Dear H. Lee,
After completing the .epb file writing part, job is terminating from the queue with error msg:
The .epb files have been correctly written
Band disentanglement is used: nbndsub = 60
Use zone-centred Wigner-Seitz cells
Number of WS vectors for electrons 279
Number of WS vectors for phonons 19
Number of WS vectors for electron-phonon 19
Maximum number of cores for efficient parallelization 912
Results may improve by using use_ws == .TRUE.
===================================================================================
= BAD TERMINATION OF ONE OF YOUR APPLICATION PROCESSES
= PID 29024 RUNNING AT n12-ib
= EXIT CODE: 9
= CLEANING UP REMAINING PROCESSES
= YOU CAN IGNORE THE BELOW CLEANUP MESSAGES
===================================================================================
Intel(R) MPI Library troubleshooting guide:
https://software.intel.com/node/561764
===================================================================================
Please have a look at the below epw input and suggest any possible fix.
&inputepw
prefix = 'xxx',
amass(1) = xx
outdir = './'
dvscf_dir = '../0-epc/save'
elph = .true.
kmaps = .false.
epbwrite = .true.
epbread = .false.
epwwrite = .true.
epwread = .false.
etf_mem = 0
nbndsub = 60
wannierize = .true.
num_iter = 50
dis_win_max = 12d0
dis_win_min = -7.5d0
dis_froz_min= 7d0
dis_froz_max= 10d0
proj(1) = 'random'
wdata(1) = 'bands_plot = .true.'
wdata(2) = 'begin kpoint_path'
wdata(3) = 'G 0.00 0.00 0.00 X 0.50 0.00 0.50'
wdata(4) = 'X 0.50 0.00 0.50 W 0.50 0.25 0.75'
wdata(5) = 'W 0.50 0.25 0.75 K 0.375 0.375 0.75'
wdata(6) = 'K 0.375 0.375 0.75 G 0.00 0.00 0.00'
wdata(7) = 'G 0.00 0.00 0.00 L 0.50 0.50 0.50'
wdata(8) = 'L 0.50 0.50 0.50 U 0.625 0.25 0.625'
wdata(9) = 'end kpoint_path'
wdata(10) = 'bands_num_points = 50'
wdata(11) = 'dis_num_iter = 50'
iverbosity = 0
elecselfen = .false.
phonselfen = .false.
a2f = .false.
fsthick = 1 ! eV
degaussw = 0.1 ! eV
nkf1 = 6 !18 !36
nkf2 = 6 !18 !36
nkf3 = 6 !18 !36
nqf1 = 2 !6 !12
nqf2 = 2 !6 !12
nqf3 = 2 !6 !12
nk1 = 6
nk2 = 6
nk3 = 6
nq1 = 2
nq2 = 2
nq3 = 2
/
4 cartesian
0.000000000 0.000000000 0.000000000
0.000000000 0.000000000 -0.622108626
0.000000000 -0.586529620 0.207369542
0.000000000 -0.586529620 -0.414739084