Double the # of cells>double the computing time?
Hi all, I have a simple question.
I am wondering if CFD methods are as simple as "TIME = constant x CELLS". I mean, if the size of the problem is doubled (twice the # of cells), then is the computation time is doubled also? (for both steady and unsteady simulation) Triple the cells, triple the computing time? Also, if you use parallel computers, two CPUs give me half the computing time? 100 or 10000 CPUs save me time, 1/100 or 1/10000? Thank you! JB 
Re: Double the # of cells>double the computing ti
I think generally the trend is nonlinear: if you double the size of the grid the CPU time is more than twice. This may vary from solver to solver.
Your second statement is generally true given the parallel implementation was done well. You should see a linear speedup: 10 processors will converge the solution in 1/10 the time. 
Re: Double the # of cells>double the computing ti
JB,
In my experience, the computation time is fairly close to linear with increasing number of cells. The code I used on a CRAY in a previous life was almost perfectly linear. On the CRAY (since you were sharing with 100 of your closest friends) it was beneficial to sneak runs into the "small job" que so that you didn't have to wait behind many people in the normal que. The "small job" que had a limit to the CPU time that could be used for jobs in that que. We would figure out how many iterations we could get in based on the calculation time per node per iteration. The estimates we would make of run time were almost perfect. Of course, I'm sure that some codes scale better than others. Also, if you run out of physical memory and have to have to use virtual memory, the computation time will go WAY up after you hit that memory threshold. Parallel speed up is a little trickier because there are more variables involved. The code we develop here is fairly efficient (I think the speedup is a factor of 7.8 on 8 machines). However that efficiency is a factor of two things  good programming for parallel computations, and good hardware. We test using a dedicated fast network switch for the machines doing the parallel computation. If you use the regular network in your office and it's being shared by emails, web surfing, etc, you may not get the same kind of performance. Alton 
Re: Double the # of cells>double the computing ti
For a 2D FEM code, 10 years ago, I observed something between n*log(n) and n^2 for the time vs # of cells law.
As for the // speed up, the Cobalt site used to provide a speed up curve, but it seems to be closed now. If I remember well, the speed up was quasilinear until 12 cpus, then it decrease drasticaly. Sylvain 
Re: Double the # of cells>double the computing ti
I think the answer depends on the applications. If you think all parameters are kept constant, your CPU time will more than just increase linearly with the number of cells. If you double your number of cells, there will be more smaller cells, which usually slow down your convergence towards a given solution. Also, for time accurate runs, the time step in explicit schemes is determined by the smallest cell, so that if you have twice as many cells, but with the smallest cell being half the size of what the previous smallest cell was, then your CPU time to reach the same result is multiplied by 4, not 2!

Re: Double the # of cells>double the computing ti
In my experience, CPU time per iteration tends to be directly proportional to the number of cells, unless you're using a trivially small number of cells (in which case the CPU's memory cache helps to speed it up) or very large number of cells (when you start using disk swapping, which is desperately slow). However, you may or may not need more iterations to achieve a converged solution. Mostly I think you need more, as others have also indicated. As far as parallel speedup is concerned, this is strongly dependent on the actual parallel implimentation.

Re: Double the # of cells>double the computing ti
I don't think it could be linear since core of the code consist in solving an algebraic system with an almost empty matrice of size N^2 (N=# of cells).
From numerical recipes, it will cost N^3 operations using LU decomposition or Gaussian elimination for a full matrice. Since a 3D FVM matrice have more or less 26 non zero elements by row or column, this power law could be decrease by one order, it will be surprising if it is by two. Any comment ? Sylvain 
Re: Double the # of cells>double the computing ti
Thanks for your opinions. So, the answer depends. Sometimes linear, sometimes not (in reality). As sylvain pointed out, if we need to solve a linear system, we cannot get O(N) easily (multigrid is O(N)?). Other algorithms may also be important to get O(N) computing time, such as search or sorting algorithms(if used) for example. So, if it is indeed linear, it must be a very good code. I just imagined that as the size of the problem gets larger, even slight deviation from linear trend could make a tremendous difference in the computing time, and actually the problem size is getting quite large these days.

Re: Double the # of cells>double the computing ti
Well, there is not much point in speculating about this too much, as it is reasonably easy to measure. Take a look at http://home.mweb.co.za/cc/ccrosby/cf...enchmarks.html These results (for what they are worth) indicate a near linear relationship when you start using realistically sized grids. My take on this is that CFD linear equation solvers tend to be hybrids (like sweeping with a TDMA type of algorithm etc.) of iterative and direct methods, but most time is probably spent on the iterative side of things, hence the near linear relationship of time per cell. However, I suspect that more iterations are needed to converge bigger grids. The picture is far from simple though, since you need those iterations in any event to handle the nonlinear nature of the problem ... Due to the asymptotic nature of convergence, it gets quite difficult to define a clearcut "point of convergence" if you're trying to benchmark grids of varying size.

Re: Double the # of cells>double the computing ti
Nice job. I agree with your comments for the biggest grid.

Re: Double the # of cells>double the computing ti
Most traditional iterative solvers are order N log (N) or maybe even order N^2. Coupled algebraic multigrid is order N, i.e., scales linearly. Unfortunately you have to assemble the matrix coefficients to give to the iterative solver, this step may not scale linearly, although, as you point out, with decent data structure design it might happen. With the exception of CFXTASClfow and CFX5, most commerical CFD codes use older order N^2 or NlogN algorithms.
Neale. 
Re: Double the # of cells>double the computing ti
I am not clear, what solving a linear system of equations with O(N) or whatever, has to do with solving a nonlinear system of coupled equations? Which is what a NS solver is. Of course, solving the linear subsystems faster during the nonlinear iterative process helps to reduce overall time, but it has nothing to do with converging the solution faster, i.e. having fewer global iterations.
I must be missing something... 
Re: Double the # of cells>double the computing ti
In their discretised form (on a discrete mesh) the Navier Stokes equations are a linear system of equations, which you throw at a linear solver. Every implicit finite volume scheme in the world does this.
Fewer global iterations is had by proper coupling, i.e. solving for u,v,w,p at the same time, in the same linear set of equations. This requires careful linearisation of the Navier Stokes equations. A better linear solver can reduce the number of global iterations as well. Simply turn off multigrid and only use a Jacobi or ILU smoother. It will take longer (i.e., more global iterations). Neale. 
Re: Double the # of cells>double the computing ti
Neal  NS is a nonlinear a coupled set of equations. You are correct saying that during iterative solution process at each iteration one solves a linear(ized) and often uncoupled set of equations. Coupling/uncoupling has very little to do with linearity/linearization. I agree that maintaining coupling similar to the original equations is good and may lead to fewer global iterations. The second parameter their is nonlinearity.
So how a better linear solver would lead to few global iterations is not clear to me. It may save time doing those linear iterations, but not reduce the number of global iterations. 
All times are GMT 4. The time now is 10:02. 