Re: [EquisMetaStock Group] Re: converting metastock formula to c#

["Roy Larsen" <rlarsen@xxxxxxxxxx>]

Hi Chellester


I don't know that my answer is better than Prestons but this is what I was writing while he was writing his.

Cum(1) accumulates 1 for each bar of a chart, starting from the first loaded bar. On the first bar the value of Cum(1) will be 1, on the second bar it will be 2 and on the 200th bar it will be 200. 

So, summing Cum(1) for the first five bars will give the total of 1+2+3+4+5 =15 on the fifth bar. Since there is insufficient data to calculate a 5-bar Sum() before the 5th bar the first 4 bars will return N/A rather than a legitimate value.

On the 6th bar the result of Sum(Cum(1),5) will be 2+3+4+5+6, which of course is 20. In other words the value of Sum(Cum(1),5) will increase by 5 on each subsequent bar.

Cum() is a very simple MS function but its simplicity should not be mistaken uselessness. It has many uses and here are just a few.

{Bar Count}
Cum(1);

{Reverse Bar Count}
LastValue(Cum(1))-Cum(1);

{Last Bar}
LastValue(Cum(1))=Cum(1);

{First Bar}
Cum(1)=1;

{Difference}
{The "Event" variable an "N" constant must be defined before attempting to use in the following expression in a real formula}
Cum(X) - ValueWhen( Nth, Event, Cum(X));

(First Valid Bar} 
{of A, B, C, etc..., variables must be defined before attempting to use in the following expression in a real formula}
Cum(IsDefined(A+B+C {etc...} ))=1;


Regards

Roy


  ----- Original Message ----- 
  From: chellester2002 
  To: equismetastock@xxxxxxxxxxxxxxx 
  Sent: Monday, March 03, 2008 12:50 AM
  Subject: [EquisMetaStock Group] Re: converting metastock formula to c#



  > And just while I'm trying to work things out - please forgive me if 
  > this is dumb but can someone please tell me if 
  > 
  > Sum(Cum(1),5) = 15? I'm just getting all mixed up and confused.
  > Thanks
  >

  Ok, I know it's not 15, but I can't figure out what it is 
  :(



   

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