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这是我的代码

program load_flow


!!!!!!!region starts 
!$acc region
p1=0;p=0
do i=2,n
    do j=1,n
        p1=(V(i)*V(j)*Yabs(i,j)*cos(angle(i,j)-d(i)+d(j)))
        p=p1+p
     end do
     P0(i-1)=p
     p=0
end do

! % % % %------Reactive Power Calculation-----%  
p=0;p1=0

do i=2,(n-m)
      do j=1,n
p1=-(V(i)*V(j)*Yabs(i,j)*sin(angle(i,j)-d(i)+d(j)))
p=p1+p
      end do
      Q0(i-1)=p
      p=0
end do

!!!!!!!!!!!mismatch factor
do i=1,(n-1)
   B1(i)=Ps(b,i)-P0(i)
  end do
do i=1,(n-m-1)
   B2(i)=Qs(b,i)-Q0(i)
end do

MF(1,1:38)=B1(1:38)
MF(1,39:67)=B2(1:29)
!!!!!!!!jacobian calculation for preddictor step
!!!!!!!!!!!!!!!!!!!!!!dia of j1
p=0;p1=0
do i=2,n
    do j=1,n
       if(j .ne. i)then
 p1=V(i)*V(j)*Yabs(i,j)*sin(angle(i,j)-d(i)+d(j))

 p=p1+p
       end if

    end do
    i=i-1
    dia1(i,i)=p

    p=0
    i=i+1
end do

!!!!!!!!!!!!!!off dia. of j1
q=0;q1=0

do k=2,n
 i=k
    do j=2,n
        if(j .ne. i)then
        q1=V(i)*V(j)*Yabs(i,j)*sin(angle(i,j)-d(i)+d(j))
        end if
        i=i-1;j=j-1
offdia1(i,j)=-q1
q1=0
i=i+1;j=j+1
    end do
end do

do i=1,38
do j=1,38
J1(i,j)=offdia1(i,j)+dia1(i,j)
end do
end do
!!!!!!!!!!!!!!!!!!!dia. of j2
p=0;p1=0
do i=2,(n-m)
    do j=1,n
       if(j .ne. i)then
         p1=V(j)*Yabs(i,j)*cos(angle(i,j)-d(i)+d(j))
 p=p1+p
       end if

    end do

    dia2(i-1,i-1)=p+(2*V(i)*Yabs(i,i)*cos(angle(i,i)))
    p=0;

end do
!!!!!!!!!!!!!!!!!!off dia. of j2
p1=0;

do k=2,n
 i=k
    do j=2,(n-m)
        if(j .ne. i)then
       p1=V(i)*Yabs(i,j)*cos(angle(i,j)-d(i)+d(j));
        end if
        i=i-1;j=j-1
offdia2(i,j)=p1
    p1=0;
    i=i+1;j=j+1
    end do
end do

do i=1,(n-m-1)

   offdia2(i,i)=dia2(i,i)

end do
J2=offdia2
!!!!!!!!!!!!!!!!!!!!dia. of j3
p=0;p1=0
do i=2,(n-m)
    do j=1,n
       if(j .ne. i)then
          p1=V(i)*V(j)*Yabs(i,j)*cos(angle(i,j)-d(i)+d(j))
 p=p1+p;
       end if
    end do
    i=i-1;
    dia3(i,i)=p
    p=0;
    i=i+1;
 end do
 !!!!!!!!!!!!!!off dia of j3
p=0;p1=0

do k=2,(n-m)
 i=k;
    do j=2,n
        if(j .ne. i)then
    p1=V(i)*V(j)*Yabs(i,j)*cos(angle(i,j)-d(i)+d(j))
        end if
     i=i-1;j=j-1
    offdia3(i,j)=-p1;
    p1=0;
    i=i+1;j=j+1
    end do
end do

do i=1,(n-m-1)
      offdia3(i,i)=dia3(i,i)

end do
J3=offdia3
!!!!!!!!!!dia of j4
p=0;p1=0
do i=2,(n-m)
    do j=1,n
       if(j .ne. i)then
           p1=V(j)*Yabs(i,j)*sin(angle(i,j)-d(i)+d(j))
 p=p1+p
       end if

    end do

    dia4(i-1,i-1)=-(2*V(i)*Yabs(i,i)*sin(angle(i,i)))-p
    p=0;p1=0
end do
 !!!!!!!!!!!!!!!off dia of j4
p1=0;p=0

do k=2,(n-m)
 i=k;
    do j=2,(n-m)
        if(j .ne. i)then
       p1=V(i)*Yabs(i,j)*sin(angle(i,j)-d(i)+d(j))
        end if
    i=i-1;j=j-1
    offdia4(i,j)=-p1
    p1=0;
    i=i+1;j=j+1
    end do
 end do

 do i=1,(n-m-1)
offdia4(i,i)=dia4(i,i);
 end do

J4=offdia4
!!!!!!!
!!!!!!!!!!!!!!!!!!!formation of final jacobian!!!!!!!!!!
Jac( 1:38, 1:38) = J1 (1:38,1:38)
Jac( 1:38,39:67) = J2 (1:38,1:29)
Jac(39:67, 1:38) = J3 (1:29,1:38)
Jac(39:67,39:67) = J4 (1:29,1:29)
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
!!!!!!!!!!region ends
!$acc end region

end program load_flow

我正在使用 PGI 加速器指令进行并行化，但它显示

“编译器无法转换加速器区域（请参阅 -Minfo 消息）：具有零步长值的感应变量”

4

2 回答 2

1

不相信这会奏效，但它应该是朝着正确方向迈出的一步：

!!!!!!!region starts 
p1=0;p=0
!$acc region do
do i=2,n
    do j=1,n
        p1=(V(i)*V(j)*Yabs(i,j)*cos(angle(i,j)-d(i)+d(j)))
        p=p1+p
     end do
     P0(i-1)=p
     p=0
end do
于 2012-03-19T03:54:55.653 回答
1

尝试更改它以在外部循环的顶部初始化 p=0：

!!!!!!!region 
!$acc region do
do i=2,n
    p=0
    do j=1,n
        p1=(V(i)*V(j)*Yabs(i,j)*cos(angle(i,j)-d(i)+d(j)))
        p=p1+p
     end do
     P0(i-1)=p
end do
于 2013-03-22T22:12:04.873 回答