I need to use Newton's method on closures.
Function f (x as Double, y as Double) as Double
f = x^3-y
End Function
I get the value of y from a cell and then I would like to find out when f is zero. In the toy example above, if the cell contains y=8, then I would expect Newton's method to find a solution close to x=2.
My solution was to make a newton_solve_f function:
Function newton_solve_f (y as Double as Double) as Double
Dim x as Double
x = 0 'initial guess for x
'do Newton's method to find x
...
newton_solve_f = x
End Function
so in effect, I copy paste my code for Newton's method (taken from here) into newton_solve_f.
The problem is that I have several such fs (some with more than two arguments), and it would be really neat if I didn't have to make a separate almost identical newton_solve_f for every one of them.
How would you solve this in VBA?
In Python, for example, it's possible to solve this problem as follows:
def f(y):
def g(x):
return x^3-y
return g
def newton_solve(f):
#do newton's method on f(x)
newton_solve(f(3))
Here f(3) is a function, a closure of one variable. (The closure example on wikipedia is almost identical to this one.)
ps. I know Newton's method also needs the (partial) derivative of f, I'm actually doing something that's more like the secant method, but that's irrelevant for what I'm asking about
Closures are not part of VBA. But you can use static variables within a method scope. They cannot be used outside the method. If you want a variable to visible outside, then you have to use global variable. Preferable declare it public in a module.
We cannot define function inside function in VB. Tried to convert the code given in the link you have mentioned. I hope it helps you. Not well versed with php, but you can see the approach below and make changes accordingly.
Sub Test()
Dim x As Double
Dim y As Double
Dim z As Double
x = Cells(1, 1).Value
y = Cells(1, 2).Value
z = NewtRap("Fun1", "dFun1", x, y)
Cells(1, 3).Value = z
End Sub
Private Function NewtRap(fname As String, dfname As String, x_guess As Double, y_value As Double) As Double
Dim cur_x As Double
Dim Maxiter As Double
Dim Eps As Double
Maxiter = 500
Eps = 0.00001
cur_x = x_guess
For i = 1 To Maxiter
If (fname = "Fun1") Then
fx = Fun1(cur_x)
ElseIf (fname = "dFun1") Then
fx = dFun1(cur_x)
ElseIf (fname = "f") Then
fx = f(cur_x, y_value)
End If
If (dfname = "Fun1") Then
fx = Fun1(cur_x)
ElseIf (dfname = "dFun1") Then
fx = dFun1(cur_x)
ElseIf (dfname = "f") Then
fx = f(cur_x, y_value)
End If
If (Abs(dx) < Eps) Then Exit For
cur_x = cur_x - (fx / dx)
Next i
NewtRap = cur_x
End Function
Function f(x As Double, y As Double) As Double
f = x ^ 3 - y
End Function
Function Fun1(x As Double) As Double
Fun1 = x ^ 2 - 7 * x + 10
End Function
Function dFun1(x As Double) As Double
dFun1 = 2 * x - 7
End Function
So to first summarise: You want to create a function that will find (using Newton-Raphson method) the roots of a function. You already have this written and working for certain functions but would like help expanding your code so it will work with a variety of functions with varying numbers of parameters?
I think you first need to think about what input functions you want it to cover. If you are only dealing with polynomials (as your example suggests), this should be fairly straightforward.
You could have general functions of:
Function fnGeneralCase (x, y, z, w, a1, a2, a3, b1, b2, b3, c1, c2, c3 as Double) as Double
fnGeneralCase = a1*x^3 + a2*x^2 + a3*x + b1*y^3 + b2*y^2 + b3*y + c1*z^3 + c2*z^2 + c3*z + w
End Function
Function fnDerivGeneralCase (x, y, z, w, a1, a2, a3, b1, b2, b3, c1, c2, c3 as Double) as Double
fnDerivGeneralCase = a1*3*x^2 + a2*2*x + a3 + b1*3*y^2 + b2*2*y + b3 + c1*3*z^2 + c2*2*z + c3
End Function
And just set the inputs to zero when you don't need them (which will be for the majority of the time).
So for your example calling:
answer = fnGeneralCase(guess, 0, 0, -8, 1, 0, 0, 0, 0, 0, 0, 0, 0)
basically gives:
function = x^3-8
If you want to include more than polynomials, this will get more complicated but you could still use the above approach...
This seems to be asking 2 related questions:
how to pass a function as an argument in vba.
how to create a closure out of an existing function.
Unfortunately neither of these are really supported, however,
for 1 you can generally work around this by passing a string function name and using 'Application.Run' to invoke the function.
2 is trickier if you have lots of functions with different numbers of parameters, but for a set number of parameters you could add extra parameters to the newton_solve function or maybe use global variables.
e.g.
Public Function f(x as Double, y as Double) as Double
f = x^3-y
End Function
Function newton_solve_f (function_name as String, y as Double) as Double
Dim x as Double
x = 0 'initial guess for x
'do Newton's method to find x
...
' invoke function_name
x = Application.Run(function_name, x, y)
...
newton_solve_f = x
End Function
Assuming f is in a module called 'Module1' you can call this with:
x = newton_solve('Module1.f', 3)
Note that the function you want to call must be public.
Related
I have written this code to split text and number in excel but whenever I run it... it doesn't work
Public Function Strip(ByVal x As String, LeaveNums As Boolean) As Variant
Dim y As String, z As String, n As Long
For n = 1 To Len(x)
y = Mid(x, n, 1)
If LeaveNums = False Then
If y Like "[A-Za-z ]" Then z = z & y 'False keeps Letters and spaces only
Else
If y Like "[0-9. ]" Then z = z & y 'True keeps Numbers and decimal points
End If
Next n
Strip = Trim(z)
End Function
You've written (copied?) a Function, not a subroutine. Functions are commonly used for repeated processes, and return (if written properly) values. So, you call a function from within a subroutine and then do something with the value the function returned. The function does something with the x and LeaveNums parameters, just like formulae do.
In fact, you can write Functions and use these as formulae in your Excel worksheets.
You can call this Function as follows (note, this is an example)
Sub Usefunction()
rslt = Strip("test this 1.01", True)
MsgBox rslt
End Sub
This will return "test this". If you set the boolean to False, the function returns "1.01" instead.
Closed. This question needs debugging details. It is not currently accepting answers.
Edit the question to include desired behavior, a specific problem or error, and the shortest code necessary to reproduce the problem. This will help others answer the question.
Closed 3 years ago.
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Here is my code for a bisection method. If I input 4 and 5 the program loops infinitely. There is a problem with it running.
Sub TheBisectionMethod1()
Dim a, b As Double 'Taking two variables, A and B
Console.Write(vbLf & "Input A: ")
a = Double.Parse(Console.ReadLine()) 'This is where user inputs and value stored for A
Console.Write(vbLf & "Input B: ")
b = Double.Parse(Console.ReadLine()) 'This is where user inputs and value stored for B
line1:
Dim c As Double
c = (a + b) / 2 'declearing variable c for the sum of half of the user entered values
If ((Math.Abs(func(c))) < 0.0001) Then 'in flow chart the value of C remians unchange so the program will not run, so it will run if i is >0.0001
Console.Write(vbLf & "C : {0}", c)
ElseIf (Math.Sign(func(c)) = Math.Sign(func(a))) Then
Console.WriteLine("Hello")
a = c
GoTo line1
ElseIf (Math.Sign(func(c)) <> Math.Sign(func(a))) Then
b = c
GoTo line1
End If
End Sub
Function func(x As Double) As Double
Dim y As Double
y = x ^ 2 - 2
Return y
End Function
Don't use a GoTo. There's no need. Also, remove the user interaction from the method that does the actual work. Read the data in one place, pass it to a method (usually a Function rather than a Sub) that does the work and returns a result, and then show the result to the user after the function ends.
That makes this question tricky, because the only result we see in the original code is writing "Hello" to the Console, and that's clearly just a debugging statement. What do you want this code to do? (I'm gonna assume you mean this)
Function Bisect(a as Double, b As Double) As Double
Dim c As Double = (a + b) / 2
Dim d As Double = func(c)
While Math.Abs(d) >= 0.0001
If Math.Sign(d) = Math.Sign(func(a)) Then
a = c
Else
b = c
End If
c = (a + b) / 2
d = func(c)
End While
Return c
End Function
Function func(x As Double) As Double
Return x ^ 2 - 2
End Function
And really, it should look like this:
Function Bisect(a as Double, b As Double, f As Function(Of Double, Double)) As Double
Dim c As Double = (a + b) / 2
Dim d As Double = f(c)
While Math.Abs(d) >= 0.0001
If Math.Sign(d) = Math.Sign(f(a)) Then
a = c
Else
b = c
End If
c = (a + b) / 2
d = f(c)
End While
Return c
End Function
and be called like this:
Bisect(a, b, Function(x) x ^ 2 - 2)
Also, the algorithm here is slightly off based the wikipedia article. This is more precise:
Function Bisect(a as Double, b As Double, f As Function(Of Double, Double)) As Double
Dim TOL As Double = 0.0001
Dim MaxSteps As Integer = 1000
Dim c As Double = (a + b) / 2
While Math.Abs(f(c)) >= TOL AndAlso (b-a)/2 >= TOL AndAlso MaxSteps > 0
If Math.Sign(f(c)) = Math.Sign(f(a)) Then
a = c
Else
b = c
End If
MaxSteps -= 1
c = (a + b) / 2
End While
If MaxSteps = 0 Then
Throw New ArgumentException("The bisection fails to converge within the allowed time for the supplied arguments.")
End If
Return c
End Function
I bring it up, because the complaint in the original question is this:
the program loops infinently[sic]
and one of the tenants of the algorithm is it's not guaranteed to converge, hence the step counter.
Finally, this looks to me like it might do well as a recursive function. Recursion can improve things here because we can rely on the call stack overflowing rather than needing to implement a step counter:
Function Bisect(a as Double, b As Double, f As Function(Of Double, Double)) As Double
Dim c As Double = (a + b) / 2
If Math.Abs(f(c)) < TOL OrElse (b-a)/2 < TOL Then Return c
If Math.Sign(f(c)) = Math.Sign(f(a)) Then
Return Bisect(c, b, f)
Else
Return Bisect(a, c, f)
End If
End Function
Of course, catching that StackOverflowException is a trick in itself, so you may still want that step counter. But I need to leave something for you to do yourself.
This also helps demonstrate part of why I recommend removing all user I/O from the Bisect() method. If this method were also responsible for talking to the end user, it would not be possible to even consider the recursive option, where the code is clearly far shorter and simpler than any of the others.
I'm working on a function to do linear interpolation:
Public Function linear_interpolation(xs As Range, ys As Range, x As Double)
Dim index As Integer
Dim x0 As Double, x1 As Double
Dim y0 As Double, y1 As Double
index = Application.WorksheetFunction.Match(x, xs)
x0 = xs(index)
y0 = ys(index)
x1 = xs(index + 1)
y1 = ys(index + 1)
linear_interpolation = ((x1 - x) * y0 + (x - x0) * y1) / (x1 - x0)
End Function
It works fine if both ranges xs and ys are fully populated, but if there is a missing value (empty cell) then it is treated as a zero, which is surprising, I was expecting a type error. If the cell contains a non-numerical value, then I get #VALUE! as expected.
What's the best way of dealing with this? Do I have to manually check to see if xs(index), ys(index), xs(index+1) and ys(index+1) empty and then return an error?
If you still need 0 in the cell but wan to flag error for NULL or "" you could easily involve a check that says
IF cell.value = VBNullString Then
'do some stuff
End If
No sense reinventing the wheel! There's a linear interpolation function on my website - I apologize for the shameless plug - that grabs the nearest values to the value you're trying to interpolate on so you don't have to worry about the empty cells registering as zeros:)
Bottom line is you want to add a check for the values equal to an empty string ""
Source: http://wellsr.com/vba/2016/excel/powerful-excel-linear-interpolation-function-vba/
I'm having two problems here. First off all I want to x to change values to x - y and if the new X is higher then 0 i want the process to repeat itself. I've worked out the code below but i'm not certiant on two things.
Am I even allowed to make the equation x = x - y or does this mess up everything? I mean in mathematical terms it would not be possible but if we take X as Hp and Y as damage I want the damage to add up. I don't want it to create an "damage HP" integer for every subtraction as I even don't know how many "Z = x - y" style equations I would have to create if I set Y to be random.
My guess is that I could create a Z integral that would copy X a moment before the subtraction would go off and then have the subtraction be X = Z - Y but I'm not sure how I would go about coding this.
I want it to go ahead and loop itself if X is higher then 0 and I'm not sure if I coded that correctly.
Here is my code:
Module Module1
Dim A As Integer
Dim B As Integer
Dim x As Integer
Dim y As Integer
Sub Main()
End Sub
Sub Maths()
A = 5
B = 4
x = 3
y = 1
Subtraction()
Console.WriteLine("You deal {0} damage to your enemy reducing it to {1} hp.", y, x)
Do Until x <= 0
Loop
End Sub
Private Sub Subtraction()
If A > B Then x = x -y
Return
End Sub
End Module
I liked this question. Here's my thoughts:
Yes, x = x - y is perfectly valid code. It's no different than if I had a string variable named myRunOnSentence and I wanted to concatenate the string that was already in the variable and another string and then store the results back in the string variable. Like this: myRunOnSentence = myRunOnSentence + "another string" Same concept, just change the datatype to an Integer. x = x + y. That programatically says: "take the value in x and the value in y, add them together, and store the result of that expression in x."
You did indeed make a mistake with the loop. You don't have any code inside the body of the loop itself.
You have nothing happening in the Main() sub of your module so this module when run will do nothing. You should just take the code from the Maths() method and put it in the Main() sub.
In your Subtraction() method, A > B will always evaluate to True because A and B are initialized with values and then never changed.
Your code should look something like this:
Module Module1
Dim A As Integer = 5
Dim B As Integer = 4
Dim x As Integer = 3
Dim y As Integer = 1
Sub Main()
Do Until x <= 0
Subtraction()
Console.WriteLine("You deal {0} damage to your enemy reducing it to {1} hp.", y, x)
Loop
End Sub
Private Sub Subtraction()
If A > B Then x = x - y 'Return statement wasn't needed.
End Sub
End Module
If this answered your question, please don't forget to mark it as the answer.
I was creating a critical value approximator of American style options. I was getting the error "#Value!" only after around 40 loops (kept track with a counter).
After some trial and error I realized it came from the part of the loop calling the BlackScholes pricing function. In theory I want to run through a range of values iteratively for the spot price while keeping the other variables fixed in a Black Scholes European price calculation. After tinkering around I reduced the issue to the fact that after the first loop it was no longer calculating Black Scholes the way it would if I just used the value on that iteration and the value I was getting was just increasing by 1, then crapping out after 40 loops of wrong values for some non obvious reason.
So below I truncated the code to a very simple skeleton which is the essence of my problem. Any help would be appreciated.
Function Looper(S As Double, K As Double, r As Double, t As Double, q As Double, Vol As Double) As Double
Dim i As Double
For i = 100 To 150 Step 1#
MsgBox i
MsgBox BS(i, K, r, t, q, Vol, "Call") 'After the first loop the values are wrong,
'What I'd like is, BS(100,...), BS(101,...),BS(102,...) which it is not.
'Not sure what it's actually calculating, since the values are way off
Next i
End Function
Public Function BS(S As Double, K As Double, r As Double, t As Double, q As Double, Vol As Double, CP As String) As Double
Dim volrootime As Double
Dim d1 As Double
Dim d2 As Double
Dim DiscF As Double
Dim DivF As Double
Dim topline1 As Double
Dim topline2 As Double
Dim topline As Double
Dim Price As Double
t = t / 365
r = r / 100
q = q / 100
DiscF = Exp(-r * t)
DivF = Exp(-q * t)
volrootime = (t ^ 0.5) * Vol
topline1 = Log(S / K)
topline2 = ((r - q) + ((Vol ^ 2) / 2)) * t
topline = topline1 + topline2
d1 = topline / volrootime
d2 = d1 - volrootime
If CP = "Call" Then
' Theta is in terms of Calendar days, changing the denominator to 252 changes it to trading days
Price = (S * DivF * Bign(d1)) - (K * DiscF * Bign(d2))
Else
' Theta is in terms of Calendar days, changing the denominator to 252 changes it to trading days
Price = K * DiscF * Bign(-d2) - S * DivF * Bign(-d1)
End If
BS = Price
End Function
The values of r, t, q change each time the BS function is called. If they must stay constant, you should use ByVal in the BS function declaration like this:
BS(S As Double, K As Double, ByVal r As Double, ByVal t As Double, ByVal q As Double, ...
By default, the parameters are passed by reference and any change in the called function are reflected in the calling function.
By the way, in this example, I wouldn't use messageboxes when debugging but instead use debug.print like this:
Debug.Print "i=" & i & vbTab & "BS=" & BS(i, K, r, t, q, Vol, "Call")
The print is made in the window opened by pressing Ctl + G (Go To).