Real Number Representation Topicscomarc/slides/lect3-3.pdf · CSE1301 Sem 2 - 2003 Lecture 30: Real Number Representation 31 30 23 22 0 Representing Numbers (cont) –23 bits for

CSE1301 Sem 2 - 2003

Lecture 30: Real Number Representation 1

1Real Number Representation

2Topics

• Terminology

• IEEE standard for floating-pointrepresentation

• Floating point arithmetic

• Limitations

3Terminology

• All digits in a number following any leading zeros are significant digits:

12.345-0.123450.00012345

CSE1301 Sem 2 - 2003


4Terminology (cont)

• The scientific notation for real numbers is:

mantissa× baseexponent

In C, the expression: 12.456e-2

means: 12.456 × 10-2

5Terminology (cont)

• The mantissa is always normalizedbetween 1 and the base (i.e., exactly one significant digitbefore the point)

Unnormalized Normalized

2997.9 × 105 2.9979 × 108

B1.39FC × 1611 B.139FC × 1612

0.010110110101 × 2-1 1.0110110101 × 2-3

6Terminology (cont)

• The precision of a number is how many digits (or bits) we use to represent it

• For example:33.143.14159263.1415926535897932384626433832795028

CSE1301 Sem 2 - 2003


7Representing Numbers

• A real number n is represented by a floating-point approximation n*

• The computer uses 32 bits (or more) to store each approximation

• It needs to store

– the mantissa

– the sign of the mantissa

– the exponent (with its sign)

831 30 02223

Representing Numbers (cont)

• The standard way to allocate 32 bits (specified by IEEE Standard 754) is:

– 23 bits for the mantissa

– 1 bit for the mantissa's sign

– 8 bits for the exponent

931 30 02223





CSE1301 Sem 2 - 2003


1031 30 02223





1131 30 02223





12• The mantissahas to be in the range 1 ≤ mantissa < base

• Therefore – If we use base 2, the digit before the point must

be a 1

– So we don't have to worry about storing itWe get 24 bits of precision using 23 bits

Representing the Mantissa

CSE1301 Sem 2 - 2003


13Representing the Mantissa (cont)

• 24 bits of precision are equivalent to a little over 7 decimal digits:

24

log210≈ 7.2

14Representing the Mantissa (cont)

• Suppose we want to represent π:3.1415926535897932384626433832795.....

• That means that we can only represent it as:3.141592 (if we truncate)3.141593 (if we round)

15Representing the Exponent

• The exponent is represented asexcess-127. E.g.,

Actual Exponent Stored Value-127 ↔ 00000000-126 ↔ 00000001

. . .0 ↔ 01111111

+1 ↔ 10000000. . .

i ↔ (i+127)2. . .

+128 ↔ 11111111

CSE1301 Sem 2 - 2003


16Representing the Exponent (cont)

• The IEEE standard restricts exponents to the range:

–126 ≤ exponent≤ +127

• The exponents –127 and +128 have special meanings:

– If exponent = –127, the stored value is 0

– If exponent = 128, the stored value is ∞

17Representing Numbers -- Example 1What is 01011011 (8-bit machine) ?

0 101 1011

sign exp mantissa

• Mantissa:1.1011

• Exponent (excess-3 format):5-3=2

1.1011 × 22⇒ 110.11

110.112 = 22 + 21 + 2-1 + 2-2

= 4 + 2 + 0.5 + 0.25 = 6.75

18Representing Numbers -- Example 2Represent -10.375 (32-bit machine)

10.37510 = 10 + 0.25 + 0.125

= 23 + 21+ 2-2 + 2-3

= 1010.0112 ⇒ 1.0100112 × 23

• Sign: 1• Mantissa:010011• Exponent (excess-127 format):

3+127 = 13010 = 100000102

1 10000010 01001100000000000000000

CSE1301 Sem 2 - 2003


19Floating Point Overflow

• Floating point representations can overflow, e.g.,

1.111111× 2127

+ 1.111111 × 2127

11.111110 × 2127

= ∞1.1111110× 2128

20Floating Point Underflow

• Floating point numbers can also get too small, e.g.,

10.010000 × 2-126

÷ 11.000000 × 20

0.110000 × 2-126

= 01.100000× 2-127

21“Normalized”“Normalized”

• Condition– exp ≠ 000…0 and exp ≠ 111 …1

• Exponent coded as biasedvalueE = Exp – Bias

• Exp : unsigned value denoted by exp

• Bias : Bias value– Single precision: 127 (Exp: 1…254, E: -126…127)– Double precision: 1023 (Exp: 1…2046, E: -1022…1023)– in general: Bias= 2e-1 - 1, where e is number of exponent bits

• Significand coded with implied leading 1M = 1.xxx …x2

• xxx …x : bits of frac

• Minimum when 000…0 (M = 1.0)

• Maximum when 111…1 (M = 2.0 –ε)

• Get extra leading bit for “free”

CSE1301 Sem 2 - 2003


22Denormalized ValuesDenormalized Values

• Condition– exp = 000…0

• Value

– Exponent value E = –Bias+ 1

– Significand value M = 0.xxx …x2

• xxx …x : bits of frac

• Cases

– exp = 000…0, frac = 000…0

• Represents value 0

• Note that have distinct values +0 and –0

– exp = 000…0, frac ≠ 000…0

• Numbers very close to 0.0

• Lose precision as get smaller

• “Gradual underflow”

23Special ValuesSpecial Values

• Condition

– exp = 111…1

• Cases

– exp = 111…1, frac = 000…0

• Represents value ∞ (infinity)

• Operation that overflows

• Both positive and negative

• E.g., 1.0/0.0 = −1.0/−0.0 = +∞, 1.0/−0.0 = −∞– exp = 111…1, frac ≠ 000…0

• Not-a-Number (NaN)

• Represents case when no numeric value can be determined

• E.g., sqrt(–1), ∞ − ∞

24Floating Point Representation

Most standard floating point representation use:1 bit for the sign (positive or negative)8 bits for the range (exponent field)

23 bits for the precision (fraction field)

S exponent fraction2381

( )( ) =××−=

≤≤××−=−

−

0 ,2.01

2541 ,2.11126

127

exponentfractionN

exponentfractionNexponentS

exponentS

CSE1301 Sem 2 - 2003




( )( ) =××−=

≤≤××−=−

−

0 ,2.01

2541 ,2.11126

127

exponentfractionN


exponentS

point? floatingin drepresente 8

56number theis How :Example −

( )( ) ( )2

22

321012

210101.1101.110

212021202121

81

21

2481

84

2485

6

×−=−=×+×+×+×+×+×−=

+++−= +++−=−

−−−

Thus the exponent is given by:

1292127 =⇒=− exponentexponent1 10000001 10101000000000000000000


(example)S exponent fraction

2381

( )( ) =××−=

≤≤××−=−

−

0 ,2.01

2541 ,2.11126

127

exponentfractionN


exponentS

00111101100000000000000000000000

What is the decimal value of the following floating point number?

exponent

exponent = 64+32+16+8+2+1=(128-8)+3=120+3=123

( )161

20.120.11 41271230 =×=××−= −−N



2381

( )( ) =××−=

≤≤××−=−

−

0 ,2.01

2541 ,2.11126

127

exponentfractionN


exponentS

01000001100101000000000000000000


exponent

exponent =128+2+1=131

( ) 24

2127131

20 1.10010200101.1200101.11 =×=××−= −N

5.182

1216222 114 =++=++= −N

CSE1301 Sem 2 - 2003




2381

( )( ) =××−=

≤≤××−=−

−

0 ,2.01

2541 ,2.11126

127

exponentfractionN


exponentS

11000001000101000000000000000000


exponent

exponent =128+2=130

( ) 23

2127130

21 01.1001200101.1200101.11 −=×−=××−= −N

( ) 25.941

18222 203 −= ++−=++−= −N

29Floating Point


( )( ) =××−=

≤≤××−=−

−

0 ,2.01

2541 ,2.11126

127

exponentfractionN


exponentS

What is the largest number that can be represented in 32 bits floatingpoint using the IEEE 754 format above?

01111111011111111111111111111111

exponentexponent =254

232221 2121....2121 −−−− ×+×++×+×=fraction

99999998807.0810241024

11

2

112121

23230 =

××−=−=×−×= −fraction

30Floating Point


( )( ) =××−=

≤≤××−=−

−

0 ,2.01

2541 ,2.11126

127

exponentfractionN


exponentS

What is the largest number that can be represented in 32 bits floatingpoint using the IEEE 754 format above?

01111111011111111111111111111111

exponentactual exponent =254-127 = 127 99999998807.0=fraction

( ) 1281270 2299999998807.11 ≈××−=N

CSE1301 Sem 2 - 2003


31Floating Point


( )( ) =××−=

≤≤××−=−

−

0 ,2.01

2541 ,2.11126

127

exponentfractionN


exponentS

What is the smallest number (closest to zero) that can be represented in 32 bits floating point using the IEEE 754 format above?

00000000000000000000000000000001

exponentactual exponent =0-126 = -126 2321 −×=fraction

( ) 149126230 2221 −−− ≈××−=N

32Special Floating Point

RepresentationsIn the 8-bit field of the exponent we can represent numbers from 0 to255. We studied how to read numbers with exponents from 0 to 254.What is the value represented when the exponent is 255 (i.e. 111111112)?

An exponent equal 255 = 111111112 in a floating point representationindicates a special value.

When the exponent is equal 255 = 111111112 and the fraction is 0,the value represented is ± infinity.

When the exponent is equal 255 = 111111112 and the fraction is non-zero, the value represented is Not a Number (NaN).

33Double Precision

32-bit floating point representation is usually called single precisionrepresentation.

A double precision floating point representation requires 64 bits. In double precision the following number of bits are used:

1 sign bit11 bits for exponent52 bits for fraction (also called significand)

CSE1301 Sem 2 - 2003


34Summary of Floating Point Real Number Encodings

NaNNaN

+∞−∞

−0

+Denorm +Normalized-Denorm-Normalized

+0

35Special Properties of

EncodingSpecial Properties of

Encoding• FP Zero Same as Integer Zero

– All bits = 0

• Can (Almost) Use Unsigned Integer Comparison

– Must first compare sign bits

– Must consider -0 = 0

– NaNs problematic

• Will be greater than any other values

• What should comparison yield?

– Otherwise OK

• Denorm vs. normalized

• Normalized vs. infinity

36Floating Point Addition

Five steps to add two floating point numbers:

1. Express the numbers with the same exponent (denormalize)

2. Add the mantissas

3. Adjust the mantissa to one digit/bit before the point (renormalize)

4. Round or truncate to required precision

5. Check for overflow/underflow

CSE1301 Sem 2 - 2003


37Floating Point Addition -- Example 1(Assume precision 4 decimal digits)

x = 9.876 × 107

y = 1.357 × 106

38Floating Point Addition -- Example 1 (cont)

(Assume precision 4 decimal digits)

1. Use the same exponents:

x = 9.876 × 107

y = 0.1357 × 107



2. Add the mantissas:

x = 9.876 × 107

y = 0.136 × 107

x+y = 10.012× 107

CSE1301 Sem 2 - 2003




3. Renormalize the sum:

x = 9.876 × 107

y = 0.136 × 107

x+y = 1.0012 × 108



4. Truncate or round:

x = 9.876 × 107

y = 0.136 × 107

x+y = 1.001 × 108



5. Check overflow and underflow:

x = 9.876 × 107

y = 0.136 × 107

x+y = 1.001 × 108

CSE1301 Sem 2 - 2003


43Floating Point Addition -- Example 2(Assume precision 4 decimal digits)

x = 3.506 × 10-5

y = -3.497 × 10-5



1. Use the same exponents:

x = 3.506 × 10-5

y = -3.497 × 10-5



2. Add the mantissas:

x = 3.506 × 10-5

y = -3.497 × 10-5

x+y = 0.009 × 10-5

CSE1301 Sem 2 - 2003




3. Renormalize the sum:

x = 3.506 × 10-5

y = -3.497 × 10-5

x+y = 9.000× 10-8



4. Truncate or round:

x = 3.506 × 10-5

y = -3.497 × 10-5

x+y = 9.000 × 10-8 (no change)



5. Check overflow and underflow:

x = 3.506 × 10-5

y = -3.497 × 10-5

x+y = 9.000 × 10-8

CSE1301 Sem 2 - 2003


49Limitations

• Floating-point representations only approximate real numbers

• The normal laws of arithmetic don't always hold, e.g., associativity is notguaranteed

50Limitations -- Example


x = 3.002 × 103

y = -3.000 × 103

z = 6.531 × 100

51Limitations -- Example (cont) (Assume precision 4 decimal digits)

x = 3.002 × 103

y = -3.000 × 103

z = 6.531 × 100

x+y = 2.000 × 100

CSE1301 Sem 2 - 2003



x = 3.002 × 103

x+y = 2.000 × 100

y = -3.000 × 103

z = 6.531 × 100

(x+y)+z = 8.531 × 100


x = 3.002 × 103

y = -3.000 × 103

z = 6.531 × 100


x = 3.002 × 103

y = -3.000 × 103

z = 6.531 × 100

y+z = -2.993 × 103

CSE1301 Sem 2 - 2003



x = 3.002 × 103

y = -3.000 × 103

y+z = -2.993 × 103

z = 6.531 × 100

x+(y+z) = 0.009 × 103


x = 3.002 × 103

x+(y+z) = 9.000 × 100

y = -3.000 × 103

y+z = -2.993 × 103

z = 6.531 × 100


x = 3.002 × 103

x+(y+z) = 9.000 × 100

y = -3.000 × 103

(x+y)+z = 8.531 × 100

z = 6.531 × 100

CSE1301 Sem 2 - 2003


58Mathematical Properties of FP AddMathematical Properties of FP Add

• Compare to those of Abelian Group–Closed under addition? YES

• But may generate infinity or NaN

–Commutative? YES

–Associative? NO• Overflow and inexactness of rounding

–0 is additive identity? YES

–Every element has additive inverseALMOST

• Except for infinities & NaNs

59Circuitry for

Addition/Subtraction

60Multiplication

• Multiply Significands

• Add Exponents

• Normalize

– Shift Significand

– Add or Subtract shift amount to exponent

• Round

– To number of bits for significand

– need to keep extra bits during computation

• Normalize again if necessary

CSE1301 Sem 2 - 2003


61For multiplication of P = X ×××× Y :

1. Compute Exponent: Exp P = ( ExpY + ExpX ) - Bias

2. Compute Product: ( 1 + Sig X ) ×××× ( 1 + SigY )Normalize if necessary; continue until most significant bit i s 1

4. Too small (e.g., 0.001xx... ) →→→→left shift result, decrement result exponent

4'. Too big (e.g., 10.1xx... ) →→→→right shift result, increment result exponent

5. If (result significand is 0) then set exponent to 0

6. if (SgnX == SgnY ) thenSgnP = positive (0)

elseSgnP = negative (1)

62FP Multiplication Algorithm

Start

1. Add the biased exponents of the two numbers, subtracting the bias from the sum to get the new biased exponent.

2. Multiply the Significands.

3. Normalize the product if necessary, shifting it right and incrementing the exponent.

4. Round the significant to the appropriate number of bits.

Overflow or underflow?

Still Normalized?

Exception

Doneyes

yes

no

no

5. Set the sign of the product to positive if the signs of the original operands are the same. If they differ, make the sign negative.

63010 1 0 1

Control

Small ALU

Big ALU

Sign Exponent Significand Sign Exponent Significand

Exponent difference

Shift right

Shift left or right

Rounding hardware

Sign Exponent Signif icand

Increment or decrement

0 10 1

Shift smaller number right Compare exponents Add

Normalize

Round

• FP ADD: Exponents are subtracted by small ALU; the difference controls the 3 MUXes;

• Shift smallerexp. to the right until exponents match;

• Significants are added in Big ALU;

• Normalization step shifts result left or right, adjusts exponents;

• Rounding and possible nornalization

CSE1301 Sem 2 - 2003


64Floating Point Multiplication

(Decimal)Assume that we only can store four digits of the significand and two digits of the exponent in a decimal f loating point representation.

How would you multiply 1.11010×1010 by 9.20010×10-5 inthis representation?

Step 1: Add the exponents: new exponent = 10 - 5 = 5

Step 2: Multiply the significands: 1.110×9.200

00000000

2220 9990

10.212000

Step 3: Normalize the product:10.21210×105 = 1.021210 ×106

Step 4: Round-off the product:1.021210×106 = 1.02110 ×106

65Math. Properties of FP MultMath. Properties of FP Mult

• Compare to Commutative Ring

– Closed under multiplication? YES

• But may generate infinity or NaN

– Multiplication Commutative? YES

– Multiplication is Associative? NO

• Possibility of overflow, inexactness of rounding

– 1 is multiplicative identity? YES

– Multiplication distributes over addition? NO

• Possibility of overflow, inexactness of rounding

• Monotonicity

– a ≥ b & c ≥ 0 ⇒ a * c ≥ b * c? ALMOST

• Except for infinities & NaNs

66Floating-Point Division

• Significands divided - exponents subtracted -bias added to difference E1-E2

• If resulting exponent out of range - overflow or underflow indication must be generated

• Resultant significand satisfies 1/β ≤ M1/M2 < β

• A single base-β shift right of significand + increase of 1 in exponent may be needed in postnormalization step - may lead to an overflow

CSE1301 Sem 2 - 2003


67Remainder in Floating-Point Division• Fixed-point remainder -R=X-QD (X, Q, D -

dividend, quotient, divisor) -|R| ≤ |D| - generated by division algorithm (restoring or nonrestoring)

• Flp division - algorithm generates quotient but not remainder -F1 REM F2 = F1-F2⋅Int(F1/F2) (Int(F1/F2)- quotient F1/F2converted to integer)

• Conversion to integer - either truncation (removing fractional part) or rounding-to-nearest

• The IEEE standard uses the round-to-nearest-evenmode -|F1 REM F2| ≤ |F2| /2

Emax-Emin68

Floating-Point Remainder • Brute-force- continue direct division algorithm for E1-E2steps

• Problem- E1-E2can be much greater than number of steps needed to generate m bits of quotient's significand - may take an arbitrary number of clock cycles

• Solution- calculate remainder in software

• Alternative- Define a REM-stepoperation - X REM F2 -performs a limited number of divide steps (e.g., limited to number of divide steps required in a regular divide operation)

• Initial X=F1, then X=remainder of previous REM-stepoperation

• REM-steprepeated until remainder ≤ F2/2

69Floating Point in CFloating Point in C

• C Guarantees Two Levelsfloat single precision

double double precision

• Conversions

– Casting between int , float , and double changes numeric values

– Double or float to int

• Truncates fractional part

• Like rounding toward zero

• Not defined when out of range

– Generally saturates to TMin or TMax– int to double

• Exact conversion, as long as int has ≤ 53 bit word size

– int to float

• Will round according to rounding mode

CSE1301 Sem 2 - 2003


70Floating Point PuzzlesFloating Point Puzzles

– For each of the following C expressions, either:• Argue that it is true for all argument values

• Explain why not true• x == (int)(float) x

• x == (int)(double) x

• f == (float)(double) f

• d == (float) d

• f == -(-f);

• 2/3 == 2/3.0

• d < 0.0 ⇒ ((d*2) < 0.0)

• d > f ⇒ -f > -d

• d * d >= 0.0

• (d+f)-d == f

int x = …;

float f = …;

double d = …;

Assume neitherd nor f is NaN

71Answers to Floating Point PuzzlesAnswers to Floating Point Puzzles

• x == (int)(float) x

• x == (int)(double) x

• f == (float)(double) f

• d == (float) d

• f == -(-f);

• 2/3 == 2/3.0

• d < 0.0 ⇒ ((d*2) < 0.0)

• d > f ⇒ -f > -d

• d * d >= 0.0

• (d+f)-d == f

int x = …;

float f = …;

double d = …;

Assume neitherd nor f is NAN

• x == (int)(float) x No: 24 bit significand

• x == (int)(double) x Yes: 53 bit significand

• f == (float)(double) f Yes: increases precision

• d == (float) d No: loses precision

• f == -(-f); Yes: Just change sign bit

• 2/3 == 2/3.0 No: 2/3 == 0

• d < 0.0 ⇒ ((d*2) < 0.0) Yes!

• d > f ⇒ -f > -d Yes!

• d * d >= 0.0 Yes!

• (d+f)-d == f No: Not associative

72MIPS Coprocessors

CPU

Registers

$0

$31

Arithmetic unit

Multiply divide

Lo Hi

Coprocessor 1 (FPU)

Registers

$0

$31

Arithmetic unit

Registers

BadVAddr

Coprocessor 0 (traps and memory)

Status

Cause

EPC

Memory

CSE1301 Sem 2 - 2003


73Floating Point in MIPS

MIPS Supports the IEEE 754 single-precision and double-precisionformats.

MIPS has a separate set of registers to store floating point operands:$f0, $f1, $f2, ...

In single precision, each individual register $f0, $f1, $f2, … contains one single precision (32-bit) value.

In double precision, each pair of registers $f0-$f1, $f2-$f3, … contains one double precision (64-bit) value.

74Floating Point in MIPS

In order to load a value in a floating point register, MIPS offers theload word coprocessor, lwcz, instructions. Because the floating pointcoprocessor is the coprocessor number 1, the instruction is lwc1.

Similarly to store the value of a floating point register into memory,MIPS offers the store word coprocessor, swc1.add.s add.d FP addition single or double sub.s sub.d FP subtraction single or double mul.s mul.d FP multiplication single or double div.s div.d FP division single or double c.x.s c.x.d FP comparison single or double (x = eq, neq. lt, le. gt, or ge) bclt FP branch true bclf FP branch false

75Floating Point Instruction in

MIPSWhat does the following assembly code do?

lwc1 $f4, 4($sp)lwc1 $f6, 8($sp)add.s $f2, $f4, $f6swc1 $f2,12($sp)

Reads two floating point values from the stack, performstheir addition and stores the result in the stack.

CSE1301 Sem 2 - 2003


76Pentium Bug

• Pentium FP Divider uses algorithm to generate multiple bits per steps

– FPU uses most significant bits of divisor & dividend/remainder to guess next 2 bits of quotient

– Guess is taken from lookup table: -2, -1,0,+1,+2 (if previous guess too large a reminder, quotient is adjusted in subsequent pass of -2)

– Guess is multiplied by divisor and subtracted from remainder to generate a new remainder

– Called SRT division after 3 people who came up with idea

• Pentium table uses 7 bits of remainder + 4 bits of divisor = 211 entries

• 5 entries of divisors omitted: 1.0001, 1.0100, 1.0111, 1.1010, 1.1101 from PLA (fix is just add 5 entries back into PLA: cost $200,000)

• Self correcting nature of SRT => string of 1s must follow error

– e.g., 1011 1111 1111 1111 1111 1011 1000 0010 0011 0111 1011 0100 (2.99999892918)

• Since indexed also by divisor/remainder bits, sometimes bug doesn’t show even with dangerous divisor value

77Pentium bug appearance

• First 11 bits to right of decimal point always correct: bits 12 to 52 where bug can occur (4th to 15th decimal digits)

• FP divisors near integers 3, 9, 15, 21, 27 are dangerous ones:

– 3.0 > d � 3.0 - 36 x 2–22 , 9.0 > d � 9.0 - 36 x 2–20

– 15.0 > d � 15.0 - 36 x 2–20 , 21.0 > d � 21.0 - 36 x 2–19

• 0.333333 x 9 could be problem

• In Microsoft Excel, try (4,195,835 / 3,145,727) * 3,145,727

– = 4,195,835 => not a Pentium with bug

– = 4,195,579 => Pentium with bug(assuming Excel doesn’t have SW bug patch)

– Rarely noticed since error in 5th significant digit

– Success of IEEE standard made discovery possible: - all computers should get same answer

78Pentium Bug Time line

• June 1994: Intel discovers bug in Pentium: takes months to make change, reverify, put into production: plans good chips in January 1995 4 to 5 million Pentiums produced with bug

• Scientist suspects errors and posts on Internet in September 1994

• Nov. 22 Intel Press release: “Can make errors in 9th digit ... Most engineers and financial analysts need only 4 of 5 digits. Theoretical mathematician should be concerned. ... So far only heard from one.”

• Intel claims happens once in 27,000 years for typical spread sheet user:

– 1000 divides/day x error rate assuming numbers random

• Dec 12: IBM claims happens once per 24 days: Bans Pentium sales

– 5000 divides/second x 15 minutes = 4,200,000divides/day

– Intel said it regards IBM's decision to halt shipments of its Pentium processor-based systems as unwarranted.

CSE1301 Sem 2 - 2003


79Pentium jokes• Q: What's another name for the "Intel Inside" sticker they put on Pentiums?

A: Warning label.

• Q: Have you heard the new name Intel has chosen for the Pentium?

A: the Intel Inacura.

• Q: According to Intel, the Pentium conforms to the IEEE standards for floating point arithmetic. If you fly in aircraft designed using a Pentium, what is the correct pronunciation of "IEEE"?

A: Aaaaaaaiiiiiiiiieeeeeeeeeeeee!

• TWO OF TOP TEN NEW INTEL SLOGANS FOR THE PENTIUM

9.9999973251 It's a FLAW, Dammit, not a Bug

7.9999414610 Nearly 300 Correct Opcodes

80Pentium conclusion: Dec. 21, 1994 $500M write-off“To owners of Pentium processor-based computers and the PC community:We at Intel wish to sincerely apologize for our handling of the recently

publicized Pentium processor flaw.The Intel Inside symbol means that your computer has a microprocessor second to none

in quality and performance. Thousands of Intel employees work very hard to ensure that this is true. But no microprocessor is ever perfect.

What Intel continues to believe is technically an extremely minor problem has taken on a life of its own. Although Intel firmly stands behind the quality of the current version of the Pentium processor, we recognize that many users have concerns.

We want to resolve these concerns.

Intel will exchange the current version of the Pentium processor for anupdated version, in which this floating-point divide flaw is corrected, forany owner who requests it, free of charge anytime during the life of theircomputer. Just call 1-800-628-8686.”

Sincerely,Andrew S. Grove Craig R. Barrett Gordon E. MoorePresident /CEO Executive Vice President Chairman of the Board

&COO

81• Pentium: Difference between bugs that board designers must know about and bugs that potentially affect all users

–Why not make public complete description of bugs in later category?

–$200,000 cost in June to repair design

–$500,000,000 loss in December in profits to replace bad parts

–How much to repair Intel’s reputation?

• What is technologists responsibility in disclosing bugs?

CSE1301 Sem 2 - 2003


82Rounding

• Why is rounding needed?

• Infinity numbers ⇒ Finite representation

• Integers only overflow

• Almost all operations need rounding

• IEEE - specifies algorithms for arithmetic

83Numbers need rounding

• Out of range:– x>2•2Emax x<1•2Emin

• Between 2 floats:– 0.110 = 0.00011001100….2 = 1.1001100…. •2-4

– 1.1001 •2-4

84Measuring Error

• ULPS (units in last place)– 1.12•10-1 Vs 0.124 : 0.4 ulps

– 1.12•10-1 Vs 0.118 : 0.2 ulps

• Relative Error– Difference/Original

– 1.12•10-1 Vs 0.124 : Err=0.004/0.124=0.032

CSE1301 Sem 2 - 2003


85Calculate Using Rounding

• Benign cancellation– Calculate 10.1-9.93 (= 0.17)

1.01 •101

0.99 •101

0.02 •101 = 2.00 •10-1

– 30 upls!

86Rounding problems

• Catastrophic cancellation– b2-4ac

– both b2 and 4ac are rounded

– the (-) exposes the error

– b=3.34 a=1.22 c=2.28

b2=11.2 4ac=11.1 b2-4ac=0.10

correct=0.0292 (70.08 upls)

87IEEE Arithmetic

• Requirement:+ - • ÷� should be EXACTLY

rounded

remainder should be EXACTLY rounded

Integer conv. should be EXACTLY rounded

• Not all (transcendental, binary to decimal)

• “Tie break” - Round to Even

CSE1301 Sem 2 - 2003


88Rounding and IEEE Rounding Modes

• When we perform math on “real” numbers, we have to worry about rounding to fit the result in the significant field.

• The FP hardware carries two extra bits of precision, and then round to get the proper value

• Rounding also occurs when converting a double to a single precision value, or converting a floating point number to an integer

Round towards +∞• ALWAYS round “up”: 2.001 → 3

• -2.001 → -2

Round towards -∞• ALWAYS round “down”: 1.999 → 1,

• -1.999 → -2

Truncate

• Just drop the last bits (round towards 0)

Round to (nearest) even

• Normal rounding, almost

89Round to Even

• Round like you learned in grade school

• Except if the value is right on the borderline, in which case weround to the nearest EVEN number

2.5 -> 2

3.5 -> 4

• Insures fairness on calculation

This way, half the time we round up on tie, the other half time we round down

• This is the default rounding mode

90Round to Even

• How will 1.005 be rounded ?– Round Up: 1.01

– Round Even: 1.00

• Why? Example:– xi=xi-1+y-y x0=1.00 y=0.125

– Round up: 1.00, 1.01, 1.02, ….

– Round even: 1.00, 1.00, 1.00, ….

CSE1301 Sem 2 - 2003


91Addition:

1.xxxxx 1.xxxxx 1.xxxxx

+ 1.xxxxx 0.001xxxxx 0.01xxxxx

1x.xxxxy 1.xxxxxyyy 1x.xxxxyyypost-normalization pre-normalization pre and post

• Guard Digits: digits to the right of the first p digits of significand to guard against loss of digits – can later be shifted left into first P places during normalization.

• Addition: carry-out shifted in

• Subtraction: borrow digit and guard

• Multiplication: carry and guard, Division requires guard

92Normalized result, but some non-zero digits to the ri ght of the significand --> the number should be rounded

E.g., B = 10, p = 3: 0 2 1.69

0 0 7.85

0 2 1.61

= 1.6900 * 10

= - .0785 * 10

= 1.6115 * 10

2-bias

2-bias

2-bias

-

one round digit must be carried to the right of the guard digit so that after a normalizing left shift, the result can be ro unded, accordingto the value of the round digit

IEEE Standard: four rounding modes: round to nearest even (default)round towards plus infinityround towards minus infinityround towards 0

round to nearest:round digit < B/2 then truncate

> B/2 then round up (add 1 to ULP: unit in last pl ace)= B/2 then round to nearest even digit

it can be shown that this strategy minimizes the mean error introduced by rounding

93Sticky Bit

Additional bit to the right of the round digit to better fine tune rounding

d0 . d1 d2 d3 . . . dp-1 0 0 00 . 0 0 X . . . X X X S

X X S+

Sticky bit: set to 1 if any 1 bits fall offthe end of the round digit

d0 . d1 d2 d3 . . . dp-1 0 0 00 . 0 0 X . . . X X X 0

X X 0-

d0 . d1 d2 d3 . . . dp-1 0 0 00 . 0 0 X . . . X X X 1-

generates a borrow

Rounding Summary

Radix 2 minimizes wobble in precision

Normal operations in +,-,*,/ require one carry /borrow bit + one guard digit

One round digit needed for correct rounding

Sticky bit needed when round digit is B/2 for max accuracy

Rounding to nearest has mean error = 0, if uniform distribution of digits are assumed

CSE1301 Sem 2 - 2003


94Floating-Point Division

• Significands divided - exponents subtracted - bias added to difference E1-E2

• If resulting exponent out of range - overflow or underflow indication must be generated

• Resultant significand satisfies 1/β ≤ M1/M2 < β

• A single base-β shift right of significand + increase of 1 in exponent may be needed in postnormalization step - may lead to an overflow

• If divisor=0 - indication of division by zerogenerated -quotient set to ±∞

• If both divisor and dividend=0 - result undefined - in the IEEE 754standard represented by NaN - not a number - also representing uninitialized variables and the result of 0 ⋅ ∞

95Remainder

• Fixed-point remainder -R=X-QD (X, Q, D - dividend, quotient, divisor) -|R| ≤ |D| - generated by division algorithm (restoring or nonrestoring)

• Flp division - algorithm generates quotient but not remainder -F1 REM F2 = F1-F2⋅Int(F1/F2) (Int(F1/F2)- quotient F1/F2converted to integer)

• Conversion to integer - either truncation (removing fractional part) or rounding-to-nearest

• The IEEE standard uses the round-to-nearest-evenmode -|F1 REM F2| ≤ |F2| /2

• Int(F1/F2)as large as β - high complexity

• Floating-point remainder calculated separately - only when required - for example, in argument reduction for periodic functions like sine and cosine

Emax-Emin96

Speeding up• Different algorithms may be used

• Result should be exact

• divide SRT algorithm in pentium– 5/2048 entries in a table

– 1/9,000,000 chance

– check:

CSE1301 Sem 2 - 2003


97MIPS R10000 arithmetic units

• Integer ALU + shifter

– All instructions take one cycle

• Integer ALU + multiplier

– Booth’s algorithm for multiplication (5-10 cycles)

– Non-restoring division (34-67 cycles)

• Floating point adder

– Carry propagate (2 cycles)

• Floating point multiplier (3 cycles)

– Booth’s algorithm

• Floating point divider (12-19 cycles)

• Floating point square root unit

• Separate unit for EA calculations

• Can start up to 5 instructions in 1 cycle

98Effect of Loss of Precision

• According to the General Accounting Office of the U.S. Government, a loss of precision in converting 24-bit integers into 24-bit floating point numbers was responsible for the failure of a Patriot anti-missile battery.

99Ariane 5

– Exploded 37 seconds after liftoff

– Cargo worth $500 million

• Why

– Computed horizontal velocity as floating point number

– Converted to 16-bit integer

– Worked OK for Ariane 4

– Overflowed for Ariane 5

• Used same software

Documents

Real Number Representation Topicscomarc/slides/lect3-3.pdf · CSE1301 Sem 2 - 2003 Lecture 30: Real Number Representation 31 30 23 22 0 Representing Numbers (cont) –23 bits for