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> -----Original Message-----
> From: Trey Johnson [mailto:dickjohnson3@xxxxxxxxxxxxxx]
> Sent: Tuesday, January 28, 2003 9:28 AM
> To: omega-list@xxxxxxxxxx
> Subject: Function Help
>
>
> I'm trying to create a function based upon a paper I read by Cynthia Kase.
> This is my first attempt at using a loop. The idea is to find the highest
> value5, which is the KSDIu, over a range of cycle lengths. I found a
> function in the archives and modified it to the following. My question is,
> does this function return the highest value for value5, which
> becomes KSDIu?
> The original function is below it.
>
> vars: mlval(0);
> mlval = 0;
>
> for value99 = 5 to 70 begin
> value5 = (high - low[value99])/(StdDev(log(C[1]/C),9)*SquareRoot(252));
> if value5> mlval then mlval = value5;
> end;
>
> KSDIu = mlval;
Looks okay, but note that (StdDev(log(C[1]/C),9)*SquareRoot(252)) is
invariant throughout the loop, and it takes 9 iterations each time to
calculate StdDev(). You can also use MAXLIST to shorten the code a bit.
Rewriting:
vars: mlval(0);
for value99 = 5 t 70 begin
mlval = maxlist(mlval, high - low[value99]);
end;
mlval = mlval / (StdDev(log(C[1]/C),9)*SquareRoot(252));
The 5, 77, and 9 values above look quite arbitrary. You might want to make
some/all of them input parameters to the function. it seemed strange to skip
the first 4 bars as well (ie, starting value99 at 5).
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