How to implement Stopping rule and other formulas

Hi All,

I'm struggling with the following, I have my code and I need to take the number that is closer than zero for every column of the matrix, I tried implementing a if argument buy it didnt work, additionally I should a a stopping rule which I'm don't really seem to understand so far how it is supposed to be. My code is the following:

Xp = Dp(:,1); % Strike price put
Vp = Dp(:,2); % Daily settlement price put
F = 9866; % Futures Dec14 price
S = 9833.07; % underlying closing price
T = 0.5; % Time to maturity to Dec14 
r = log(F./S)./T; % interest rate calculated using the futures price
q = 0; % No divided yield
t = 0; % Initial day
flag=0; % To calculate for a put
sg = [0.001:0.001:1]'; % Volatility range

ImpVol =zeros(length(sg),length(Vp));
for i =1:length(sg)
   for  j =1:length(Vp)
   ImpVol(i,j) = BSM(S,Xp(j),r,q,sg(i),t,T,flag)- Vp(j);
   end 
end

So I need to take the closest number to 0 for every column in ImpVol and add a stopping rule, I suppose in case the interval is too long and the jump from step to step is too small.

I would really appreciate your help!!!

Regards,

Laura R

[Geoff Hayes]

Laura - for every column in ImpVol, so after the above code has been executed, do you want to find that element that is the closest to zero, using something like

 [smallVal,indexVal] = min(abs(ImpVol));

which would return the smallest values, smallVal, from every column that is closest to zero (since we use abs). That smallest value would be found in the rows given by indexVal.

As for the stopping rule, could you provide more details on what you mean by this?

[Laura Ramirez]

Thank for the reply. But this means as well that it could take one of the negative values as the minimum and for the purpose of the exercise is that not possible. The stopping rule means that I should put lets say maximum of iterations that the loop does I guess. Do you happen to have any idea of how to do this ?

[Laura Ramirez]

I tried doing the following but I'm doing something wrong because the program does not take the whole matrix

if ImpVol > 0
    [smallVal,indexVal] = min(abs(ImpVol));
else 
    ImpVol = NaN;
end

[Pierre Benoit]

If you don't want to return the negative value use this :

[smallVal,indexVal] = min(ImpVol(ImpVol>0));

But I don't understand your stopping rule either. Could you give a concrete example in which such a rule should be used ?

[Laura Ramirez]

I tried using your command but it returns the minimum value of the entire ImpVol matrix not per Column. The stopping rule is because i need to establish a maximum of iterations to the program, and if this maximum is reached then to show and error or something. Thanks!!!

[Pierre Benoit]

Oops, my bad.

You could use another variable to store all positive values of ImpVol and NaN otherwise

dummy_ImpVol = Imp_Vol;
dummy_ImpVol(Imp_Vol<=0) = NaN;
[smallVal,indexVal] = min(dummy_ImpVol);

For your stopping rule, you could use the function error with a if.

[Geoff Hayes]

@Laura - I typically use a stopping rule whenever I have a while loop and have some condition that may always evaluate true so I just add a second condition to ensure that the maximum number of iterations has not yet been reached

 maxNumIters   = 100;
 atIter        = 1;
 someCondition = true;

 while someCondition && atIter<=maxNumIters

     % increment to the next iteration
     atIter = atIter + 1;

     % do stuff

     % re-evaluate someCondition

 end

 if atIter>maxNumIters
     % maximum number of iterations has been surpassed so write error
     error('maximum number of iterations in my function have been met');
 end

I'm not sure how this could be migrated to your code where you have two for loops, unless there is some code that you have not shown?