# Computing time vs order of accuracyΒΆ

Using a higher order implies using smaller time steps and using more basis functions. The expected increase in computed time can then be estimated as follow:

Decrease of the time steps width by a factor: \((2N+1)/(2n+1)\)

Increase of the number of basis functions by a factor:
\((N+1)(N+2)(N+3)/((n+1)(n+2)(n+3))\)

with:

n: initial order of accuracy -1

N: new order of accuracy -1

Here is an example of the theoretical increase of the compute time relative to order 3

3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | |
---|---|---|---|---|---|---|---|---|

Increase due to time steps | 1 | 1.4 | 1.8 | 2.2 | 2.6 | 3 | 3.4 | 3.8 |

Increase due to number of basis functions | 1 | 2 | 3.5 | 5.6 | 8.4 | 12 | 16.5 | 22 |

theoretical increase relative to order 3 | 1 | 2.8 | 6.3 | 12.32 | 21.84 | 36 | 56.1 | 83.6 |

observed increase (on SM2) | 1 | 2.0 | 4.6 | 11.5 |

The last line shows the observed time increase on SuperMUC Phase 2 on a small run with dynamic rupture and LTS-DR. Low order calculation (up to order 5 included) are memory bounds, and are then less efficient. As a consequence, the higher-order simulations cost less than expected in comparison with order 3.