I have come across a issue while working in VBA . I'm supposed to write program that is Numerical integration of trapeze method (I'm not sure if It is how it's called in English) of function 100*x^99 lower limit = 0 upper limit = 1 . Cells (j,5) contains numbers (10,30,100,300,1000,3000,10000) - amount of point splits . Code seems to work but given wrong results , for amount of splits it should be around
10 - 5.000295129200607
30 - 1.786588019299606
100 - 1.0812206997600746
300 - 1.0091505687770146
1000 - 1.0008248693208752
3000 - 1.0000916650530287
10000 - 1.000008249986933
Function F(x)
F = 100 * (x ^ 99)
End Function
Sub calka()
Dim n As Single
Dim xp As Single
Dim dx As Single
Dim xk As Single
Dim ip As Single
Dim pole As Single
xp = 0
xk = 1
For j = 5 To 11
n = Cells(j, 5)
dx = (xk - xp) / n
pole = 0
For i = 1 To n - 1
pole = pole + F(xp + i * dx)
Next i
pole = pole + ((F(xp) + F(xk)) / 2)
pole = pole * dx
Worksheets("Arkusz1").Cells(j, 7) = pole
Next j
End Sub
I tried to implement same code in java and c++ and it worked flawlessly but VBA always gives me wrong results , I'm not sure if it's rounds at some point and I can disable in settings or my code is just not written right .
Apologies for low clarity It's hard for me to translate mathematic to English.
Use Doubles rather than Singles
http://www.techrepublic.com/article/comparing-double-vs-single-data-types-in-vb6/
I have a calculation like this: 3 * 12300 / 160. The result is: 230.625. But I just want the integer part, 230.
In C, this can be done using something like this: int MyVar = (int)3*12300/160;
Is there a way in VBA (With MS-Access) for force the result to be an integer?
You can round down using the Int or Fix functions.
Since you know the result you want is a whole number, you should store the result in a variable of type Long (or Integer if you're absolutely certain it will always be smaller than 32768).
Dim l As Long
l = Int(3 / 160 * 12300) ' <~~~~ Yes, I switched the numbers around on purpose!*
MsgBox "l = " & l
* Why did I switch the numbers around? Because the expression 3 * 12300 / 160 will throw an error in VBA. Read here why: Overflow when multiplying Integers and assigning to Long
Im trying to create Monte-Carlo simulation that can be used to derive estimates for integration problems (summing up the area under
a curve). Have no idea what to do now and i am stuck
"to solve this problem we generate a number (say n) of random number pairs for x and y between 0 and 1, for each pair we see if the point (x,y) falls above or below the line. We count the number of times this happens (say c). The area under the curve is computed as c/n"
Really confused please help thank you
Function MonteCarlo()
Dim a As Integer
Dim b As Integer
Dim x As Double
Dim func As Double
Dim total As Double
Dim result As Double
Dim j As Integer
Dim N As Integer
Console.WriteLine("Enter a")
a = Console.ReadLine()
Console.WriteLine("Enter b")
b = Console.ReadLine()
Console.WriteLine("Enter n")
N = Console.ReadLine()
For j = 1 To N
'Generate a new number between a and b
x = (b - a) * Rnd()
'Evaluate function at new number
func = (x ^ 2) + (2 * x) + 1
'Add to previous value
total = total + func
Next j
result = (total / N) * (b - a)
Console.WriteLine(result)
Console.ReadLine()
Return result
End Function
You are using the rejection method for MC area under the curve.
Do this:
Divide the range of x into, say, 100 equally-spaced, non-overlapping bins.
For your function y = f(x) = (x ^ 2) + (2 * x) + 1, generate e.g. 10,000 values of y for 10,000 values of x = (b - a) * Rnd().
Count the number of y-values in each bin, and divide by 10,000 to get a "bin probability." --> p(x).
Next, the proper way to randomly simulate your function is to use the rejection method, which goes as follows:
4a. Draw a random x-value using x = (b - a) * Rnd()
4b. Draw a random uniform U(0,1). If U(0,1) is less than p(x) add a count to the bin.
4c. Continue steps 4a-4b 10000 times.
You will now be able to simulate your y=f(x) function using the rejection method.
Overall, you need to master these approaches before you do what you want since it sounds like you have little experience in bin counts, simulation, etc. Area under the curve is always one using this approach, so just be creative for integrating using MC.
Look at some good textbooks on MC integration.
How do I use Excel VBA to find the minimum value of an equation?
For example, if I have the equation y = 2x^2 + 14, and I want to make a loop that will slowly increase/decrease the value of x until it can find the smallest value possible for y, and then let me know what the corresponding value of x is, how would I go about doing that?
Is there a method that would work for much more complicated equations?
Thank you for your help!
Edit: more details
I'm trying to design a program that will find a certain constant needed to graph a nuclear decay. This constant is a part of an equation that gets me a calculated decay. I'm comparing this calculated decay against a measured decay. However, the constant changes very slightly as the decay happens, which means I have to use something called a residual-square to find the best constant to use that will fit the entire decay best to make my calculated decay as accurate as possible.
It works by doing (Measured Decay - Calculated Decay) ^2
You do that for the decay at several times, and add them all up. What I need my program to do is to slowly increase and decrease this constant until I can find a minimum value for the value I get when I add up the residual-squared results for all the times using this decay. The residual-squared that has the smallest value has the value of the constant that I want.
I already drafted a program that does all the calculations and such. I'm just not sure how to find this minimum value. I'm sure if a method works for something like y = x^2 + 1, I can adapt it to work for my needs.
Test the output while looping to look for the smallest output result.
Here's an Example:
Sub FormulaLoop()
Dim x As Double
Dim y As Double
Dim yBest As Double
x = 1
y = (x ^ 2) + 14
yBest = y
For x = 2 To 100
y = (x ^ 2) + 14
If y < yBest Then
yBest = y
End If
Next x
MsgBox "The smallest output of y was: " & yBest
End Sub
If you want to loop through all the possibilities of two variables that make up x then I'd recommend looping in this format:
Sub FormulaLoop_v2()
Dim MeasuredDecay As Double
Dim CalculatedDecay As Double
Dim y As Double
Dim yBest As Double
MeasuredDecay = 1
CalculatedDecay = 1
y = ((MeasuredDecay - CalculatedDecay) ^ 2) + 14
yBest = y
For MeasuredDecay = 2 To 100
For CalculatedDecay = 2 To 100
y = ((MeasuredDecay - CalculatedDecay) ^ 2) + 14
If y < yBest Then
yBest = y
End If
Next CalculatedDecay
Next MeasuredDecay
MsgBox "The smallest output of y was: " & yBest
End Sub
Given numbers like 499, 73433, 2348 what VBA can I use to round to the nearest 5 or 10? or an arbitrary number?
By 5:
499 -> 500
2348 -> 2350
7343 -> 7345
By 10:
499 -> 500
2348 -> 2350
7343 -> 7340
etc.
It's simple math. Given a number X and a rounding factor N, the formula would be:
round(X / N)*N
Integrated Answer
X = 1234 'number to round
N = 5 'rounding factor
round(X/N)*N 'result is 1235
For floating point to integer, 1234.564 to 1235, (this is VB specific, most other languages simply truncate) do:
int(1234.564) 'result is 1235
Beware: VB uses Bankers Rounding, to the nearest even number, which can be surprising if you're not aware of it:
msgbox round(1.5) 'result to 2
msgbox round(2.5) 'yes, result to 2 too
Thank you everyone.
To round to the nearest X (without being VBA specific)
N = X * int(N / X + 0.5)
Where int(...) returns the next lowest whole number.
If your available rounding function already rounds to the nearest whole number then omit the addition of 0.5
In VB, math.round has additional arguments to specify number of decimal places and rounding method. Math.Round(10.665, 2, MidpointRounding.AwayFromZero) will return 10.67 . If the number is a decimal or single data type, math.round returns a decimal data type. If it is double, it returns double data type. That might be important if option strict is on.
The result of (10.665).ToString("n2") rounds away from zero to give "10.67". without additional arguments math.round returns 10.66, which could lead to unwanted discrepancies.
'Example: Round 499 to nearest 5. You would use the ROUND() FUNCTION.
a = inputbox("number to be rounded")
b = inputbox("Round to nearest _______ ")
strc = Round(A/B)
strd = strc*B
msgbox( a & ", Rounded to the nearest " & b & ", is" & vbnewline & strd)
For a strict Visual Basic approach, you can convert the floating-point value to an integer to round to said integer. VB is one of the rare languages that rounds on type conversion (most others simply truncate.)
Multiples of 5 or x can be done simply by dividing before and multiplying after the round.
If you want to round and keep decimal places, Math.round(n, d) would work.
Here is our solution:
Public Enum RoundingDirection
Nearest
Up
Down
End Enum
Public Shared Function GetRoundedNumber(ByVal number As Decimal, ByVal multiplier As Decimal, ByVal direction As RoundingDirection) As Decimal
Dim nearestValue As Decimal = (CInt(number / multiplier) * multiplier)
Select Case direction
Case RoundingDirection.Nearest
Return nearestValue
Case RoundingDirection.Up
If nearestValue >= number Then
Return nearestValue
Else
Return nearestValue + multiplier
End If
Case RoundingDirection.Down
If nearestValue <= number Then
Return nearestValue
Else
Return nearestValue - multiplier
End If
End Select
End Function
Usage:
dim decTotal as Decimal = GetRoundedNumber(CDec(499), CDec(0.05), RoundingDirection.Up)
Simply ROUND(x/5)*5 should do the job.
I cannot add comment so I will use this
in a vbs run that and have fun figuring out why the 2 give a result of 2
you can't trust round
msgbox round(1.5) 'result to 2
msgbox round(2.5) 'yes, result to 2 too
something like that?
'nearest
n = 5
'n = 10
'value
v = 496
'v = 499
'v = 2348
'v = 7343
'mod
m = (v \ n) * n
'diff between mod and the val
i = v-m
if i >= (n/2) then
msgbox m+n
else
msgbox m
end if
Try this function
--------------start----------------
Function Round_Up(ByVal d As Double) As Integer
Dim result As Integer
result = Math.Round(d)
If result >= d Then
Round_Up = result
Else
Round_Up = result + 1
End If
End Function
-------------end ------------
I slightly updated the function provided by the "community wiki" (the best answer), just to round to the nearest 5 (or anything you like), with this exception : the rounded number will NEVER be superior to the original number.
This is useful in cases when it is needed to say that "a company is alive for 47 years" : I want the web page to display "is alive for more than 45 years", while avoiding lying in stating "is alive for more than 50 years".
So when you feed this function with 47, it will not return 50, but will return 45 instead.
'Rounds a number to the nearest unit, never exceeding the actual value
function RoundToNearestOrBelow(num, r)
'#param num Long/Integer/Double The number to be rounded
'#param r Long The rounding value
'#return OUT Long The rounded value
'Example usage :
' Round 47 to the nearest 5 : it will return 45
' Response.Write RoundToNearestBelow(47, 5)
Dim OUT : OUT = num
Dim rounded : rounded = Round((((num)) / r), 0) * r
if (rounded =< num) then
OUT = rounded
else
OUT = rounded - r
end if
'Return
RoundToNearestOrBelow = OUT
end function 'RoundToNearestOrBelow
To mimic in Visual Basic the way the round function works in Excel, you just have to use:
WorksheetFunction.Round(number, decimals)
This way the banking or accounting rounding don't do the rounding.