1. Floating Point Representations CDA 3101 Discussion Session 02

2. Question 1 Converting the binary number 1010 0100 1001 0010 0100 1001 0010 0100 2 to decimal, if the binary is Unsigned? 2 ’ s complement? Single precision floating-point?

3. Question 1.1 Converting bin (unsigned) to dec 1010 0100 1001 0010 0100 1001 0010 0100 2 1*2 31 + 1*2 29 + … + 1*2 8 + 1*2 5 + 1*2 2 = 2761050404

4. Question 1.2 Converting bin (2 ’ s complement) to dec 1010 0100 1001 0010 0100 1001 0010 0100 2 -1*2 31 + 1*2 29 + … + 1*2 8 + 1*2 5 + 1*2 2 = -1533916892

5. Question 1.3 Converting bin (Single precision FP) to dec 1010 0100 1001 0010 0100 1001 0010 0100 2 Sign bit : 1 Exponent : 01001001 = 73 Fraction : 00100100100100100100100 =1*2 -3 + 1*2 -6 + … + 1*2 -15 + 1*2 -18 + 1*2 -21 =0.142857074 (-1) S * (1.Fraction) * 2 (Exponent - 127) = (-1) 1 * (1.142857074) * 2 (73 - 127) = -1.142857074 * 2 -54 = -6.344131187 * 10 -17 S(1)Biased Exponent(8)Fraction (23)

6. Question 2 Show the IEEE 754 binary representation for the floating-point number 0.1 10 in single ­ precision and double ­ precision

7. Question 2.1 Converting 0.1 10 to single-precision FP Step1: Covert fraction 0.1 to binary (multiplying by 2) 0.1*2 = 0.2, 0.2*2 = 0.4, 0.4*2 = 0.8, 0.8*2 = 1.6, 0.6*2 = 1.2, 0.2*2 = 0.4, 0.4*2 = 0.8, 0.8*2 = 1.6, 0.6*2 = 1.2, … 000110011 … 1.10011 … * 2 -4 Step2: Express in single precision format (-1) S * (1.Fraction) * 2 (Exponent +127) = (-1) 0 * (1.10011001100110011001100) * 2 (-4+127) 00111101110011001100110011001100

8. Question 2.2 Converting 0.1 10 to double-precision FP Step1: Covert fraction 0.1 to binary (multiplying by 2) 0.1*2 = 0.2, 0.2*2 = 0.4, 0.4*2 = 0.8, 0.8*2 = 1.6, 0.6*2 = 1.2, 0.2*2 = 0.4, 0.4*2 = 0.8, 0.8*2 = 1.6, 0.6*2 = 1.2, … 000110011 … 1.10011 … * 2 -4 Step2: Express in double precision format (-1) S * (1.Fraction) * 2 (Exponent +1023) = (-1) 0 * (1.1001100110011001100110) * 2 (-4+1023) 0011111110111001100110011001100110011001100110011001100110011001

9. Question 3 Convert the following single-precision numbers into decimal a. 0 11111111 0000000000000000000000 b. 0 00000000 0000000000000000000010

10. Question 3.1 Converting bin (Single precision FP) to dec 0 11111111 00000000000000000000000 2 Sign bit : 0 Exponent : 11111111 = Infinity Fraction : 00000000000000000000000 = 0 Infinity S(1)Biased Exponent(8)Fraction (23)

11. Question 3.2 Converting bin (Single precision FP) to dec 0 00000000 00000000000000000000010 2 Sign bit : 0 Exponent : 00000000 = 0 Fraction : 00000000000000000000010 =1*2 -22 =0.000000238 (-1) S * (0.Fraction) * 2 -126 = (-1) 0 * (0.000000238) * 2 -126 = 2.797676555 * 10 -45 S(1)Biased Exponent(8)Fraction (23)

12. Question 4 Consider the 80-bit extended-precision IEEE 754 floating point standard that uses 1 bit for the sign, 16 bits for the biased exponent and 63 bits for the fraction (f). Then, write (i) the 80- bit extended-precision floating point representation in binary and (ii) the corresponding value in base-10 positional (decimal) system of a.the third smallest positive normalized number b.the largest (farthest from zero) negative normalized number c.the third smallest positive denormalized number that can be represented.

13. Question 4.1 The third smallest positive normalized number Bias: 2 15 -1 = 32767 Sign: 0 Biased Exponent: 0000 0000 0000 0001 Fraction (f): 61 zeros followed by 10 Decimal Value: (-1) 0 *2 (1-32767) *(1+2 -62 ) = 2 -32766 +2 -32828

14. Question 4.2 The largest (farthest from zero) negative normalized number Sign: 1 Biased Exponent: 1111 1111 1111 1110 Fraction: 63 ones Decimal Value: (-1) 1 *2 (65534-32767) *(1+2 -1 +2 -2 +…+2 -63 ) = -2 32767 (2 64 -1)2 -63 = -2 32768 (approx.)

15. Question 4.3 The third smallest positive denormalized number Sign: 0 Biased Exponent: 0000 0000 0000 0000 Fraction: 61 zeros followed by 11 Decimal Value: (-1) 0 *2 -32766 *(2 -62 +2 -63 ) = 3*2 -32829