On 21/10/2011 3:25 PM, Chandan Choudhury wrote:
Dear gmx users,
A simulation was performed for 50 ns with
; RUN CONTROL
integrator = md
nsteps = 50000000
dt = 0.001
; OUTPUT CONTROL
nstxout = 1000
nstvout = 1000
nstxtcout = 0
nstlog = 100
nstenergy = 100
The output produced were 0-50.edr, 0-50.trr, 0-50.log, 0-50.pdb
state.cpt, state_prev.cpt. The simulation completed normally.
After 50 ns, I intend to extend for 4 more ns, write the trajectory
file frequently with
Use tpbconv -extend. See
http://www.gromacs.org/Documentation/How-tos/Extending_Simulations
; RUN CONTROL
integrator = md
nsteps = 4000000
dt = 0.001
; OUTPUT CONTROL
nstxout = 00
nstvout = 00
nstxtcout = 10
nstlog = 500
nstenergy = 10
$grompp -f md.mdp -p topol.top -o md50-54.tpr -c 0-50.pdb -n index.ndx
$gmxdump -s md50-54.tpr | more
md50-54.tpr:
inputrec:
integrator = md
nsteps = 4000000
init_step = 0
ns_type = Grid
nstlist = 10
ndelta = 2
nstcomm = 1
comm_mode = Linear
nstlog = 500
nstxout = 0
nstvout = 0
nstfout = 0
nstenergy = 10
nstxtcout = 10
init_t = 0
delta_t = 0.001
xtcprec = 1000
nkx = 22
nky = 22
nkz = 40
pme_order = 4
ewald_rtol = 1e-05
ewald_geometry = 0
epsilon_surface = 0
optimize_fft = FALSE
ePBC = xyz
bPeriodicMols = FALSE
bContinuation = FALSE
bShakeSOR = FALSE
etc = V-rescale
epc = Berendsen
epctype = Isotropic
tau_p = 1
$mdrun_mpi-4.5.5 -s md50-54.tpr -c 50-54.pdb -x 50-54.xtc -e 50-54.edr
-g 50-54.log -cpi state.cpt -nice 0
state.cpt is the output of the initial 50ns run. I have used state.cpt
to incorporate the velocities frm the initial run.
So, I would expect my 2nd simulation to get complete after running for
4 ns. But, that was not the case. I have checked the log file. The
time clearly exceeds 4ns.
$ tail -15 50-54.log
DD step 54500989 load imb.: force 5.6% pme mesh/force 0.429
DD step 54500999 load imb.: force 6.0% pme mesh/force 0.428
Step Time Lambda
54501000 54501.00259 0.00000
Energies (kJ/mol)
Bond Angle Proper Dih. Ryckaert-Bell.
LJ-14
1.82780e+02 3.37147e+02 8.77788e+01 1.92877e+01
3.90071e+01
Coulomb-14 LJ (SR) Coulomb (SR) Coul. recip.
Potential
5.96624e+02 7.91553e+04 -5.73437e+05 -2.73830e+04
-5.20402e+05
Kinetic En. Total Energy Temperature Pressure (bar)
9.75908e+04 -4.22811e+05 2.97855e+02 -5.82198e+01
Kindly suggest what wrong am I doing. and how do I incorporate the
velocity from my earlier run.
On the information given, I can't explain this observation.
Mark
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