Excel Linear Interpolation VBA - vba

this function interpolates/extrapolates a table of known x,y
For example,
x y
1 10
2 15
3 20
Linterp(A1:B3, -1) = 0
However, this code can only do two adjacent arrays.
I would like to modify this code so that I can
select two separate arrays, for example N106:N109,P106:P109.
How can I make this adjustment in this code?
Function Linterp(r As Range, x As Double) As Double
' linear interpolator / extrapolator
' R is a two-column range containing known x, known y
Dim lR As Long, l1 As Long, l2 As Long
Dim nR As Long
'If x = 1.5 Then Stop
nR = r.Rows.Count
If nR < 2 Then Exit Function
If x < r(1, 1) Then ' x < xmin, extrapolate
l1 = 1: l2 = 2: GoTo Interp
ElseIf x > r(nR, 1) Then ' x > xmax, extrapolate
l1 = nR - 1: l2 = nR: GoTo Interp
Else
' a binary search would be better here
For lR = 1 To nR
If r(lR, 1) = x Then ' x is exact from table
Linterp = r(lR, 2)
Exit Function
ElseIf r(lR, 1) > x Then ' x is between tabulated values, interpolate
l1 = lR: l2 = lR - 1: GoTo Interp
End If
Next
End If
Interp:
Linterp = r(l1, 2) _
+ (r(l2, 2) - r(l1, 2)) _
* (x - r(l1, 1)) _
/ (r(l2, 1) - r(l1, 1))
End Function

one very simple way is having the function accepting two ranges in input, one for X values (say rX) and one for Y ones (say rY), and then changing every occurrence of r(foo,1) to rX(foo) and r(foo,2) to rY(foo)
like follows
Option Explicit
Function Linterp2(rX As Range, rY As Range, x As Double) As Double
' linear interpolator / extrapolator
' R is a two-column range containing known x, known y
Dim lR As Long, l1 As Long, l2 As Long
Dim nR As Long
'If x = 1.5 Then Stop
nR = rX.Rows.Count
If nR < 2 Then Exit Function
If x < rX(1) Then ' x < xmin, extrapolate
l1 = 1: l2 = 2: GoTo Interp
ElseIf x > rX(nR) Then ' x > xmax, extrapolate
l1 = nR - 1: l2 = nR: GoTo Interp
Else
' a binary search would be better here
For lR = 1 To nR
If rX(lR) = x Then ' x is exact from table
Linterp2 = rY(lR)
Exit Function
ElseIf rX(lR) > x Then ' x is between tabulated values, interpolate
l1 = lR: l2 = lR - 1: GoTo Interp
End If
Next
End If
Interp:
Linterp2 = rY(l1) _
+ (rY(l2) - rY(l1)) _
* (x - rX(l1)) _
/ (rX(l2) - rX(l1))
End Function
but you must implement code to check for consistency of the two ranges, like being both of one column each and with the same number of rows

use this function :
Public Function lineare_iterpolation(x As Variant, x1 As Variant, x2 As Variant, y1 As Variant, y2 As Variant) As Variant
If x = x1 Then
lineare_iterpolation = y1
Exit Function
End If
If x = x2 Then
lineare_iterpolation = y2
Exit Function
End If
lineare_iterpolation = y1 + (x - x1) * (y2 - y1) / (x2 - x1)
Exit Function
End Function

Related

Closest distance between lat/longs in large dataset in excel vba

Beginner looper here...I am working on this well spacing project that looks at lat/longs and determines the next closest well. I think I may be creating an infinite loop or the program is just taking forever to run (It's looping through 15,000 rows). My main struggle has been trying to make sure each location is compared to every location in the dataset. From there I take the 2nd lowest distance (since the lowest will be zero when it compares to itself).
Sub WellSpacing()
Dim r As Integer, c As Integer, L As Integer, lastrow As Integer
Dim lat1 As Double, lat2 As Double, long1 As Double, long2 As Double
Dim distance As Double, d1 As Double, d2 As Double, d3 As Double
Dim PI As Double
PI = Application.WorksheetFunction.PI()
L = 2
r = 3
c = 10
lastrow = Sheets("Test").Cells(Rows.Count, "J").End(xlUp).Row
For L = 2 To lastrow
For r = 2 To lastrow
lat1 = Sheets("Test").Cells(L, c)
long1 = Sheets("Test").Cells(L, c + 1)
lat2 = Sheets("Test").Cells(r, c)
long2 = Sheets("Test").Cells(r, c + 1)
d1 = Sin((Abs((lat2 - lat1)) * PI / 180 / 2)) ^ 2 + Cos(lat1 * PI / 180) * Cos(lat2 * PI / 180) * Sin(Abs(long2 - long1) * PI / 180 / 2) ^ 2
d2 = 2 * Application.WorksheetFunction.Atan2(Sqr(1 - d1), Sqr(d1))
d3 = 6371 * d2 * 3280.84
Sheets("Working").Cells(r - 1, c - 9) = d3
Next r
Sheet2.Activate
Range("A:A").Sort Key1:=Range("A1"), Order1:=xlAscending
distance = Sheet2.Range("A2")
Sheets("Test").Cells(L, c + 2) = distance
Sheet2.Range("A:A").Clear
Sheet1.Activate
Next L
End Sub
I've been working with geo-location math (aka, coordinate geometry) a lot lately and wrote a sub to do pretty much the same thing you're seeking.
Your code probably isn't creating an infinite loop, but calculating distances between thousands of coordinates can be very processor-intensive and even minor changes to your code can have a huge impact on processing time.
Calculating closest coordinate pair: Brute Force Method
There are a number of algorithms for determining closest points however the easiest to code (therefore possibly best for one-time use) is known as the Brute Force Method.
For p1 = 1 to numPoints
For p2 = p1 + 1 to numPoints
...calculate {distance}
...if {distance} < minDistance then minDist = {distance}
Next p2
Next p1
Using this method, distance will be calculated between x * ( n - 1 ) / 2 pairs of points.
For example, a list of 5 points would require 10 comparisons:
Point 1 ↔ Point 2
Point 1 ↔ Point 3
Point 1 ↔ Point 4
Point 1 ↔ Point 5
Point 2 ↔ Point 3
Point 2 ↔ Point 4
Point 2 ↔ Point 5
Point 3 ↔ Point 4
Point 3 ↔ Point 5
Point 4 ↔ Point 5
Since additional points will increase execution time exponentially, this method can create some lengthy processing times, especially on a slower machine or with an excessive number of points.
The methods I use for calculating distances between points and for comparing distances between lists of points are far from the [code-heavier] most-efficient alternatives, but they work for my "one-off" needs.
Depending on my purposes, I'll switch (almost identical code) between Excel & Access, but Access is much faster, so you may want to move your list into a table and do it that way.
One of the lists of points I compare has 252 items, which requires 31,628 individual comparisons using this "easy-code" method. In Excel, the process completes in 1.12 seconds, which is Access it only takes 0.16 seconds.
This may not seem like a big difference until we starting working with longer lists of points: another list of mine (closer to the size of yours) has about 12,000 points, which requires 71,994,000 calculations using the Brute Force method. In Access, the process completes in 8.6 minutes, so I estimate it would take about an hour in Excel.
Of course, all of these times are based on my operating system, processing power, Office version, etc. VBA isn't ideal for this level of computation, and everything you can do to reduce length of code will make a big difference, including commenting-out the status bar updates, immediate-window output, turn off screen updates, etc.
This code is a little messy & un-commented since I slapped it together for my own purposes, but it works for me. Let me know if you have any questions about how it works. All calculations are in metric but can be easily converted.
Sub findShortestDist_Excel()
Const colLatitude = "C" ' Col.C = Lat, Col.D = Lon
Dim pointList As Range, pointCount As Long, c As Range, _
arrCoords(), x As Long, y As Long
Dim thisDist As Double, minDist As Double, minDist_txt As String
Dim cntCurr As Long, cntTotal As Long, timerStart As Single
timerStart = Timer
Set pointList = Sheets("Stops").UsedRange.Columns(colLatitude)
pointCount = WorksheetFunction.Count(pointList.Columns(1))
'build array of numbers found in Column C/D
ReDim arrCoords(1 To 3, 1 To pointCount)
For Each c In pointList.Columns(1).Cells
If IsNumeric(c.Value) And Not IsEmpty(c.Value) Then
x = x + 1
arrCoords(1, x) = c.Value
arrCoords(2, x) = c.Offset(0, 1).Value
End If
Next c
minDist = -1
cntTotal = pointCount * (pointCount + 1) / 2
'loop through array
For x = 1 To pointCount
For y = x + 1 To pointCount
If (arrCoords(1, x) & arrCoords(2, x)) <> (arrCoords(1, y) & arrCoords(2, y)) Then
cntCurr = cntCurr + 1
thisDist = Distance(arrCoords(1, x), arrCoords(2, x), _
arrCoords(1, y), arrCoords(2, y))
'check if this distance is the smallest yet
If ((thisDist < minDist) Or (minDist = -1)) And thisDist > 0 Then
minDist = thisDist
'minDist_txt = arrCoords(1, x) & "," & arrCoords(2, x) & " -> " & arrCoords(1, y) & "," & arrCoords(2, y)
End If
'Application.StatusBar = "Calculating Distances: " & Format(cntCurr / cntTotal, "0.0%")
End If
Next y
'DoEvents
Next x
Debug.Print "Minimum distance: " & minDist_txt & " = " & minDist & " meters"
Debug.Print "(" & Round(Timer - timerStart, 2) & "sec)"
Application.StatusBar = "Finished. Minimum distance: " & minDist_txt & " = " & minDist & "m"
End Sub
Note that the procedure above is dependent on the following (which has slightly different versions for Access vs. Excel):
Excel: Calculate distance between points
Public Function Distance(ByVal lat1 As Double, ByVal lon1 As Double, _
ByVal lat2 As Double, ByVal lon2 As Double) As Double
'returns Meters distance in Excel (straight-line)
Dim theta As Double: theta = lon1 - lon2
Dim Dist As Double: Dist = Math.Sin(deg2rad(lat1)) * Math.Sin(deg2rad(lat2)) + Math.Cos(deg2rad(lat1)) * Math.Cos(deg2rad(lat2)) * Math.Cos(deg2rad(theta))
Dist = rad2deg(WorksheetFunction.Acos(Dist))
Distance = Dist * 60 * 1.1515 * 1.609344 * 1000
End Function
Function deg2rad(ByVal deg As Double) As Double
deg2rad = (deg * WorksheetFunction.PI / 180#)
End Function
Function rad2deg(ByVal rad As Double) As Double
rad2deg = rad / WorksheetFunction.PI * 180#
End Function
...and alternative code, for Microsoft Access:
Access: Shortest Distance
Sub findShortestDist_Access()
Const tableName = "Stops"
Dim pointCount As Long, arrCoords(), x As Long, y As Long
Dim thisDist As Double, minDist As Double
Dim cntCurr As Long, cntTotal As Long, timerStart As Single
Dim rs As Recordset
timerStart = Timer
Set rs = CurrentDb.OpenRecordset("SELECT * FROM " & tableName)
With rs
.MoveLast
.MoveFirst
pointCount = .RecordCount
'build array of numbers found in Column C/D
ReDim arrCoords(1 To 2, 1 To pointCount)
Do While Not .EOF
x = x + 1
arrCoords(1, x) = !stop_lat
arrCoords(2, x) = !stop_lon
.MoveNext
Loop
.Close
End With
minDist = -1
cntTotal = pointCount * (pointCount + 1) / 2
SysCmd acSysCmdInitMeter, "Calculating Distances:", cntTotal
'loop through array
For x = 1 To pointCount
For y = x + 1 To pointCount
cntCurr = cntCurr + 1
thisDist = Distance(arrCoords(1, x), arrCoords(2, x), _
arrCoords(1, y), arrCoords(2, y))
'check if this distance is the smallest yet
If ((thisDist < minDist) Or (minDist = -1)) And thisDist > 0 Then
minDist = thisDist
End If
SysCmd acSysCmdUpdateMeter, cntCurr
Next y
DoEvents
Next x
SysCmd acSysCmdRemoveMeter
Debug.Print "Minimum distance: " & minDist_txt & " = " & minDist & " meters"
Debug.Print "(" & Round(Timer - timerStart, 2) & "sec)"
End Sub
Note that the procedure above is dependent on the following... (Access may handle mass-calculations more quickly, but we have to build some functions ourselves that are built-in to Excel)
Access: Calculate distance between points
Const pi As Double = 3.14159265358979
Public Function Distance(ByVal lat1 As Double, ByVal lon1 As Double, _
ByVal lat2 As Double, ByVal lon2 As Double) As Double
'returns Meters distance in Access (straight-line)
Dim theta As Double: theta = lon1 - lon2
Dim dist As Double
dist = Math.Sin(deg2rad(lat1)) * Math.Sin(deg2rad(lat2)) + Math.Cos(deg2rad(lat1)) _
* Math.Cos(deg2rad(lat2)) * Math.Cos(deg2rad(theta))
dist = rad2deg(aCos(dist))
Distance = dist * 60 * 1.1515 * 1.609344 * 1000
End Function
Function deg2rad(ByVal deg As Double) As Double
deg2rad = (deg * pi / 180#)
End Function
Function rad2deg(ByVal rad As Double) As Double
rad2deg = rad / pi * 180#
End Function
Function aTan2(x As Double, y As Double) As Double
aTan2 = Atn(y / x)
End Function
Function aCos(x As Double) As Double
On Error GoTo aErr
If x = 0 Or Abs(x) = 1 Then
aCos = 0
Else
aCos = Atn(-x / Sqr(-x * x + 1)) + 2 * Atn(1)
End If
Exit Function
aErr:
aCos = 0
End Function
Planar Case
Another method of calculating closer points is called Planar Case. I haven't seen any ready-to-use code samples and I don't need it bad enough to bother coding it, but the gist of it is this:
Read about this and more about the Closest pair of points problem on Wikipedia.
I would recommend using arrays as #Qharr said. I would also look to speed up the process by including some logic steps that avoid doing the complex math on every set of points.
What I mean is that you can do a Rough Estimate first to see whether or not to bother doing the actual calculations. I went with looking at whether or not either the Lat or Long of the current position is closer than the last closest point, but you could do anything you wanted.
I would change your code to something like:
Sub WellSpacing()
Dim R As Integer, C As Integer, L As Integer, LastRow As Integer, Shortest() As Integer
Dim Lats() As Double, Longs() As Double, Distances() As Double
Dim Distance As Double, D1 As Double, D2 As Double, D3 As Double
Dim PI As Double
On Error Resume Next
PI = Application.WorksheetFunction.PI()
L = 2
R = 3
C = 10
LastRow = Sheets("Test").Cells(Rows.Count, 10).End(xlUp).Row
ReDim Lats(1 To (LastRow - 1)) As Double
ReDim Longs(1 To (LastRow - 1)) As Double
ReDim Distances(1 To (LastRow - 1)) As Double
ReDim Shortest(1 To (LastRow - 1)) As Integer
For L = 2 To LastRow
Lats(L - 1) = Sheets("Test").Range("J" & L).Value
Longs(L - 1) = Sheets("Test").Range("K" & L).Value
Next L
For L = 1 To (LastRow - 1)
'This is a method of setting an initial value that can't be obtained through the caclucations (so you will know if any calcs have been done or not).
Distances(L) = -1
For R = 1 To (LastRow - 1)
'This minimises your calculations by 15,000 to begin with
If R = L Then GoTo Skip_This_R
'This skips checking the previous distances if it is the first calculation being checked.
If Distances(L) = -1 Then GoTo Skip_Check
'If there has already been a distance calculated, this does a rough check of whether the Lat or Long is closer. If neither
'the Lat or Long are closer than the current closest, then it will skip it. This reduces the code by 7 lines for most pairs.
If Abs(Lats(L) - Lats(R)) < Abs(Lats(L) - Lats(Shortest(L))) Or Abs(Longs(L) - Longs(R)) < Abs(Longs(L) - Longs(Shortest(L))) Then
Skip_Check:
D1 = Sin((Abs((Lats(R) - Lats(L))) * PI / 180 / 2)) ^ 2 + Cos(Lats(L) * PI / 180) * Cos(Lats(R) * PI / 180) * Sin(Abs(Longs(R) - Longs(L)) * PI / 180 / 2) ^ 2
D2 = 2 * Application.WorksheetFunction.Atan2(Sqr(1 - D1), Sqr(D1))
D3 = 6371 * D2 * 3280.84
If D3 < Distances(L) Or Distances(L) = -1 Then
Distances(L) = D3
'This stores the index value in the array of the closest Lat/Long point so far.
Shortest(L) = R
End If
End If
Skip_This_R:
Next R
'This puts the resulting closest distance into the corresponding cell.
Sheets("Test").Range("L" & (L + 1)).Value = Distances(L)
'This clears any previous comments on the cell.
Sheets("Test").Range("L" & (L + 1)).Comments.Delete
'This adds a nice comment to let you know which Lat/Long position it is closest to.
Sheets("Test").Range("L" & (L + 1)).AddComment "Matched to Row " & (Shortest(L) + 1)
Next L
End Sub

Find the indices for the minimum values in a multi dimensional array in VBA

In the code below I have an n x n x n array of values. I need to identify the indices that contain the minimum, second to minimum, third to minimum, ..., and put them into their own array to be used later on in the code. CC is currently defined as a 11 x 11 x 11 array and I need to identify the minimums. Below is the setup of my array CC that contains the values. n is defined as the length of the array h2s, which happens to be 11 in this case. h2st is the sum of the values in h2s.
h2s = [1.099, 0.988, 0.7, 0.8, 0.5, 0.432, 0.8, 1.12, 0.93, 0.77, 0.658]
h2st = 0
n = Ubound(h2s) - Lbound(h2s) + 1
For i = 1 to n
h2st = h2st + h2s(i)
Next i
For i = 1 To n
For j = i + 1 To n
For k = j + 1 To n
CC(i, j, k) = Abs(h2st - ((h2s(i) + h2s(j) + h2s(k)) * (n / 3)))
Next k
Next j
Next i
You can use this function that takes a multidimensional array and returns an array of its n minimum values, where n is a parameter. Importantly, the elements in the returned array are a data structure of Type Point, containing the coordinates and the value of each found point.
You can easily adjust it for finding the n max values, just by changing two characters in the code, as indicated in comments (the initialization and the comparison)
Option Explicit
Type Point
X As Long
Y As Long
Z As Long
value As Double
End Type
Function minVals(ar() As Double, nVals As Long) As Point()
Dim i As Long, j As Long, k As Long, m As Long, n As Long, pt As Point
'Initialize returned array with max values.
pt.value = 9999999# ' <-------- change to -9999999# for finding max
ReDim ret(1 To nVals) As Point
For i = LBound(ret) To UBound(ret): ret(i) = pt: Next
For i = LBound(ar, 1) To UBound(ar, 1)
For j = LBound(ar, 2) To UBound(ar, 2)
For k = LBound(ar, 3) To UBound(ar, 3)
' Find first element greater than this value in the return array
For m = LBound(ret) To UBound(ret)
If ar(i, j, k) < ret(m).value Then ' <------- change to > for finding max
' shift the elements on this position and insert the current value
For n = UBound(ret) To m + 1 Step -1: ret(n) = ret(n - 1): Next n
pt.X = i: pt.Y = j: pt.Z = k: pt.value = ar(i, j, k)
ret(m) = pt
Exit For
End If
Next m
Next k
Next j
Next i
minVals = ret
End Function
Sub Test()
Dim i As Long, j As Long, k As Long, pt As Point
Const n As Long = 11
ReDim CC(1 To n, 1 To n, 1 To n) As Double
For i = 1 To n
For j = 1 To n
For k = 1 To n
CC(i, j, k) = Application.RandBetween(100, 100000)
Next k
Next j
Next i
' Testing the function: get the smalles 5 values and their coordinates
Dim mins() As Point: mins = minVals(CC, 5)
' Printing the results
For i = LBound(mins) To UBound(mins)
Debug.Print mins(i).value, mins(i).X, mins(i).Y, mins(i).Z
Next
End Sub

VBA - Modular exponentiation

I've been trying to do Modular exponentiation in VBA for use in MS excel, but there seems to be a logical error which crashes Excel everytime i try to use the formula.
Function expmod(ax As Integer, bx As Integer, cx As Integer)
' Declare a, b, and c
Dim a As Integer
Dim b As Integer
Dim c As Integer
' Declare new values
Dim a1 As Integer
Dim p As Integer
' Set variables
a = ax
b = bx
c = cx
a1 = a Mod c
p = 1
' Loop to use Modular exponentiation
While b > 0
a = ax
If (b Mod 2 <> 0) Then
p = p * a1
b = b / 2
End If
a1 = (a1 * a1) Mod c
Wend
expmod = a1
End Function
I used the pseudocode which was provided here.
Here is an implementation I wrote a while back. Using Long rather than Integer enables it to handle higher exponents:
Function mod_exp(alpha As Long, exponent As Long, modulus As Long) As Long
Dim y As Long, z As Long, n As Long
y = 1
z = alpha Mod modulus
n = exponent
'Main Loop:
Do While n > 0
If n Mod 2 = 1 Then y = (y * z) Mod modulus
n = Int(n / 2)
If n > 0 Then z = (z * z) Mod modulus
Loop
mod_exp = y
End Function

The results of my functions when I call a spline function gives wrong values

I have a function that only call the spline function when something happens..in this case when a division is less than zero..the inputs for the function is the same that for the spline function(called CUBIC), the spline was tested and works well when I call it direct! someone can help me?...follows a party of the code
Function NDF6(T As Variant, dias As Variant, taxas As Variant)
If T <= dias(1) Then
NDF6 = taxas(1)
Exit Function
End If
If T >= dias(tam) Then
NDF6 = taxas(tam)
Exit Function
End If
For i = 1 To tam
If T <= dias(i) Then
If taxas(i) / taxas(i - 1) < 0 Then
Call CUBIC(T, dias, taxas)
Else
i0 = ((taxas(i - 1) * dias(i - 1)) / 360) + 1
i1 = ((taxas(i - 1) * dias(i - 1)) / 360) + 1
irel = i1 / i0
i2 = irel ^ ((T - dias(i - 1)) / (dias(i) - dias(i - 1)))
i2rel = i2 * i0
i2real = i2rel - 1
NDF6 = i2real * (360 / T)
End If
Public Function CUBIC(x As Variant, input_column As Variant, output_column As Variant)
The function returns a zero value when I call the cubic function. The inputs are a cell with a value with a value equivalent a day and two arrays(DUONOFF and ONOFF) equivalent a days and rates, I call the function like:
NDF6(512,DUONOFF,ONOFF)
follows the CUBIC function
Public Function CUBIC(x As Variant, input_column As Variant, output_column As Variant)
'Purpose: Given a data set consisting of a list of x values
' and y values, this function will smoothly interpolate
' a resulting output (y) value from a given input (x) value
' This counts how many points are in "input" and "output" set of data
Dim input_count As Integer
Dim output_count As Integer
input_count = input_column.Rows.Count
output_count = output_column.Rows.Count
Next check to be sure that "input" # points = "output" # points
If input_count <> output_count Then
CUBIC = "Something's messed up! The number of indeces number of output_columnues don't match!"
GoTo out
End If
ReDim xin(input_count) As Single
ReDim yin(input_count) As Single
Dim c As Integer
For c = 1 To input_count
xin(c) = input_column(c)
yin(c) = output_column(c)
Next c
values are populated
Dim N As Integer 'n=input_count
Dim i, k As Integer 'these are loop counting integers
Dim p, qn, sig, un As Single
ReDim u(input_count - 1) As Single
ReDim yt(input_count) As Single 'these are the 2nd deriv values
N = input_count
yt(1) = 0
u(1) = 0
For i = 2 To N - 1
sig = (xin(i) - xin(i - 1)) / (xin(i + 1) - xin(i - 1))
p = sig * yt(i - 1) + 2
yt(i) = (sig - 1) / p
u(i) = (yin(i + 1) - yin(i)) / (xin(i + 1) - xin(i)) - (yin(i) - yin(i - 1)) / (xin(i) - xin(i - _1))
u(i) = (6 * u(i) / (xin(i + 1) - xin(i - 1)) - sig * u(i - 1)) / p
Next i
qn = 0
un = 0
yt(N) = (un - qn * u(N - 1)) / (qn * yt(N - 1) + 1)
For k = N - 1 To 1 Step -1
yt(k) = yt(k) * yt(k + 1) + u(k)
Next k
now eval spline at one point
Dim klo, khi As Integer
Dim h, b, a As Single
first find correct interval
klo = 1
khi = N
Do
k = khi - klo
If xin(k) > x Then
khi = k
Else
klo = k
End If
k = khi - klo
Loop While k > 1
h = xin(khi) - xin(klo)
a = (xin(khi) - x) / h
b = (x - xin(klo)) / h
y = a * yin(klo) + b * yin(khi) + ((a ^ 3 - a) * yt(klo) + (b ^ 3 - b) * yt(khi)) * (h ^ 2) _/ 6
CUBIC = y
out:
End Function

I'm having overflow issues in this two-variable optimization program

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.