| 查看: 388 | 回复: 1 | |||
| 当前主题已经存档。 | |||
| 【有奖交流】积极回复本帖子，参与交流，就有机会分得作者 futurehero 的 100 个金币 | |||
futurehero新虫 (正式写手)
|
[交流]
【求助】诚求助并行计算中的一个问题，谢谢大家！
|
||
|
这个程序大致如下，我想用并行计算解决“do循环中包含积分”的计算问题。如果将并行的线程设为1，则和串行计算的结果一直，但是线程大于1时，计算的结果就不正确了。小弟刚刚接触并行计算两天，做了很多种方案都发现线程大于1时计算的结果都不对，请诸位大侠帮助，我想得到争取的并行计算结果。如果将程序中的call omp_set_num_threads(4 )线程改为call omp_set_num_threads(1 )就和串行结果是一致的。(可能您只需要看一下并行区就可以发现问题所在了) 程序如下： program main implicit none integer:: Num, i, p, j, k ,var, r ,Y real(kind=8):: D, epsl21, epsl22, T, eta, Gama, GamaL, GamaR, epsl0, VL, VR parameter(D=1.0_8, Num=10001, Gama=0.01_8, GamaL=0.005_8, GamaR=0.005_8, epsl0=-0.1_8, VL=0.0_8, VR= 0.0_8) real(kind=8):: pi, Dc, Hc, fL(Num), fR(Num), f0(Num), epsl(Num), epsl11, epsl12, dIdV(Num) complex(kind=8):: sigma0, FG11(Num), FG12(Num), FG21(Num), FG22(Num) complex(kind=8):: G011(Num), G012(Num), G021(Num), G022(Num), G11(Num), G12(Num), G21(Num), G22(Num) complex(kind=8):: DDOS(Num), DOS, D11, D12, D21, D22 complex(kind=8):: n11, n12, n21, n22 complex(kind=8):: delta1211(Num), delta2111(Num), delta2211(Num) complex(kind=8):: delta1112(Num), delta2112(Num), delta2212(Num) complex(kind=8):: delta1121(Num), delta1221(Num), delta2221(Num) complex(kind=8):: delta1122(Num), delta1222(Num), delta2122(Num) complex(kind=8):: FZB11(Num), FZB12(Num), FZB21(Num), FZB22(Num) complex(kind=8):: B1211(Num), B2111(Num), B2211(Num) complex(kind=8):: FB1211(Num), FB2111(Num), FB2211(Num) complex(kind=8):: B1112(Num), B2112(Num), B2212(Num) complex(kind=8):: FB1112(Num), FB2112(Num), FB2212(Num) complex(kind=8):: B1121(Num), B1221(Num), B2221(Num) complex(kind=8):: FB1121(Num), FB1221(Num), FB2221(Num) complex(kind=8):: B1222(Num), B2122(Num), B1122(Num) complex(kind=8):: FB1222(Num), FB2122(Num), FB1122(Num) real(kind=8):: m, Field, SOC, Tk Dc=2*D / (Num-1) eta =2.0*Dc pi=4.0_8*datan(1.0_8) m=0.00_8 T=0.0_8 SOC=0.00000_8 Field=0.00000_8 epsl11=epsl0 - SOC/2.0_8 - m*Field - Field epsl12=epsl0 + SOC/2.0_8 + m*Field - Field epsl21=epsl0 + SOC/2.0_8 - m*Field + Field epsl22=epsl0 - SOC/2.0_8 + m*Field + Field do i=1, Num, 1 epsl(i)=-D + (i-1) *Dc end do do i=1, Num, 1 if( epsl(i) < VL ) fL(i)=1.0_8 if( epsl(i) > VL ) fL(i)=0.0_8 if( dabs(epsl(i)-VL) < 0.01_8*Dc ) fL(i)=0.5_8 if( epsl(i) < VR ) fR(i)=1.0_8 if( epsl(i) > VR ) fR(i)=0.0_8 if( dabs(epsl(i)-VR) < 0.01_8*Dc ) fR(i)=0.5_8 end do f0=( fL*GamaL + fR*GamaR ) /Gama delta1211=dcmplx( epsl12 - epsl11, eta ) - epsl delta2111=dcmplx( epsl21 - epsl11, eta ) - epsl delta2211=dcmplx( epsl22 - epsl11, eta ) - epsl delta1112=dcmplx( epsl11 - epsl12, eta ) - epsl delta2112=dcmplx( epsl21 - epsl12, eta ) - epsl delta2212=dcmplx( epsl22 - epsl12, eta ) - epsl delta1121=dcmplx( epsl11 - epsl21, eta ) - epsl delta1221=dcmplx( epsl12 - epsl21, eta ) - epsl delta2221=dcmplx( epsl22 - epsl21, eta ) - epsl delta1122=dcmplx( epsl11 - epsl22, eta ) - epsl delta1222=dcmplx( epsl12 - epsl22, eta ) - epsl delta2122=dcmplx( epsl21 - epsl22, eta ) - epsl !================================ sigma0=dcmplx(0.0_8, -Gama) G011 =epsl-epsl11-sigma0 G012 =epsl-epsl12-sigma0 G021 =epsl-epsl21-sigma0 G022 =epsl-epsl22-sigma0 !******************************************** call omp_set_num_threads(4 ) !$OMP parallel !$OMP do private (Y) do Y=1, Num, 1 FB1211=f0/( epsl(Y) + delta1211 ) FB2111=f0/( epsl(Y) + delta2111 ) FB2211=f0/( epsl(Y) + delta2211 ) FB1112=f0/( epsl(Y) + delta1112 ) FB2112=f0/( epsl(Y) + delta2112 ) FB2212=f0/( epsl(Y) + delta2212 ) FB1121=f0/( epsl(Y) + delta1121 ) FB1221=f0/( epsl(Y) + delta1221 ) FB2221=f0/( epsl(Y) + delta2221 ) FB1122=f0/( epsl(Y) + delta1122 ) FB1222=f0/( epsl(Y) + delta1222 ) FB2122=f0/( epsl(Y) + delta2122 ) call INTEGRAL(Num, Dc, FB1211, B1211(Y)) call INTEGRAL(Num, Dc, FB2111, B2111(Y)) call INTEGRAL(Num, Dc, FB2211, B2211(Y)) call INTEGRAL(Num, Dc, FB1112, B1112(Y)) call INTEGRAL(Num, Dc, FB2112, B2112(Y)) call INTEGRAL(Num, Dc, FB2212, B2212(Y)) call INTEGRAL(Num, Dc, FB1121, B1121(Y)) call INTEGRAL(Num, Dc, FB1221, B1221(Y)) call INTEGRAL(Num, Dc, FB2221, B2221(Y)) call INTEGRAL(Num, Dc, FB1122, B1122(Y)) call INTEGRAL(Num, Dc, FB1222, B1222(Y)) call INTEGRAL(Num, Dc, FB2122, B2122(Y)) end do !$OMP end do !$OMP end parallel !****************************************** B1211=B1211*Gama/pi B2111=B2111*Gama/pi B2211=B2211*Gama/pi B1112=B1112*Gama/pi B2112=B2112*Gama/pi B2212=B2212*Gama/pi B1121=B1121*Gama/pi B1221=B1221*Gama/pi B2221=B2221*Gama/pi B1122=B1122*Gama/pi B1222=B1222*Gama/pi B2122=B2122*Gama/pi G11= ( 1.0_8 - 3.0_8*0.245_8 )/( G011 - B1211-B2111-B2211 ) G12= ( 1.0_8 - 3.0_8*0.245_8 )/( G012 - B1112-B2112-B2212 ) G21= ( 1.0_8 - 3.0_8*0.245_8 )/( G021 - B1121-B1221-B2221 ) G22= ( 1.0_8 - 3.0_8*0.245_8 )/( G022 - B1122-B1222-B2122 ) FG11=-f0*G11/pi FG12=-f0*G12/pi FG21=-f0*G21/pi FG22=-f0*G22/pi call INTEGRAL(Num, Dc, FG11, n11) call INTEGRAL(Num, Dc, FG12, n12) call INTEGRAL(Num, Dc, FG21, n21) call INTEGRAL(Num, Dc, FG22, n22) write(*,*) "n11=", dimag(n11), "n12=", dimag(n12) write(*,*) "n21=", dimag(n21), "n22=", dimag(n22) stop end program main subroutine INTEGRAL(Num, Dc, Y, answer) implicit none integer:: Num real(kind=8):: Dc complex(kind=8):: Y(Num), answer integer:: i complex:: s1 s1=(Y(1)+ Y(Num))/2 do i=2, Num-1, 1 s1=s1+ Y(i) end do answer=s1*Dc return end subroutine INTEGRAL stop end program main [ Last edited by futurehero on 2010-3-4 at 23:09 ] |
回复此楼» 猜你喜欢
请问有评职称，把科研教学业绩算分排序的高校吗
已经有6人回复
-
2025冷门绝学什么时候出结果
已经有6人回复
-
Bioresource Technology期刊，第一次返修的时候被退回好几次了
已经有7人回复
-
真诚求助：手里的省社科项目结项要求主持人一篇中文核心，有什么渠道能发核心吗
已经有8人回复
-
寻求一种能扛住强氧化性腐蚀性的容器密封件
已经有5人回复
-
请问哪里可以有青B申请的本子可以借鉴一下。
已经有4人回复
-
请问下大家为什么这个铃木偶联几乎不反应呢
已经有5人回复
-
天津工业大学郑柳春团队欢迎化学化工、高分子化学或有机合成方向的博士生和硕士生加入
已经有4人回复
-
康复大学泰山学者周祺惠团队招收博士研究生
已经有6人回复
-
AI论文写作工具：是科研加速器还是学术作弊器？
已经有3人回复
futurehero
新虫 (正式写手)
- 应助: 0 (幼儿园)
- 金币: 22
- 散金: 464
- 红花: 1
- 帖子: 350
- 在线: 47.7小时
- 虫号: 250574
- 注册: 2006-05-13
- 性别: GG
- 专业: 凝聚态物性 II ：电子结构
2楼2010-03-20 16:08:46