# Computing time vs order of accuracy

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.