1. 1 IEEE Floating Point Revision Guide for Phase Test Week 5

2. 2 Floating Point 15900000000000000 could be represented as 14 15.9 * 10 15 1.59 * 10 16 A calculator might display 159 E14 159 * 10 14 MantissaExponent

3. 3 Binary The value of real binary numbers… Scientific2 2121 2020.2 -1 2 -2 2 -3 Fractions. ½¼¾ Decimal421..5.25.125 101.101 = 4+1+1/2+1/8 = 4+1+.5+.125= 5.625 = 5 ⅝ 1 0 1. 1 0 1

4. 4 Binary Fractions The value of real binary numbers… Scientific2 2121 2020.2 -1 2 -2 2 -3 Fractions. ½¼⅛ Decimal421..5.25.125 101.101 = 4+1+1/2+1/8 = 4+1+.5+.125= 5.625 = 5 ⅝ 1 0 1. 1 0 1

5. 5 Binary Fractions The value of real binary numbers… Scientific2 2121 2020.2 -1 2 -2 2 -3 Fractions. ½¼⅛ Decimal421..5.25.125 101.101 = 4+1+1/2+1/8 = 4+1+.5+.125= 5.625 = 5 ⅝ 1 0 1. 1 0 1

6. 6 IEEE Single Precision The number will occupy 32 bits The first bit represents the sign of the number; 1= negative 0= positive. The next 8 bits will specify the exponent stored in biased 127 form. The remaining 23 bits will carry the mantissa normalised to be between 1 and 2. i.e. 1<= mantissa < 2

7. 7 Basic Conversion Converting a decimal number to a floating point number. 1.Take the integer part of the number and generate the binary equivalent. 2.Take the fractional part and generate a binary fraction 3.Then place the two parts together and normalise.

8. 8 IEEE – Example 1 Convert 6.75 to 32 bit IEEE format. 1.The Mantissa. The Integer first. 6 / 2 = 3 r 0 3 / 2= 1 r 1 1 / 2= 0 r 1 2. Fraction next..75 * 2 = 1.5.5 * 2 = 1.0 3. put the two parts together… 110.11 Now normalise 1.1011 * 2 2 = 110 2 = 0.11 2

9. 9 IEEE – Example 1 Convert 6.75 to 32 bit IEEE format. 1.The Mantissa. The Integer first. 6 / 2 = 3 r 0 3 / 2= 1 r 1 1 / 2= 0 r 1 2. Fraction next..75 * 2 = 1.5.5 * 2 = 1.0 3. put the two parts together… 110.11 Now normalise 1.1011 * 2 2 = 110 2

10. 10 IEEE – Example 1 Convert 6.75 to 32 bit IEEE format. 1.The Mantissa. The Integer first. 6 / 2 = 3 r 0 3 / 2= 1 r 1 1 / 2= 0 r 1 2. Fraction next..75 * 2 = 1.5.5 * 2 = 1.0 3. put the two parts together… 110.11 Now normalise 1.1011 * 2 2 = 110 2 = 0.11 2

11. 11 IEEE Biased 127 Exponent To generate a biased 127 exponent Take the value of the signed exponent and add 127. Example. 2 16 then 2 127+16 = 2 143 and my value for the exponent would be 143 = 10001111 2 So it is simply now an unsigned value....

12. 12 Possible Representations of an Exponent

13. 13 Why Biased ? The smallest exponent 00000000 Only one exponent zero 01111111 The highest exponent is 11111111 To increase the exponent by one simply add 1 to the present pattern.

14. 14 Back to the example Our original example revisited…. 1.1011 * 2 2 Exponent is 2+127 =129 or 10000001 in binary. NOTE: Mantissa always ends up with a value of ‘1’ before the Dot. This is a waste of storage therefore it is implied but not actually stored. 1.1000 is stored.1000 6.75 in 32 bit floating point IEEE representation:- 0 10000001 10110000000000000000000 sign(1) exponent(8) mantissa(23)

15. 15 Special cases 0 + Infinity and - infinity.  Zero is a pattern that only contains ‘0’s 00000000000000000000000000000000  Positive Infinity is the pattern 011111111….  Negative Infinity is the pattern 111111111….

16. 16 Truncation and Rounding Following arithmetic operations on a floating point number we may have increased the number of mantissa bits. Since we will have a fixed storage (23 places) for the mantissa we require to limit these bits. The simplest approach is to truncate the result prior to storage Example0.1101101 stored in 4 bits stored in 4 bits => 0.1101 ( loss 0.0000101 )

17. 17 Rounding If lost digit is > ½ then add 1 to LSB Example – in 4 bits 0.1101101 <- 0.1101 + 0.0001 = 0.1110 ( rounded UP) 0.1101011 <- 0.1101 ( rounded DOWN) NOTE: Rounding is always preferred to truncation partly because it is intrinsically more accurate, and because we end up with a FAIR error.

18. 18 Other Considerations Truncation always undervalues the result, and can lead to a systematic error situation. Rounding has one major disadvantage since it requires up to two further arithmetic operations. Note. When we use floating point care has to be taken when comparing the size of numbers because we are generating binary fractions of a predefined length. There is always going to be the chance of recurring numbers etc like 1/3 in decimal 0.333333333333333333333 etc..

19. 19 From Floating Point Binary to Decimal Example 1 01111011 11100000100000000000000 Sign = 1 therefore this number is a negative number. Exponent 01111011 = 64+32+16+8+2+1 = 123 subtract the 127 = - 4 Mantissa = 1.111000001 1.111000001 * 2 - 4 -ve 0.0001111000001 1/16 + 1/32 +1/64+1/128+1/8192 or - 0.1173095703125

20. 20 Floating Point Maths Floating point addition and subtraction. 1. Make sure that the two numbers are of the same magnitude. Their Exponents have to be equal. 2. We then add or subtract the mantissas 3. Starting with the existing exponent re-normalise if needed.

21. 21 Example Example 1.1* 2 3 + 1.1 * 2 2 Select the smaller number and make the mantissa smaller by moving the point whilst increasing the exponent until the exponents match. 1.1 * 2 2 0.11 * 2 3 Add the mantissas Re-normalise.

22. 22 Example 1.1* 2 3 001.1 2 3 +1.1 * 2 2 000.11 2 3 010.01 2 3 Re normalise 010.01 * 2 3 = 1.001 * 2 4

23. 23 FP math Floating Point Multiplication Assume two numbers a x 2 m b x 2 n Result (a x 2 m ) x (b x 2 n ) = ( a x b ) x ( 2 m+n ) Floating Point Division Assume two number a x 2 m and b x 2 n Result (a x 2 m ) / (b x 2 n ) = (a/b ) x 2 m-n