I wrote user-defined function called myaverage, which calculates weighted average of 3 numbers:
Function myaverage(x, y, z) As Double
Dim a As Single
Dim x As Double
Dim y As Double
Dim z As Double
a = 0.4 * x + 0.5 * y + 0.1 * z
myaverage = a
End Function
But if I type =myaverage(A1:A3), I cannot see the result.
Simplify your function to something like this:
Function myaverage(x As Double, y As Double, z As Double) As Double
myaverage = 0.4 * x + 0.5 * y + 0.1 * z
End Function
In Excel cell, use this:
= myaverage(A1, B1, C1)
You had double declarations. I mean, x was alreaday declared in the signature and you re-declared it in the function body. So this will work:
Public Function myaverage(x As Double, y As Double, z As Double) As Double
myaverage = 0.4 * x + 0.5 * y + 0.1 * z
End Function
Do not forget the function must be placed in Module (e.g. Module1). See Step 2 in this tutorial.
If you change the code, always re-test your function for language errors using menu Debug > Compile. If errors are shown, fix them.
If you want to use range A1:C1 as the input argument then UDF has to be written accordingly:
Public Function RangeAverage(rRef As Range) As Double
With rRef
RangeAverage = 0.4 * .Cells(1).Value2 + 0.5 * .Cells(2).Value2 + 0.1 * .Cells(3).Value2
End With
End Function
Related
Hey I have no idea why I get an error "run time error 13 type mismatch". Thats my code and the place where I get an error:
EDIT: That is my code:
Function payoff(S_T, K, CallPut As String)
If CallPut = "call" Then
omega = 1
Else: omega = -1
End If
payoff = WorksheetFunction.Max(omega * (S_T - K), 0)
End Function
Function BS_trajektoria(S_0 As Double, T As Double, r As Double, q As Double, sigma As Double, N As Long) As Double()
Randomize
Dim S() As Double
Dim delta_t As Double
Dim i As Long
ReDim S(N)
S(0) = S_0
delta_t = T / N
For i = 1 To N
S(i) = S(i - 1) * Exp((r - q - 0.5 * sigma ^ 2) * delta_t + sigma * delta_t ^ 0.5 * Application.NormSInv(Rnd))
Next i
BS_trajektoria = S
End Function
Function barrier_MC(S_0 As Double, K As Double, T As Double, r As Double, q As Double, sigma As Double, _
B As Double, N As Long, num_of_sim As Long, CallPut As String, BarType As String) As Double
Randomize
Dim max_value As Double
Dim suma_wyplat As Double
Dim wyplata As Double
Dim i As Long
Dim S() As Double
suma_wyplat = 0
If (BarType = "DO" Or BarType = "DI") And B > S_0 Then
MsgBox "Too high barrier!"
Exit Function
ElseIf (BarType = "UO" Or BarType = "UI") And B < S_0 Then
MsgBox "Too low barrier!"
Exit Function
End If
With WorksheetFunction
For i = 1 To num_of_sim
S = BS_trajektoria(S_0, T, r, q, sigma, N)
max_value = .Max(S)
If max_value >= B Then
wyplata = 0
Else
wyplata = payoff(S(N), K, CallPut)
End If
suma_wyplat = suma_wyplat + wyplata
Next i
End With
barrier_MC = Exp(-r * T) * suma_wyplat / num_of_sim
End Function
Sub test3()
MsgBox barrier_MC(100, 100, 1, 0.05, 0.02, 0.2, 120, 1000, 1000000, "call", "UO")
End Sub
Anyone know where is the problem? For smaller value of N and num_of_sim everything works fine, the problem is when I use bigger values for these variables.
If you declare a new Double variable called rand and modify the main loop so that it looks like:
For i = 1 To N
rand = Rnd
S(i) = S(i - 1) * Exp((r - q - 0.5 * sigma ^ 2) * delta_t + sigma * delta_t ^ 0.5 * Application.NormSInv(rand))
Next i
you will see that the problem always happens when rand = 0. Why it throws that particular error is a bit of a mystery, but it is what it is. As a fix, what you could do is to keep the code as modified above with the following twist:
For i = 1 To N
rand = Rnd
If rand = 0 Then rand = 0.0000001
S(i) = S(i - 1) * Exp((r - q - 0.5 * sigma ^ 2) * delta_t + sigma * delta_t ^ 0.5 * Application.NormSInv(rand))
Next i
Then the code will run without error. It is still somewhat slow, but optimizing it (if possible) would be for a different question.
I'm trying to calculate distance in kilometers between two geographical coordinates using the haversine formula.
Code:
Dim dbl_dLat As Double
Dim dbl_dLon As Double
Dim dbl_a As Double
dbl_P = WorksheetFunction.Pi / 180
dbl_dLat = dbl_P * (dbl_Latitude2 - dbl_Latitude1)
dbl_dLon = dbl_P * (dbl_Longitude2 - dbl_Longitude1)
dbl_a = Sin(dbl_dLat / 2) * Sin(dbl_dLat / 2) + Cos(dbl_Latitude1 * dbl_P) * Cos(dbl_Latitude2 * dbl_P) * Sin(dbl_dLon / 2) * Sin(dbl_dLon / 2)
dbl_Distance_KM = 6371 * 2 * WorksheetFunction.Atan2(Sqr(dbl_a), Sqr(1 - dbl_a))
I'm testing with these coordinates:
dbl_Longitude1 = 55.629178
dbl_Longitude2 = 29.846686
dbl_Latitude1 = 37.659466
dbl_Latitude2 = 30.24441
And the code returns 20015.09, which is obviously wrong. It should be 642 km according to Yandex maps.
Where am I wrong? Are the longitude and latitude in wrong format?
As far as I can tell, the issue is that the order of arguments to atan2() varies by language. The following works* for me:
Option Explicit
Public Sub Distance()
Dim dbl_Longitude1 As Double, dbl_Longitude2 As Double, dbl_Latitude1 As Double, dbl_Latitude2 As Double
dbl_Longitude1 = 55.629178
dbl_Longitude2 = 29.846686
dbl_Latitude1 = 37.659466
dbl_Latitude2 = 30.24441
Dim dbl_dLat As Double
Dim dbl_dLon As Double
Dim dbl_a As Double
Dim dbl_P As Double
dbl_P = WorksheetFunction.Pi / 180
dbl_dLat = dbl_P * (dbl_Latitude2 - dbl_Latitude1) 'to radians
dbl_dLon = dbl_P * (dbl_Longitude2 - dbl_Longitude1) 'to radians
dbl_a = Sin(dbl_dLat / 2) * Sin(dbl_dLat / 2) + _
Cos(dbl_Latitude1 * dbl_P) * Cos(dbl_Latitude2 * dbl_P) * Sin(dbl_dLon / 2) * Sin(dbl_dLon / 2)
Dim c As Double
Dim dbl_Distance_KM As Double
c = 2 * WorksheetFunction.Atan2(Sqr(1 - dbl_a), Sqr(dbl_a)) ' *** swapped arguments to Atan2
dbl_Distance_KM = 6371 * c
Debug.Print dbl_Distance_KM
End Sub
*Output: 2507.26205401321, although gcmap.com says the answer is 2512 km. This might be a precision issue --- I think it's close enough to count as working. (Edit it might also be that gcmap uses local earth radii rather than the mean radius; I am not sure.)
Explanation
I found this description of the haversine formula for great-circle distance, which is what you are implementing. The JavaScript implementation on that page gives this computation for c:
var c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a));
In JavaScript, atan2() takes parameters y, x. However, in Excel VBA, WorksheetFunction.Atan2 takes parameters x, y. Your original code passed Sqr(dbl_a) as the first parameter, as it would be in JavaScript. However, Sqr(dbl_a) needs to be the second parameter in Excel VBA.
A comment on naming
Building on #JohnColeman's point, there are lots of ways to name variables. In this case, I would recommend using the prefixes for unit rather than for type: e.g., deg_Latitude1, RadPerDeg = Pi/180, and rad_dLat = RadPerDeg * (deg_Latitude2 - deg_Latitude1). I personally think that helps avoid unit-conversion mishaps.
My VBA code that returns the answer in feet; However 'd' is the answer in kilometers.
Imports System.Math
Module Haversine
Public Function GlobalAddressDistance(sLat1 As String, sLon1 As String, sLat2 As String, sLon2 As String) As String
Const R As Integer = 6371
Const cMetersToFeet As Single = 3.2808399
Const cKiloMetersToMeters As Integer = 1000
Dim a As Double = 0, c As Double = 0, d As Double = 0
'Convert strings to numberic double values
Dim dLat1 As Double = Val(sLat1)
Dim dLat2 As Double = Val(sLat2)
Dim dLatDiff As Double = DegreesToRadians(CDbl(sLat2) - CDbl(sLat1))
Dim dLonDiff As Double = DegreesToRadians(CDbl(sLon2) - CDbl(sLon1))
a = Pow(Sin(dLatDiff / 2), 2) + Cos(DegreesToRadians(dLat1)) * Cos(DegreesToRadians(dLat2)) * Pow(Sin(dLonDiff / 2), 2)
c = 2 * Atan2(Sqrt(a), Sqrt(1 - a))
d = R * c
'Convert kilometers to feet
Return Format((d * cKiloMetersToMeters * cMetersToFeet), "0.##").ToString
End Function
Private Function DegreesToRadians(ByVal dDegrees As Double) As Double
Return (dDegrees * PI) / 180
End Function
End Module
It would be nice if someone could explain what causes function above return #value error.
Public Function papild(x)
Dim Sum As Double, A As Double, pi As Double,
Sum = 0.5 - (x - pi / 4)
A = -(x - pi / 4)
pi = Application.WorksheetFunction.pi()
Dim k As Integer, i As Integer
k = 2
i = 0
Do While Abs(A) > 0.0001
A = -A * 4 * A * A / (k + i) * (k + i + 1)
Sum = Sum + A
k = k + 1
i = i + 1
Loop
paplid = Sum
End Function
Function takes x value from MS Excel cell and it's equal = -1.5708 (=-PI()/2 #Formula Bar)
In lines 3 and 4 you work with variable pi before setting it in line 5...
Could there be some brackets missing in your formula. It basically says:
A = -4A^3 * (k+i+1)/(k+1)
This obviously drifts to +/- infinite so your loop cannot end.
Also there is a comma too much in the second line and a spelling error in the last line (paplid instead of papild).
Have you tried debugging the code?
When I run the code I get an overflow error # the 6th iteration of the while loop starting with x = -1.5708. Number gets to large to fit inside variable
.Other than that there are some minor things:
missing As Double
Public Function papild(x) As Double
and unnecessary comma at the end
Dim Sum As Double, A As Double, pi As Double,
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^(1.5sin(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.