Decreasing Volume Scan

[apaul015]

Hey guys,

Can anyone help me out in building a Scan idea I had? I pretty much want to scan for potential textbook consolidation patterns based on Volume and not Price. I don't really care if the consolidation pattern is a flag, channel, or triangle.

I pretty much want to scan for at least 3 consecutive volume bars that are decreasing based on my criteria. I would use this on 5 and 15 minute Timeframes. So say for example Bar 1 has a Volume of 100,000, Bar 2 has a Volume of 90,000, Bar 3 has a Volume of 80,000, etc. I'm trying to set real time alerts as soon as Bar 3 has closed, but I'd also want to filter out patterns that have occurred earlier in the day or week as well.

I don't really care about Price, since I'm planning on using this scan in combination with other scans to filter out the stocks I'm watching during the day.

Any tips are appreciated!!

 

[halcyonguy]

Hey guys,

Can anyone help me out in building a Scan idea I had? I pretty much want to scan for potential textbook consolidation patterns based on Volume and not Price. I don't really care if the consolidation pattern is a flag, channel, or triangle.

I pretty much want to scan for at least 3 consecutive volume bars that are decreasing based on my criteria. I would use this on 5 and 15 minute Timeframes. So say for example Bar 1 has a Volume of 100,000, Bar 2 has a Volume of 90,000, Bar 3 has a Volume of 80,000, etc. I'm trying to set real time alerts as soon as Bar 3 has closed, but I'd also want to filter out patterns that have occurred earlier in the day or week as well.

I don't really care about Price, since I'm planning on using this scan in combination with other scans to filter out the stocks I'm watching during the day.

Any tips are appreciated!!

here are a couple versions that i think will work

# test if volume is less than the previous volume.
def v = volume;
plot z = ( v[2] > v[1] and v[1] > v[0] );

# find the % change in volume and compare it to a constant.
def v = volume;
# enter a %, positivecfor increasing volume, negative for decreasing.
input diffper = -1.0;
# calc % change of volume
def per2 = round((( v[1] - v[2] ) / v[2] ) *100,1);
def per1 = round((( v[0] - v[1] ) / v[1] ) *100,1);
# is the chg > diffper ?
def x2;
def x1;
if diffper > 0 then {
 x2 = ( per2 > diffper);
 x1 = ( per1 > diffper);
} else {
 x2 = ( per2 < diffper);
 x1 = ( per1 < diffper);
};
plot z = (x2 and x1);

 

[apaul015]

Thanks! I ended up figuring something else out, but never thought of using volume average decreasing. I think that works much better.