S7-1200 SCL Bubble Sort - I'm Stuck!

adfox

Member
adfox
Join Date
Jun 2006
Location
Poland
Posts
334
Hi, im creating compressor station project with S7-1200.

Project requires that 4 compressors should work in cascade - that's already done, and they should start always the one with smallest hour counter.

So i've done bubble sort in scl. Function takes time in hours from counter
and should output the number of compressor in descending hours order.

Code: 
//------------------------------------------------------------------------------
//    Bubble sort 
//------------------------------------------------------------------------------

// --- Make an array from inputs
#Array1[1]:=#Input1;
#Array1[2]:=#Input2;
#Array1[3]:=#Input3;
#Array1[4]:=#Input4;

// --- Run bubble sort on array
REPEAT #swap:=false;
  
  FOR #i:=4 TO 2 BY -1 DO
    IF #Array1[#i-1] > #Array1[#i] THEN
    
        #aux:=#Array1[#i];
        #Array1[#i] :=#Array1[#i-1];
        #Array1[#i-1] :=#aux;
        #swap:=true;
        
    END_IF;
  END_FOR;
UNTIL NOT #swap END_REPEAT;

// --- Assign positions to values
FOR #i:=1 TO 4 BY 1 DO

  IF #Array1[#i] = #Input1 THEN #Array1[#i] := 1; END_IF;
  IF #Array1[#i] = #Input2 THEN #Array1[#i]  := 2; END_IF;
  IF #Array1[#i] = #Input3 THEN #Array1[#i]  := 3; END_IF;
  IF #Array1[#i] = #Input4 THEN #Array1[#i]  := 4; END_IF;

END_FOR;




#Out1 := #Array1[1];
#Out2 := #Array1[2];
#Out3 := #Array1[3];
#Out4 := #Array1[4];


Problem is that when if 2 or more Inputs got the same value for example i've got the same numbers on outputs.

For example 4 2 2 1

I've really stuck as it my first scl function - could You post some hints how to avoid duplicated output ?

Regards.
 
Problem starts when more compressors will be added - and still they must make cascade - or replace one of faulted with higher priority... it's confusing how customer may complicate things ;)
 
Use a second array initialised with the pump numbers and perform the same swap operations inside your bubble sort.
 
adfox said:
Problem starts when more compressors will be added - and still they must make cascade - or replace one of faulted with higher priority... it's confusing how customer may complicate things ;)
Click to expand...

If there are N compressors you need N-1 compares to find the one with the minimum hours. One you find the compressor with the minimum hours there is no need to sort the rest. This becomes more important as N gets larger. Use one FOR..END_FOR loop.
 
adfox said:
Won't it give me the same result - duplicated numbers ?
Click to expand...

No. As the PumpNo array is allocated the numbers 1,2,3,4 there cannot be duplicates.

Code: 
#Out1 := #PumpNo[1];
#Out2 := #PumpNo[2];
#Out3 := #PumpNo[3];
#Out4 := #PumpNo[4];
 
quick edit to your code:

Code: 
//------------------------------------------------------------------------------
//    Bubble sort 
//------------------------------------------------------------------------------
// --- Make an array from inputs
#Array1[1]:=#Input1;
#Array1[2]:=#Input2;
#Array1[3]:=#Input3;
#Array1[4]:=#Input4;
#PumpNo[1]:=1;
#PumpNo[2]:=2;
#PumpNo[3]:=3;
#PumpNo[4]:=4;
// --- Run bubble sort on array
REPEAT #swap:=false;
  
  FOR #i:=4 TO 2 BY -1 DO
    IF #Array1[#i-1] > #Array1[#i] THEN
    
        #aux:=#Array1[#i];
        #Array1[#i] :=#Array1[#i-1];
        #Array1[#i-1] :=#aux;
        #swap:=true;
        #aux:=#PumpNo[#i];
        #PumpNo[#i]:=#PumpNo[#i-1];
        #PumpNo[#i-1]:=#aux;          
    END_IF;
  END_FOR;
UNTIL NOT #swap END_REPEAT;
// --- Assign positions to values
#Out1 := #PumpNo[1];
#Out2 := #PumpNo[2];
#Out3 := #PumpNo[3];
#Out4 := #PumpNo[4];
 
Does it matter if 2 or more compressors have the same running hours? Just start the first one and then that compressor will accumulate more running hours than the second one so next time to run the second will have less hours than the first one.
 
Calistodwt said:
Does it matter if 2 or more compressors have the same running hours? Just start the first one and then that compressor will accumulate more running hours than the second one so next time to run the second will have less hours than the first one.
Click to expand...

For customer - it does :) He wants to see sorted list - don't ask why :)

L D[AR2,P#0.0] - thank You! I've been battling this issue for a 3rd day now.
Thanks again!
 
Hello friends,

I brought this matter back to the agenda :)

I have a question.

For example; I want 3 runs with the least running, 2 other runs to stop working.

How can I do it ?

Thank you.
 
adfox said:
For customer - it does :) He wants to see sorted list - don't ask why :)
Click to expand...

I see a new PO in your future to change this to be done with a static list by ascending or descending order...

I don't have a crystal ball, but can see people servicing compressor 2 when it was compressor 3 which was the second listed from the top that had enough hours for service time. Particularly if the pumps are set up in a way for a while and then one day they change.

Why is a client defining this? Aren't you the expert? ;)
 

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