Question

The following algorithm (attributed to Clever Moler) estimates machine precision (eps): a = 4.0/3.0; b = a − 1.0; c = b + b + b; eps = abs(c − 1.0). Implement the program twice with single and double precision variable types and report the value of eps for both single and double precision.

Answer 1

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ANSWER:

CODE:

// EPS.java

public class EPS {

      public static void main(String[] args) {

            // finding eps using single precision (float) variables

            float a = (float) (4.0 / 3.0);

            float b = (float) (a - 1.0);

            float c = b + b + b;

            float result = (float) (c - 1.0);

            float eps1 = Math.abs(result);

            System.out.println("eps for single precision variables: " + eps1);

            // finding eps using double precision (float) variables

            double x = 4.0 / 3.0;

            double y = x - 1.0;

            double z = y + y + y;

            double eps2 = Math.abs(z - 1.0);

            System.out.println("eps for double precision variables: " + eps2);

      }

}

/*OUTPUT*/

eps for single precision variables: 1.1920929E-7

eps for double precision variables: 2.220446049250313E-16

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