View Issue Details
| ID | Project | Category | View Status | Date Submitted | Last Update |
|---|---|---|---|---|---|
| 0002522 | OpenFOAM | Bug | public | 2017-04-06 12:33 | 2017-04-12 19:10 |
| Reporter | Shorty | Assigned To | henry | ||
| Priority | normal | Severity | minor | Reproducibility | always |
| Status | closed | Resolution | no change required | ||
| Platform | GNU/Linux | OS | Ubuntu | OS Version | 14.04 |
| Product Version | dev | ||||
| Summary | 0002522: Equation relaxation for 1 will lead to a modified matrix | ||||
| Description | Hey, if we build a vector matrix and relax it with the factor of 1, the relaxed matrix is different to the original one. Based on a lot of textbooks, I guessed that relax factor of 1 should not influence the matrix but it does based on the boundary treatment. First we add the MAXIMUM component, then we relax and finally we remove the MINIMUM. It is easy to see that this can influence the matrix system. In my case the solution with out any relaxation factor took 0.7s while using a relaxation factor of 1, lead to 32s. After that I went into the code to see what is going on and I realized the maximum/minimum addition/removal. For scalar matrix it is fine but in my displacement matrix I focused on that different behavior. But I think you know it much better and know the reason why we add the maximum first and remove the minimum at the end. I just wanted to let you know. Maybe it is really a simple mistake. | ||||
| Tags | equation, fvVectorMatrix, relaxation | ||||
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> if we build a vector matrix and relax it with the factor of 1, the relaxed matrix is different to the original one. Correct, relaxation is applied to a matrix which is first made at least diagonally equal, otherwise it would not improve diagonal dominance. Thus relaxation with a factor of 1 makes the matrix at least diagonally equal but no more. |
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> Correct, relaxation is applied to a matrix which is first made at least diagonally equal, otherwise it would not improve diagonal dominance. Thus relaxation with a factor of 1 makes the matrix at least diagonally equal but no more. Okay, thus, the new vector matrix with relaxation factor of 1 does not has to be equal to the matrix we had before, because the new one can be diagonal improved (thats what I got from the sources and you confirmed now). If so, I am wondering why I need more outer corrections than not setting any relaxation factor. I would expect solving a improved diagonal matrix will speed up the solution. Thanks Henry. |
| Date Modified | Username | Field | Change |
|---|---|---|---|
| 2017-04-06 12:33 | Shorty | New Issue | |
| 2017-04-06 12:35 | Shorty | Tag Attached: equation | |
| 2017-04-06 12:35 | Shorty | Tag Attached: fvVectorMatrix | |
| 2017-04-06 12:35 | Shorty | Tag Attached: relaxation | |
| 2017-04-06 13:27 | henry | Note Added: 0008020 | |
| 2017-04-06 14:22 | Shorty | Note Added: 0008021 | |
| 2017-04-06 14:52 | henry | Assigned To | => henry |
| 2017-04-06 14:52 | henry | Status | new => closed |
| 2017-04-06 14:52 | henry | Resolution | open => no change required |
