How to round decimal value up to nearest 0.05 value?

Question

Is there any way to round up decimal value to its nearest 0.05 value in .Net?

Ex:

7.125 -> 7.15

6.66 -> 6.7

If its now available can anyone

Answer 1

Use this:

Math.Round(mydecimal / 0.05m, 0) * 0.05m;

The same logic can be used in T-SQL:

ROUND(@mydecimal / 0.05, 0) * 0.05

I prefer this approach to the selected answer simply because you can directly see the precision used.

Answer 2

I could not get proper rounding in most of the formulas

This one "rounds" to nearest

float roundFloat(float value, float toNearest) {
            float divVal = (1 / (toNearest == 0 ? 1 : toNearest));
            return ((float)(Math.Round(value * divVal)) / divVal);
        }

Result:

roundFloat(2, 0.125F); -> 2
roundFloat(2.11, 0.125F); -> 2.125
roundFloat(2.22, 0.125F); -> 2.25
roundFloat(2.33, 0.125F); -> 2.375
roundFloat(2.44, 0.125F); -> 2.5
roundFloat(2.549999, 0.125F); -> 2.5
roundFloat(2.659999, 0.125F); -> 2.625
roundFloat(2.769999, 0.125F); -> 2.75
roundFloat(2.879999, 0.125F); -> 2.875
roundFloat(2.989999, 0.125F); -> 3

Example 0.125 nearest rounding

 2.000 
 2.125 
 2.250 
 2.375 
 2.500 
 2.625 
 2.750 
 2.875 
 3.000 

Answer 3

Duplicated here and here for ruby and python. It shouldn't be too different.

Answer 4

How about:

Math.Ceiling(myValue * 20) / 20

Answer 5

Something like this should work for any step, not just 0.05:

private decimal RoundUp (decimal value, decimal step)
{
    var multiplicand = Math.Ceiling (value / step);
    return step * multiplicand;
}

Answer 6

Math..::.Round Method (Decimal, Int32, MidpointRounding)

Rounds a double-precision floating-point value to the specified number of fractional digits. A parameter specifies how to round the value if it is midway between two other numbers.

   Math.Round(1.489,2,MidpointRounding.AwayFromZero)