LU分解FORTRANプログラムNO.3
2003/08/18 日立製作所 & 早稲田大学 後 保範 ( Ushiro Yasunori )
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1. 概要
実密行列を係数とする連立一次方程式Ax=bをガウス消去法でLU分解し、
LUx=bの数値解を前進後退代入で求める。
部分軸交換付きで軸交換対象行は全て交換するプログラム。
2. プログラム
C=================================================================C
SUBROUTINE GLU3(A,B,N,ND,EPS,IP,IER)
C=================================================================C
C Real-Dense LU Decomposition by Gauss Elimination C
C and Solve Ax=b by Substitution C
C A ---> LU Decomposition with Partial Pivoting C
C Type-1 Partial Pivoting (Changing all rows) C
C-----------------------------------------------------------------C
C A(ND,N) R*8, I/O, A Coefficient Matrix C
C B(N) R*8, I/O, A right-hand vector(b) and solution(x) C
C N I*4, In, Matrix Size of A C
C ND I*4, In, Array Size of A ( ND >= N ) C
C EPS R*8, In, Value for Singularity Check C
C IP(N) I*4, Out, Pivot Number C
C IER I*4, Out, 0 : Normal Execution C
C 1 : Singular Stop C
C 2 ; Parameter Error C
C-----------------------------------------------------------------C
C Written by Yasunori Ushiro, 2003/08/17 C
C ( Hitachi Ltd. and Waseda University ) C
C=================================================================C
IMPLICIT REAL*8(A-H,O-Z)
DIMENSION A(ND,N), B(N), IP(N)
C----- Gauss Elimination Step ------------
C Check Parameter
IER = 0
if(ND.lt.N) then
IER = 2
go to 100
end if
C Gauss Elimination
do k=1,N
C Search Maximum Value in k's column
KPIV = k
PIV = abs(A(k,k))
do i=k+1,N
if(abs(A(i,k)).gt.PIV) then
KPIV = i
PIV = abs(A(i,k))
end if
end do
C Check Singularity
if(PIV.lt.EPS) then
IER = 1
go to 100
end if
IP(k) = KPIV
C Change A(k,*) <--> A(KPIV,*), All rows
if(KPIV.ne.k) then
do j=1,N
AKJ = A(k,j)
A(k,j) = A(KPIV,j)
A(KPIV,j) = AKJ
end do
end if
C Pivot Value
DPIV = 1.0/A(k,k)
A(k,k) = DPIV
C A(*,k)=A(*,k)/A(k,k)
do i=k+1,N
A(i,k) = A(i,k)*DPIV
end do
C Main Elimination
do j=k+1,N
AKJ = A(k,j)
do i=k+1,N
A(i,j) = A(i,j) - AKJ*A(i,k)
end do
end do
end do
C---- Solve LUx=b by Substitution -------------
C Interchange Entries of B
do k=1,N-1
i = IP(k)
BW = B(k)
B(k) = B(i)
B(i) = BW
end do
C Forward Substitution
do j=1,N-1
do i=j+1,N
B(i) = B(i) - B(j)*A(i,j)
end do
end do
C Back Substitution
do j=N,1,-1
B(j) = B(j)*A(j,j)
do i=1,j-1
B(i) = B(i) - A(i,j)*B(j)
end do
end do
C
100 continue
C
RETURN
END