Roy: I think you were right with "i" in the date() after all. I didn't understand what you were doing with "b" and "b1." You have "b" as the previous day and "b1" as the day of the date, or date(i) -- in other words, the day of the hit, when the condition is true. So for DELL, if I put "b" and "b1" in the output, the value of "b1" on 7/01 is .65. B is the value on the previous day.
DELL ,07/01/1998, 92.88, B: 0.61, B1: 0.65
Of course, that's why you have if b<b1 when you're looking for slope moving up. B is the first value, and B1 is the "next j."
So the scan should be the way you had it, with q<0 and date(i). I've just changed the output to show B1:
println symbol,",", date(i),",",close(0):6:2,","," B: ", b:4:2,","," B1: ", b1:4:2;
Here's the whole thing again:
//Slope formula by Bob Jagow
//Back-testing version by Roy Yorgensen
output = "linregslope.lst";
input="portfoli.lst";
issuetype common;
//Daystoload = 50; //needed for R2 bug!!!
integer i,j, S, Sx, Sxx;
float b,b1, Sxy, Sy;
b1:=0; // only initialized first time
for j = 0 to -10 step -1 do
// reset all variables at stat of next loop
S := 21;
Sx := 0;
Sxx := 0;
Sxy := 0;
Sy := 0;
for i = j+(1 - S) to j do
Sx := Sx + i;
Sy := Sy + close(i); //kills println wo d2ld
Sxx := Sxx + i*i;
Sxy := Sxy + i*close(i); //kills println wo d2ld
//Sxy := Sxy + i*close(0);
next i;
b := (S*Sxy - Sx*Sy)/(S*Sxx - Sx*Sx);
if
j<0 and // eliminate first pass thru loop
b< b1 then
//println symbol,",", date(i),",",close(0):8:2,",",b:8:4;
println symbol,",", date(i),",",close(0):6:2,","," B: ", b:4:2,","," B1: ", b1:4:2;
b1:=b; // save current for next test
else
b1:=b; // must alway set ot last day's val
endif;
next j;
Brooke |
