VB Nonlinear Programming Example

Imports System

Imports CenterSpace.NMath.Core


Namespace CenterSpace.NMath.Examples.VisualBasic

  A .NET example in Visual Basic
  Module NonlinearProgrammingExample

    Sub Main()

      Dim Objective As DoubleFunctional = New ObjectiveFunction()
      Dim Problem As New NonlinearProgrammingProblem(Objective)

      Dimensionality of the objective and constraint function domains
      Dim xDimension As Integer = 2

      Add constraint x0*x1 - x0 -x1 <= -1.5
      Problem.AddUpperBoundConstraint(xDimension, AddressOf Constraint1, -1.5)

      Add constraint x0*x1 >= -10
      Problem.AddLowerBoundConstraint(xDimension, AddressOf Constraint2, -10.0)

      Set some options on the solver by creating an Options object and setting
      some properties, then use it to construct the solver instance.
      Dim solveroptions As New ActiveSetLineSearchSQP.Options()
      solveroptions.StepDirectionTolerance = 0.00000001
      solveroptions.FunctionChangeTolerance = 0.000001
      Since our constraints are nearly linear (one quadratic term) well
      use the simpler constant step size.
      solveroptions.StepSizeCalculator = New ConstantSQPStepSize(1)

      Create the solver with the above options and use it to solve the problem.
      Dim Solver As New ActiveSetLineSearchSQP(solveroptions)

      Initial solution guess
      Dim x0 As New DoubleVector(-1.0, 1.0)
      Dim success As Boolean = Solver.Solve(Problem, x0)

      Console.WriteLine()

      Console.WriteLine(If(success, "Solver was successful", "Solver encountered a problem."))
      Console.WriteLine("Termination status = " & Solver.SolverTerminationStatus.ToString())
      Console.WriteLine("X = " & Solver.OptimalX.ToString())
      Console.WriteLine("f(x) = " & Solver.OptimalObjectiveFunctionValue)
      Console.WriteLine("Iterations = " & Solver.Iterations)

      Console.WriteLine()
      Console.WriteLine("Press Enter Key")
      Console.Read()

    End Sub

    Function Constraint1(ByVal X As DoubleVector) As Double
      Return X(0) * X(1) - X(0) - X(1)
    End Function

    Function Constraint2(ByVal x As DoubleVector) As Double
      Return x(0) * x(1)
    End Function

  End Module

  <summary>
  f(x) = exp(x0)*(4*x0^2 + 2*x1^2 + 4*x0*x1 + 2*x1 + 1)
  </summary>
  Class ObjectiveFunction
    Inherits DoubleFunctional

    <summary>
    Constructor. Must initialize the base Module with dimension
    of the domain, 2 in this case.
    </summary>
    Public Sub New()
      MyBase.New(2)
    End Sub

    <summary>
    Evaluate the objective function at a point x.
    </summary>
    <param name="x">The point at which to evaluate the function.</param>
    <returns>The value of the objective function.</returns>
    Public Overrides Function Evaluate(ByVal X As DoubleVector) As Double
      Dim x0 As Double = X(0)
      Dim x1 As Double = X(1)
      Return Math.Exp(x0) * (4 * x0 * x0 + 2 * x1 * x1 + 4 * x0 * x1 + 2 * x1 + 1)
    End Function

    <summary>
    Returns the gradient of the objective function at a point x.
    </summary>
    <param name="x">The point at which to evaluate the gradient.</param>
    <param name="grad">Vector into which to place the gradient values.</param>
    Public Overrides Sub Gradient(ByVal X As DoubleVector, ByVal grad As DoubleVector)
      Dim x0 As Double = X(0)
      Dim x1 As Double = X(1)
      Dim ex0 As Double = Math.Exp(x0)
      grad(0) = ex0 * (4 * x0 * x0 + 2 * x1 * x1 + 4 * x0 * x1 + 2 * x1 + 1) + ex0 * (8 * x0 + 4 * x1)
      grad(1) = ex0 * (4 * x0 + 4 * x1 + 2)
    End Sub

  End Class
End Namespace