# How to find the optimal SSOR parameter

The symmetric successive overrelaxation method features the iteration matrix

$$P=\left(\frac{D}{\omega}+L\right)\frac{\omega}{2-\omega}D^{-1}\left(\frac{D}{\omega}+U\right)$$

Either as a stationary method or as a preconditioner, a system in that matrix is solved at every iteration.

The method depends on a parameter $$\omega \in (0,2)$$. While it converges for every $$\omega$$ in that parameter range, there is little information in standard textbooks on how to find the optimal parameter in practice. Often, $$\omega = 1$$ is a choice without further discussion.

What are practical ways of picking a relaxation parameter? Or is picking the parameter a purely theoretical issue?

• the book by Axelsson & Barker has a discussion. It comes down to the fact that for SSOR the optimal parameter is much less important than for SOR. Sep 21 at 19:27

In fact, $$\omega=1.2$$ is often used in practice when using SSOR as a preconditioner in an outer iteration. The way to find which value is optimal is simply one of trying many different values and recording how many iterations you need to solve a linear system to a certain accuracy. It's a one-dimensional optimization problem (the parameter is $$\omega$$, the objective function is the number of iterations), so easy enough to solve by simply sampling many different values of $$\omega$$.

• So in practice it comes down to trial and error? That sounds bleak. Sep 24 at 9:25
• It's a 1d optimization problem, that really isn't all that hard. Or you could just accept the value of 1.2 that others have found as gospel. Sep 25 at 3:04

Corless/Fillion book gives a formula for "optimal" $$\omega$$ of SOR but doesn't say where it comes from. Here's the relevant section: Not sure if this helps in some way, but in the book Numerical Recipes, the authors have this to say about the SOR parameter $$\omega$$ and how it can be iteratively updated: where Please note that the context where these conclusions were drawn in the book comes from considerations of a 2D SOR scheme, and they mention nothing about the impact of this update scheme on the SSOR. Also, the intermediate calculation step, denoted with bracketed indices $$1/2$$ and $$n+1/2$$ come from the 2D red-black SOR scheme, where first even-even and odd-odd i,j coordinates are solved, and then even-odd and odd-even i,j are solved. i,j here are indices of the grid points along the $$x$$ and $$y$$ axes, respectively

New contributor
Željko_JL is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
• As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center.
– Community Bot
Sep 27 at 19:50