sptrfs()

subroutine sptrfs	(	integer	n,
		integer	nrhs,
		real, dimension( * )	d,
		real, dimension( * )	e,
		real, dimension( * )	df,
		real, dimension( * )	ef,
		real, dimension( ldb, * )	b,
		integer	ldb,
		real, dimension( ldx, * )	x,
		integer	ldx,
		real, dimension( * )	ferr,
		real, dimension( * )	berr,
		real, dimension( * )	work,
		integer	info
	)

SPTRFS

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Purpose:

 SPTRFS improves the computed solution to a system of linear
 equations when the coefficient matrix is symmetric positive definite
 and tridiagonal, and provides error bounds and backward error
 estimates for the solution.

Parameters

[in]	N	N is INTEGER The order of the matrix A. N >= 0.
[in]	NRHS	NRHS is INTEGER The number of right hand sides, i.e., the number of columns of the matrix B. NRHS >= 0.
[in]	D	D is REAL array, dimension (N) The n diagonal elements of the tridiagonal matrix A.
[in]	E	E is REAL array, dimension (N-1) The (n-1) subdiagonal elements of the tridiagonal matrix A.
[in]	DF	DF is REAL array, dimension (N) The n diagonal elements of the diagonal matrix D from the factorization computed by SPTTRF.
[in]	EF	EF is REAL array, dimension (N-1) The (n-1) subdiagonal elements of the unit bidiagonal factor L from the factorization computed by SPTTRF.
[in]	B	B is REAL array, dimension (LDB,NRHS) The right hand side matrix B.
[in]	LDB	LDB is INTEGER The leading dimension of the array B. LDB >= max(1,N).
[in,out]	X	X is REAL array, dimension (LDX,NRHS) On entry, the solution matrix X, as computed by SPTTRS. On exit, the improved solution matrix X.
[in]	LDX	LDX is INTEGER The leading dimension of the array X. LDX >= max(1,N).
[out]	FERR	FERR is REAL array, dimension (NRHS) The forward error bound for each solution vector X(j) (the j-th column of the solution matrix X). If XTRUE is the true solution corresponding to X(j), FERR(j) is an estimated upper bound for the magnitude of the largest element in (X(j) - XTRUE) divided by the magnitude of the largest element in X(j).
[out]	BERR	BERR is REAL array, dimension (NRHS) The componentwise relative backward error of each solution vector X(j) (i.e., the smallest relative change in any element of A or B that makes X(j) an exact solution).
[out]	WORK	WORK is REAL array, dimension (2*N)
[out]	INFO	INFO is INTEGER = 0: successful exit < 0: if INFO = -i, the i-th argument had an illegal value

Internal Parameters:

  ITMAX is the maximum number of steps of iterative refinement.

Author

Univ. of Tennessee

Univ. of California Berkeley

Univ. of Colorado Denver

NAG Ltd.

Definition at line 161 of file sptrfs.f.

*
*  -- LAPACK computational routine --
*  -- LAPACK is a software package provided by Univ. of Tennessee,    --
*  -- Univ. of California Berkeley, Univ. of Colorado Denver and NAG Ltd..--
*
*     .. Scalar Arguments ..
      INTEGER            INFO, LDB, LDX, N, NRHS
*     ..
*     .. Array Arguments ..
      REAL               B( LDB, * ), BERR( * ), D( * ), DF( * ),
     $                   E( * ), EF( * ), FERR( * ), WORK( * ),
     $                   X( LDX, * )
*     ..
*
*  =====================================================================
*
*     .. Parameters ..
      INTEGER            ITMAX
      parameter( itmax = 5 )
      REAL               ZERO
      parameter( zero = 0.0e+0 )
      REAL               ONE
      parameter( one = 1.0e+0 )
      REAL               TWO
      parameter( two = 2.0e+0 )
      REAL               THREE
      parameter( three = 3.0e+0 )
*     ..
*     .. Local Scalars ..
      INTEGER            COUNT, I, IX, J, NZ
      REAL               BI, CX, DX, EPS, EX, LSTRES, S, SAFE1, SAFE2,
     $                   SAFMIN
*     ..
*     .. External Subroutines ..
      EXTERNAL           saxpy, spttrs, xerbla
*     ..
*     .. Intrinsic Functions ..
      INTRINSIC          abs, max
*     ..
*     .. External Functions ..
      INTEGER            ISAMAX
      REAL               SLAMCH
      EXTERNAL           isamax, slamch
*     ..
*     .. Executable Statements ..
*
*     Test the input parameters.
*
      info = 0
      IF( n.LT.0 ) THEN
         info = -1
      ELSE IF( nrhs.LT.0 ) THEN
         info = -2
      ELSE IF( ldb.LT.max( 1, n ) ) THEN
         info = -8
      ELSE IF( ldx.LT.max( 1, n ) ) THEN
         info = -10
      END IF
      IF( info.NE.0 ) THEN
         CALL xerbla( 'SPTRFS', -info )
         RETURN
      END IF
*
*     Quick return if possible
*
      IF( n.EQ.0 .OR. nrhs.EQ.0 ) THEN
         DO 10 j = 1, nrhs
            ferr( j ) = zero
            berr( j ) = zero
   10    CONTINUE
         RETURN
      END IF
*
*     NZ = maximum number of nonzero elements in each row of A, plus 1
*
      nz = 4
      eps = slamch( 'Epsilon' )
      safmin = slamch( 'Safe minimum' )
      safe1 = nz*safmin
      safe2 = safe1 / eps
*
*     Do for each right hand side
*
      DO 90 j = 1, nrhs
*
         count = 1
         lstres = three
   20    CONTINUE
*
*        Loop until stopping criterion is satisfied.
*
*        Compute residual R = B - A * X.  Also compute
*        abs(A)*abs(x) + abs(b) for use in the backward error bound.
*
         IF( n.EQ.1 ) THEN
            bi = b( 1, j )
            dx = d( 1 )*x( 1, j )
            work( n+1 ) = bi - dx
            work( 1 ) = abs( bi ) + abs( dx )
         ELSE
            bi = b( 1, j )
            dx = d( 1 )*x( 1, j )
            ex = e( 1 )*x( 2, j )
            work( n+1 ) = bi - dx - ex
            work( 1 ) = abs( bi ) + abs( dx ) + abs( ex )
            DO 30 i = 2, n - 1
               bi = b( i, j )
               cx = e( i-1 )*x( i-1, j )
               dx = d( i )*x( i, j )
               ex = e( i )*x( i+1, j )
               work( n+i ) = bi - cx - dx - ex
               work( i ) = abs( bi ) + abs( cx ) + abs( dx ) + abs( ex )
   30       CONTINUE
            bi = b( n, j )
            cx = e( n-1 )*x( n-1, j )
            dx = d( n )*x( n, j )
            work( n+n ) = bi - cx - dx
            work( n ) = abs( bi ) + abs( cx ) + abs( dx )
         END IF
*
*        Compute componentwise relative backward error from formula
*
*        max(i) ( abs(R(i)) / ( abs(A)*abs(X) + abs(B) )(i) )
*
*        where abs(Z) is the componentwise absolute value of the matrix
*        or vector Z.  If the i-th component of the denominator is less
*        than SAFE2, then SAFE1 is added to the i-th components of the
*        numerator and denominator before dividing.
*
         s = zero
         DO 40 i = 1, n
            IF( work( i ).GT.safe2 ) THEN
               s = max( s, abs( work( n+i ) ) / work( i ) )
            ELSE
               s = max( s, ( abs( work( n+i ) )+safe1 ) /
     $             ( work( i )+safe1 ) )
            END IF
   40    CONTINUE
         berr( j ) = s
*
*        Test stopping criterion. Continue iterating if
*           1) The residual BERR(J) is larger than machine epsilon, and
*           2) BERR(J) decreased by at least a factor of 2 during the
*              last iteration, and
*           3) At most ITMAX iterations tried.
*
         IF( berr( j ).GT.eps .AND. two*berr( j ).LE.lstres .AND.
     $       count.LE.itmax ) THEN
*
*           Update solution and try again.
*
            CALL spttrs( n, 1, df, ef, work( n+1 ), n, info )
            CALL saxpy( n, one, work( n+1 ), 1, x( 1, j ), 1 )
            lstres = berr( j )
            count = count + 1
            GO TO 20
         END IF
*
*        Bound error from formula
*
*        norm(X - XTRUE) / norm(X) .le. FERR =
*        norm( abs(inv(A))*
*           ( abs(R) + NZ*EPS*( abs(A)*abs(X)+abs(B) ))) / norm(X)
*
*        where
*          norm(Z) is the magnitude of the largest component of Z
*          inv(A) is the inverse of A
*          abs(Z) is the componentwise absolute value of the matrix or
*             vector Z
*          NZ is the maximum number of nonzeros in any row of A, plus 1
*          EPS is machine epsilon
*
*        The i-th component of abs(R)+NZ*EPS*(abs(A)*abs(X)+abs(B))
*        is incremented by SAFE1 if the i-th component of
*        abs(A)*abs(X) + abs(B) is less than SAFE2.
*
         DO 50 i = 1, n
            IF( work( i ).GT.safe2 ) THEN
               work( i ) = abs( work( n+i ) ) + nz*eps*work( i )
            ELSE
               work( i ) = abs( work( n+i ) ) + nz*eps*work( i ) + safe1
            END IF
   50    CONTINUE
         ix = isamax( n, work, 1 )
         ferr( j ) = work( ix )
*
*        Estimate the norm of inv(A).
*
*        Solve M(A) * x = e, where M(A) = (m(i,j)) is given by
*
*           m(i,j) =  abs(A(i,j)), i = j,
*           m(i,j) = -abs(A(i,j)), i .ne. j,
*
*        and e = [ 1, 1, ..., 1 ]**T.  Note M(A) = M(L)*D*M(L)**T.
*
*        Solve M(L) * x = e.
*
         work( 1 ) = one
         DO 60 i = 2, n
            work( i ) = one + work( i-1 )*abs( ef( i-1 ) )
   60    CONTINUE
*
*        Solve D * M(L)**T * x = b.
*
         work( n ) = work( n ) / df( n )
         DO 70 i = n - 1, 1, -1
            work( i ) = work( i ) / df( i ) + work( i+1 )*abs( ef( i ) )
   70    CONTINUE
*
*        Compute norm(inv(A)) = max(x(i)), 1<=i<=n.
*
         ix = isamax( n, work, 1 )
         ferr( j ) = ferr( j )*abs( work( ix ) )
*
*        Normalize error.
*
         lstres = zero
         DO 80 i = 1, n
            lstres = max( lstres, abs( x( i, j ) ) )
   80    CONTINUE
         IF( lstres.NE.zero )
     $      ferr( j ) = ferr( j ) / lstres
*
   90 CONTINUE
*
      RETURN
*
*     End of SPTRFS
*

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