SscIc0 SscIcSolve SscSsorSolve SsrIc0 SsrIcSolve SsrSsorSolve

SscSsorSolve() Sub SscSsorSolve ( Uplo As  String, N As  Long, Omega As  Double, Val() As  Double, Colptr() As  Long, Rowind() As  Long, D() As  Double, B() As  Double, X() As  Double, Optional Info As  Long, Optional Base As  Long = -1  ) Symmetric successive over-relaxation (SSOR) preconditioner (Symmetric matrix) (CSC) Purpose This routine is the symmetric successive over-relaxation (SSOR) preconditioner for the symmetric coefficient matrix A of the sparse linear equations. It solves the equation M*x = b or M^T*x = b, where M is the preconditioner matrix. Only the upper or lower trianglar part of A is referred for calculation. Parameters [in] Uplo = "U": Matrix A is stored in upper triangle. = "L": Matrix A is stored in lower triangle. [in] N Dimension of preconditioner matrix. (N >= 0) (if N = 0, returns without computation) [in] Omega The relaxation parameter ω. (0 < ω < 2) [in] Val() Array Val(LVal - 1) (LVal >= Nnz) Values of non-zero elements of matrix A. (Nnz is number of non-zero elements) [in] Colptr() Array Colptr(LColptr - 1) (LColptr >= N + 1) Column pointers of matrix A. [in] Rowind() Array Rowind(LRowind - 1) (LRowind >= Nnz) Row indices of matrix A (where Nnz is the number of nonzero elements). [in] D() Array D(LD) (LD >= N) Diagonal elements of preconditioner matrix M obtained by CSC_SSOR(). [in] B() Array B(LB - 1) (LB >= N) Right hand side vector b. [out] X() Array X(LX - 1) (LX >= N) Solution vector x. [out] Info (Optional) = 0: Successful exit. = i < 0: The (-i)-th argument is invalid. = j > 0: Matrix is singular (j-th diagonal element is zero). [in] Base (Optional) Indexing of Colptr() and Rowind(). = 0: Zero-based (C style) indexing: Starting index is 0. = 1: One-based (Fortran style) indexing: Starting index is 1. (default: Assumes 1 if Colptr(0) = 1, 0 otherwise) Example Program Solve the system of linear equations Ax = B by CG method with SSOR preconditioner, where ( 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_Cg_Ssor_Ssc() Const N = 3, Nnz = 6, Omega = 0.9, Tol = 0.0000000001 '1.0e-10 Dim A(Nnz - 1) As Double, Ia(N) As Long, Ja(Nnz - 1) As Long Dim B(N - 1) As Double, X(N - 1) As Double Dim XX(N - 1) As Double, YY(N - 1) As Double Dim Iter As Long, Res As Double, IRev As Long, Info As Long A(0) = 2.2: A(1) = -0.11: A(2) = -0.32: A(3) = 2.93: A(4) = 0.81: A(5) = 2.37 Ia(0) = 0: Ia(1) = 3: Ia(2) = 5: Ia(3) = 6 Ja(0) = 0: Ja(1) = 1: Ja(2) = 2: Ja(3) = 1: Ja(4) = 2: Ja(5) = 2 B(0) = -1.566: B(1) = -2.8425: B(2) = -1.1765 Dim D(Nnz - 1) As Double Call CsxSsor(N, Omega, A(), Ia(), Ja(), D(), Info) If Info <> 0 Then Debug.Print "Ssor Info =" + Str(Info) IRev = 0 Do Call Cg_r(N, B(), X(), Info, XX(), YY(), IRev, Iter, Res) If IRev = 1 Then '- Matvec Call SscDusmv("L", N, 1, A(), Ia(), Ja(), XX(), 0, YY()) ElseIf IRev = 3 Then '- Psolve Call SscSsorSolve("L", N, Omega, A(), Ia(), Ja(), D(), YY(), XX(), Info) If Info <> 0 Then Debug.Print "SsorSolve Info =" + Str(Info) ElseIf IRev = 10 Then '- Check convergence If Res < Tol Then IRev = 11 End If Loop While IRev <> 0 Debug.Print "X =", X(0), X(1), X(2) Debug.Print "Iter = " + CStr(Iter) + ", Res = " + CStr(Res) + ", Info = " + CStr(Info) End Sub Example Results X = -0.8 -0.92 -0.29 Iter = 3, Res = 6.67060401143409E-19, Info = 0