[Scilab-users] Avoiding a loop

[Tim Wescott]

Depending on how often you switch between battery and generator, and
how icky-picky you're willing to be, there may be a way to reduce
computation.

It looks like the term P(n,2) * (P(n+1,1) - P(n,1)) is always there,
and you're either adding it to 'gen' (is it energy production?) or
subtracting it from 'batt'.

If you really want to go there, you can vectorize "if" statements by
using boolean expressions on vectors and the "find" function, which
returns indexes of true results.  Then you can use "cumsum" on your
P(n,2) * (P(n+1,1) - P(n,1)) term to find where the battery state of
charge (I assume that's what 'batt' is) hits 800.

It'll be complicated.  It'll be prone to error.  Proving that it's
correct will be a pain.  But when you get it working, it'll be
considerably faster.

On Thu, 2017-05-11 at 09:17 +0200, Frieder Nikolaisen wrote:
> Thanks for all the answers. 
> 
> I feared that there is no way around a loop. During the process batt
> (Battery) is charged and discharged. In my example, it is only
> discharged. I will code the entire problem with a loop, maybe
> somebody knows something to speed up the process with the full
> problem. (Tim: I am not a programming pro, a C-function might not be
> a solution. )
> Why do I try avoidng a loop? I do have txt-document with 50 000 to
> 100 000 lines about a (hybrid-)locomotive shunting process. I do need
> to optimize the energy managment. Because I am not mathemtic student,
> I have to solve the problem empirical (try and error). The programm
> has to run a few hundred times. With a matrix thats no problem, but
> with matrixes only, I can only calculate the diesel usage without any
> battery energy storage. 
> Thanks for the checking my code anyway. 
> 
> Am 10.05.2017 um 20:53 schrieb Amanda Osvaldo:
> > What it's the equation you need to compute ?
> > Perhaps I can help.
> > 
> > I think it's possible to compute with something in this way:
> > 
> > map = find (P(:,2) > 100 );
> > if batt > 800 then
> >     batt = batt - P(map,2) * (P(map+1,1) - P(map,1));
> > end
> > 
> > 
> > On Wed, 2017-05-10 at 17:23 +0200, Frieder Nikolaisen wrote:
> > > Hello,
> > > I did write an example code, but I do not like the time consuming
> > > way I solved the problem. With 50 000 lines in the matrix, it
> > > wouldn't be fun.
> > > How can I avoid using the for-loop?
> > > 10,     80;
> > > 11,     200
> > > 15,     0];
> > > 
> > > batt = 1000;
> > > gen = 0;
> > > 
> > > n = 1
> > > for n=1:5
> > > 
> > >     if P(n,2) > 100 then
> > >         if batt > 800 then batt = batt - P(n,2) * (P(n+1,1) -
> > > P(n,1))
> > >         else
> > >         gen = gen + P(n,2) * (P(n+1,1) - P(n,1))
> > >         end
> > >     
> > >     else
> > >     batt = batt - P(n,2) * (P(n+1,1) - P(n,1))
> > >     end
> > > disp('n ' + string(n))
> > > disp('batt ' + string(batt))
> > > disp('gen ' + string(gen))
> > > end
> > > Thanks alot!
> > > 
> > > Best regards
> > > Frieder 
> > >  
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-- 

Tim Wescott
www.wescottdesign.com
Control & Communications systems, circuit & software design.
Phone: 503.631.7815
Cell:  503.349.8432