cbc team mailing list archive

[Anders Logg <logg@xxxxxxxxx>]

On Tue, Apr 13, 2010 at 03:10:50PM +0200, Harish Narayanan wrote:
> On 4/13/10 3:08 PM, Anders Logg wrote:
> > On Tue, Apr 13, 2010 at 03:02:08PM +0200, Harish Narayanan wrote:
> >> On 4/13/10 2:47 PM, Anders Logg wrote:
> >>> On Tue, Apr 13, 2010 at 02:42:34PM +0200, Harish Narayanan wrote:
> >>>> On 4/13/10 1:32 PM, Anders Logg wrote:
> >>>>> What if T = 10 and dt = 0.9999999999. Then we get
> >>>>>
> >>>>>   ceil(T / dt) = 11.
> >>>>
> >>>> I am not sure I understand what is wrong with this.
> >>>>
> >>>> If dt = 1.0, then we get 10 time steps. If dt < 1.0, we get more time steps.
> >>>>
> >>>> Harish
> >>>
> >>> Isn't there a chance that one might specify the time step to be k and
> >>> the end time to be n*k and get T / k = n + eps?
> >>
> >> This, in my mind, is a common case. But the same problem would have
> >> existed in the earlier code as well. All I was doing was to replace the
> >> earlier code which always added one time step. Now:
> >>
> >> If T / k is an integer, dt = k.
> >> If T / k is not, then dt = T /(int(T / k) + 1)
> >>
> >> The earlier code was always using the second definition.
> >
> > I'm not saying the old code was correct, but there should be some
> > clever solution that does the right thing. How about this:
> >
> > n = ceil(T / dt - DOLFIN_EPS)
>
> Will think about it. I am guessing it will need an intelligent +/-
> DOLFIN_EPS.
>
> Harish

It should probably not be DOLFIN_EPS, but something much
larger. Something like this should be ok

  n = ceil(0.99 * T / dt)

--
Anders

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