No, it's not an input parameter of QE. Try to run your job on a different
machine.
PG
On Mon, Feb 3, 2020 at 8:27 AM Pooja Vyas
wrote:
> I even tried reducing my k-points from 11*11*11 to 3*3*3. but still the
> run gets terminated with signal 9 error.
>
> On Mon, Feb 3, 2020 at 12:55 PM Pooja V
I even tried reducing my k-points from 11*11*11 to 3*3*3. but still the run
gets terminated with signal 9 error.
On Mon, Feb 3, 2020 at 12:55 PM Pooja Vyas
wrote:
> Respected sir,
> I don't have any idea about the time limit which you are talking about. Is
> it any parameter of the QE input file
Respected sir,
I don't have any idea about the time limit which you are talking about. Is
it any parameter of the QE input file?
On Mon, Feb 3, 2020 at 12:53 PM Paolo Giannozzi
wrote:
> Yes, but it is much larger than that (115 days). I was referring to a
> maximum limit set by your batch queue
Yes, but it is much larger than that (115 days). I was referring to a
maximum limit set by your batch queue system, or by your operating system.
Paolo
On Mon, Feb 3, 2020 at 5:32 AM Pooja Vyas
wrote:
> Respected sir,
> Is there any default value of time after which the run can terminate by
> it
Respected sir,
Is there any default value of time after which the run can terminate by
itself?
On Mon, Feb 3, 2020 at 3:29 AM Paolo Giannozzi
wrote:
> On Sat, Feb 1, 2020 at 11:12 AM Pooja Vyas
> wrote:
>
> total cpu time spent up to now is72841.4 secs
>> [...]
>>
> Primary job termin
On Sat, Feb 1, 2020 at 11:12 AM Pooja Vyas
wrote:
total cpu time spent up to now is72841.4 secs
> [...]
>
Primary job terminated normally, but 1 process returned
> a non-zero exit code. Per user-direction, the job has been aborted.
>
[...]
> What could be the reason for this termination
While computing energy of CaO with 216 atoms, the run gets terminated by
itself after 6 iterations with the following message,
iteration # 6 ecut= 100.00 Ry beta= 0.70
Davidson diagonalization with overlap
ethr = 8.13E-07, avg # of iterations = 3.0
negative rho (up,
Thank you sir
On Tue, 21 Jan 2020, 11:52 am Premkumar Thirumalaisamy,
wrote:
> Hi,
> Always compare the energy per formula unit, if you do that in your
> case there is a small different in energy which may be due to different
> k-grid and energy cutoff.
>I think it is better to optimize
Hi,
Always compare the energy per formula unit, if you do that in your case
there is a small different in energy which may be due to different k-grid
and energy cutoff.
I think it is better to optimize the kpoints and energy cutoff for
particular unit cell or supercell.
On Sun, Jan 19, 2020
Hi Pooja,
I think CaO is a cubic crystal. You can use VESTA to generate a supercell.
It can take .cif file. Play with it. I believe extending lattice
parameter 2X2X2 generate a 36 atom supercell.
HTH
Manu
(McMaster University)
On Sat, Jan 18, 2020 at 12:38 AM Pooja Vyas
wrote:
> Respected sir/ma
Respected sir/madam,
Referring a paper on calculation of energy with vacancy in CaO, I want to
re-calculate the energy with the same number of atoms used in the paper.
They have 36 atoms in their supercell, 3 x 3 x 2 monk horst pack grid. From
this information, is it possible to know what could be
Well, maybe because in your first calculation you had two atoms, and 64
in the second one.You should try to see if multiplying -107.10 by
64/2 improve the comparison
L.
On 16/01/2020 06:23, Pooja Vyas wrote:
Initially I had run my input script with ecut=100Ry and k-points= 11
11 11 1
Initially I had run my input script with ecut=100Ry and k-points= 11 11 11
1 1 1. At that time my energy was -107.10Ry. During this run, I had
specified only two atomic positions (Ca: 0 0 0, O: 0.5 0.5 0.5)
For the same system but with ecut=60Ry and k-points= 3 3 3 1 1 1 the
calculated energy was -
Please, I would rather suggest to spy the suggested ecut in your
pseudopotentials files and from there you could have an idea on how you
might conduct your test for checking and others..
Best Wishes!
Bruno, IMSP-Benin
On Fri, Jan 10, 2020 at 8:53 AM Tone Kokalj wrote:
> On 2020-01-10 04:18, Po
On 2020-01-10 04:18, Pooja Vyas wrote:
The process gets terminated by itself.
Insufficient memory?
Can it be because of large cutoff and large kpoints?
Yes, reduce the cutoff. Try using something like:
ecutwfc = 30.0
ecutrho = 240.0
and
K_POINTS (gamma)
But you should make a test which
>
> The process gets terminated by itself. May be due to long time it takes.
> Can it be because of large cutoff and large kpoints?
>
Try taking small kinetic energy cutoff (ecut), charge density cutoff, and
less kpts; and see whether it proceeds to completion. Also, keep an eye on
RAM as suggeste
The process gets terminated by itself. May be due to long time it takes.
Can it be because of large cutoff and large kpoints?
On Thu, 9 Jan 2020, 5:44 pm Paolo Giannozzi, wrote:
> On Thu, Jan 9, 2020 at 10:45 AM Pooja Vyas
> wrote:
>
> But the iteration does not start and the calculation doesn'
On Thu, Jan 9, 2020 at 10:45 AM Pooja Vyas
wrote:
But the iteration does not start and the calculation doesn't proceed and
> gets stuck at the following point:
>
the calculation proceeds, if you have enough memory:
Estimated total dynamical RAM > 31.02 GB
>
but it takes time. With yo
Respected sir,
I understand completely that 2*2*2 takes time, but in my case, the
iteration doesn't even start.
Though I shall perform convergence, but can you suggest any specific
numbers for both these parameters (cutoff and k-point) suitable for my
input?
On Thu, Jan 9, 2020 at 4:40 PM Tone Kok
Dear Pooja,
All seems fine. Note that calculation of (2x2x2) supercell takes
considerably longer than that of the (1x1x1) unit cell.
Are you sure you need that high cutoff energy and that dense k-point
grid? If you reduce these two then the calculation will run faster.
Best regards,
Tone Kokalj
Following is my input file:
&control
calculation = 'scf',
prefix = '9.1334'
tstress= .true.
tprnfor= .true.
outdir = '/home/user/cao.oct/'
pseudo_dir = '/home/user/cao.oct/pseudo/'
/
&system
ibrav = 0,
celldm(1)=9.1334,
nat = 64,
ntyp = 2,
ecutwfc = 1
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