The following program calculates the minimum point of a multi-variable function using Newton's method. This method is implemented in a basic form without any enhancements.
Click here to download a ZIP file containing the project files for this program.
The program prompts you to either use the predefined default input values or to enter the following:
1. The values for the initial set of variables
2. The values for the tolerances for each variable.
3. The function tolerance
4. The maximum number of iterations
In case you choose the default input values, the program displays these values and proceeds to find the optimum point. In the case you select being prompted, the program displays the name of each input variable along with its default value. You can then either enter a new value or simply press Enter to use the default value. This approach allows you to quickly and efficiently change only a few input values if you so desire.
The program displays the following final results:
1. The coordinates of the minimum value.
2. The fine step sizes for each variable.
3. The minimum function value.
4. The number of iterations
The current code finds the minimum for the following function:
f(x1,x2) = x1 - x2 + 2 * x1 ^ 2 + 2 * x1 * x2 + x2 ^ 2
Using, for each variable, an initial value of 0, initial step size of 0.1, minimum step size of 1e-7, and using a function tolerance of 1e-7. Here is the sample console screen:
Here is the listing for the main module. The module contains several test functions:
Module Module1
Sub Main()
Dim nNumVars As Integer = 2
Dim fX() As Double = {0, 0}
Dim fParam() As Double = {0, 0}
Dim fToler() As Double = {0.00001, 0.00001}
Dim nIter As Integer = 0
Dim nMaxIter As Integer = 100
Dim fEpsFx As Double = 0.0000001
Dim I As Integer
Dim fBestF As Double
Dim sAnswer As String, sErrorMsg As String = ""
Dim oOpt As COptimNewton1
Dim MyFx As MyFxDelegate = AddressOf Fx3
Dim SayFx As SayFxDelegate = AddressOf SayFx3
oOpt = New COptimNewton1
Console.WriteLine("Newton's Method for Optimization (basic version)")
Console.WriteLine("Finding the minimum of function:")
Console.WriteLine(SayFx())
Console.Write("Use default input values? (Y/N) ")
sAnswer = Console.ReadLine()
If sAnswer.ToUpper() = "Y" Then
For I = 0 To nNumVars - 1
Console.WriteLine("X({0}) = {1}", I + 1, fX(I))
Console.WriteLine("Tolerance({0}) = {1}", I + 1, fToler(I))
Next
Console.WriteLine("Function tolerance = {0}", fEpsFx)
Console.WriteLine("Maxumum cycles = {0}", nMaxIter)
Else
For I = 0 To nNumVars - 1
fX(I) = GetIndexedDblInput("X", I + 1, fX(I))
fToler(I) = GetIndexedDblInput("Tolerance", I + 1, fToler(I))
Next
fEpsFx = GetDblInput("Function tolerance", fEpsFx)
nMaxIter = GetIntInput("Maxumum cycles", nMaxIter)
End If
Console.WriteLine("******** FINAL RESULTS *************")
fBestF = oOpt.CalcOptim(nNumVars, fX, fParam, fToler, fEpsFx, nMaxIter, nIter, sErrorMsg, MyFx)
If sErrorMsg.Length > 0 Then
Console.WriteLine("** NOTE: {0} ***", sErrorMsg)
End If
Console.WriteLine("Optimum at")
For I = 0 To nNumVars - 1
Console.WriteLine("X({0}) = {1}", I + 1, fX(I))
Next
Console.WriteLine("Function value = {0}", fBestF)
Console.WriteLine("Number of iterations = {0}", nIter)
Console.WriteLine()
Console.Write("Press Enter to end the program ...")
Console.ReadLine()
End Sub
Function GetDblInput(ByVal sPrompt As String, ByVal fDefInput As Double) As Double
Dim sInput As String
Console.Write("{0}? ({1}): ", sPrompt, fDefInput)
sInput = Console.ReadLine()
If sInput.Trim().Length > 0 Then
Return Double.Parse(sInput)
Else
Return fDefInput
End If
End Function
Function GetIntInput(ByVal sPrompt As String, ByVal nDefInput As Integer) As Integer
Dim sInput As String
Console.Write("{0}? ({1}): ", sPrompt, nDefInput)
sInput = Console.ReadLine()
If sInput.Trim().Length > 0 Then
Return Integer.Parse(sInput)
Else
Return nDefInput
End If
End Function
Function GetIndexedDblInput(ByVal sPrompt As String, ByVal nIndex As Integer, ByVal fDefInput As Double) As Double
Dim sInput As String
Console.Write("{0}({1})? ({2}): ", sPrompt, nIndex, fDefInput)
sInput = Console.ReadLine()
If sInput.Trim().Length > 0 Then
Return Double.Parse(sInput)
Else
Return fDefInput
End If
End Function
Function GetIndexedIntInput(ByVal sPrompt As String, ByVal nIndex As Integer, ByVal nDefInput As Integer) As Integer
Dim sInput As String
Console.Write("{0}({1})? ({2}): ", sPrompt, nIndex, nDefInput)
sInput = Console.ReadLine()
If sInput.Trim().Length > 0 Then
Return Integer.Parse(sInput)
Else
Return nDefInput
End If
End Function
Function SayFx1() As String
Return "F(X) = 10 + (X(1) - 2) ^ 2 + (X(2) + 5) ^ 2"
End Function
Function Fx1(ByVal N As Integer, ByRef X() As Double, ByRef fParam() As Double) As Double
Return 10 + (X(0) - 2) ^ 2 + (X(1) + 5) ^ 2
End Function
Function SayFx2() As String
Return "F(X) = 100 * (X(1) - X(2) ^ 2) ^ 2 + (X(2) - 1) ^ 2"
End Function
Function Fx2(ByVal N As Integer, ByRef X() As Double, ByRef fParam() As Double) As Double
Return 100 * (X(0) - X(1) ^ 2) ^ 2 + (X(1) - 1) ^ 2
End Function
Function SayFx3() As String
Return "F(X) = X(1) - X(2) + 2 * X(1) ^ 2 + 2 * X(1) * X(2) + X(2) ^ 2"
End Function
Function Fx3(ByVal N As Integer, ByRef X() As Double, ByRef fParam() As Double) As Double
Return X(0) - X(1) + 2 * X(0) ^ 2 + 2 * X(0) * X(1) + X(1) ^ 2
End Function
End Module
Notice that the user-defined functions have accompanying helper functions to display the mathematical expression of the function being optimized. For example, function Fx1 has the helper function SayFx1 to list the function optimized in Fx1. Please observe the following rules::
The program uses the following class to optimize the objective function:
Public Delegate Function MyFxDelegate(ByVal nNumVars As Integer, ByRef fX() As Double, ByRef fParam() As Double) As Double
Public Delegate Function SayFxDelegate() As String
Public Class COptimNewton1
Dim m_MyFx As MyFxDelegate
Protected Function MyFxEx(ByVal nNumVars As Integer, _
ByRef fX() As Double, ByRef fParam() As Double, _
ByRef fDeltaX() As Double, ByVal fLambda As Double) As Double
Dim I As Integer
Dim fXX(nNumVars) As Double
For I = 0 To nNumVars - 1
fXX(I) = fX(I) + fLambda * fDeltaX(I)
Next I
MyFxEx = m_MyFx(nNumVars, fXX, fParam)
End Function
Protected Function FirstDeriv(ByVal nNumVars As Integer, _
ByRef fX() As Double, ByRef fParam() As Double, _
ByVal nIdxI As Integer) As Double
Dim fXt, h, Fp, Fm As Double
fXt = fX(nIdxI)
h = 0.01 * (1 + Math.Abs(fXt))
fX(nIdxI) = fXt + h
Fp = m_MyFx(nNumVars, fX, fParam)
fX(nIdxI) = fXt - h
Fm = m_MyFx(nNumVars, fX, fParam)
fX(nIdxI) = fXt
FirstDeriv = (Fp - Fm) / 2 / h
End Function
Protected Function SecondDeriv(ByVal nNumVars As Integer, _
ByRef fX() As Double, ByRef fParam() As Double, _
ByVal nIdxI As Integer, ByVal nIdxJ As Integer) As Double
Dim fXt, fYt, fHX, fHY, F0, Fp, Fm As Double
Dim Fpp, Fmm, Fpm, Fmp, fResult As Double
' calculate second derivative?
If nIdxI = nIdxJ Then
F0 = m_MyFx(nNumVars, fX, fParam)
fXt = fX(nIdxI)
fHX = 0.01 * (1 + Math.Abs(fXt))
fX(nIdxI) = fXt + fHX
Fp = m_MyFx(nNumVars, fX, fParam)
fX(nIdxI) = fXt - fHX
Fm = m_MyFx(nNumVars, fX, fParam)
fX(nIdxI) = fXt
fResult = (Fp - 2 * F0 + Fm) / fHX ^ 2
Else
fXt = fX(nIdxI)
fYt = fX(nIdxJ)
fHX = 0.01 * (1 + Math.Abs(fXt))
fHY = 0.01 * (1 + Math.Abs(fYt))
' calculate Fpp
fX(nIdxI) = fXt + fHX
fX(nIdxJ) = fYt + fHY
Fpp = m_MyFx(nNumVars, fX, fParam)
' calculate Fmm
fX(nIdxI) = fXt - fHX
fX(nIdxJ) = fYt - fHY
Fmm = m_MyFx(nNumVars, fX, fParam)
' calculate Fpm
fX(nIdxI) = fXt + fHX
fX(nIdxJ) = fYt - fHY
Fpm = m_MyFx(nNumVars, fX, fParam)
' calculate Fmp
fX(nIdxI) = fXt - fHX
fX(nIdxJ) = fYt + fHY
Fmp = m_MyFx(nNumVars, fX, fParam)
fX(nIdxI) = fXt
fX(nIdxJ) = fYt
fResult = (Fpp - Fmp - Fpm + Fmm) / (4 * fHX * fHY)
End If
Return fResult
End Function
Protected Sub GetFirstDerives(ByVal nNumVars As Integer, _
ByRef fX() As Double, ByRef fParam() As Double, _
ByRef fFirstDerivX() As Double)
Dim I As Integer
For I = 0 To nNumVars - 1
fFirstDerivX(I) = FirstDeriv(nNumVars, fX, fParam, I)
Next I
End Sub
Protected Sub GetSecondDerives(ByVal nNumVars As Integer, _
ByRef fX() As Double, ByRef fParam() As Double, _
ByRef fSecondDerivX(,) As Double)
Dim I, J As Integer
For I = 0 To nNumVars - 1
For J = 0 To nNumVars - 1
fSecondDerivX(I, J) = SecondDeriv(nNumVars, fX, fParam, I, J)
Next J
Next I
End Sub
Public Function CalcOptim(ByVal nNumVars As Integer, ByRef fX() As Double, ByRef fParam() As Double, _
ByRef fToler() As Double, ByVal fEpsFx As Double, ByVal nMaxIter As Integer, _
ByRef nIter As Integer, ByRef sErrorMsg As String, _
ByVal MyFx As MyFxDelegate) As Double
Dim I As Integer
Dim fNorm As Double
Dim g(nNumVars) As Double, Index(nNumVars) As Integer
Dim DeltaX(nNumVars) As Double, J(nNumVars, nNumVars) As Double
Dim bStop As Boolean
m_MyFx = MyFx
nIter = 1
Do
nIter += 1
If nIter > nMaxIter Then
sErrorMsg = "Reached maximum iterations limit"
Exit Do
End If
GetFirstDerives(nNumVars, fX, fParam, g)
' test if gradient is shallow enough
fNorm = MatrixLibVb.VectNorm(g)
If fNorm < fEpsFx Then Exit Do
GetSecondDerives(nNumVars, fX, fParam, J)
MatrixLibVb.MV_LUDecomp(J, Index, nNumVars)
MatrixLibVb.MV_LUBackSubst(J, Index, nNumVars, g)
For I = 0 To nNumVars - 1
DeltaX(I) = g(I)
fX(I) -= DeltaX(I)
Next I
bStop = True
For I = 0 To nNumVars - 1
If Math.Abs(DeltaX(I)) > fToler(I) Then
bStop = False
Exit For
End If
Next I
Loop Until bStop
Return MyFx(nNumVars, fX, fParam)
End Function
End Class
Copyright (c) Namir Shammas. All rights reserved.