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No of cells and iterations to converge

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Old   May 8, 2007, 07:10
Default No of cells and iterations to converge
  #1
Grigery
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can we relate the number iteration for the solution to converge to the numbers cells? i.e, if my case with 1 million cells converged (residuals upto 10^-5) within 2000 iterations, then can I guess the same domain with 2 million cells and same boundary conditions will take more than 2000 iterations to reach the same convergence level? (the maximum skewness, y+, etc are also the same in both the cases)

Grig
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Old   May 8, 2007, 08:59
Default Re: No of cells and iterations to converge
  #2
George
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I think I have read somewhere (I don't remember where) that there is almost a linear relation between these parameters (2 million cells-> 4000 iterations). But I haven't checked it myself for its validity.
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Old   May 8, 2007, 20:05
Default Re: No of cells and iterations to converge
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Bak_Flow
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Dear Grigery,

this is a good question and depends on a number of things:

1. what solver algorithm 2. coupling of problem physics 3. linearity/non-linearity of problem physics 4. Multigrid or some other acceleration...almost a given these days...????

I could give examples but it is probably better if you think about them yourself.....things like a coupled vs segregated algorithm for coupled physics and a linear problem???? Get it?

It is generally accepted .... but not always proven that coupled algorithms will give linear scalablity ie # iterations is constant regardless of number of cells.

A good reveiw reference is the Handbook of Computational Fluid Dynamics, ed Roger Peyret Ch 2. P65 Section E: Strongly Coupled Algorithms.

Here they quote: Moreover, with suitable acceleration techniques, the increase in computational effort with number N of pointes per direction is at most the order of N, whereas segregated methods yield an increase greater than N^2...Im not sure if that fits into your calc...are you solving 2D or 3D?

Hope this is of help.

Best Regards,

Bak_Flow
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Old   May 9, 2007, 00:31
Default Re: No of cells and iterations to converge
  #4
Grigery
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dear bak_flow

Thank you for detailed answer.. It was really helpful. I will definitely have a look at that book. I would've rather posted my doubt in a different way. All that I wanted to know was, given correct boundary conditions and other parameters, for same geometry of a 3-D domain (coupled solver, k-e turbulence model), will the solver take more iterations to converge if i increase ONLY the grid points?, And as a corrollary of this, Can I expect a domain of 2 million cells to converge within 100 iterations? My idea is, this is possible if the correct boundary conditions were given and the number of iterations to reach the required residual level depend ONLY on the BC (for any conditions of 2-D or 3-D type of solver, physics, linearity, etc). Is that true?

thanks again for your help

Grig
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Old   May 9, 2007, 11:20
Default Re: No of cells and iterations to converge
  #5
Jonas Holdeman
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I have been doing a convergence study for some divergence-free finite elements applied to the 2D lid-driven cavity problem, using simple iteration to solve the nonlinearity problem. This is a stream function-velocity method, the pressure being decoupled and eliminated. I find that the nonlinear convergence rates are essentially independent of mesh size (maybe 1% faster for finer meshes). Comparing 3rd, 4th and 5th order elements, the nonlinear convergence rates of the higher order elements are faster, but require more work per iteration. Nonlinear convergence rates of the methods are about the same for small Reynolds number (Re<100), and all are slower (if at all) for larger Re. The convergence-limiting Re (beyond which the simple iteration does not converge) is higher for higher-order elements. Using optimal under-relaxation, the convergence rates are similar at higher Re, and the convergence limits are extended. For low-enough Re, nonlinear convergence rates may be increased by over-relaxation.

This does not address the accuracy, which is higher for the higher-order methods on a given mesh, and (accuracy) convergence rate under mesh refinement which is, of course, faster for higher order elements.
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Old   May 9, 2007, 12:05
Default Re: No of cells and iterations to converge
  #6
Mani
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An important factor here (that was mentioned before) is multigrid. If each iteration entails a full multigrid cycle down to the same coarseness, then using a higher grid resolution will mean more work per iteration. The number of iterations will then not strongly depend on the grid size. You could actually say that that's the point of using multigrid: to take care of errors on all spatial scales within each iteration. That's the theory, anyway.
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Old   May 13, 2007, 12:18
Default Re: No of cells and iterations to converge
  #7
OPS
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Hi,

Yes, number of iterations taken for a solution to converge is strongly depend on the BCs and Cells. If you correct BC, the solution will converge much faster (less iterations) than the incorrect BC (more iterations for convergence). Here I quote from a book (Patanker, page 134). I hope you are aware about the SIMPLE and SIMPLER algorithm.

"If the given velocity field happens to be the correct velocity field, then the pressure equation in the SIMPLER will produce the correct pressure field, and there will be no need for an further iterations." (IT MEANS SOLUTION COVERGED).

Regarding number of cells, yes, the number of cells effect the correct solution (i.e a converged solution). Coarse cells do not represent actual model and hence require more iterations to get the correct solution.

Regards OPS

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Old   May 13, 2007, 21:11
Default Re: No of cells and iterations to converge
  #8
Grigery
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thanx OPS...this is what I asked exactly

Grig
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