Bicg Bicg1 Bicg_r Bicg_s Cgs Cgs_r Cgs_s Diom Diom_r Diom_s Dqgmres Dqgmres_r Dqgmres_s Fgmres Fgmres_r Fgmres_s Fom Fom_r Fom_s Gcr Gcr_r Gcr_s Gpbicg Gpbicg_r Gpbicg_s Orthomin Orthomin_r Orthomin_s Qmr Qmr_r Qmr_s Sor Sor_r Sor_s Tfqmr Tfqmr_r Tfqmr_s

Bicg1() Sub Bicg1 ( N As  Long, Val() As  Double, Rowptr() As  Long, Colind() As  Long, B() As  Double, X() As  Double, Optional Info As  Long, Optional Iter As  Long, Optional Res As  Double, Optional MaxIter As  Long = 500, Optional Tol As  Double = 1.0E-10  ) Solution of linear system Ax = b using bi-conjugate gradient (BICG) method (Simple driver) Purpose This routine solves the linear system Ax = b using the bi-conjugate gradient (BICG) iterative method. Matrix A is represented in CSR format. Parameters [in] N Dimension of the matrix A. (N >= 0) (if N = 0, returns without computation) [in] Val() Array Val(LVal - 1) (LVal >= Nnz) (Nnz is the number of nonzero elements of the matrix A) Values of nonzero elements of the matrix A. [in] Rowptr() Array Rowptr(LRowptr - 1) (LRowptr >= N + 1) Row pointers of the matrix A. [in] Colind() Array Colind(LColind - 1) (LColind >= Nnz) Column indices of the matrix A. [in] B() Array B(LB - 1) (LB >= N) Right hand side vector b. [in,out] X() Array X(LX - 1) (LX >= N) [in] Initial guess of solution. [out] Obtained approximate solution. [out] Info (Optional) = 0: Successful exit = i < 0: The (-i)-th argument is invalid. = 11: Maximum number of iterations exceeded. = 12: Breakdown occurred. [out] Iter (Optional) Actual number of iterations performed for convergence. [out] Res (Optional) Final residual norm value norm(b - A*x). [in] MaxIter (Optional) Maximum number of iterations. (MaxIter > 0) (default = 500) [in] Tol (Optional) Tolerance for convergence test. (default = 1.0e-10) Assumed to be converged if norm(b - A*x) <= Tol*norm(b). If Tol < eps (machine epsilon), Tol = eps is assumed. Note Indexing of Rowptr() and Ind() is assumed to be zero-based (C style) if Rowptr(0) = 1, or one-based (Fortran style) otherwise. Example Program Solve the system of linear equations Ax = B, where ( 0.2 -0.11 -0.93 ) ( -0.3727 ) A = ( -0.32 0.81 0.37 ), B = ( 0.4319 ) ( -0.8 -0.92 -0.29 ) ( -1.4247 ) Sub Ex_Bicg1() Const N = 3, Nnz = N * N 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 Iter As Long, Res As Double, Info As Long A(0) = 0.2: A(1) = -0.11: A(2) = -0.93: A(3) = -0.32: A(4) = 0.81: A(5) = 0.37: A(6) = -0.8: A(7) = -0.92: A(8) = -0.29 Ia(0) = 0: Ia(1) = 3: Ia(2) = 6: Ia(3) = 9 Ja(0) = 0: Ja(1) = 1: Ja(2) = 2: Ja(3) = 0: Ja(4) = 1: Ja(5) = 2: Ja(6) = 0: Ja(7) = 1: Ja(8) = 2 B(0) = -0.3727: B(1) = 0.4319: B(2) = -1.4247 Call Bicg1(N, A(), Ia(), Ja(), B(), X(), Info, Iter, Res) 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 = 0.86 0.640000000000001 0.509999999999996 Iter = 3, Res = 1.86243126342707E-15, Info = 0