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 eg 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

我会尝试乘以 20，四舍五入到最接近的整数，然后除以 20。这是一个技巧，但应该会得到正确的答案。

Answer 3

I wrote this in Java a few years ago: https://github.com/marcolopes/dma/blob/master/org.dma.java/src/org/dma/java/math/BusinessRules.java

/**
 * Rounds the number to the nearest<br>
 * Numbers can be with or without decimals<br>
 */
public static BigDecimal round(BigDecimal value, BigDecimal rounding, RoundingMode roundingMode){

    return rounding.signum()==0 ? value :
        (value.divide(rounding,0,roundingMode)).multiply(rounding);

}


/**
 * Rounds the number to the nearest<br>
 * Numbers can be with or without decimals<br>
 * Example: 5, 10 = 10
 *<p>
 * HALF_UP<br>
 * Rounding mode to round towards "nearest neighbor" unless
 * both neighbors are equidistant, in which case round up.
 * Behaves as for RoundingMode.UP if the discarded fraction is >= 0.5;
 * otherwise, behaves as for RoundingMode.DOWN.
 * Note that this is the rounding mode commonly taught at school.
 */
public static BigDecimal roundUp(BigDecimal value, BigDecimal rounding){

    return round(value, rounding, RoundingMode.HALF_UP);

}


/**
 * Rounds the number to the nearest<br>
 * Numbers can be with or without decimals<br>
 * Example: 5, 10 = 0
 *<p>
 * HALF_DOWN<br>
 * Rounding mode to round towards "nearest neighbor" unless
 * both neighbors are equidistant, in which case round down.
 * Behaves as for RoundingMode.UP if the discarded fraction is > 0.5;
 * otherwise, behaves as for RoundingMode.DOWN.
 */
public static BigDecimal roundDown(BigDecimal value, BigDecimal rounding){

    return round(value, rounding, RoundingMode.HALF_DOWN);

}

Answer 4

Here are a couple of very simple methods in c# I wrote to always round up or down to any value passed.

public static Double RoundUpToNearest(Double passednumber, Double roundto)
    {

        // 105.5 up to nearest 1 = 106
        // 105.5 up to nearest 10 = 110
        // 105.5 up to nearest 7 = 112
        // 105.5 up to nearest 100 = 200
        // 105.5 up to nearest 0.2 = 105.6
        // 105.5 up to nearest 0.3 = 105.6

        //if no rounto then just pass original number back
        if (roundto == 0)
        {
            return passednumber;
        }
        else
        {
            return Math.Ceiling(passednumber / roundto) * roundto;
        }
    }
    public static Double RoundDownToNearest(Double passednumber, Double roundto)
    {

        // 105.5 down to nearest 1 = 105
        // 105.5 down to nearest 10 = 100
        // 105.5 down to nearest 7 = 105
        // 105.5 down to nearest 100 = 100
        // 105.5 down to nearest 0.2 = 105.4
        // 105.5 down to nearest 0.3 = 105.3

        //if no rounto then just pass original number back
        if (roundto == 0)
        {
            return passednumber;
        }
        else
        {
            return Math.Floor(passednumber / roundto) * roundto;
        }
    }

Answer 5

In Scala I did the following (Java below)

import scala.math.BigDecimal.RoundingMode

def toFive(
   v: BigDecimal,
   digits: Int,
   roundType: RoundingMode.Value= RoundingMode.HALF_UP
):BigDecimal = BigDecimal((2*v).setScale(digits-1, roundType).toString)/2

And in Java

import java.math.BigDecimal;
import java.math.RoundingMode;

public static BigDecimal toFive(BigDecimal v){
    return new BigDecimal("2").multiply(v).setScale(1, RoundingMode.HALF_UP).divide(new BigDecimal("2"));
}

Answer 6

Based on your edit, another possible solution would be:

BigDecimal twenty = new BigDecimal(20);
BigDecimal amount = new BigDecimal(990.49)

// To round to the nearest .05, multiply by 20, round to the nearest integer, then divide by 20
BigDecimal result =  new BigDecimal(amount.multiply(twenty)
                                          .add(new BigDecimal("0.5"))
                                          .toBigInteger()).divide(twenty);

This has the advantage, of being guaranteed not to lose precision, although it could potentially be slower of course...

And the scala test log:

scala> var twenty = new java.math.BigDecimal(20) 
twenty: java.math.BigDecimal = 20

scala> var amount = new java.math.BigDecimal("990.49");
amount: java.math.BigDecimal = 990.49

scala> new BigDecimal(amount.multiply(twenty).add(new BigDecimal("0.5")).toBigInteger()).divide(twenty)
res31: java.math.BigDecimal = 990.5

Answer 7

For this test to pass :

assertEquals(bd("1.00"), round(bd("1.00")));
assertEquals(bd("1.00"), round(bd("1.01")));
assertEquals(bd("1.00"), round(bd("1.02")));
assertEquals(bd("1.00"), round(bd("1.024")));
assertEquals(bd("1.05"), round(bd("1.025")));
assertEquals(bd("1.05"), round(bd("1.026")));
assertEquals(bd("1.05"), round(bd("1.049")));

assertEquals(bd("-1.00"), round(bd("-1.00")));
assertEquals(bd("-1.00"), round(bd("-1.01")));
assertEquals(bd("-1.00"), round(bd("-1.02")));
assertEquals(bd("-1.00"), round(bd("-1.024")));
assertEquals(bd("-1.00"), round(bd("-1.0245")));
assertEquals(bd("-1.05"), round(bd("-1.025")));
assertEquals(bd("-1.05"), round(bd("-1.026")));
assertEquals(bd("-1.05"), round(bd("-1.049")));

Change ROUND_UP in ROUND_HALF_UP :

private static final BigDecimal INCREMENT_INVERTED = new BigDecimal("20");
public BigDecimal round(BigDecimal toRound) {
    BigDecimal divided = toRound.multiply(INCREMENT_INVERTED)
                                .setScale(0, BigDecimal.ROUND_HALF_UP);
    BigDecimal result = divided.divide(INCREMENT_INVERTED)
                               .setScale(2, BigDecimal.ROUND_HALF_UP);
    return result;
}

Answer 8

  public static BigDecimal roundTo5Cents(BigDecimal amount)
  {
    amount = amount.multiply(new BigDecimal("2"));
    amount = amount.setScale(1, RoundingMode.HALF_UP);
    // preferred scale after rounding to 5 cents: 2 decimal places
    amount = amount.divide(new BigDecimal("2"), 2, RoundingMode.HALF_UP);
    return amount;
  }

Note that this is basically the same answer as John's .

Answer 9

public static void roundUp()
{
    try
    {
        System.out.println("Enter the currency : $");
        Scanner keyboard = new Scanner(System.in);
        String myint = keyboard.next();
        if (!isEmptyOrBlank(myint).booleanValue())
        {
            BigDecimal d = new BigDecimal(myint);
            System.out.println("Enter the round up factor: $");
            String roundUpFactor = keyboard.next();
            if (!isEmptyOrBlank(roundUpFactor).booleanValue())
            {
                BigDecimal scale = new BigDecimal(roundUpFactor);
                BigDecimal y = d.divide(scale, MathContext.DECIMAL128);
                BigDecimal q = y.setScale(0, 0);
                BigDecimal z = q.multiply(scale);
                System.out.println("Final price after rounding up to " + roundUpFactor + " is : $" + z);
                System.out.println("Want to try with other price Y/N :");

                String exit = keyboard.next();
                if ((!isEmptyOrBlank(exit).booleanValue()) && ("y".equalsIgnoreCase(exit))) {
                    roundUp();
                } else {
                    System.out.println("See you take care");
                }
            }
        }
        else
        {
            System.out.println("Please be serious u r dealing with critical Tx Pricing");
        }
    }
    catch (Exception e)
    {
        System.out.println("Please be serious u r dealing with critical Tx Pricing enter correct rounding off value");
    }
}

Answer 10

Tom has the right idea, but you need to use BigDecimal methods, since you ostensibly are using BigDecimal because your values are not amenable to a primitive datatype. Something like:

BigDecimal num = new BigDecimal(0.23);
BigDecimal twenty = new BigDecimal(20);
//Might want to use RoundingMode.UP instead,
//depending on desired behavior for negative values of num.
BigDecimal numTimesTwenty = num.multiply(twenty, new MathContext(0, RoundingMode.CEILING)); 
BigDecimal numRoundedUpToNearestFiveCents
  = numTimesTwenty.divide(twenty, new MathContext(2, RoundingMode.UNNECESSARY));

Answer 11

You can use plain double to do this.

double amount = 990.49;
double rounded = ((double) (long) (amount * 20 + 0.5)) / 20;

EDIT: for negative numbers you need to subtract 0.5