I am trying to find the x intercept of a 4th degree function by incrementing the x value. I feel like this way doesnt work always and isnt the most efficient way to do this, is there another way I am missing?
My code is:
Sub Findintercept()
Dim equation As Double, x As Double, A As Double, B As Double, C As Double, D As Double, E As Double
A = 0.000200878
B = -0.002203704
C = 0.0086
D = -0.02333
E = 0.02033
x = 0
equation = A * x ^ 4 + B * x ^ 3 + C * x ^ 2 + D * x + E
While (equation > 0.00001 Or equation < -0.00001)
If (x > 5) Then
MsgBox "Could not find intercept"
equation = 0
Else
x = x + 0.0001
equation = A * x ^ 4 + B * x ^ 3 + C * x ^ 2 + D * x + E
End If
Wend
MsgBox x
End Sub
Sometimes it fails to find the intercept hence the IF condition in the while loop. (Im always expecting the intercept to be less than 5!
Your method suffers from two problems:
You assume a step size to change x. The step could be too large, causing you to "walk past" the value your are looking for. To deal with this, you make a small step size, which can mean an excessively large number of iterations are needed to find the solution.
You always assume the same direction to change x. Even with seemingly small values for your step size, you could "walk past" the solution, and have no means to change direction. Or, your initial guess may be on the wrong side of the solution, and you never find an answer.
The Newton-Raphson method handles both of these issues neatly. You do still need to choose your initial guess somewhat close to the root you are looking for.
This method does have potential problems, but for polynomials such as the one you are dealing with, it is quite good.
Below is a simple VBA sub that implements this method. It solves your problem in 4 iterations. I recommend adjusting the initial guess (xii) a lot to see how it impacts the solution you get.
Sub SimpleNewtonRaphson()
Const Tol As Double = 1E-06
Const MaxIter As Long = 50
Dim xi As Double, xii As Double, deriv As Double
Dim IterCount As Long
' Initialize
xi = 0#
xii = 1#
IterCount = 0
' Method
Do While IterCount < MaxIter And Abs(xii - xi) > Tol
xi = xii
deriv = myDeriv(xi)
If deriv = 0# Then Exit Do
xii = xi - myFunc(xi) / deriv
IterCount = IterCount + 1
Loop
' Results
If deriv = 0 Then MsgBox "Ran into a 0 derivative, modify initial guess"
If IterCount >= MaxIter Then MsgBox "MaxIterations reached"
If Abs(xii - xi) <= Tol Then MsgBox "Solution found #" & vbCrLf & "F(" & xii & ") = " & myFunc(xii)
End Sub
... and two VBA functions for your equation and it's derivative ...
Function myFunc(x As Double) As Double
Const A As Double = 0.000200878
Const B As Double = -0.002203704
Const C As Double = 0.0086
Const D As Double = -0.02333
Const E = 0.02033
myFunc = A * x ^ 4 + B * x ^ 3 + C * x ^ 2 + D * x + E
End Function
Function myDeriv(x As Double) As Double
Const A As Double = 0.000200878
Const B As Double = -0.002203704
Const C As Double = 0.0086
Const D As Double = -0.02333
myDeriv = 4 * A * x ^ 3 + 3 * B * x ^ 2 + 2 * C * x + D
End Function
Related
I'm trying to create a code that uses the false position method to find the roots of an equation. The equation is as follows:
y = x^(1.5sinā”(x)) * e^(-x/7) + e^(x/10) - 4
I used a calculator to find the roots, and they are 6.9025, 8.8719, and 12.8079.
My VBA code is as follows:
Option Explicit
Function Func(x)
Func = (x ^ (1.5 * Sin(x))) * Exp(-x / 7) + Exp(x / 10) - 4
End Function
Function FalsePos(Guess1, Guess2)
Dim a, b, c As Single
Dim i As Integer
a = Guess1
b = Guess2
For i = 0 To 1000
c = a - Func(a) * (b - a) / (Func(b) - Func(a))
If (Func(c) < 0.00001) Then
i = 1001
Else
If Func(a) * Func(c) < 0 Then
b = c
Else
a = c
End If
End If
Next
FalsePos = c
End Function
My problem is that when I call the function and use for example 4 and 8 as my two guesses, the number it returns is 5.29 instead of the root between 4 and 8 which is 6.9025.
Is there something wrong with my code or am I just not understanding the false position method correctly?
You should use Double for precision with Maths problems. Three other notes about coding that you may not be aware of:
dim a, b, c as Single
will dim a and b as Variants, and c as a Single, and you can use Exit For to escape from a for loop, rather than setting the control variable out of the bounds. Finally, you should define the outputs of a Function by specifying As ... after the closing parenthesis.
You should use breakpoints (press F9 with the carrot in a line of code to breakpoint that line), then step through the code by pressing F8 to advance line-by-line to see what is happening, and keep your eye on the Locals window (Go to View > Locals)
This is the code with the above changes:
Function Func(x As Double) As Double
Func = (x ^ (1.5 * Sin(x))) * Exp(-x / 7) + Exp(x / 10) - 4
End Function
Function FalsePos(Guess1 As Double, Guess2 As Double) As Double
Dim a As Double, b As Double, c As Double
Dim i As Integer
a = Guess1
b = Guess2
For i = 0 To 1000
c = a - Func(a) * (b - a) / (Func(b) - Func(a))
If (Func(c) < 0.00001) Then
Exit For
Else
If Func(a) * Func(c) < 0 Then
b = c
Else
a = c
End If
End If
Next
FalsePos = c
End Function
If I have a convex curve, and want to find the minimum point (x,y) using a for or while loop. I am thinking of something like
dim y as double
dim LastY as double = 0
for i = 0 to a large number
y=computefunction(i)
if lasty > y then exit for
next
how can I that minimum point? (x is always > 0 and integer)
Very Close
you just need to
dim y as double
dim smallestY as double = computefunction(0)
for i = 0 to aLargeNumber as integer
y=computefunction(i)
if smallestY > y then smallestY=y
next
'now that the loop has finished, smallestY should contain the lowest value of Y
If this code takes a long time to run, you could quite easily turn it into a multi-threaded loop using parallel.For - for example
dim y as Double
dim smallestY as double = computefunction(0)
Parallel.For(0, aLargeNumber, Sub(i As Integer)
y=computefunction(i)
if smallestY > y then smallestY=y
End Sub)
This would automatically create separate threads for each iteration of the loop.
For a sample function:
y = 0.01 * (x - 50) ^ 2 - 5
or properly written like this:
A minimum is mathematically obvious at x = 50 and y = -5, you can verify with google:
Below VB.NET console application, converted from python, finds a minimum at x=50.0000703584199, y=-4.9999999999505, which is correct for the specified tolerance of 0.0001:
Module Module1
Sub Main()
Dim result As Double = GoldenSectionSearch(AddressOf ComputeFunction, 0, 100)
Dim resultString As String = "x=" & result.ToString + ", y=" & ComputeFunction(result).ToString
Console.WriteLine(resultString) 'prints x=50.0000703584199, y=-4.9999999999505
End Sub
Function GoldenSectionSearch(f As Func(Of Double, Double), xStart As Double, xEnd As Double, Optional tol As Double = 0.0001) As Double
Dim gr As Double = (Math.Sqrt(5) - 1) / 2
Dim c As Double = xEnd - gr * (xEnd - xStart)
Dim d As Double = xStart + gr * (xEnd - xStart)
While Math.Abs(c - d) > tol
Dim fc As Double = f(c)
Dim fd As Double = f(d)
If fc < fd Then
xEnd = d
d = c
c = xEnd - gr * (xEnd - xStart)
Else
xStart = c
c = d
d = xStart + gr * (xEnd - xStart)
End If
End While
Return (xEnd + xStart) / 2
End Function
Function ComputeFunction(x As Double)
Return 0.01 * (x - 50) ^ 2 - 5
End Function
End Module
Side note: your initial attempt to find minimum is assuming a function is discrete, which is very unlikely in real life. What you would get with a simple for loop is a very rough estimate, and a long time to find it, as linear search is least efficient among other methods.
I have a list of distances that I would like to display like you would read off a tape measure, for example 144.125 would display as 144 1/8". I have the following formula
=TEXT(A1,"0"&IF(ABS(A1-ROUND(A1,0))>1/32,"0/"&CHOOSE(ROUND(MOD(A1,1)*16,0),16,8,16,4,16,8,16,2,16,8,16,4,16,8,16),""))&""""
I'd like to simplify it to a 1 argument function (for A1) so I could use it throughout the workbook, but the amount of " quotes and vba keywords is causing problems. Is there an easier way to get a UDF to insert a complicated formula?
If you want to use a UDF with visual basic then try this:
Public Function Fraction(ByVal x As Double, Optional ByVal tol As Double = 1 / 64#) As String
Dim s As Long, w As Long, d As Long, n As Long, f As Double
s = Sgn(x): x = Abs(x)
If s = 0 Then
Fraction = "0"
Exit Function
End If
w = CInt(WorksheetFunction.Floor_Precise(x)): f = x - w
d = CInt(WorksheetFunction.Floor_Precise(1 / tol)): n = WorksheetFunction.Round(f * d, 0)
Dim g As Long
Do
g = WorksheetFunction.Gcd(n, d)
n = n / g
d = d / g
Loop While Abs(g) > 1
Fraction = Trim(IIf(s < 0, "-", vbNullString) + CStr(w) + IIf(n > 0, " " + CStr(n) + "/" + CStr(d), vbNullString))
End Function
With results:
The TEXT function can do this directly:
A B
1 144,1250 144 1/8 "
Formula in B1:
=TEXT(A1;"# ??/??\""")
Greetings
Axel
First off, here is what I have so far:
Option Explicit
Dim y As Variant
Dim yforx As Variant
Dim yfork As Variant
Dim ynew As Variant
Dim ymin As Variant
Dim x As Variant
Dim xmin As Variant
Dim k As Variant
Dim kmin As Variant
Dim s As Variant
Dim Z As Variant
Dim Track As Variant
Sub PracticeProgram()
'Selects the right sheet
Sheets("PracticeProgram").Select
'y = k ^ 2 * (x ^ 2 + 2 * x * k - 6) / (x + k) ^ 2
'these are the bounds we are stepping through
Track = 0
x = 1
xmin = 1
k = 1
kmin = 1
y = 100000000
yforx = 100000
yfork = 1000000000
Do
y = 100000000
For x = 0 To 1000 Step 0.1
ynew = kmin ^ 2 * (x ^ 2 + 2 * x * kmin - 6) / (x + kmin) ^ 2
'This checks the new y-value against an absurdly high y-value we know is wrong. if it is less than this y-value, we keep the x-value that corresponds with it.
If ynew < y Then
xmin = x
y = ynew
yforx = y
xmin = Application.Evaluate("=Round(" & xmin & ", 3)")
Else
End If
Next
MsgBox (yforx)
For k = 0 To 1000 Step 0.1
y = k ^ 2 * (xmin ^ 2 + 2 * xmin * k - 6) / (xmin + k) ^ 2
If ynew < y Then
kmin = k
y = ynew
yfork = y
kmin = Application.Evaluate("=Round(" & kmin & ",3)")
Else
End If
Next
MsgBox (yfork)
Loop Until (Abs(yforx - yfork) < 10)
End Sub
This program is supposed to find the values of x and k in order to minimize the value of y. This is a practice for a much more complicated program that will use this same concept. In my actual program y, k, and x will all be greater than zero no matter what, but since it was hard to think of a simple equation whose results would be in the shape of a parabola opening up, I decided to allow negative answers for this practice program.
Basically, it should bounce back and forth between the equations finding the ideal values for x and k until finally it has a minimal answer for y using ideal answers for both x and k. I'm not sure what the actual answer is, so I'm letting it stop within a range of 10. If it works, I'll make it smaller, but I don't want the program going for forever, just in case.
MY PROBLEM: I keep getting overflow errors! I'm trying to round the values for xmin and kmin to three figures after the decimal, but it doesn't seem to be helping. Am I using them wrong? Can someone help me get this program working?
You're doing a division by zero. xmin = 0, k = 0, (xmin + k) ^ 2 = 0. (I'm not sure why it isn't reporting division by zero.)
A suggestion: use the Locals pane to see the value of local variables. You can also use the Watch pane to see the value of expressions you want to monitor.
During some iterative optimization, the following VBA code for the computation of the bivariate normal CDF sometimes throws an Overflow error on the line with z = hx * hy * c inside the while loop of the upper function.
I debugged the code and the overflow occurs when the numbers being multiplied result in a number bigger than what a double can hold.
Can you show me how to handle the problem by ignoring the iterations of the loop with such high values - I guess that's the only feasible solution (?). I tried myself with a On Error Goto nextiteration line before the multiplication and placing the nextiteration jump point before the Wend, but the error persists.
Function tetrachoric(x As Double, y As Double, rho As Double) As Double
Const FACCURACY As Double = 0.0000000000001
Const MinStopK As Integer = 20
Dim k As Integer
Dim c As Double
Dim z As Double
Dim s As Double
Dim hx As Double
Dim hx1 As Double
Dim hx2 As Double
Dim hy As Double
Dim hy1 As Double
Dim hy2 As Double
Dim CheckPass As Integer
hx = 1
hy = 1
hx1 = 0
hy1 = 0
k = 0
c = rho
z = c
s = z
CheckPass = 0
While CheckPass < MinStopK
k = k + 1
hx2 = hx1
hy2 = hy1
hx1 = hx
hy1 = hy
hx = x * hx1 - (k - 1) * hx2
hy = y * hy1 - (k - 1) * hy2
c = c * rho / (k + 1)
z = hx * hy * c
s = s + z
If Abs(z / s) < FACCURACY Then
CheckPass = CheckPass + 1
Else
CheckPass = 0
End If
Wend
tetrachoric = s
End Function
Public Function bivnor(x As Double, y As Double, rho As Double) As Double
'
' bivnor function
' Calculates bivariat normal CDF F(x,y,rho) for a pair of standard normal
' random variables with correlation RHO
'
If rho = 0 Then
bivnor = Application.WorksheetFunction.NormSDist(x) * _
Application.WorksheetFunction.NormSDist(y)
Else
bivnor = Application.WorksheetFunction.NormSDist(x) * _
Application.WorksheetFunction.NormSDist(y) + _
Application.WorksheetFunction.NormDist(x, 0, 1, False) * _
Application.WorksheetFunction.NormDist(y, 0, 1, False) * _
tetrachoric(x, y, rho)
End If
End Function
Source: Available for download at http://michael.marginalq.com/
you're hitting on the limits of the computer architecture. Many complex algorithms can't be implemented 1:1 with their mathematical representation because of performance reasons and/or erroneous behavior when overflowing. There's an exceptionally good blog about these issues - John D. Cook.
Please take a look here for a better implementation.
You can also try binding an external library, that gives you arbitrary precision number handling, of course implemented using very expensive (in terms of CPU time) software algorithms. More can be found here.
Updated code using On Error Resume Next instead of On Error Goto:
While CheckPass < MinStopK
k = k + 1
hx2 = hx1
hy2 = hy1
hx1 = hx
hy1 = hy
hx = x * hx1 - (k - 1) * hx2
hy = y * hy1 - (k - 1) * hy2
c = c * rho / (k + 1)
On Error Resume Next
z = hx * hy * c
If Err.Number = 0 Then
s = s + z
If Abs(z / s) < FACCURACY Then
CheckPass = CheckPass + 1
Else
CheckPass = 0
End If
Else
Err.Clear
End If
Wend
http://www.codeproject.com/KB/recipes/float_point.aspx treats how to "Use Logarithms to Avoid Overflow and Underflow", which is a simple but quite effective way of working around overflow problems. In fact, it's so simple yet logical, why haven't we thought of that solution ourselves? ;)