Is my Clayburg adaptive stop Interpretation Correct?

[Ernie Bonugli <ebonugli@xxxxxxxx>]

Hello omega-list,

I am not sure whether I've interpreted ClayBurgs Adaptive stop.  Here
is what he wrote in the Active Trader Feb 2002 issue, page 76:

"Our next stop routine makes use of the average range of recent bars to
set trailing stops.  This calculation is performed by subtracting a
multiple of the average range of the last several bars from the low of
the bar on which a long position was established.
   For reference, we will use the average range of the last 21 bars
multiplied by 3 as a stop level.  Therefore, the value subtracted from
the initial low will be 3 times the average range of the last 21 bars.
The resulting value is the initial stop to be used for the new long
position.
  On each successive bar the calculations are repeated, using the new
value for the 21 -bar average range.  If the market has not made a new
higher low since the last bar, the new value is subtracted from the old
highest  low.  If, on the other hand, a new higher low was established
on the previous bar, three times the 21-bar average range is
subtracted from the new higher low. "

So, here is my EL code.  I just don't know whether I've interpreted the
above correctly.

{Volatility stop by Clayburg}
var: xLOW(0),TR(0);
TR = avgtruerange(21);
if MP <> 1 or xLOW  < low then  xLOW = low;
exitlong("lxVola") at xLOW-TR*3 stop;

var: xhigh(0);
if mp <> -1 or xHIGH > high then xHIGH = high;
exitshort("sxVola") at xHIGH+TR*3 stop;

{end of Volatility Stop}


Here are some results on one of my systems:

        variable parabolic exit      Clayburg Volatility Stop

%profitable =      56.66                58.06
ProfitFactor =     3.31                 2.9
Sharpe Ratio =     4.39                 3.79
Trades       =     532                  496



Best regards,
 Ernie                          mailto:ebonugli@xxxxxxxx