LFISF

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Uses iterative refinement to improve the solution of a real symmetric system of linear equations.
Required Arguments
AN by N matrix containing the coefficient matrix of the symmetric linear system. (Input)
Only the upper triangle of A is referenced
FACTN by N matrix containing the factorization of the coefficient matrix A as output from routine LFCSF/DLFCSF or LFTSF/DLFTSF. (Input)
Only the upper triangle of FACT is used.
IPVT — Vector of length N containing the pivoting information for the factorization of A as output from routine LFCSF/DLFCSF or LFTSF/DLFTSF. (Input)
B — Vector of length N containing the right-hand side of the linear system. (Input)
X — Vector of length N containing the solution to the linear system. (Output)
If B is not needed, B and X can share the same storage locations.
RES — Vector of length N containing the residual vector at the improved solution. (Output)
Optional Arguments
N — Number of equations. (Input)
Default: N = size (A,2).
LDA — Leading dimension of A exactly as specified in the dimension statement of the calling program. (Input)
Default: LDA = size (A,1).
LDFACT — Leading dimension of FACT exactly as specified in the dimension statement of the calling program. (Input)
Default: LDFACT = size (FACT,1).
FORTRAN 90 Interface
Generic: CALL LFISF (A, FACT, IPVT, B, X, RES [, …])
Specific: The specific interface names are S_LFISF and D_LFISF.
FORTRAN 77 Interface
Single: CALL LFISF (N, A, LDA, FACT, LDFACT, IPVT, B, X, RES)
Double: The double precision name is DLFISF.
Description
Routine LFISF computes the solution of a system of linear algebraic equations having a real symmetric indefinite coefficient matrix. Iterative refinement is performed on the solution vector to improve the accuracy. Usually almost all of the digits in the solution are accurate, even if the matrix is somewhat ill-conditioned.
To compute the solution, the coefficient matrix must first undergo a U DUT factorization. This may be done by calling either LFCSF or LFTSF.
Iterative refinement fails only if the matrix is very ill-conditioned.
LFISF and LFSSF both solve a linear system given its U DUT factorization. LFISF generally takes more time and produces a more accurate answer than LFSSF. Each iteration of the iterative refinement algorithm used by LFISF calls LFSSF.
Comments
Informational error
Type
Code
Description
3
2
The input matrix is too ill-conditioned for iterative refinement to be effective.
Example
A set of linear systems is solved successively. The right-hand-side vector is perturbed after solving the system each of the first two times by adding 0.2 to the second element.
 
USE LFISF_INT
USE UMACH_INT
USE LFCSF_INT
USE WRRRN_INT
! Declare variables
PARAMETER (LDA=3, N=3)
INTEGER IPVT(N), NOUT
REAL A(LDA,LDA), B(N), X(N), FACT(LDA,LDA), RES(N), RCOND
!
! Set values for A and B
! A = ( 1.0 -2.0 1.0)
! ( -2.0 3.0 -2.0)
! ( 1.0 -2.0 3.0)
!
! B = ( 4.1 -4.7 6.5)
!
DATA A/1.0, -2.0, 1.0, -2.0, 3.0, -2.0, 1.0, -2.0, 3.0/
DATA B/4.1, -4.7, 6.5/
! Factor A and compute the estimate
! of the reciprocal condition number
CALL LFCSF (A, FACT, IPVT, RCOND)
! Print condition number
CALL UMACH (2, NOUT)
WRITE (NOUT,99999) RCOND, 1.0E0/RCOND
! Solve, then perturb right-hand side
DO 10 I=1, 3
CALL LFISF (A, FACT, IPVT, B, X, RES)
! Print results
CALL WRRRN (’X’, X, 1, N, 1)
CALL WRRRN (’RES’, RES, 1, N, 1)
B(2) = B(2) + .20E0
10 CONTINUE
!
99999 FORMAT (’ RCOND = ’,F5.3,/,’ L1 Condition number = ’,F6.3)
END
Output
 
RCOND < 0.035
L1 Condition number < 40.0
 
X
1 2 3
-4.100 -3.500 1.200
RES
1 2 3
-2.384E-07 -2.384E-07 0.000E+00
X
1 2 3
-4.500 -3.700 1.200
RES
1 2 3
-2.384E-07 -2.384E-07 0.000E+00
X
1 2 3
-4.900 -3.900 1.200
RES
1 2 3
-2.384E-07 -2.384E-07 0.000E+00
Published date: 03/19/2020
Last modified date: 03/19/2020