round BigDecimal to nearest 5 cents

Question

I'm trying to figure out how to round a monetary amount upwards to the nearest 5 cents. The following shows my expected results

1.03     => 1.05
1.051    => 1.10
1.05     => 1.05
1.900001 => 1.10

I need the result to be have a precision of 2 (as shown above).

Update

Following the advice below, the best I could do is this

    BigDecimal amount = new BigDecimal(990.49)

    // To round to the nearest .05, multiply by 20, round to the nearest integer, then divide by 20
   def result =  new BigDecimal(Math.ceil(amount.doubleValue() * 20) / 20)
   result.setScale(2, RoundingMode.HALF_UP)

I'm not convinced this is 100% kosher - I'm concerned precision could be lost when converting to and from doubles. However, it's the best I've come up with so far and seems to work.

Answer 1

Using BigDecimal without any doubles (improved on the answer from marcolopes):

public static BigDecimal round(BigDecimal value, BigDecimal increment,
                               RoundingMode roundingMode) {
    if (increment.signum() == 0) {
        // 0 increment does not make much sense, but prevent division by 0
        return value;
    } else {
        BigDecimal divided = value.divide(increment, 0, roundingMode);
        BigDecimal result = divided.multiply(increment);
        return result;
    }
}

The rounding mode is e.g. RoundingMode.HALF_UP. For your examples, you actually want RoundingMode.UP (bd is a helper which just returns new BigDecimal(input)):

assertEquals(bd("1.05"), round(bd("1.03"), bd("0.05"), RoundingMode.UP));
assertEquals(bd("1.10"), round(bd("1.051"), bd("0.05"), RoundingMode.UP));
assertEquals(bd("1.05"), round(bd("1.05"), bd("0.05"), RoundingMode.UP));
assertEquals(bd("1.95"), round(bd("1.900001"), bd("0.05"), RoundingMode.UP));

Also note that there is a mistake in your last example (rounding 1.900001 to 1.10).

Answer 2

I'd try multiplying by 20, rounding to the nearest integer, then dividing by 20. It's a hack, but should get you the right answer.