Mnf Mnf_r Mng Mng_r Mnh Mnh_r Optif0 Optif0_r Optif9 Optif9_r Subplex Subplex_r

Optif0() Sub Optif0 ( N As  Long, X() As  Double, F As  LongPtr, Xpls() As  Double, Fpls As  Double, Info As  Long  ) Minimum of a multivariable nonlinear function (quasi-Newton method) (simple driver) Purpose This routine finds the local minimum point (xs1, xs2, ..., xsn) of general nonlinear function f(x1, x2, ..., xn) (a twice continuously differentiable real-valued function). The first order finite difference is used to compute the gradients, and the secant method (BFGS update) is used to compute the Hessian. The steps are computed by the line search. Optif0 is equivalent to using Optif9 with setting default parameters. So that the problem is solved by the quasi-Newton method (BFGS method). Parameters [in] N The order or dimension of the problem. (N > 1) [in] X() Array X(LX - 1) (LX >= N) Initial approximation of the solution vector. [in] F The user-supplied subroutine which calculates the function f(x1, x2, ..., xn) defined as follows. Sub F(N As Long, X() As Double, Fval As Double) Calculate the function value from given N and X() and return in Fval. The other variables should not be altered. End Sub [out] Xpls() Array Xpls(LXpls - 1) (LXpls >= N) Local minimum. [out] Fpls Function value at local minimum Xpls. [out] Info = 0: Successful exit. = -1: The argument N had an illegal value. (N <= 1) = -2: The argument X() is invalid. = -4: The argument Xpls() is invalid. = 1: Last global step failed to locate a point lower than Xpls. Either Xpls is an approximate local minimum of the function, the function is too nonlinear for this algorithm, or Steptl is too large. = 2: Iteration limit (200) exceeded. = 3: Maximum step size Stepmx exceeded five consecutive times. Either the function is unbounded below, becomes asymptotic to a finite value from above in some direction, or Stepmx is too small. Reference CMLIB Example Program Find the minimum point of the following function (Rosenbrock function). f(x1, x2) = 100(x2 - x1^2)^2 + (1 - x1)^2 The initial approximation is (x1, x2) = (-1.2, 1). Sub FOptif(N As Long, X() As Double, F As Double) F = 100 * (X(1) - X(0) ^ 2) ^ 2 + (1 - X(0)) ^ 2 End Sub Sub Ex_Optif0() Const N As Long = 2 Dim X(N - 1) As Double, Xp(N - 1) As Double, Fp As Double Dim Info As Long X(0) = -1.2: X(1) = 1 Call Optif0(N, X(), AddressOf FOptif, Xp(), Fp, Info) Debug.Print "X1, X2 =", Xp(0), Xp(1) Debug.Print "Info =", Info End Sub Example Results X1, X2 = 0.999990556479338 0.999981114653297 Info = 0