Dpocon Dpotrf Dpotri Dpotrs Dppcon Dpptrf Dpptri Dpptrs Dsposv

Dsposv() Sub Dsposv ( Uplo As  String, N As  Long, A() As  Double, B() As  Double, X() As  Double, Iter As  Long, Info As  Long, Optional Nrhs As  Long = 1  ) (Simple driver) Solution to system of linear equations AX = B for a symmetric positive definite matrix (mixed precision with iterative refinement) Purpose This routine computes the solution to a real system of linear equations A * X = B, where A is an n x n symmetric positive definite matrix and X and B are n x nrhs matrices. Dsposv first attempts to factorize the matrix in single precision and use this factorization within an iterative refinement procedure to produce a solution with double precision normwise backward error quality (see below). If the approach fails the method switches to a double precision factorization and solve. The iterative refinement is not going to be a winning strategy if the ratio single precision performance over double precision performance is too small. A reasonable strategy should take the number of right-hand sides and the size of the matrix into account. Up to now, we always try iterative refinement. The iterative refinement process is stopped if   iter > itermax or for all the rhs we have:   rnrm < sqrt(n)*xnrm*anrm*eps*bwdmax where   iter is the number of the current iteration in the iterative refinement process   rnrm is the infinity-norm of the residual   xnrm is the infinity-norm of the solution   anrm is the infinity-operator-norm of the matrix A   eps is the machine epsilon returned by dlamch('E') The value itermax and bwdmax are fixed to 30 and 1.0D+00 respectively. Parameters [in] N Number of linear equations, i.e., order of the matrix A. (N >= 0) (If N = 0, returns without computation) [in,out] A() Array A(LA1 - 1, LA2 - 1) (LA1 >= N, LA2 >= N) [in] N x N symmetric positove definite matrix A. The upper or lower triangular part is to be referenced in accordance with Uplo. [out] If iterative refinement has been successfully used (Info = 0 and Iter >= 0, see description below), then A() is unchanged. If double precision factorization has been used (Info = 0 and Iter < 0, see description below), then the array A() contains the factor U or L from the Cholesky factorization A = U^T*U or A = L*L^T. [in] B() Array B(LB1 - 1, LB2 - 1) (LB1 >= max(1, N), LB2 >= Nrhs) (2D array) or B(LB - 1) (LB >= max(1, N), Nrhs = 1) (1D array) N x Nrhs matrix of right hand side matrix B. [out] X() Array X(LX1 - 1, LX2 - 1) (LX1 >= max(1, N), LX2 >= Nrhs) (2D array) or X(LX - 1) (LX >= max(1, N), Nrhs = 1) (1D array) If Info = 0, the N x Nrhs solution matrix X. [out] Iter < 0: Iterative refinement has failed, double precision factorization has been performed.   = -1: The routine fell back to full precision for implementation- or machine-specific reasons.   = -2: Narrowing the precision induced an overflow, the routine fell back to full precision.   = -3: Failure of Sgetrf.   = -31: Stop the iterative refinement after the 30th iterations. > 0: Iterative refinement has been successfully used. Returns the number of iterations. [out] Info = 0: Successful exit. = -2: The argument N had an illegal value. (N < 0) = -3: The argument A() is invalid. = -4: The argument B() is invalid. = -5: The argument X() is invalid. = -8: The argument Nrhs had an illegal value. (Nrhs < 0) = i > 0: The i-th diagonal element of the factor U computed in double precision is exactly zero. The factorization has been completed, but the factor U is exactly singular, so the solution could not be computed. [in] Nrhs (Optional) Number of right hand sides, i.e., number of columns of the matrix B. (Nrhs >= 0) (If Nrhs = 0, returns without computation) (default = 1) Reference LAPACK Example Program Solve the system of linear equations Ax = B , where A is symmetric positive definite and ( 2.2 -0.11 -0.32 ) ( -1.566 ) A = ( -0.11 2.93 0.81 ), B = ( -2.8425 ) ( -0.32 0.81 2.37 ) ( -1.1765 ) Sub Ex_Dsposv() Const N = 3 Dim A(N - 1, N - 1) As Double, B(N - 1) As Double, X(N - 1) As Double Dim Iter As Long, Info As Long A(0, 0) = 2.2 A(1, 0) = -0.11: A(1, 1) = 2.93 A(2, 0) = -0.32: A(2, 1) = 0.81: A(2, 2) = 2.37: B(0) = -1.566: B(1) = -2.8425: B(2) = -1.1765 Call Dsposv("L", N, A(), B(), X(), Iter, Info) Debug.Print "X =", X(0), X(1), X(2) Debug.Print "Iter =", Iter, "Info =", Info End Sub Example Results X = -0.8 -0.92 -0.29 Iter = 2 Info = 0