Re: [foxboro] HLBL Array Question

Peter,

Don't move the data. Change a variable that tells you where to insert
the next value instead.


Consider this array:

[1]
[2]
[3]
NF=3D1

With a next free spot (NF) index set to 1 initially.

First sample goes into NF, increment NF. So, we have:

[1] is 'a'
[2]
[3]
NF=3D2

Get next sample


[1] is 'a'
[2] is 'b'
[3]
NF=3D3

Get next sample

[1] is 'a'
[2] is 'b'
[3] is 'c'
NF=3D1

Now that we've reached the end of the array, NF wraps around to 1.

Get next sample

[1] is 'd'
[2] is 'b'
[3] is 'c'
NF=3D2

And so on.

Make sense? You increment NF until it hits the limit and wrap it to the
beginning.

In some situations, you also want a Next To Use pointer, but that's a
different lesson.


Regards,
 =

Alex Johnson
Invensys Process Systems
10900 Equity Drive
Houston, TX 77041
713 329 8472 (voice)
713 329 1600 (switchboard)
713 329 1944 (fax)
alex.johnson@xxxxxxxxxxxxxxxx
http://ips.invensys.com =


-----Original Message-----
From: foxboro-bounce@xxxxxxxxxxxxx [mailto:foxboro-bounce@xxxxxxxxxxxxx]
On Behalf Of Bruley, Peter T
Sent: Thursday, May 15, 2008 1:46 PM
To: 'Foxboro DCS Mail List' ('Foxboro DCS Mail List')
Subject: [foxboro] HLBL Array Question

Gentlemen (and Ladies)

        I have a large array size is 1440 (holds 24 hours worth of 1 min
sa=3D
mples)

        Every minute I want to update the 1 of the sample value and
throw o=3D
ut the oldest value.

Problem 1)

        Currently I am shuffling each value 1 at a time using this loop:
----------------------------
      x :=3D3D 1;
      SN01 :=3D3D "Preparing Array for new value ....";
      WHILE (x < 1440) DO
        ARHR[x] :=3D3D ARHR[x+1]; {* throw out arry position 1 and shuffle
*}
        x :=3D3D x + 1;
      ENDWHILE;
      ARHR[1440] :=3D3D RI01;     {* Load newest value into array position
14=3D
40 *}
------------------------------


        This works however just to shuffle the values takes almost 30
secon=3D
ds.

        Is there a quicker way to do this ?
      Is there a way to copy array position [2] through [1440] to
another a=3D
rray using a single command ?

Problem 2)

        During the next step of my sequence I have to find the average
for =3D
the last 24 hours so I need to find the sum of the array then divide by
144=3D
0. I use the following loop to find the sum:

----------------------------
   y :=3D3D 1;
   SUM :=3D3D 0;
   WHILE ( y < 1441 ) DO
     SUM :=3D3D SUM + ARHR[ y ];
     y :=3D3D y + 1;
   ENDWHILE;
   RO01 :=3D3D SUM/1440; {* 24 Hour rolling average *}
----------------------------

        Again it takes about 30 seconds just to find the sum of the
array.

        Is there a quicker way to find the sum of an array?


Thank You,

Peter Bruley



 =

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