asked on

Multiplication accurary

Hi,

working on a simple problem and wanted to get some ideas of the best way around it. Essentially I want to solve the following problem:

$0.75 * 120%

The answer is $0.90.

I have tried the following:

		BigDecimal bd1 = new BigDecimal(.75);
		BigDecimal bd2 = new BigDecimal(.2);
		BigDecimal bd3 = new BigDecimal(1);
		
		System.out.println( bd1.multiply(bd2.add(bd3)));
		
		System.out.println( 0.75 * ( 1 + 0.20) );

		System.out.println( .75d *  1.2d );
		
		System.out.println( .75d / 5d * 6d);
		
		System.out.println( 75d * 1.2d /100d);
		
		System.out.println( .75d * (( 1d + 0.2d) * 100d)/100d );

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With the following output:

0.90000000000000000832667268468867405317723751068115234375
0.8999999999999999
0.8999999999999999
0.8999999999999999
0.9
0.9

for_yan

Look at this recent question with solution:

https://www.experts-exchange.com/questions/27240925/how-to-round.html?sfQueryTermInfo=1+10+30+bigdecim+yan

  BigDecimal federalTaxableEarnings =  new BigDecimal(40.0276);
 federalTaxableEarnings = federalTaxableEarnings.setScale(2, RoundingMode.HALF_UP);
        System.out.println("earning: " + federalTaxableEarnings);

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for_yan

I think it should be like that for your code:

BigDecimal bd4 = bd1.multiply(bd2.add(bd3));
bd4 = bd4.setScale(2,RoundingMode.HALF_UP);
System.out.println("bd4: " + bd4);

Let me test.

for_yan

Yes, this worked for me:

BigDecimal bd1 = new BigDecimal(.75);
		BigDecimal bd2 = new BigDecimal(.2);
		BigDecimal bd3 = new BigDecimal(1);

		
        BigDecimal bd4 = bd1.multiply(bd2.add(bd3));
        bd4 = bd4.setScale(2,RoundingMode.HALF_UP);
         System.out.println("bd4: " + bd4);

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Output:

bd4: 0.90

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ASKER CERTIFIED SOLUTION

Frosty555

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SOLUTION

CEHJ

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infinidem

ASKER

The answer is given to two parts as the explanation that a decimal can not be accurately represented via binary and that the constructors for BidDecimal should be with more accuracy. Thanks Everyone.