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Memory and Run Time Scaling

It seems intuitive and obvious that the amount of memory used (M) should scale with the total number of cells . For a 3-D problem,

N = m_x~m_y~m_z

dx = \frac{L_x}{m_x} = \frac{L_y}{m_y} = \frac{L_z}{m_z}

m_x \propto m_y \propto m_z \propto \frac{1}{dx}

So if the total memory M is proportional to N, then it follows from the above equations that

M \propto dx^{-3}

You might also think that the run time t_{run} scales in the same way, but that is not quite correct. Memory usage is a function of how much information you have and run time is a function of how many calculations you have to do on that information. The single most important factor determining run time is how many time steps (N_{steps}) are required to update each cell. So it now makes more sense to write an expression like this:

t_{run} \propto dx^{-3}~N_{steps}

The CFL condition is needed to calculate how the number of time steps scales:

dt = C \frac{dx}{v_{max}}

where C is the CFL number and v_{max} is the maximum wave speed. C and v_{max} can be ignored since they are constants, and then we can write

dt \propto dx

So if we have a run where we are trying to get to some final time t_{final}, we have to take a certain number of time steps.

N_{steps} = \frac{t_{final}}{dt} \propto \frac{1}{dx}

Combining this equation with our previous equation for t_{run}, we get the final result:

t_{run} \propto dx^{-4}

Below is a very simple example that illustrates how these equations work. For simplicity, the 1 x 1 x 1 simulation has all of its values explicitly set to 1.

Grid	N	dx	dt	t_{final}	N_{steps}	t_{run}
1 x 1 x 1	1	1	1	1	1	1
2 x 2 x 2	8	½	½	1	2	16
4 x 4 x 4	64	¼	¼	1	4	256

The scaling equations can also be generalized to problems of any dimension ndim.

astrobear

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Memory and Run Time Scaling

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