[gmx-users] question about continuation using tpbconv and mdrun -cpi

[Baofu Qiao]

Hi all,

Recently I made some tests on the continuation of gmx 4.0.5. Firstly, I
run a complete simulation as the reference. In the full.mdp file,
gen_seed=123456 & continuation=yes are used to build the same initial
structure.  In all the simulations, "mdrun -dlb no" is used because I
want to reproduce exactly the same potential energy.

1. Using tpbconv.  (part1.mdp is the same as full.mdp except the "nsteps")
  1a) grompp .... -s part1
      mpirun -np 32 mdrun -deffnm part1 -dlb no
      tpbconv -s part1.tpr -f part1.trr -e part1.edr -extend 100 -o part2
      mpirun -np 32 mdrun -deffnm part2 -dlb no
      eneconv -f part1.edr part2.edr -o part_all.edr
  1b) tpbconv -s part1.tpr -extend 100 -o part2.tpr
      mpirun -np 32 mdrun -deffnm part2 -cpi part1.cpt -dlb no
      eneconv -f part1.edr part2.edr -o part_all.edr
The potential energy is generated from g_energy, and compared. In this
method, I met two problems: 
     Q.a) The potential energies of part2  in both 1a and 1b are not
exactly the same as the reference potential!  And also the potential of
part2 in 1a is different from that in 1b. (Both potentials of part1 from
1a and 1b are identical to the reference.)
     Q.b) The RMSD of the potential energies of part_all.edr is very big
(1 order of magnitude bigger than the corresponding one from the
separate .edr files)

2) using mdrun -cpi -append
        grompp -s full2
        mpirun -np 32 mdrun -deffnm full2 -dlb no  (stop the job at some
time point, then run the following)
        mpirun -np 32 mdrun -deffnm full2 -dlb no  -cpi full2_prev.cpt
-append
The second (-append) section of the potential energy is also different
from the result of the reference potential, even though I am using the
same full.mdp file.  (the potential in the former section is identical
to the reference)

Then, how to reproduce exactly the potential?  It is said that the
continuation from "mdrun -cpi" is binary identical. However, it seems
not in my tests. What's the problem of the very big RMSD of the
potential from eneconv?

Thanks in advance!

regards,
Baofu Qiao