[Malthe Eisum]

I'm doing DNS simulations of a turbulent wave boundary layer, using a slightly modified version of pimpleFOAM (a body force is added to drive the flow).

To estimate the optimal number of processors I want to do a small study of the execution time with increasing no. of processors.

I have therefore two questions:
1) The listed execution time is the accumulative time eg. from code example:

ExecutionTime = 1.31 s

at the second time step indicates that from the start of time loop to the end of the second time step there have elapsed 1.31 s CPU time. Also meaning that from time step one to two

1.31-1.24=0.07 s

CPU time elapsed.

2) Is the listed execution time an estimate for one processor or a sum of all eg. (for 8 processors) the sum of CPU time is eather

1.31*8

or

1.31/8

since CPU time is close to clock time I would consider that the first case is more likely. But I would like to be confirmed.

Starting time loop

Courant Number mean: 0 max: 0 velocity magnitude: 0
deltaT = 0.00107403
Time = 0.00107403

dU0dt = dU0dt [0 1 -2 0 0 0 0] (3.51258e-06 0 0)
DILUPBiCG:  Solving for Ux, Initial residual = 1, Final residual = 0.0074636, No Iterations 1
DILUPBiCG:  Solving for Uy, Initial residual = 0, Final residual = 0, No Iterations 0
GAMG:  Solving for p, Initial residual = 3.27377e-09, Final residual = 3.27377e-09, No Iterations 0
time step continuity errors : sum local = 1.65251e-28, global = 3.78285e-29, cumulative = 3.78285e-29
GAMG:  Solving for p, Initial residual = 3.15544e-09, Final residual = 3.15544e-09, No Iterations 0
time step continuity errors : sum local = 1.59278e-28, global = 4.38014e-29, cumulative = 8.16298e-29
dU0dt = dU0dt [0 1 -2 0 0 0 0] (3.51258e-06 0 0)
DILUPBiCG:  Solving for Ux, Initial residual = 0.086718, Final residual = 1.17036e-10, No Iterations 2
DILUPBiCG:  Solving for Uy, Initial residual = 7.4425e-11, Final residual = 7.4425e-11, No Iterations 0
GAMG:  Solving for p, Initial residual = 3.48381e-09, Final residual = 3.48381e-09, No Iterations 0
time step continuity errors : sum local = 1.75853e-28, global = 4.38014e-29, cumulative = 1.25431e-28
GAMG:  Solving for p, Initial residual = 3.59952e-09, Final residual = 3.59952e-09, No Iterations 0
time step continuity errors : sum local = 1.81694e-28, global = 5.37562e-29, cumulative = 1.79187e-28
ExecutionTime = 1.24 s  ClockTime = 1 s

Courant Number mean: 2.75088e-29 max: 8.27559e-12 velocity magnitude: 4.59998e-10
deltaT = 0.0012803
Time = 0.00235434

dU0dt = dU0dt [0 1 -2 0 0 0 0] (3.52066e-06 0 0)
DILUPBiCG:  Solving for Ux, Initial residual = 0.999568, Final residual = 0.00876511, No Iterations 1
DILUPBiCG:  Solving for Uy, Initial residual = 9.62229e-11, Final residual = 9.62229e-11, No Iterations 0
GAMG:  Solving for p, Initial residual = 2.84384e-08, Final residual = 2.84384e-08, No Iterations 0
time step continuity errors : sum local = 1.71118e-27, global = 1.55691e-27, cumulative = 1.7361e-27
GAMG:  Solving for p, Initial residual = 3.60149e-08, Final residual = 3.60149e-08, No Iterations 0
time step continuity errors : sum local = 2.16706e-27, global = 1.8607e-27, cumulative = 3.5968e-27
dU0dt = dU0dt [0 1 -2 0 0 0 0] (3.52066e-06 0 0)
DILUPBiCG:  Solving for Ux, Initial residual = 0.0487339, Final residual = 1.07318e-10, No Iterations 2
DILUPBiCG:  Solving for Uy, Initial residual = 1.39588e-10, Final residual = 1.39588e-10, No Iterations 0
GAMG:  Solving for p, Initial residual = 3.72728e-08, Final residual = 3.72728e-08, No Iterations 0
time step continuity errors : sum local = 2.24275e-27, global = 1.8607e-27, cumulative = 5.4575e-27
GAMG:  Solving for p, Initial residual = 3.09594e-08, Final residual = 3.09594e-08, No Iterations 0
time step continuity errors : sum local = 1.86287e-27, global = 1.70881e-27, cumulative = 7.16631e-27
ExecutionTime = 1.31 s  ClockTime = 1 s

[TobM]

Hi,

clock time is written out in full seconds, when you look it up for larger times you will see that it is always larger than the execution time. Execution time is the time from the beginning of the simulation until the last processor finished its calculation. Both times don't differ much from another.