# ECE331: Hardware Organization and Design

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1 ECE331: Hardware Organization and Design Lecture 10: Multiplication & Floating Point Representation Adapted from Computer Organization and Design, Patterson & Hennessy, UCB

2 MIPS Division Two 32-bit registers for product HI: most-significant 32 bits LO: least-significant 32-bits Instructions div rs, rt / divu rs, rt 32-bit result in LO, 32-bit remainder in HI mfhi rd / mflo rd Moves data from hi and lo registers Example mfhi \$s0 -> move remainder to register \$s0 mflo \$s0 -> move result of division to register \$s0 ECE331: Floating Point 2

3 MULTIPLY Paper and pencil example (unsigned): Multiplicand 1000 Multiplier Product m bits x n bits = m+n bit product Issues to note: The number of partial products is equal to the number of bits in the multiplier The partial products are added together A total of n-1 adders could be used to generate the result ECE331: Floating Point 3

4 MULTIPLY Paper and pencil example (unsigned): Multiplicand 1000 Multiplier Product Need one adder ECE331: Floating Point 4

5 MULTIPLY Paper and pencil example (unsigned): Multiplicand 1000 Multiplier Product Need one adder ECE331: Floating Point 5

6 MULTIPLY Paper and pencil example Multiplicand 1000 Multiplier three additions 1000 Product B 3 B 2 B 1 B 0 -> Multiplicand A 3 A 2 A 1 A 0 -> Multiplier ECE331: Floating Point 6

7 What is the Delay Through the Multiplier? x3,x2,x1,x0=1111 y3,y2,y1,y0=0001 ΔFA - operation time - delay Assuming equal delays for sum and carry-out Longest carry propagation chain when adding two 4-bit numbers In synchronous arithmetic units - time allowed for adder's operation is worst-case delay - nδfa ECE331: Floating Point 7

9 Multiplier Delay Critical paths C0 delay = 0 C1 delay = ΔFA C2 delay = 3ΔFA C3 delay = 5ΔFA C4 delay = 6ΔFA C5 delay = 7ΔFA C6 delay = 8ΔFA C7 delay = 8ΔFA What s the delay for n bits? (n-2) * (2ΔFA) + nδfa = (3n-4)ΔFA ECE331: Floating Point 9 Critical path for c7

10 Floating Point Representation for non-integral numbers Including very small and very large numbers Like scientific notation In binary ±1.xxxxxxx 2 2 yyyy Types float and double in C normalized ECE331: Floating Point 10

11 Floating Point Numbers The largest 32 bit unsigned integer number is = 4,294,967,295 What if we want to encode the approx. age of the earth? 4,600,000,000 or 4.6 x 10 9 or the weight in kg of one a.m.u. (atomic mass unit) or 1.6 x There is no way we can encode either of the above in a 32- bit integer. ECE331: Floating Point 11

12 Exponential Notation The following are equivalent representations of 1, ,400.0 x ,340.0 x ,234.0 x x x x x x 10 5 The representations differ in that the decimal place the point - floats to the left or right (with the appropriate adjustment in the exponent). ECE331: Floating Point 12

13 Parts of a Floating Point Number x 10-3 Exponent Sign of mantissa Location of decimal point Mantissa Sign of exponent Base Mantissa is also called Significand ECE331: Floating Point 13

14 Single Precision Format Note that the exponent has no explicit sign bit Base? 32 bits M: Mantissa (23 bits) E: Exponent (8 bits) S: Sign of mantissa (1 bit) ECE331: Floating Point 14

15 Normalization The mantissa M is a normalized fraction Has an implied decimal place on left Has an implied (hidden) 1 on left of the decimal place E.g., Fraction Represents = The significand=1.f is in the range [1, 2-ulp] ulp unit in the last position (what remains to reach a whole number when all bits are set to one) F ECE331: Floating Point 15 S = ( 1) 1. f 2 E Bias Value of exponent (unsigned integer) Bias value (known; set by convention)

16 Binary Fractions To convert binary fractions to floating point = 1*(0.5) + 1*(0.25) + 1*(0.125) + 0*(0.0625) + 0*( ) + 0*( ) + 1*( ) + 0*( ) = ECE331: Floating Point 16

17 Binary Fractions To convert floating point to binary whole number fraction 9 à *(0.5) = *(0.25) = neg. num *(0.125) = 0 X à = x 2 4 = x 2 3 Note that we can shift positions left and right of the decimal point by multiplying by different powers of 2 ECE331: Floating Point 17

18 Working with normalization (single precision) S {0,1} (1-bit) 1 E 254 (8-bits unsigned integer); Bias = f base-2 < Formula provides for the full range of possible floating point numbers. F S = ( 1) 1. f 2 E Bias ECE331: Floating Point 18

19 IEEE Floating-Point Format single: 8 bits double: 11 bits S Exponent single: 23 bits double: 52 bits Fraction x = ( 1) S (1+ Fraction) 2 (Exponent Bias) S: sign bit (0 non-negative, 1 negative) Normalize significand: 1.0 significand < 2.0 Always has a leading pre-binary-point 1 bit, so no need to represent it explicitly (hidden bit) Significand is Fraction with the 1. restored Exponent: excess representation: actual exponent + Bias Ensures exponent is unsigned Single: Bias = 127; Double: Bias = 1023 ECE331: Floating Point 19

20 Single-Precision Range Exponents and reserved Smallest value Exponent: actual exponent = = 126 Fraction: significand = 1.0 ± ± Largest value exponent: actual exponent = = +127 Fraction: significand 2.0 ± ± ECE331: Floating Point 20

21 Floating-Point Example To convert floating point number to binary Represent = = ( 1) S = 1 Fraction = Exponent = 1 + Bias Single: = 126 = Double: = 1022 = Single: fraction sign exponent and bias Double: ECE331: Floating Point 21

22 Floating-Point Example To convert from binary to floating point What number is represented by the single-precision float? Identify the components S = 1 Fraction = Exponent = = 129 Calculate the value x = ( 1) 1 ( ) 2 ( ) = ( 1) = 5.0 ECE331: Floating Point 22

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