Floating Point Representations CDA 3101 Discussion Session 02

Question 1 Converting the binary number 1100 0000 1101 1001 1001 1001 1001 10102 to decimal, if the binary is single precision floating-point?

Question 1 Converting bin (Single precision FP) to decimal 1100 0000 1101 1001 1001 1001 1001 10102 Sign bit : 1 Exponent : 10000001 = 129 Fraction : 10110011001100110011010 =1*2-1 + 1*2-3 + … + 1*2-19 + 1*2-20 + 1*2-22 =0.7000000476837158 (-1)S * (1.Fraction) * 2(Exponent - 127) =(-1)1 * (1.7000000476837158) * 2(129 - 127) =- 1.7000000476837158 * 2(129 - 127) =-6.80000019073486328125 S(1) Biased Exponent(8) Fraction (23)

Question 2 Show the IEEE 754 binary representation for the floating-point number 0.110 in single­precision and double­precision

Question 2.1 Converting 0.110 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) 01111011 10011001100110011001100

Question 2.2 Converting 0.110 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) 01111111011 1001100110011001100110011001100110011001100110011001

Question 3 Convert the following single-precision numbers into decimal

Question 3.1 Converting bin (Single precision FP) to dec 0 11111111 000000000000000000000002 Sign bit : 0 Exponent : 11111111 = Infinity Fraction : 00000000000000000000000 = 0 Infinity S(1) Biased Exponent(8) Fraction (23)

Question 3.2 Converting bin (Single precision FP) to dec 0 00000000 000000000000000000000102 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)

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 the third smallest positive normalized number the largest (farthest from zero) negative normalized number the third smallest positive denormalized number that can be represented.

Question 4.1 The third smallest positive normalized number Bias: 215-1 = 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

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) = -232767(264-1)2-63 = -232768 (approx.)

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