Round to increments of 2.5? - vb.net

Question

I need to round a value up to the nearest multiple of 2.5.
For example:
6 --> 7.5
7.6 --> 10
etc.
This seems like the best way to do this?
Function RoundToIncrement(ByVal originalNumber As Decimal, ByVal increment As Decimal) As Decimal
Dim num = Math.Round(originalNumber / increment, MidpointRounding.AwayFromZero) * increment
If originalNumber Mod increment <> 0 And num < originalNumber Then
num += increment
End If
Return num
End Function

Answer 1

Divide the number by 2.5, round up to the nearest integer, then multiply the result by 2.5.
You're close.
Function RoundToIncrement(ByVal orignialNumber As Decimal, ByVal increment As Decimal) As Decimal
Return Math.Ceiling( orignialNumber / increment ) * increment
End Function
Math.Ceiling will always round non-integers up, so you don't need the post-adjustment.

Answer 2

Divide the number by 2.5. Round to nearest 1. Multiply by 2.5.
Beware of cumulative errors, and you're all set.
-Adam

Answer 3

/*
This will round up (Math.Ceiling) or down (Math.Floor) based on the midpoint of the increment.
The other examples use Math.Ceiling and therefore always round up.
Assume the increment is 2.5 in this example and the number is 6.13
*/
var halfOfIncrement = Increment / 2; // 2.5 / 2 = 1.25
var floorResult = Math.Floor(originalNumber / Increment); //Math.Floor(6.13 / 2.5) = Math.Floor(2.452) = 2
var roundingThreshold = (floorResult * Increment) + halfOfIncrement; //(2 * 2.5) = 5 + 1.25 = 6.25
if (originalNumber >= roundingThreshold) //6.13 >= 6.25 == false therefore take Math.Floor(6.13/2.5) = Math.Floor(2.452) = 2 * 2.5 = 5
result = Math.Ceiling(originalNumber / Increment) * Increment;
else
result = Math.Floor(originalNumber / Increment) * Increment;