I'm trying to calculate a cross currency rate by simply summing the forex rates for A/B and B/C and multiplying the means to find the A/C rate and I keep getting 0 in return. This is the code:
Function forex(audData As Range, euData As Range)
a = Application.Count(audData)
e = Application.Count(euData)
'Counts how many values are in the data
aSum = 0
eSum = 0
aMean = 0
eMean = 0
For i = 1 To aud ' This sums the 1st forex rate and finds the mean
aSum = aSum + audData(i)
Next i
aMean = aSum / a
For i = 1 To eu ' This sums the 2nd forex rate and finds the mean
eSum = eSum + euData(i)
Next i
eMean = eSum / e
forex = (aMean * eMean)
End Function
You could use the Average() function:
Function forex(audData As Range, euData As Range)
With WorksheetFunction
forex = .Average(audData) * .Average(euData)
End With
End Function
You seem to be a victim of undeclared variables. You have two loops,
For i = 1 To aud
and (later on)
For i = 1 To eu
where neither aud nor eu are declared. Thus, they default to variants with an implicit value of 0, hence neither of these loops ever execute and all your variables stay at 0.
You really should get in the habit of using Option Explicit at the top of all of your modules. This can be done automatically by enabling the option Require Variable Declarations in the VBA editor options. In the long run, it will save you hours of debugging time.
I can't test your code, but if you declare your variables and replace aud and eu by what I think you meant you would get:
Function forex(audData As Range, euData As Range) As Double
Dim a As Long, e As Long, aSum As Double, eSum As Double, aMean As Double, eMean As Double, i As Long
a = Application.Count(audData)
e = Application.Count(euData)
'Counts how many values are in the data
For i = 1 To a ' This sums the 1st forex rate and finds the mean
aSum = aSum + audData(i)
Next i
aMean = aSum / a
For i = 1 To e ' This sums the 2nd forex rate and finds the mean
eSum = eSum + euData(i)
Next i
eMean = eSum / e
forex = (aMean * eMean)
End Function
I skipped the lines like aSum = 0 since properly declared VBA variables have reasonable default values.
Related
Suppose in column Z with 200 rows, are my optimal averages.
Now I want a macro that generates n random integers between a and b inclusive (n <= 20) so that difference between the average of numbers generated with optimal average is in (-0.15,+0.15).
Example:
Z1:optimal average1=5.5
Z2:optimal average2=5.3
Z200:optimal average200=6.3
n=8
a=1; b=10
numbers of generated:
A1:H1)5-9-4-3-7-4-9-3
A2:H2)10-7-3-2-5-4-3-9
.
.
.
A200:H200)4-8-9-6-6-6-10-2
Here is a hit-or-miss approach (which is often the only viable way to get random numbers which satisfy additional constraints in an unbiased way):
Function RandIntVect(n As Long, a As Long, b As Long, mean As Double, tol As Double, Optional maxTries As Long = 1000) As Variant
'Uses a hit-or-miss approach to generate a vector of n random ints in a,b inclusive whose mean is
'within the tolerance tol of the given target mean
'The function raises an error if maxTries misses occur without a hit
Dim sum As Long, i As Long, j As Long
Dim lowTarget As Double, highTarget As Double 'targets for *sums*
Dim vect As Variant
lowTarget = n * (mean - tol)
highTarget = n * (mean + tol)
For i = 1 To maxTries
ReDim vect(1 To n)
sum = 0
j = 0
Do While j < n And sum + a * (n - j) <= highTarget And sum + b * (n - j) >= lowTarget
j = j + 1
vect(j) = Application.WorksheetFunction.RandBetween(a, b)
sum = sum + vect(j)
Loop
If j = n And lowTarget <= sum And sum <= highTarget Then
'Debug.Print i 'uncomment this line to see how many tries required
RandIntVect = vect
Exit Function
End If
Next i
'error if we get to here
RandIntVect = CVErr(xlErrValue)
End Function
This could be used as a worksheet array formula. The target means were in column I and in A2:H2 I entered =RandIntVect(8,1,10,I2,0.15) (with ctrl+shift+enter as an array formula) and then copied down:
Note that array formulas are volatile, so these numbers would be recalculated every time the worksheet is. You could use the function in VBA to place the numbers directly in the ranges rather than using the function as a worksheet formula. Something like:
Sub test()
Dim i As Long
For i = 1 To 3
Range(Cells(i + 1, 1), Cells(i + 1, 8)).Value = RandIntVect(8, 1, 10, Cells(i + 1, 9).Value, 0.15)
Next i
End Sub
enter image description here
The difference between two means is not within range (0.15+, 0.15-)
I am trying to use VBA function to add up a number's all power result together.
For example if I have 6 as revenue, and have first=1 and last=5, diff=5-1=4, then the result should be like total=6+6^2+6^3+6^4.
Below is my VBA code and not sure why it is not working and always give me result as 0:
Function AC(last, first, revenue)
diff = last - firs
For i = 1 To diff
Count = revenue ^ i
Total = Total + Count
Next i
End Function
Thanks if anyone can help me
You need to assign a value to the Function for it to return a value. You also have a spelling error on diff = last - firs :
Function AC(last, first, revenue)
diff = last - first
For i = 1 To diff
Count = revenue ^ i
Total = Total + Count
Next i
AC = Total
End Function
To be safe one should also always declare the variables, and we can do the addition right to the function:
Option Explicit
Function AC(ByVal last As Long, ByVal first As Long, ByVal revenue As Double) As Double
Dim diff As Long
Dim i as long
Dim Count as Double
diff = last - first
For i = 1 To diff
Count = revenue ^ i
AC = AC + Count
Next i
End Function
With array formula AC = [sum(6 ^ row(1:4))] ( # is short for As Double) :
Function AC#(last#, first#, revenue#)
AC = Evaluate("sum(" & revenue & " ^ row(1:" & last - first & "))")
End Function
With Excel Formula:
=SUMPRODUCT(revenue ^ ROW(INDIRECT("1:" & last - first)))
Here's a loop-less implementation:
Function AC(last, first, revenue)
If revenue = 1 then AC = last - first: Exit Function
AC = revenue*((revenue^(last - first) - 1) / (revenue-1))
End Function
I'm trying to write a simple function in VBA that will test a real value and output a string result if it's a perfect cube. Here's my code:
Function PerfectCubeTest(x as Double)
If (x) ^ (1 / 3) = Int(x) Then
PerfectCubeTest = "Perfect"
Else
PerfectCubeTest = "Flawed"
End If
End Function
As you can see, I'm using a simple if statement to test if the cube root of a value is equal to its integer portion (i.e. no remainder). I tried testing the function with some perfect cubes (1, 8, 27, 64, 125), but it only works for the number 1. Any other value spits out the "Flawed" case. Any idea what's wrong here?
You are testing whether the cube is equal to the double supplied.
So for 8 you would be testing whether 2 = 8.
EDIT: Also found a floating point issue. To resolve we will round the decimals a little to try and overcome the issue.
Change to the following:
Function PerfectCubeTest(x As Double)
If Round((x) ^ (1 / 3), 10) = Round((x) ^ (1 / 3), 0) Then
PerfectCubeTest = "Perfect"
Else
PerfectCubeTest = "Flawed"
End If
End Function
Or (Thanks to Ron)
Function PerfectCubeTest(x As Double)
If CDec(x ^ (1 / 3)) = Int(CDec(x ^ (1 / 3))) Then
PerfectCubeTest = "Perfect"
Else
PerfectCubeTest = "Flawed"
End If
End Function
#ScottCraner correctly explains why you were getting incorrect results, but there are a couple other things to point out here. First, I'm assuming that you are taking a Double as input because the range of acceptable numbers is higher. However, by your implied definition of a perfect cube only numbers with an integer cube root (i.e. it would exclude 3.375) need to be evaluated. I'd just test for this up front to allow an early exit.
The next issue you run into is that 1 / 3 can't be represented exactly by a Double. Since you're raising to the inverse power to get your cube root you're also compounding the floating point error. There's a really easy way to avoid this - take the cube root, cube it, and see if it matches the input. You get around the rest of the floating point errors by going back to your definition of a perfect cube as an integer value - just round the cube root to both the next higher and next lower integer before you re-cube it:
Public Function IsPerfectCube(test As Double) As Boolean
'By your definition, no non-integer can be a perfect cube.
Dim rounded As Double
rounded = Fix(test)
If rounded <> test Then Exit Function
Dim cubeRoot As Double
cubeRoot = rounded ^ (1 / 3)
'Round both ways, then test the cube for equity.
If Fix(cubeRoot) ^ 3 = rounded Then
IsPerfectCube = True
ElseIf (Fix(cubeRoot) + 1) ^ 3 = rounded Then
IsPerfectCube = True
End If
End Function
This returned the correct result up to 1E+27 (1 billion cubed) when I tested it. I stopped going higher at that point because the test was taking so long to run and by that point you're probably outside of the range that you would reasonably need it to be accurate.
For fun, here is an implementation of a number-theory based method described here . It defines a Boolean-valued (rather than string-valued) function called PerfectCube() that tests if an integer input (represented as a Long) is a perfect cube. It first runs a quick test which throws away many numbers. If the quick test fails to classify it, it invokes a factoring-based method. Factor the number and check if the multiplicity of each prime factor is a multiple of 3. I could probably optimize this stage by not bothering to find the complete factorization when a bad factor is found, but I had a VBA factoring algorithm already lying around:
Function DigitalRoot(n As Long) As Long
'assumes that n >= 0
Dim sum As Long, digits As String, i As Long
If n < 10 Then
DigitalRoot = n
Exit Function
Else
digits = Trim(Str(n))
For i = 1 To Len(digits)
sum = sum + Mid(digits, i, 1)
Next i
DigitalRoot = DigitalRoot(sum)
End If
End Function
Sub HelperFactor(ByVal n As Long, ByVal p As Long, factors As Collection)
'Takes a passed collection and adds to it an array of the form
'(q,k) where q >= p is the smallest prime divisor of n
'p is assumed to be odd
'The function is called in such a way that
'the first divisor found is automatically prime
Dim q As Long, k As Long
q = p
Do While q <= Sqr(n)
If n Mod q = 0 Then
k = 1
Do While n Mod q ^ k = 0
k = k + 1
Loop
k = k - 1 'went 1 step too far
factors.Add Array(q, k)
n = n / q ^ k
If n > 1 Then HelperFactor n, q + 2, factors
Exit Sub
End If
q = q + 2
Loop
'if we get here then n is prime - add it as a factor
factors.Add Array(n, 1)
End Sub
Function factor(ByVal n As Long) As Collection
Dim factors As New Collection
Dim k As Long
Do While n Mod 2 ^ k = 0
k = k + 1
Loop
k = k - 1
If k > 0 Then
n = n / 2 ^ k
factors.Add Array(2, k)
End If
If n > 1 Then HelperFactor n, 3, factors
Set factor = factors
End Function
Function PerfectCubeByFactors(n As Long) As Boolean
Dim factors As Collection
Dim f As Variant
Set factors = factor(n)
For Each f In factors
If f(1) Mod 3 > 0 Then
PerfectCubeByFactors = False
Exit Function
End If
Next f
'if we get here:
PerfectCubeByFactors = True
End Function
Function PerfectCube(n As Long) As Boolean
Dim d As Long
d = DigitalRoot(n)
If d = 0 Or d = 1 Or d = 8 Or d = 9 Then
PerfectCube = PerfectCubeByFactors(n)
Else
PerfectCube = False
End If
End Function
Fixed the integer division error thanks to #Comintern. Seems to be correct up to 208064 ^ 3 - 2
Function isPerfectCube(n As Double) As Boolean
n = Abs(n)
isPerfectCube = n = Int(n ^ (1 / 3) - (n > 27)) ^ 3
End Function
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 calculate how many layers a commodity will be stacked in. I have a variable quantity (iQty), a given width for the loadbed (dRTW), a width per unit for the commodity (dWidth) and a quantity per layer (iLayerQty).
The quantity per layer is calculated as iLayerQty = Int(dRTW/dWidth)
Now I need to divide the total quantity by the quantity per layer and round up. In an Excel formula it would be easy, but I'm trying to avoid WorksheetFunction calls to minimise A1/R1C1 confusion. At the moment I'm approximating it with this:
(Number of layers) = ((Int(iQty / iLayerQty) + 1)
And that works fine most of the time - except when the numbers give an integer (a cargo width of 0.5 m, for instance, fitting onto a 2.5 m rolltrailer). In those instances, of course, adding the one ruins the result.
Is there any handy way of tweaking that formula to get a better upward rounding?
I don't see any reason to avoid WorksheetFunction; I don't see any confusion here.
Number_of_layers = WorksheetFunction.RoundUp(iQty / iLayerQty, 0)
You could also roll your own function:
Function RoundUp(ByVal Value As Double)
If Int(Value) = Value Then
RoundUp = Value
Else
RoundUp = Int(Value) + 1
End If
End Function
Call it like this:
Number_of_layers = RoundUp(iQty / iLayerQty)
If using a WorksheetFunction object to access a ROUNDUP or CEILING function is off the table then the same can be accomplished with some maths.
Number of layers = Int(iQty / iLayerQty) - CBool(Int(iQty / iLayerQty) <> Round(iQty / iLayerQty, 14))
A VBA True is the equivalent of (-1) when used mathematically. The VBA Round is there to avoid 15 digit floating point errors.
I use -int(-x) to get the ceiling.
?-int(-1.1) ' get ceil(1.1)
2
?-int(1.1) ' get ceil(-1.1)
-1
?-int(-5) ' get ceil(5)
5
These are the functions I put together for this purpose.
Function RoundUp(ByVal value As Double) as Integer
Dim intVal As Integer
Dim delta As Double
intVal = CInt(value)
delta = intVal - value
If delta < 0 Then
RoundUp = intVal + 1
Else
RoundUp = intVal
End If
End Function
Function RoundDown(ByVal value As Double) as Integer
Dim intVal As Integer
Dim delta As Double
intVal = CInt(value)
delta = intVal - value
If delta <= 0 Then
RoundDown = intVal
ElseIf delta > 0 Then
RoundDown = intVal - 1
End If
End Function
This is my Ceiling in VBA.
Function Ceiling(ByVal Number As Double, ByVal Significance As Double) As Double
Dim intVal As Long
Dim delta As Double
Dim RoundValue As Double
Dim PreReturn As Double
If Significance = 0 Then
RoundValue = 1
Else
RoundValue = 1 / Significance
End If
Number = Number * RoundValue
intVal = CLng(Number)
delta = intVal - Number
If delta < 0 Then
PreReturn = intVal + 1
Else
PreReturn = intVal
End If
Ceiling = PreReturn / RoundValue
End Function