I'm running the following program, and get an overflow error at the first iteration once the program gets to a. From my understanding this shouldn't happen because I'm using Doubles which have a ridiculous amount of capacity, and a Long, but because it's only going up to 100 currently it shouldn't matter, and the code fails long before then. Here is my original inputs:
h0 = 1000, v0 = 0, a0 = g = -9.80665, dt = .01, m = 752.2528, b = .287875
and here's the code:
Sub drag()
Dim h0 As Double
Dim v0 As Double
Dim a0 As Double
Dim dt As Double
Dim m As Double
Dim b As Double
Dim g As Double
Dim i As Long
Dim h As Double
Dim v As Double
Dim a As Double
h0 = Worksheets("Drag").Cells(2, 8).Value
v0 = Worksheets("Drag").Cells(2, 9).Value
a0 = Worksheets("Drag").Cells(2, 10).Value
g = Worksheets("Drag").Cells(2, 10).Value
dt = Worksheets("Drag").Cells(2, 7).Value
m = Worksheets("Drag").Cells(2, 4).Value
b = Worksheets("Drag").Cells(2, 5).Value
Debug.Print h0 & v0 & a0 & dt & m & b
For i = 1 To 100
v = v0 + a0 * dt
h = 0.5 * a0 * (dt ^ 2) + v0 * dt + h0
a = m * g - b * (v ^ 2) 'Line where overflow occurs
v0 = v
h0 = h
a0 = a
Cells(i + 2, 8) = h0
Cells(i + 2, 9) = v0
Cells(i + 2, 10) = a0
Next i
Debug.Print h0 & v0 & a0 & dt & m & b
End Sub
Hope this is an easy fix.
Your overflow happens when i=14 (14th pass through the loop) and when v = -1.689 x 10^209. I don't know what you are trying to do, but v is blowing up. It would help if you describe what you are trying to do. – John Coleman 17 mins ago
#John Coleman I see the problem now, I'm doing the drag equation and I forgot to divide the a term by m, thanks. – Anthrochange 9 mins ago
You have already identified what needs to be done.
Now for the explanation on why the overflow on m * g - b * (v ^ 2). I added a watch as shown below
Consider the before and after screenshot
Before v = v0 + a0 * dt is calculated
After v = v0 + a0 * dt is calculated
What you see here is a very peculiar behaviour. A Type Double changing to an Integer
This is not normal and has been experienced before. Unfortunately this problem exists in Excel for a long time and unfortunately is also present in Excel 2016. It occurs only when you are using very large numbers. These kind of bugs are very rare, but yes, they do exist.
One can have a similar kind of experience when using MOD as mentioned in the below link
The MOD Function
The variable in question that is getting an overflow is v. You defined it as a double, which means you can only put numbers in the range of 1.79769313486232 * 10^308 to 4.94065645841247 * 10^–324. VBA uses 8 bytes to hold the double data. Any number outside of this huge range causes an overflow. (see here for more information about DBA data types)
How big of a number are you expecting? A rough estimate fo the number of atoms in the universe is 1 * 10^82, and the double value range goes to almost 2 *10^308. If you need to work with numbers that large, you're going to have to use a different method because it simply won't fit in the standard VBA data types.
Related
I am trying to populate and then perform calculations based on a rather large array. This is to populate a binomial tree to calculate option price, so after populating the array I would need to perform repeated calculation so I'd prefer having 3 dimensions in my array for easy reference further down in the code. First dimension is the number of periods that have passed, second dimension is the number of price increases and the third is the number of price decreases.
Dim arr() As Double
Dim periods As Integer
Dim p As Long, i As Long
Dim u As Double, d As Double
Dim iniprice As Double
Let periods = 400
Let iniprice = 100
Let u = 1.1
Let d = 0.9
ReDim arr(0 To periods, 0 To periods, 0 To periods)
Let arr(0, 0, 0) = iniprice
For p = 1 To UBound(arr, 1)
For i = 0 To p
arr(p, i, p - i) = WorksheetFunction.RoundDown(arr(0, 0, 0) * u ^ i * d ^ (p - i), 2)
Next i
Next p
Is this a limitation stemming from the amount of RAM available on my PC (currently having 8Gb) or is this a limitation of VBA itself? Since one period is usually one day, a periods value of 1000 is normal (252 days = 1 trading year).
I also noticed that I have a lot of unused values, because I want to populate only values that have this format arr(p, i, p-i), values such as arr(10,10,10) will be 0. I'd greatly appreciate a workaround to this.
You can try using a single lookup column to represent the 3D array (not the best way but helps with the memory errors):
Dim periods As Integer
Dim p As Long, i As Long
Dim u As Double, d As Double
Dim iniprice As Double
Dim d1 As Long, d2 As Long, d3 As Long, someRow As Long
Dim fndRange As Range
Let periods = 400
Let iniprice = 100
Let u = 1.1
Let d = 0.9
' Col A will be your lookup in format "X-X-X", Col B will hold values
With Sheets("some sheet")
.Range("B1") = iniprice
For d1 = 0 To periods
For d2 = 0 To periods
For d3 = 0 To periods
.Range("A" & someRow).Value = d1 & "-" & d2 & "-" & d3
someRow = someRow + 1
Next
Next
Next
End With
For p = 1 To periods
For i = 0 To p
Set fndRange = Sheets("some sheet").Columns(1).Find(p & "-" & i & "-" & (p-1))
fndRange.Offset(0,1).Value = WorksheetFunction.RoundDown(iniprice * u ^ i * d ^ (p - i), 2)
Next i
Next p
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
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
I use vba on excel 2007, OS: windows vista, to make calculation using kinematic wave equation in finite difference scheme. But, when it runs the run-time 5 (invalid procedure call or arguments) message appears. I really don't what is going wrong. Anyone can help?
Sub kwave()
Dim u(500, 500), yy(500, 500), alpha, dt, dx, m, n, so, r, f, X, L, K As Single
Dim i, j As Integer
dx = 0.1
dt = 0.01
L = 10
m = 5 / 3
r = 1
f = 0.5
n = 0.025
so = 0.1 'this is slope
alpha = 1 / n * so ^ 0.5
X = 0
For i = 0 To 100
Cells(i + 1, 1) = X
u(i, 1) = L - so * X
X = X + dx
Cells(i + 1, 2) = u(i, 1)
Next i
For j = 0 To 100
For i = 1 To 100
'predictor step
u(i, j + 1) = u(i, j) - alpha * dt / dx * (u(i + 1, j) ^ m - u(i, j) ^ m) + (r - f) * dt
'corrector step
K = u(i, j + 1) ^ m - u(i - 1, j + 1) ^ m '<<<<----- RUNTIME ERROR 5 HAPPENS AT THIS LINE
yy(i, j + 1) = 0.5 * ((yy(i, j) + u(i, j + 1)) - alpha * dt / dx * K + (r - f) * dt)
Next i
Next j
End Sub
You are declaring the variables wrong- the array should store a double/single but it is defaulting to a variant. See this article.
http://www.cpearson.com/excel/declaringvariables.aspx -
"Pay Attention To Variables Declared With One Dim Statement
VBA allows declaring more than one variable with a single Dim
statement. I don't like this for stylistic reasons, but others do
prefer it. However, it is important to remember how variables will be
typed. Consider the following code:
Dim J, K, L As Long You may think that all three variables are
declared as Long types. This is not the case. Only L is typed as a
Long. The variables J and K are typed as Variant. This declaration is
functionally equivalent to the following:
Dim J As Variant, K As Variant, L As Long You should use the As Type
modifier for each variable declared with the Dim statement:
Dim J As Long, K As Long, L As Long "
Additionally, when i = 99 and j = 10, u(99,11), which is j+1, produces a negative number. Note that this does not fully cause the problem though, because you can raise negative numbers to exponents. Ex, -5^3 = -125
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.