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
Related
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,
how can I make sure that the results dont remove the last 0 for example, this produces "1.454" instead of "1.4540"
Dim test1 As Decimal = 14540
Debug.Print(TransformDecimal(test1, 1))
Private Shared Function TransformDecimal(value As Decimal, numberOfPlaces As Integer) As Decimal
Dim min = CDec(Math.Pow(10, numberOfPlaces - 1))
Dim max = CDec(Math.Pow(10, numberOfPlaces))
If (value >= max) Then
While value >= max
value /= 10
End While
ElseIf (value < min) Then
While value < min
value *= 10
End While
End If
Return value
End Function
You really must put Option Explicit On.
When you don't and you have, for example, this code Dim x As Decimal = 1.0 you are creating a Double and then converting it to Decimal. With Option Strict On this code won't compile.
When you write this code Dim x As Decimal = 1.0d you are immediately creating a Decimal- and that preserves the decimal places.
So, if you now write your method like this:
Private Shared Function TransformDecimal(value As Decimal, numberOfPlaces As Integer) As Decimal
Dim min = CDec(Math.Pow(10, numberOfPlaces - 1))
Dim max = CDec(Math.Pow(10, numberOfPlaces))
If (value >= max) Then
While value >= max
Dim bits = Decimal.GetBits(value)
bits(3) = ((bits(3) \ 65536) + 1) * 65536
value = New Decimal(bits)
End While
ElseIf (value < min) Then
While value < min
Dim bits = Decimal.GetBits(value)
bits(3) = ((bits(3) \ 65536) - 1) * 65536
value = New Decimal(bits)
End While
End If
Return value
End Function
And call it like this:
Dim test1 As Decimal = 14540d
Debug.Print(TransformDecimal(test1, 1).ToString())
You'll find that you get the result you want:
1.4540
The crux of this code is the lines bits(3) = ((bits(3) \ 65536) + 1) * 65536 & bits(3) = ((bits(3) \ 65536) - 1) * 65536 which shift the exponent to change the decimal by multiples of 10.
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.
I am trying to run some mathematical formula which has large values, so I am using the Double datatype. But still I am getting the answer as NaN if I input large values. So how can I solve this?
For example, if I pass Varx=3 and countx=1230 I get the result as NaN.
Public Function EulerForPro(ByVal Varx As Integer, ByVal Countx As Integer) As Double
Dim Result1 As Double = 1
Dim Result2 As Double = Varx
Dim Result As Double = 0
For i = 1 To Countx
Result1 = Result1 + (Math.Pow(-1, i) * Math.Pow(Varx, (2 * i)) / factx(2 * i))
Next
For i = 1 To Countx
Result2 = Result2 + (Math.Pow(-1, i) * Math.Pow(Varx, ((2 * i) + 1)) / factx((2 * i) + 1))
Next
Result = Result1 + Result2
Label2.Text = Result1
Label3.Text = Result2
Label4.Text = Result
End Function
Use the logarithm of the results. It is standard practice for large numbers.
So instead of multiplying a*b (which causes overflow), use
Y = log(a) + log(b)
You could try using the Decimal data type instead. It can hold an integer value of approximately 7,9 x 10^28 (the exact number is 2^96 - 1, since it can use 96 bits of its 128 for storing the integer part of a value). However, it's a complex data type and you should carefully consider how to use it in your code. Not all implicit conversions work, especially when using Double.
See MSDN on decimal for exact specifications of Decimal.