CSI 312 Lecture 5, Fall 2011

Computer Architecture Lecture 9 - Floating Point 2011 Reading: 3.6-3.9 Homework: 3.30, 3.35, 3.37, 3.38, 3.40, 3.44 Why did the Ariane 5 Explode? (image source: java.sun.com) 2 Outline - Floating Point  Motivation and Key Ideas   IEEE 754 Floating Point Format  Floating Point Arithmetic  MIPS Floating Point Instructions  Summary 3 Floating Point - Motivation  Review: n-bit integer representations  Unsigned: 0 to 2n-1  Signed Two’s Complement: - 2n-1 to 2n-1-1  Biased (excess-b): -b to 2n-b  Problem: how do we represent:  Very large numbers 9,345,524,282,135,672, 2354  Very small numbers 0.00000000000000005216, 2-100  Rational numbers 2/3  Irrational numbers sqrt(2)  Transcendental numbers e, π 4 Fixed Point Representation  Idea: fixed-point numbers with fractions  Decimal point (binary point) marks start of fraction  Decimal: 1.2503 = 1 X 100 + 2 X 10-1 + 5 X 10-2 + 3 X 10-4  Binary: 1.0100001 = 1 X 20 + 1 X 2-2 + 1 X 2-7  Problems  Limited locations for “decimal point” (binary point”)  Won’t work for very small or very larger numbers 5 Another Approach: Scientific Notation  Represent a number as a combination of  Mantissa (significand): Normalized number AND  Exponent (base 10)  Example: 6.02 X 1023 Significand (mantissa) Radix (base) Exponent 6 Floating Point  Key idea: adapt scientific notation to binary  Fixed-width binary number for significand  Fixed-width binary number for exponent (base 2)  Idea: represent a number as 1.xxxxxxxtwo X 2yyyy Significand (mantissa) Radix (2) Exponent Leading ‘1’ (Implicit) Important Points: This is a tradeoff between precision and range Arithmetic is approximate - error is inevitable! 7 Outline - Floating Point  Motivation and Key Ideas  IEEE 754 Floating Point Format   Floating Point Arithmetic  MIPS Floating Point Instructions  Summary 8 IEEE 754 Floating Point  Single precision (C/C++/Java float type) Value N = (-1)S X 1.F X 2E-127  Double precision (C/C++/Java double type) Value N = (-1)S X 1.F X 2E-1023 S E Exponent F Significand 1 bit 8 bits 23 bits S E Exponent F Significand F Significand (continued - 52 bits total) 1 bit 11 bits 20 bits 32 bits Bias Bias The exponent represents both positive and negative values. To do this, a bias is added to the actual exponent in order to get the stored exponent. For IEEE SP , this value is 127 to represent 0, thus, a stored value of 200 indicated an exponent of (200 – 127) = 73. an exponent of 3 is stored as 130. an exponent of -2 is stored as (-2 +127) = 125. For DP bias = 1024. 9 Floating Point Examples  8.75ten = 1 X 23 + 1 X 2-1 + 1 X 2-2 = 1.00011 X 23  Single Precision: • Significand: 1.00011000…. (note leading 1 is implied) • Exponent: 3 + 127 = 130 = 10000010two  Double Precision: • Significand: 1.00011000… • Exponent: 3 + 1023 = 1026 = 10000000010two 0 E Exponent F Significand S 1 0 0 0 0 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 S E Exponent F Significand F Significand (continued - 52 bits total) 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 10 Floating Point Examples  -0.375ten = 1 X 2-2 + 1 X 2-3 = 1. 1 X 2-2  Single Precision: • Significand: 1.1000…. • Exponent: -2 + 127 = 125 = 01111101two  Double Precision: • Significand: 1.1000… • Exponent: -2 + 1023 = 1021 = 01111111101two 1 E Exponent F Significand S 0 1 1 1 1 1 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 S E Exponent F Significand F Significand (continued - 52 bits total) 1 0 1 1 1 1 1 1 1 1 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 11 Floating Point Examples  Q: What is the value of the following single- precision word?  Significand = 1 + 2-1 + 2-4 + 2-8 + 2-10 + 2-12  Exponent = 8 - 127 = -119  Final Result = (1 + 2-1 + 2-4 + 2-8 + 2-10 + 2-12) X 2-119 = 2.36 X 10-36 0 E Exponent F Significand S 0 0 0 0 1 0 0 0 1 0 0 1 0 0 0 1 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 12 Outline - Floating Point  Motivation and Key Ideas  IEEE 754 Floating Point Format  Floating Point Arithmetic   MIPS Floating Point Instructions  Rounding & Errors  Summary Floating Point Addition 13 Add: 9.999 x 101 to 1.610 x 10-1 Step 1: Normalize the smaller number to fit with the larger: 0.0161 x 101 If we can represent only 4 digit number so it becomes: 0.016 x 101 Step 2: 9.999 0.016 10.015 (not in normalized scientific notation Step 3: 10.015 x 101 = 1.0015 x 102 14 Floating Point Addition (Fig. 3.16) 1. Align binary point to number with larger exponent 2. Add significands 3. Normalize result and adjust exponent 4. If overflow/underflow throw exception 5. Round result (go to 3 if normalization needed again) A 1.11 X 20 1.11 X 20 1.75 + B + 1.00 X 2-2 + 0.01 X 20 0.25 10.00 X 20 (Normalize) 1.00 X 21 2.00 15 Floating Point Multiplication (Fig. 3.18) 1. Add 2 exponents together to get new exponent (subtract 127 to get proper biased value) 2. Multiply significands 3. Normalize result if necessary (shift right) & adjust exponent 4. If overflow/underflow throw exception 5. Round result (go to 3 if normalization needed again) 6. Set sign of result using sign of X, Y 16 Outline - Floating Point  Motivation and Key Ideas  IEEE 754 Floating Point Format  Floating Point Arithmetic  MIPS Floating Point Instructions   Summary 17 MIPS Floating Point Instructions  Organized as a coprocessor  Separate registers $f0-$f31  Separate operations  Separate data transfer (to same memory)  Basic operations  add.s - single add.d - double  sub.s - single sub.d - double  mul.s - single mul.d - double  div.s - single div.d - double 18 MIPS Floating Point Instructions (cont’d)  Data transfer  lwc1, swcl (l.s, s.s) - load/store float to fp reg  l.d, s.d - load/store double to fp reg pair  Testing / branching  c.lt.s, c.lt.d, c.eq.s, c.eq.d, … compare and set condition bit if true  bclt - branch if condition true  bclf - branch if condition false 19 Outline - Floating Point  Motivation and Key Ideas  IEEE 754 Floating Point Format  Floating Point Arithmetic  MIPS Floating Point Instructions  Summary 20 Addendum - Why Did the Ariane 5 Explode?  In 1996 Ariane 5 Flight 501 exploded after launch.  Estimated cost of accident: $500 million 21 Addendum - Why Did the Ariane 5 Explode?  The cause was traced to the Inertial reference system (SRI).  Both the main and backup SRI failed.  Both units failed due to an out-of-range conversion Input: double precision floating point Output: 16-bit integer for “horizontal bias” (BH)  Careful analysis during software design had indicated that BH would “fit” in 16 bits  So, why didn’t it fit? 22 Addendum - Why did the Ariane 5 Explode?  Careful analysis during software design had indicated that BH would “fit” in 16 bits  BUT, all analysis had been done for the Ariane 4, the predecessor of Ariane 5 - software was reused  Since Ariane 5 was a larger rocket, the values for BH were higher than anticipated  AND, there was no handler to deal with the exception!  For more information:  http://www.ima.umn.edu/~arnold/disasters/ariane.html  Or, Google “Ariane 5 Flight 501” 23 Summary - Chapter 3  Important Topics  Signed & Unsigned Numbers (3.2)  Addition and Subtraction (3.3)  Constructing an ALU (B.5)  Multiplication and Division (3.4, 4.5)  Floating Point (3.6)  Coming Up:  Performance (Chapter 4)