canny georgina cannysiska at gmail.com
Thu Jul 12 16:33:48 CEST 2018

```well, thank you for the explanation before about how the ETA is computed by
this formula:

start_time + work_to_do * (now - start_time) / work_done.

However, I've read it (the formula) again but can't seem to catch the idea
of what the ETA works for.
But, my real intention is not actually bug reporting.
What I wanted to find out is about how to calculate the speed of the
sieving computation process.
Then I happen to see the ETA information printed there when I use CADO-NFS.
But, I am firstly confused by the XX:XX:XX digit for ETA format there, not
sure whether it's showing a duration, or time.
Would you kindly explain to me how can I somewhat "measure"/"calculate" the
speed of the sieving (e.g: like how many percent per minute or per secs,
etc) using the information printed by ETA there?

Thank you

2018-07-10 16:18 GMT+07:00 Emmanuel ThomÃ© <Emmanuel.Thome at inria.fr>:

> On Tue, Jul 10, 2018 at 04:01:22PM +0700, canny georgina wrote:
> > when i tried to somewhat compute what time it took from (for example 1.3%
> > to 1.4%), im confused since what the ETA shows is weird (for me).
> > when I collect them all just like this:
> > > ok (*1.3% => ETA Tue Jul 17 01:40:07 2018*)
> > > ok *(1.3% => ETA Tue Jul 17 07:43:21 2018)*
> > > ok (*1.4% => ETA Tue Jul 17 01:58:11 2018*)
> > > ok (*1.4% => ETA Tue Jul 17 04:24:17 2018)*
> >  i'm confused about how to compute it when it's *01:40:07 ** going
> forward
> > to this hour **( 07:43:21) and backward at this  hour ( 01:58:11) and
> going
> > forward again at this hour ( 04:24:17)*
>
> No bug here.
>
> The ETA is computed as
>
>     start_time + work_to_do * (now - start_time) / work_done.
>
> At the beginning of the computation, you have enough irregularity in the
> speed per workunit to account for jitter that adds up to hours or days
> when multiplied by "work_to_do/work_done".
>
> E.
>
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