Please help in convergence of the array

 for l = 1:6
 c11 (l)=opt11{l};

 end

 output = c11 =

Columns 1 through 5

   89.0475   98.7932   98.7932   98.7932   98.7932

Column 6

   98.7932

I want to break the loop "l" in program if c11(i+1)<=c11(i). please help me.

Answer 1

[Guillaume]

Well, you already nearly wrote the answer yourself!

for ...
  ... do something
  if condition
     break; %stop the loop
  end
end

So, in your case, probably:

for l = 1:6
   c11(l) = opt11{l}
   if l>1 && c11(l) <= c11(l-1)
        break;
   end
end

Of course, the above could be just written without the loop:

c11 = cell2mat(opt11(1:6));
c11 = c11(logical(cumprod(diff(c11) > 0)));

[Triveni]

Thanks sir, you have given me perfect answer which i want. Sir please help me once again, if i want, c11(l+1)<= or ==c11(l+4) continuously for three times, then i want to break or defined 1,2,3 or 4 then how can i break? or,

 Columns 1 through 4

    [93.6568]    [98.9636]    [93.4562]    [103.8119]

Columns 5 through 8

    [101.4730]    [103.8119]    [101.4730]    [103.8119]

Columns 9 through 12

    [101.4730]    [103.8119]    [101.4730]    [103.8119]

Columns 13 through 16

    [101.4730]    [103.8119]    [101.4730]    [103.8119]

Columns 17 through 20

    [101.4730]    [103.8119]    [101.4730]    [103.8119]

 %
 %After column 4, I want to avoid this type of repetition.

[Stephen23]

@Triveni: it sounds like you are really trying achieve some task which you have not told us about. Rather than getting us to fix some broken code, it would be better if you actually described what you are trying to achieve:

http://xyproblem.info/

http://meta.stackexchange.com/questions/66377/what-is-the-xy-problem

[Triveni]

Actually if i post whole code then you need to pay attention at-least for 5 to 10 minutes to analyze my code. My question is not so difficult, i have to stop only loop in different circumstances which i have told you.

     [93.6568]    [98.9636]    [93.4562]    [103.8119]

Columns 5 through 8

    [101.4730]    [103.8119]

In column 1, 2, 3, 4, and 5 value increase and then decrease, then value is continuously increase and decrease. If increase and decrease is steady then i have to break my loop. I have to find out maximum value from c11. When value is not maximizing and it's going steady then i need not to solve it.