[Wien] A question on the lapw2
Gavin Abo
gsabo at crimson.ua.edu
Tue Apr 9 21:50:03 CEST 2013
Are you using Wien2k 12.1? This error might be caused by the 'maxindex'
bug in an older Wien2k version:
http://zeus.theochem.tuwien.ac.at/pipermail/wien/2012-October/017854.html
http://zeus.theochem.tuwien.ac.at/pipermail/wien/2012-October/017853.html
On 4/9/2013 1:14 PM, Hu, Wenhao wrote:
> Hi, all:
>
> I met a problem in the calculation of Ni doped diamond supercell.
> Explicitly, what I'm trying to calculate is a 2x2x2 diamond supercell
> doped with a Nickel atom in the center of the supercell. Part of the
> structure files is pasted as follows:
>
> Ni_Diamond
> P 10
> RELA
> 13.492650 13.492650 13.492650 90.000000 90.000000 90.000000
> ATOM 1: X=0.00000000 Y=0.00000000 Z=0.00000000
> MULT= 1 ISPLIT= 2
> C 1 NPT= 781 R0=0.00010000 RMT= 1.3600 Z: 6.0
> LOCAL ROT MATRIX: 1.0000000 0.0000000 0.0000000
> 0.0000000 1.0000000 0.0000000
> 0.0000000 0.0000000 1.0000000
> ATOM -2: X=0.50000000 Y=0.00000000 Z=0.00000000
> MULT= 3 ISPLIT=-2
> -2: X=0.00000000 Y=0.50000000 Z=0.00000000
> -2: X=0.00000000 Y=0.00000000 Z=0.50000000
> C 2 NPT= 781 R0=0.00010000 RMT= 1.3600 Z: 6.0
> LOCAL ROT MATRIX: 0.0000000 0.0000000 1.0000000
> 1.0000000 0.0000000 0.0000000
> 0.0000000 1.0000000 0.0000000
> ATOM -3: X=0.50000000 Y=0.50000000 Z=0.00000000
> MULT= 3 ISPLIT=-2
> -3: X=0.50000000 Y=0.00000000 Z=0.50000000
> -3: X=0.00000000 Y=0.50000000 Z=0.50000000
> C 3 NPT= 781 R0=0.00010000 RMT= 1.3600 Z: 6.0
> LOCAL ROT MATRIX: 1.0000000 0.0000000 0.0000000
> 0.0000000 1.0000000 0.0000000
> 0.0000000 0.0000000 1.0000000
> ATOM 4: X=0.50000000 Y=0.50000000 Z=0.50000000
> MULT= 1 ISPLIT= 2
> Ni4 NPT= 781 R0=0.00005000 RMT= 1.7500 Z: 28.0
> LOCAL ROT MATRIX: 1.0000000 0.0000000 0.0000000
> 0.0000000 1.0000000 0.0000000
> 0.0000000 0.0000000 1.0000000
> ATOM -5: X=0.25000000 Y=0.25000000 Z=0.00000000
> MULT=12 ISPLIT= 8
> -5: X=0.75000000 Y=0.25000000 Z=0.00000000
> -5: X=0.25000000 Y=0.75000000 Z=0.00000000
> -5: X=0.75000000 Y=0.75000000 Z=0.00000000
> -5: X=0.25000000 Y=0.00000000 Z=0.25000000
>
> I used the parallel calculation mode and the whole program just
> crashed in the lapw2 step. In the output file, I have the following
> messages:
>
> LAPW1 END
> LAPW1 END
> LAPW2 - FERMI; weighs written
> forrtl: severe (174): SIGSEGV, segmentation fault occurred
> Image PC Routine Line Source
> forrtl: severe (174): SIGSEGV, segmentation fault occurred
> Image PC Routine Line Source
> lapw2c 00000001000652A1 _l2main_ 893 l2main_tmp_.F
> lapw2c 00000001000652A1 _l2main_ 893 l2main_tmp_.F
> lapw2c 00000001000798E3 _MAIN__ 564 lapw2_tmp_.F
> lapw2c 000000010000141C Unknown Unknown Unknown
> lapw2c 00000001000798E3 _MAIN__ 564 lapw2_tmp_.F
> lapw2c 00000001000013D4 Unknown Unknown Unknown
> lapw2c 000000010000141C Unknown Unknown Unknown
> lapw2c 00000001000013D4 Unknown Unknown Unknown
> forrtl: severe (174): SIGSEGV, segmentation fault occurred
> Image PC Routine Line Source
> lapw2c 00000001000652A1 _l2main_ 893 l2main_tmp_.F
> lapw2c 00000001000798E3 _MAIN__ 564 lapw2_tmp_.F
> lapw2c 000000010000141C Unknown Unknown Unknown
> lapw2c 00000001000013D4 Unknown Unknown Unknown
> forrtl: severe (174): SIGSEGV, segmentation fault occurred
> Image PC Routine Line Source
> lapw2c 00000001000652A1 _l2main_ 893 l2main_tmp_.F
> forrtl: severe (174): SIGSEGV, segmentation fault occurred
> lapw2c 00000001000798E3 _MAIN__ 564 lapw2_tmp_.F
> Image PC Routine Line Source
> lapw2c 000000010000141C Unknown Unknown Unknown
> lapw2c 00000001000652A1 _l2main_ 893 l2main_tmp_.F
> lapw2c 00000001000013D4 Unknown Unknown Unknown
> lapw2c 00000001000798E3 _MAIN__ 564 lapw2_tmp_.F
> lapw2c 000000010000141C Unknown Unknown Unknown
> lapw2c 00000001000013D4 Unknown Unknown Unknown
> forrtl: severe (174): SIGSEGV, segmentation fault occurred
> ........................................................
>
> I checked the mailist and tried many possible solutions like
> decreasing the value of RMT*KMax. But they just doesn't work. I can
> paste the other configuration files if necessary. I really appreciate
> any suggestions or comments. Thank you in advance.
>
> Best,
> Wenhao
>
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