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4 Commits
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Author SHA1 Message Date
gabrielkheisa
80f1f66052 fix 2022-11-27 12:37:54 +07:00
gabrielkheisa
45293c6962 fix old fortran syntax writing 2022-11-27 12:23:27 +07:00
gabrielkheisa
115bcdb4c8 fix typo 2022-11-27 12:11:34 +07:00
gabrielkheisa
436d13ab14 add pdf and zip 2022-11-16 20:15:04 +07:00
34 changed files with 304 additions and 332 deletions
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<p>
Gabriel Possenti Kheisa Drianasta<br>
19/442374/PA/19123
</p>
<h1>Hasil Tugas OpenMP</h1>
<h3>Pengujian terhadap 1, 2, 4, 8, 16, 32, hingga 64 thread</h3>
<h2>Hasil</h2>
<p>Matrix size: 4096</p>
<ul>
<li>1 Thread: 227.454 sec</li>
<li>2 Thread: 115.605 sec</li>
<li>4 Thread: 60.744 sec</li>
<li>8 Thread: 30.498 sec</li>
<li>16 Thread: 26.556 sec</li>
<li>32 Thread: 24.960 sec</li>
<li>64 Thread: 24.707 sec</li>
</ul>
<br>
<h3>Waktu (detik)</h3>
<img src="https://media.discordapp.net/attachments/1003173519879847966/1041940259719618560/image.png">
<h3>Selisih waktu terhadap jumlah core sebelumnya (detik)</h3>
<img src="https://cdn.discordapp.com/attachments/1003173519879847966/1042339831071657994/image.png">
<br><br>
<h1>Pertanyaan</h1>
<h3>1. What is the maximum speed up compared to single thread
execution?</h3>
<p>Dengan menghitung perbandingan selisih waktu yang di dapat antara 1 thread dengan 2 thread, yakni ( 227 / 115 ) detik, maka speedup yang di dapat adalah <b>1.97</b> kali. Dengan menggunakan rumus perhitungan speedup berikut:</p>
<img src="https://cdn.discordapp.com/attachments/1003173519879847966/1042413697986994176/image.png">
<p>Maka program bersifat <b>98.4%</b> parallel. Dengan persentase berikut,maka, perbandingan performa dengan program yang bersifat parallel sempurna adalah sebagai berikut:</p>
<pre>
Parallel 100% = [1, 2, 4, 8, 16, 32, 64]
Parallel 98.4% = [1, 1.97, 3.8, 7.19, 12.9, 21.4, 31.8]
</pre>
<img src="https://media.discordapp.net/attachments/1003173519879847966/1042418071593304117/image.png">
<p>Selisih speedup terbesar tampak pada jumlah thread <b>8 ke 16</b>, sama seperti hasil percobaan sebelumnya, sehingga penambahan jumlah core dari 8, ke 16, ke 32, dan seterusnya menjadi relatif lebih tidak efektif.</p>
<h3>2. As we increase the number of threads, the execution time starts to
saturate at one point (cannot get any faster). Explain why this
happens.</h3>
<p>Karena program tidak bersifat parallel sempurna (100%). Program bersifat 98.4% parallel, dimana <b>1.6%</b> bersifat serial. Bagian serial ini merupakan bagian yang tidak dapat dikerjakan secara parallel karena berbagai faktor. Meskipun angka tersebut kecil, selisih speedup dengan program yang dapat dijalankan secara parallel sempurna akan semakin terlihat dengan jumlah core atau thread yang semakin banyak.</p>
<img src="https://media.discordapp.net/attachments/1003173519879847966/1042420674385416202/image.png">
<img src="https://media.discordapp.net/attachments/1003173519879847966/1042421381591203871/image.png">
<h3>3. Find the exact number of threads where saturation begins. What is its relationship with the number of physical cores of the system?
Explain.</h3>
<p>Berdasarkan hasil yang terlampir, baik hasil eksperimen maupun perhitungan dengan Hukum Amdahl, <b>8 thread</b> merupakan jumlah core terbaik karena selisih waktu eksekusi (hasil percobaan) tidak begitu besar, sedangkan selisih speedup (perhitungan menggunakan Hukum Amdahl) sangat besar. Dengan kata lain, 8 thread ke 16 thread dan seterusnya relatif tidak efektif. Hubungan antara jumlah thread program dengan jumlah core adalah berbanding lurus dalam hal speedup dan berbanding terbalik dengan waktu eksekusi, dengan program yang bersifat 100% parallel.</p>
<br>
<p>Lampiran program dan hasil: https://repo.gabrielkheisa.xyz/gabrielkheisa/tugas3-openMP</p>

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<!DOCTYPE html>
<html>
<script src="https://cdnjs.cloudflare.com/ajax/libs/Chart.js/2.5.0/Chart.min.js"></script>
<body>
<canvas id="myChart" style="width:100%;max-width:600px"></canvas>
<script>
var xValues = [1,2,4,8,16,32,64];
var data = [227.454, 115.605, 60.774, 30.498, 26.556, 24.960, 24.707];
var d_data = [];
var speedup = [1, 1.97, 3.8, 7.19, 12.9, 21.4, 31.8];
var i = 1;
while(i < data.length){
d_data[i] = data[i+1] - data[i];
i++;
}
/*
1 Thread: 227.454 sec
2 Thread: 115.605 sec
4 Thread: 60.744 sec
8 Thread: 30.498 sec
16 Thread: 26.556 sec
32 Thread: 24.960 sec
64 Thread: 24.707 sec
*/
new Chart("myChart", {
type: "line",
data: {
labels: xValues,
datasets: [{
data: data,
borderColor: "black",
fill: false
}]
},
options: {
legend: {display: false}
}
});
</script>

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mahasiswa2
komputasi06
mimax= 129 mjmax= 65 mkmax= 65
imax= 128 jmax= 64 kmax= 64
Start rehearsal measurement process.
Measure the performance in 10000 times.
MFLOPS: 6017.17725 time(s): 27.3678112 8.79942896E-10

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mahasiswa2
komputasi06
mimax= 129 mjmax= 65 mkmax= 65
imax= 128 jmax= 64 kmax= 64
Start rehearsal measurement process.
Measure the performance in 10000 times.
MFLOPS: 724.292114 time(s): 227.362640 8.79942896E-10

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program summ
implicit none
integer :: sum = 0
integer :: n
!$OMP parallel do
do n = 0, 1000
sum = sum + n
print*, n, " ", sum
end do
!$OMP end parallel do
print*, " "
print*, " "
print*, " "
print*, "hasilnya adalah ", sum
end program summ

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#!/bin/bash
#SBATCH --nodes=1
#SBATCH --time=00:01:10
#SBATCH --job-name=Gabriel
whoami
hostname
gfortran -fopenmp no_reduction.f90 -o no_reduction.x
export OMP_NUM_THREADS=4
./no_reduction.x

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<p>
Gabriel Possenti Kheisa Drianasta<br>
19/442374/PA/19123
</p>
<h1>Hasil Performance tuning</h1>
<h3>Sebelum parallel</h3>
<pre>
mahasiswa2
komputasi06
mimax= 129 mjmax= 65 mkmax= 65
imax= 128 jmax= 64 kmax= 64
Start rehearsal measurement process.
Measure the performance in 10000 times.
MFLOPS: 724.292114 time(s): 227.362640 8.79942896E-10
</pre>
<h3>Sesudah parallel</h3>
<pre>
mahasiswa2
komputasi06
mimax= 129 mjmax= 65 mkmax= 65
imax= 128 jmax= 64 kmax= 64
Start rehearsal measurement process.
Measure the performance in 10000 times.
MFLOPS: 6017.17725 time(s): 27.3678112 8.79942896E-10
</pre>
<h1>Pembahasan</h1>
<h3>Program OMP do end do disispkan pada bagian berikut pada program <a href="https://elok.ugm.ac.id/pluginfile.php/2173252/mod_resource/content/1/sample4.f">himeno</a>:</h3>
<p>https://repo.gabrielkheisa.xyz/gabrielkheisa/tugas3-openMP/commit/1f11c6330793ba22afb953d114078c87e65d522c</p>
<img src="https://cdn.discordapp.com/attachments/1003173519879847966/1046321839607136297/image.png">
<img src="https://media.discordapp.net/attachments/1003173519879847966/1046321891406786580/image.png">
<h3>Kemudian dilanjutkan dengan penambahan $OMP parallel private() dan do reduction untuk deklarasi variabel parallel:</h3>
<p>https://repo.gabrielkheisa.xyz/gabrielkheisa/tugas3-openMP/commit/8babfcc27e3b0dc524cf99a00cbd47645b9d0273</p>
<img src="https://media.discordapp.net/attachments/1003173519879847966/1046322875797352518/image.png">
<p>Hasilnya adalah peningkatan performa dari <b>724MFLOPS</b> menjadi <b>6017MFLOPS</b> atau sebesar <b>8.3 kali</b> untuk <b>20 threads</b></p>
<br>
<p>Lampiran source code dan dokumentasi: https://repo.gabrielkheisa.xyz/gabrielkheisa/tugas3-openMP/src/branch/master2</p>

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mahasiswa2
komputasi06
Matrix Size = 4096
Execution Time = 227.454 sec A(n,n) = 0.102092747142180D+04

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mahasiswa2
komputasi06
Matrix Size = 4096
Execution Time = 115.605 sec A(n,n) = 0.102437460360345D+04

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mahasiswa2
komputasi06
Matrix Size = 4096
Execution Time = 60.744 sec A(n,n) = 0.102843459928075D+04

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mahasiswa2
komputasi06
Matrix Size = 4096
Execution Time = 30.498 sec A(n,n) = 0.100725125084073D+04

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mahasiswa2
komputasi06
Matrix Size = 4096
Execution Time = 26.556 sec A(n,n) = 0.100253761446337D+04

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mahasiswa2
komputasi03
Matrix Size = 4096
Execution Time = 24.960 sec A(n,n) = 0.101898724293035D+04

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mahasiswa2
komputasi03
Matrix Size = 4096
Execution Time = 24.707 sec A(n,n) = 0.101964571498883D+04

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C*********************************************************************
C
C This benchmark test program is measuring a cpu performance
C of floating point operation by a Poisson equation solver.
CC
C If you have any question, please ask me via email.
C written by Ryutaro HIMENO, November 26, 2001.
C Version 3.0
C ----------------------------------------------
C Ryutaro Himeno, Dr. of Eng.
C Head of Computer Information Division,
C RIKEN (The Institute of Pysical and Chemical Research)
C Email : himeno@postman.riken.go.jp
C ---------------------------------------------------------------
C You can adjust the size of this benchmark code to fit your target
C computer. In that case, please chose following sets of
C (mimax,mjmax,mkmax):
C small : 65,33,33
C small : 129,65,65
C midium: 257,129,129
C large : 513,257,257
C ext.large: 1025,513,513
C This program is to measure a computer performance in MFLOPS
C by using a kernel which appears in a linear solver of pressure
C Poisson eq. which appears in an incompressible Navier-Stokes solver.
C A point-Jacobi method is employed in this solver as this method can
C be easyly vectrized and be parallelized.
C ------------------
C Finite-difference method, curvilinear coodinate system
C Vectorizable and parallelizable on each grid point
C No. of grid points : imax x jmax x kmax including boundaries
C ------------------
C A,B,C:coefficient matrix, wrk1: source term of Poisson equation
C wrk2 : working area, OMEGA : relaxation parameter
C BND:control variable for boundaries and objects ( = 0 or 1)
C P: pressure
C -------------------
C -------------------
C "use portlib" statement on the next line is for Visual fortran
C to use UNIX libraries. Please remove it if your system is UNIX.
C -------------------
! use portlib
use omp_lib
IMPLICIT REAL*4(a-h,o-z)
real*8 t1,t2
C
C PARAMETER (mimax=513,mjmax=257,mkmax=257)
C PARAMETER (mimax=257,mjmax=129,mkmax=129)
PARAMETER (mimax=129,mjmax=65,mkmax=65)
C PARAMETER (mimax=65,mjmax=33,mkmax=33)
C
C ttarget specifys the measuring period in sec
PARAMETER (ttarget=60.0)
CC Arrey
common /pres/ p(mimax,mjmax,mkmax)
common /mtrx/ a(mimax,mjmax,mkmax,4),
+ b(mimax,mjmax,mkmax,3),c(mimax,mjmax,mkmax,3)
common /bound/ bnd(mimax,mjmax,mkmax)
common /work/ wrk1(mimax,mjmax,mkmax),wrk2(mimax,mjmax,mkmax)
CC Other constants
common /others/ imax,jmax,kmax,omega
C
dimension time0(2),time1(2)
C
omega=0.8
imax=mimax-1
jmax=mjmax-1
kmax=mkmax-1
CC Initializing matrixes
call initmt
write(*,*) ' mimax=',mimax,' mjmax=',mjmax,' mkmax=',mkmax
write(*,*) ' imax=',imax,' jmax=',jmax,' kmax=',kmax
CC Start measuring
C
nn=10000
write(*,*) ' Start rehearsal measurement process.'
write(*,*) ' Measure the performance in 10000 times.'
C
! cpu0=dtime(time0)
t1 = omp_get_wtime()
C
C Jacobi iteration
call jacobi(nn,gosa)
C
! cpu1= dtime(time1)
t2 = omp_get_wtime()
! cpu = cpu1
cpu = t2-t1
flop=real(kmax-2)*real(jmax-2)*real(imax-2)*34.0*real(nn)
xmflops2=flop/cpu*1.0e-6
write(*,*) ' MFLOPS:',xmflops2,' time(s):',cpu,gosa
C
C end the test loop
! nn=ifix(ttarget/(cpu/3.0))
! write(*,*) 'Now, start the actual measurement process.'
! write(*,*) 'The loop will be excuted in',nn,' times.'
! write(*,*) 'This will take about one minute.'
! write(*,*) 'Wait for a while.'
C
C Jacobi iteration
! cpu0=dtime(time0)
! call jacobi(nn,gosa)
C
! cpu1= dtime(time1)
! cpu = cpu1
! flop=real(kmax-2)*real(jmax-2)*real(imax-2)*34.0*real(nn)
! xmflops2=flop*1.0e-6/cpu
C
CCC xmflops2=nflop/cpu*1.0e-6*float(nn)
C
! write(*,*) ' Loop executed for ',nn,' times'
! write(*,*) ' Gosa :',gosa
! write(*,*) ' MFLOPS:',xmflops2, ' time(s):',cpu
! score=xmflops2/82.84
! write(*,*) ' Score based on Pentium III 600MHz :',score
C
! pause
stop
END
C
C
C**************************************************************
subroutine initmt
C**************************************************************
IMPLICIT REAL*4(a-h,o-z)
C
C PARAMETER (mimax=513,mjmax=257,mkmax=257)
C PARAMETER (mimax=257,mjmax=129,mkmax=129)
PARAMETER (mimax=129,mjmax=65,mkmax=65)
C PARAMETER (mimax=65,mjmax=33,mkmax=33)
C
CC Arrey
common /pres/ p(mimax,mjmax,mkmax)
common /mtrx/ a(mimax,mjmax,mkmax,4),
+ b(mimax,mjmax,mkmax,3),c(mimax,mjmax,mkmax,3)
common /bound/ bnd(mimax,mjmax,mkmax)
common /work/ wrk1(mimax,mjmax,mkmax),wrk2(mimax,mjmax,mkmax)
CC other constants
common /others/ imax,jmax,kmax,omega
C
!$OMP parallel private(k,j,i)
!$OMP do
do k=1,mkmax
do j=1,mjmax
do i=1,mimax
a(i,j,k,1)=0.0
a(i,j,k,2)=0.0
a(i,j,k,3)=0.0
a(i,j,k,4)=0.0
b(i,j,k,1)=0.0
b(i,j,k,2)=0.0
b(i,j,k,3)=0.0
c(i,j,k,1)=0.0
c(i,j,k,2)=0.0
c(i,j,k,3)=0.0
p(i,j,k) =0.0
wrk1(i,j,k)=0.0
bnd(i,j,k)=0.0
enddo
enddo
enddo
!$OMP end do
C
!$OMP do
do k=1,kmax
do j=1,jmax
do i=1,imax
a(i,j,k,1)=1.0
a(i,j,k,2)=1.0
a(i,j,k,3)=1.0
a(i,j,k,4)=1.0/6.0
b(i,j,k,1)=0.0
b(i,j,k,2)=0.0
b(i,j,k,3)=0.0
c(i,j,k,1)=1.0
c(i,j,k,2)=1.0
c(i,j,k,3)=1.0
p(i,j,k) =float((k-1)*(k-1))/float((kmax-1)*(kmax-1))
wrk1(i,j,k)=0.0
bnd(i,j,k)=1.0
enddo
enddo
enddo
!$OMP end do
!$OMP end parallel
C
return
end
C
C*************************************************************
subroutine jacobi(nn,gosa)
C*************************************************************
IMPLICIT REAL*4(a-h,o-z)
C
C PARAMETER (mimax=513,mjmax=257,mkmax=257)
C PARAMETER (mimax=257,mjmax=129,mkmax=129)
PARAMETER (mimax=129,mjmax=65,mkmax=65)
C PARAMETER (mimax=65,mjmax=33,mkmax=33)
C
CC Arrey
common /pres/ p(mimax,mjmax,mkmax)
common /mtrx/ a(mimax,mjmax,mkmax,4),
+ b(mimax,mjmax,mkmax,3),c(mimax,mjmax,mkmax,3)
common /bound/ bnd(mimax,mjmax,mkmax)
common /work/ wrk1(mimax,mjmax,mkmax),wrk2(mimax,mjmax,mkmax)
CC other constants
common /others/ imax,jmax,kmax,omega
C
C
DO loop=1,nn
gosa=0.0
!$OMP parallel private(K,J,I,S0,SS,wrk2)
!$OMP do reduction(+:GOSA)
DO K=2,kmax-1
DO J=2,jmax-1
DO I=2,imax-1
S0=a(I,J,K,1)*p(I+1,J,K)+a(I,J,K,2)*p(I,J+1,K)
1 +a(I,J,K,3)*p(I,J,K+1)
2 +b(I,J,K,1)*(p(I+1,J+1,K)-p(I+1,J-1,K)
3 -p(I-1,J+1,K)+p(I-1,J-1,K))
4 +b(I,J,K,2)*(p(I,J+1,K+1)-p(I,J-1,K+1)
5 -p(I,J+1,K-1)+p(I,J-1,K-1))
6 +b(I,J,K,3)*(p(I+1,J,K+1)-p(I-1,J,K+1)
7 -p(I+1,J,K-1)+p(I-1,J,K-1))
8 +c(I,J,K,1)*p(I-1,J,K)+c(I,J,K,2)*p(I,J-1,K)
9 +c(I,J,K,3)*p(I,J,K-1)+wrk1(I,J,K)
SS=(S0*a(I,J,K,4)-p(I,J,K))*bnd(I,J,K)
GOSA=GOSA+SS*SS
wrk2(I,J,K)=p(I,J,K)+OMEGA *SS
enddo
enddo
enddo
!$OMP end do
C
!$OMP do
DO K=2,kmax-1
DO J=2,jmax-1
DO I=2,imax-1
p(I,J,K)=wrk2(I,J,K)
enddo
enddo
enddo
!$OMP end do
!$OMP end parallel
C
enddo
CC End of iteration
return
end

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@ -78,11 +78,13 @@ C
C
! cpu0=dtime(time0)
t1 = omp_get_wtime()
!$OMP PARALLEL DO
C
C Jacobi iteration
call jacobi(nn,gosa)
C
! cpu1= dtime(time1)
!$OMP END PARALLEL DO
t2 = omp_get_wtime()
! cpu = cpu1
cpu = t2-t1
@ -140,7 +142,6 @@ CC Arrey
CC other constants
common /others/ imax,jmax,kmax,omega
C
!$OMP do
do k=1,mkmax
do j=1,mjmax
do i=1,mimax
@ -160,10 +161,7 @@ C
enddo
enddo
enddo
!$OMP end do
C
!$OMP do
do k=1,kmax
do j=1,jmax
do i=1,imax
@ -183,7 +181,6 @@ C
enddo
enddo
enddo
!$OMP end do
C
return
end
@ -211,7 +208,6 @@ C
C
DO loop=1,nn
gosa=0.0
!$OMP do reduction(+:GOSA)
DO K=2,kmax-1
DO J=2,jmax-1
DO I=2,imax-1
@ -231,9 +227,7 @@ C
enddo
enddo
enddo
!$OMP end do
C
!$OMP do
DO K=2,kmax-1
DO J=2,jmax-1
DO I=2,imax-1
@ -241,7 +235,6 @@ C
enddo
enddo
enddo
!$OMP end do
C
enddo
CC End of iteration

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program sample3
use omp_lib
implicit real(8)(a-h,o-z)
parameter (n=4096)
real(8) a(n,n), c(n,n)
real(4) b(n,n)
real*8 t1, t2
a=0.0d0
call random_number(b)
call random_number(c)
write(6,50) ' Matrix Size = ', n
50 format(1x,a,i5)
t1 = omp_get_wtime()
!$OMP PARALLEL DO
do j=1,n
do k=1,n
do i=1,n
a(i,j)=a(i,j)+b(i,k)*c(k,j)
end do
end do
end do
!$OMP END PARALLEL DO
t2 = omp_get_wtime()
write(6, 60) ' Execution Time = ',t2-t1,' sec',' A(n,n) = ',a(n,n)
60 format(1x,a,f10.3,a,1x,a,d24.15)
stop
end

13
tugas3.sh Normal file
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#!/bin/bash
#SBATCH --nodes=1
#SBATCH --time=00:01:10
#SBATCH --job-name=Gabriel
whoami
hostname
gfortran -fopenmp tugas3.f90
export OMP_NUM_THREADS=512
ulimit -s unlimited
./a.out

6
script-serial.sh → tugas3_16thread.sh
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@ -1,15 +1,13 @@
#!/bin/bash
#SBATCH --nodes=1
#SBATCH --node-list=komputasi06
#SBATCH --time=00:05:00
#SBATCH --job-name=Gabriel
whoami
hostname
gfortran -fopenmp tugas-serial.f
export OMP_NUM_THREADS=20
export OMP_STACKSIZE=32m
gfortran -fopenmp tugas3.f90
export OMP_NUM_THREADS=16
ulimit -s unlimited
./a.out

6
script-parallel.sh → tugas3_1thread.sh
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@ -1,15 +1,13 @@
#!/bin/bash
#SBATCH --nodes=1
#SBATCH --node-list=komputasi06
#SBATCH --time=00:05:00
#SBATCH --job-name=Gabriel
whoami
hostname
gfortran -fopenmp tugas-parallel.f
export OMP_NUM_THREADS=20
export OMP_STACKSIZE=32m
gfortran -fopenmp tugas3.f90
export OMP_NUM_THREADS=1
ulimit -s unlimited
./a.out

13
tugas3_2thread.sh Normal file
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@ -0,0 +1,13 @@
#!/bin/bash
#SBATCH --nodes=1
#SBATCH --time=00:05:00
#SBATCH --job-name=Gabriel
whoami
hostname
gfortran -fopenmp tugas3.f90
export OMP_NUM_THREADS=2
ulimit -s unlimited
./a.out

13
tugas3_32thread.sh Normal file
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@ -0,0 +1,13 @@
#!/bin/bash
#SBATCH --nodes=1
#SBATCH --time=00:05:00
#SBATCH --job-name=Gabriel
whoami
hostname
gfortran -fopenmp tugas3.f90
export OMP_NUM_THREADS=32
ulimit -s unlimited
./a.out

13
tugas3_4thread.sh Normal file
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@ -0,0 +1,13 @@
#!/bin/bash
#SBATCH --nodes=1
#SBATCH --time=00:05:00
#SBATCH --job-name=Gabriel
whoami
hostname
gfortran -fopenmp tugas3.f90
export OMP_NUM_THREADS=4
ulimit -s unlimited
./a.out

13
tugas3_64thread.sh Normal file
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@ -0,0 +1,13 @@
#!/bin/bash
#SBATCH --nodes=1
#SBATCH --time=00:05:00
#SBATCH --job-name=Gabriel
whoami
hostname
gfortran -fopenmp tugas3.f90
export OMP_NUM_THREADS=64
ulimit -s unlimited
./a.out

13
tugas3_8thread.sh Normal file
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@ -0,0 +1,13 @@
#!/bin/bash
#SBATCH --nodes=1
#SBATCH --time=00:05:00
#SBATCH --job-name=Gabriel
whoami
hostname
gfortran -fopenmp tugas3.f90
export OMP_NUM_THREADS=8
ulimit -s unlimited
./a.out

15
with_reduction.f90 Normal file
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program summ
implicit none
integer :: sum = 0
integer :: n
!$OMP parallel do
do n = 0, 1000
sum = sum + n
print*, n, " ", sum
end do
!$OMP end parallel do
print*, " "
print*, " "
print*, " "
print*, "hasilnya adalah ", sum
end program summ

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with_reduction.sh Normal file
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#!/bin/bash
#SBATCH --nodes=1
#SBATCH --time=00:01:10
#SBATCH --job-name=Gabriel
whoami
hostname
gfortran -fopenmp with_reduction.f90 -o with_reduction.x
export OMP_NUM_THREADS=4
./with_reduction.x

BIN
with_reduction.x Executable file
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Binary file not shown.