option is still running after a minute.