GitLab now enforces expiry dates on tokens that originally had no set expiration date. Those tokens were given an expiration date of one year later. Please review your personal access tokens, project access tokens, and group access tokens to ensure you are aware of upcoming expirations. Administrators of GitLab can find more information on how to identify and mitigate interruption in our documentation.
@@ -65,20 +65,20 @@ Optionally, adjusting `maxIter` for every run (when the decomposition method or
# Methodology
This section describes how the cases were setup. It serves for informative purposes only and is not a part of the microbenchmark.
* The three cases were first run up to t=0.5s. For each case the following procedure was performed.
* Find out averaged initial residual $r_c$ at the end of the run: [0.4s, 0.5s].
* The initial residual reaches a plateau (not decreasing anymore) at some point before t=0.4s. Find out this point in time by finding at which time $1.2 r_c$ of the initial residual is reached (e.g. for the 1M case $t_c$=0.34s). Put differently, it is assumed that the
* Calculate the averaged iteration number $iter_{run}$ from $t$ up to $t_c$.
* Calculate the averaged iteration number $iter_{start}$ from the first *15* time steps.
* Build a factor $a_{case}=iter_{start}/iter_{run}$.
* An average factor $a$ was build based on the three cases. It is used for the extrapolation from *fixedTol* setup to *fixedIter* setup giving a mean iteration number, which is representative for the whole run of the initial transient. In this way there is no need to run a case for the complete duration of 0.5s.
* Run the cases with the fixed iteration number $iter_{run}$ for 1000 time steps.
* Analyze the wall clock time per time step $dt_{wc}$.
* Compute its average $dt_{wcAv}$ and standard deviation $dt_{wcStd}$
* Compute the sample size of time steps needed for 95% confidence level with 1% error and given $dt_{wcAv}$, $dt_{wcStd}$ according to this formula:
$n_{dt} = 1.96^2*dt_{wcStd}^2/(dt_{wcAv}*0.01)^2$
* Several runs to evaluate $n_{dt}$ on both a workstation and on a HPC were performed. The highest $n_{dt}$ was found to be 13. Finally, it is set to 15 to be aligned with the *fixedTol* setup.
* Find out averaged initial residual $`r_c`$ at the end of the run: [0.4s, 0.5s].
* The initial residual reaches a plateau (not decreasing anymore) at some point before t=0.4s. Find out this point in time by finding at which time $`1.2 r_c`$ of the initial residual is reached (e.g. for the 1M case $`t_c`$=0.34s). Put differently, it is assumed that the
* Calculate the averaged iteration number $`iter_{run}`$ from $`t`$ up to $`t_c`$.
* Calculate the averaged iteration number $`iter_{start}`$ from the first *15* time steps.
* Build a factor $`a_{case}=iter_{start}/iter_{run}`$.
* An average factor $`a`$ was build based on the three cases. It is used for the extrapolation from *fixedTol* setup to *fixedIter* setup giving a mean iteration number, which is representative for the whole run of the initial transient. In this way there is no need to run a case for the complete duration of 0.5s.
* Run the cases with the fixed iteration number $`iter_{run}`$ for 1000 time steps.
* Analyze the wall clock time per time step $`dt_{wc}`$.
* Compute its average $`dt_{wcAv}`$ and standard deviation $`dt_{wcStd}`$
* Compute the sample size of time steps needed for 95% confidence level with 1% error and given $`dt_{wcAv}`$, $`dt_{wcStd}`$ according to this formula:
* Several runs to evaluate $`n_{dt}`$ on both a workstation and on a HPC were performed. The highest $`n_{dt}`$ was found to be 13. Finally, it is set to 15 to be aligned with the *fixedTol* setup.
# Remarks
# General remarks
* Fixing the iteration number per time step n_iter is not based on a fixed tolerance run up to the stationary state (t=1s), which may be argued to be the actual representative case. During the run with fixed tolerance, n_iter decreases and the average value n_iterAv of the complete run would be relatively low. If the same case is run with a fixed iteration number of n_iterAv, which is too low to achieve the needed convergence, the run might end in a failure.
* CFL number of 1 is too high for the cases to depict the start-up process of the flow correctly, which may be seen when mean Co of two cases with different mesh sizes is analyzed. Co=1 is chosen because this is closer to the industrial setups.
