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To: Bob Jagow who wrote (5244)7/7/1998 4:52:00 PM
From: TechTrader42  Read Replies (1) | Respond to of 11149
 
Slope enthusiasts (philanslopists?): Bob Jagow has solved the dizzying d by z problem in the slope formula when percentage gain is added as a condition. Here's his new formulation of the formula (for moolah?):

output = "linregslope.lst";
input="portfoli.lst";
//issuetype common;
integer first, i, j, S, Sx;
float b,b1, Sxx, Sxy, Sy, pctgain;
first:= -10;
Daystoload = 50 -first;
DaysRequired = 22 - first;
S := 21; // set here
b1 := 0; // only initialized first time
for j = 0 to first step -1 do //why 0 to -10?
// reset all variables at start of loop
Sx := 0; Sxx := 0; Sxy := 0; Sy := 0;
for i = j+(1 - S) to j do
Sx := Sx + i;
Sy := Sy + close(i);
Sxx := Sxx + i*i;
Sxy := Sxy + i*close(i);
next i;
b := .000001+(S*Sxy - Sx*Sy)/(S*Sxx - Sx*Sx);
if b = 0 then print symbol,"b was zero with j: ",j; endif;
pctgain := abs((b1/b) -1); //gives div by 0 -- strange?
if j < 0 // eliminate first pass thru loop -- clumsy
and b < b1 then
//pctgain := abs((b1/b)-1); // not % and constrained to be > 0 ;)
if pctgain > 1 then
println symbol,",", date(j),",",close(0):6:3,","," B: ",
b:4:3,","," B1: ", b1:4:3,","," Pctgain: ", pctgain:4:2;
endif; endif;
b1:=b; // must alway set to last day's val
next j;

Note this line:
b := .000001+(S*Sxy - Sx*Sy)/(S*Sxx - Sx*Sx);

Kills z bugs every time.

If you put "i" in date() instead of "j," you get the date of the move up (B1). With j in the date, you get the previous date (the date of B). B1 or B: That is now the question.

You can change this line:

if pctgain > 1 then

to get the percentage change you want. Of course, you're going to get greater percentage changes if you use a shorter period for slope. Instead of using 21 days for the period, for example, you could use 3 days. Change this line to do that:

S := 21;

What more could a slopephiliac desire?

Brooke


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