Coupled Cluster Scaling Question

  • #1

Main Question or Discussion Point

Hi,

I have a relatively small system. It takes 10 minutes to perform a CCSDT calculation. I know scaling is notoriously bad, but does anyone have a rough idea of how long a CCSDTQ calculation would take on the same system? Are we talking days or months?

Thomas
 

Answers and Replies

  • #2
cgk
Science Advisor
521
42
First: Depends on the program. For these kinds of things you should use Kallay's MRCC. It will even give you a coarse estimation of the time per iteration. Like in CCSD(T) (which you probably want to do in Molpro or CFOUR), different programs can differ by one or two orders of magnitude in calculation time. (While producing exactly the same number to 12 digits precision)

Second: Depends on the system (symmetries, ratio of occupied to virtual orbitals). So it's hard to tell. In principle, CCSDT is N^8 and CCSDTQ is N^10, but that does not translate well into the factors for *the same system*. It is very hard to estimate.

Also note that CCSDT is generally not a very useful method, because in a large majority of cases it gives worse results than CCSD(T) (see e.g., Jan Martin's W4 paper). Generally, the next thing better than CCSD(T) is CCSDT(Q) (N^9) or CCSDTQ; so unless you are doing the CCSDT as part of a basis set extrapolation series, it is better to not do it at all.
 
  • #3
First: Depends on the program. For these kinds of things you should use Kallay's MRCC. It will even give you a coarse estimation of the time per iteration. Like in CCSD(T) (which you probably want to do in Molpro or CFOUR), different programs can differ by one or two orders of magnitude in calculation time. (While producing exactly the same number to 12 digits precision)

Second: Depends on the system (symmetries, ratio of occupied to virtual orbitals). So it's hard to tell. In principle, CCSDT is N^8 and CCSDTQ is N^10, but that does not translate well into the factors for *the same system*. It is very hard to estimate.

Also note that CCSDT is generally not a very useful method, because in a large majority of cases it gives worse results than CCSD(T) (see e.g., Jan Martin's W4 paper). Generally, the next thing better than CCSD(T) is CCSDT(Q) (N^9) or CCSDTQ; so unless you are doing the CCSDT as part of a basis set extrapolation series, it is better to not do it at all.
You have no idea how useful this is to me. Tearing my hair out over explaining CCSDT discrepancies when the answer was right in the W4 paper. Thanks!
 

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