Root Bracketing Solvers in Visual Basic QuickStart Sample
Illustrates the use of the root bracketing solvers for solving equations in one variable in Visual Basic.
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Option Infer On
' The RootBracketingSolver and derived classes reside in the
' Extreme.Mathematics.EquationSolvers namespace.
Imports Extreme.Mathematics.EquationSolvers
' Function delegates reside in the Extreme.Mathematics
' namespace.
Imports Extreme.Mathematics.Algorithms
Imports Extreme.Mathematics
' Illustrates the use of the root bracketing solvers
' in the Extreme.Mathematics.EquationSolvers namespace of Extreme Numerics.NET.
Module RootBracketingSolvers
Sub Main()
' The license is verified at runtime. We're using
' a demo license here. For more information, see
' https://numerics.net/trial-key
Extreme.License.Verify("Demo license")
' Root bracketing solvers are used to solve
' non-linear equations in one variable.
'
' Root bracketing solvers start with an interval
' which is known to contain a root. This interval
' is made smaller and smaller in successive
' iterations until a certain tolerance is reached,
' or the maximum number of iterations has been
' exceeded.
'
' The properties and methods that give you control
' over the iteration are shared by all classes
' that implement iterative algorithms.
'
' Target function
'
' The function we are trying to solve must be
' provided as a Func(Of Double, Double). For more
' information about this delegate, see the
' Functions QuickStart sample.
Dim f As Func(Of Double, Double) = AddressOf Math.Cos
' All root bracketing solvers inherit from
' RootBracketingSolver, an abstract base class.
Dim solver As RootBracketingSolver
'
' Bisection method
'
' The bisection method halves the interval during
' each iteration. It is implemented by the
' BisectionSolver class.
Console.WriteLine("BisectionSolver: cos(x) = 0 over [1,2]")
solver = New BisectionSolver()
solver.LowerBound = 1
solver.UpperBound = 2
solver.TargetFunction = f
Dim result As Double = solver.Solve()
' The Status property indicates
' the result of running the algorithm.
Console.WriteLine(" Result: {0}",
solver.Status)
' The result is also available through the
' Result property.
Console.WriteLine(" Solution: {0}", solver.Result)
' You can find out the estimated error of the result
' through the EstimatedError property:
Console.WriteLine(" Estimated error: {0}",
solver.EstimatedError)
Console.WriteLine(" # iterations: {0}",
solver.IterationsNeeded)
'
' Regula Falsi method
'
' The Regula Falsi method optimizes the selection
' of the next interval. Unfortunately, the
' optimization breaks down in some cases.
' Here is an example:
Console.WriteLine("RegulaFalsiSolver: cos(x) = 0 over [1,2]")
solver = New RegulaFalsiSolver()
solver.LowerBound = 1
solver.UpperBound = 2
solver.MaxIterations = 1000
solver.TargetFunction = f
result = solver.Solve()
Console.WriteLine(" Result: {0}",
solver.Status)
Console.WriteLine(" Solution: {0}", solver.Result)
Console.WriteLine(" Estimated error: {0}",
solver.EstimatedError)
Console.WriteLine(" # iterations: {0}",
solver.IterationsNeeded)
' However, for sin(x) = 0, everything is fine:
Console.WriteLine("RegulaFalsiSolver: sin(x) = 0 over [-0.5,1]")
solver = New RegulaFalsiSolver()
solver.LowerBound = -0.5
solver.UpperBound = 1
solver.TargetFunction = AddressOf Math.Sin
result = solver.Solve()
Console.WriteLine(" Result: {0}",
solver.Status)
Console.WriteLine(" Solution: {0}", solver.Result)
Console.WriteLine(" Estimated error: {0}",
solver.EstimatedError)
Console.WriteLine(" # iterations: {0}",
solver.IterationsNeeded)
'
' Dekker-Brent method
'
' The Dekker-Brent method combines the best of
' both worlds. It is the most robust and, on average,
' the fastest method.
Console.WriteLine("DekkerBrentSolver: cos(x) = 0 over [1,2]")
solver = New DekkerBrentSolver()
solver.LowerBound = 1
solver.UpperBound = 2
solver.TargetFunction = f
result = solver.Solve()
Console.WriteLine(" Result: {0}",
solver.Status)
Console.WriteLine(" Solution: {0}", solver.Result)
Console.WriteLine(" Estimated error: {0}",
solver.EstimatedError)
Console.WriteLine(" # iterations: {0}",
solver.IterationsNeeded)
'
' Controlling the process
'
Console.WriteLine("Same with modified parameters:")
' You can set the maximum # of iterations:
' If the solution cannot be found in time, the
' Status will return a value of
' IterationStatus.IterationLimitExceeded
solver.MaxIterations = 20
' You can specify how convergence is to be tested
' through the ConvergenceCriterion property:
solver.ConvergenceCriterion =
ConvergenceCriterion.WithinRelativeTolerance
' And, of course, you can set the absolute or
' relative tolerance.
solver.RelativeTolerance = 0.00001
' In this example, the absolute tolerance will be
' ignored.
solver.AbsoluteTolerance = 1.0E-24
solver.LowerBound = 157081
solver.UpperBound = 157082
solver.TargetFunction = f
result = solver.Solve()
Console.WriteLine(" Result: {0}",
solver.Status)
Console.WriteLine(" Solution: {0}", solver.Result)
' The estimated error will be less than 0.157
Console.WriteLine(" Estimated error: {0}",
solver.EstimatedError)
Console.WriteLine(" # iterations: {0}",
solver.IterationsNeeded)
Console.Write("Press Enter key to exit...")
Console.ReadLine()
End Sub
End Module