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Computer Arithmetic Chapter 3PowerPoint Presentation

Computer Arithmetic Chapter 3

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Computer Arithmetic Chapter 3. Vishwani D. Agrawal James J. Danaher Professor Department of Electrical and Computer Engineering Auburn University http://www.eng.auburn.edu/~vagrawal [email protected] What Goes on Inside ALU?. Machine instr.: add $t1, $s1, $s2

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### Computer ArithmeticChapter 3

Vishwani D. Agrawal

James J. Danaher Professor

Department of Electrical and Computer Engineering

Auburn University

http://www.eng.auburn.edu/~vagrawal

ELEC 5200-001/6200-001 Lecture 7

What Goes on Inside ALU?

- Machine instr.: add $t1, $s1, $s2
- What it means to computer:
000000 10001 10010 01000 00000 100000

Arithmetic Logic Unit

(ALU)

Control unit

Flags

Registers

Registers

ELEC 5200-001/6200-001 Lecture 7

Basic Idea

- Hardware can only deal with binary digits, 0 and 1.
- Must represent all numbers, integers or floating point, positive or negative, by binary digits, called bits.
- Devise electronic circuits to perform arithmetic operations, add, subtract, multiply and divide, on binary numbers.

ELEC 5200-001/6200-001 Lecture 7

Positive Integers

- Decimal system: made of 10 digits, {0,1,2, . . . , 9}
41 = 4×10 + 1

255 = 2×102 + 5×101 + 5×100

- Binary system: made of two digits, {0,1}
00101001 = 0×27 + 0×26 + 1×25 + 0×24 +1×23

+0×22 + 0×21 + 1×20

=32 + 8 +1 = 41

11111111 = 255, largest number with 8 binary digits

28-1

ELEC 5200-001/6200-001 Lecture 7

Base or Radix

- For decimal system, 10 is called the base or radix.
- Decimal 41 is also written as 4110 or 41ten
- Base (radix) for binary system is 2.
- Thus, 41ten = 1010012 or 101001two
- Also, 111ten = 1101111two
and 111two = 7ten

ELEC 5200-001/6200-001 Lecture 7

Signed Integers – How not to Do

- Use fixed length binary representation
- Use left-most bit (called most significant bit or MSB) for sign
- 0 for positive
- 1 for negative

- Example: +18ten = 00010010two
- 18ten = 10010010two

ELEC 5200-001/6200-001 Lecture 7

Why Not Signed-Integers

- Sign and magnitude bits should be differently treated in arithmetic operations.
- Addition and subtraction require different logic circuits.
- Overflow is difficult to detect.
- “Zero” has two representations:
- +0ten = 00000000two
- - 0ten = 10000000two

- Signed-integers are not used in modern computers.

ELEC 5200-001/6200-001 Lecture 7

Signed Integers – Other Ways

- Use fixed-length representation, but no sign bit
- One’s complement: To form a negative number, complement each bit in the positive magnitude.
- Two’s complement: To form a negative number, start with positive magnitude, subtract one, and then complement each bit, or
first complement each bit in the positive magnitude, and then add one.

- Two’s complement is the most preferred representation.

ELEC 5200-001/6200-001 Lecture 7

Three Representations

Sign-magnitude

000 = +0

001 = +1

010 = +2

011 = +3

100 = - 0

101 = - 1

110 = - 2

111 = - 3

One’s complement

000 = +0

001 = +1

010 = +2

011 = +3

100 = - 3

101 = - 2

110 = - 1

111 = - 0

Two’s complement

000 = +0

001 = +1

010 = +2

011 = +3

100 = - 4

101 = - 3

110 = - 2

111 = - 1

(Preferred)

ELEC 5200-001/6200-001 Lecture 7

001

0

111

1

-1

2

010

110

-2

-3

3

011

-4

101

100

Two’s Complement NumbersELEC 5200-001/6200-001 Lecture 7

Two’s Complement n-bit Numbers

- Range: -2n-1 through 2n-1- 1
- Unique zero: 00000000 . . . . . 0
- Negation rule: see slide 8.
- Expansion of bit length: stretch the left-most bit all the way, e.g., 11111101 is still – 3.
- Overflow rule: If two numbers with the same sign bit (both positive or both negative) are added, the overflow occurs if and only if the result has the opposite sign.
- Subtraction rule: for A-B, add –B to A.

ELEC 5200-001/6200-001 Lecture 7

Converting Two’s Compliment to Decimal

n-1

an-1an-2 . . . a1a0 = -2n-1an-1 + Σ 2i ai

i=0

8-bit conversion box

-128 64 32 16 8 4 2 1

-128 64 32 16 8 4 2 1

1 1 1 1 1 1 0 1

Example

-128+64+32+16+8+4+1 = -128 + 125 = -3

ELEC 5200-001/6200-001 Lecture 7

MIPS

- MIPS architecture uses 32-bit numbers. What is the range of integers (positive and negative) that can be represented?
Positive integers: 0 to 2,147,483,647

Negative integers: -1 to -2,147,483,648

- What are the binary representations of the extreme positive and negative integers?
0111 1111 1111 1111 1111 1111 1111 1111 = 231-1 = 2,147,483,647

1000 0000 0000 0000 0000 0000 0000 0000 = -231 = -2,147,483,648

- What is the binary representation of zero?
0000 0000 0000 0000 0000 0000 0000 0000

ELEC 5200-001/6200-001 Lecture 7

Addition

- Adding bits:
- 0 + 0 = 0
- 0 + 1 = 1
- 1 + 0 = 1
- 1 + 1 = (1)0

- Adding integers:

carry

1 1 0

0 0 0 . . . . . . 0 1 1 1 two = 7ten

+ 0 0 0 . . . . . . 0 1 1 0 two = 6ten

= 0 0 0 . . . . . . 1 (1) 1 (1) 0 (0) 1 two = 13ten

ELEC 5200-001/6200-001 Lecture 7

Subtraction

- Direct subtraction
- Two’s complement subtraction

0 0 0 . . . . . . 0 1 1 1 two = 7ten

- 0 0 0 . . . . . . 0 1 1 0 two = 6ten

= 0 0 0 . . . . . . 0 0 0 1two = 1ten

1 1 0

0 0 0 . . . . . . 0 1 1 1 two = 7ten

+ 1 1 1 . . . . . . 1 0 1 0 two = - 6ten

= 0 0 0 . . . . . . 0 (1) 0 (1) 0 (0)1 two = 1ten

ELEC 5200-001/6200-001 Lecture 7

Overflow: An Error

- Examples: Addition of 3-bit integers (range -4 to +3)
- -2-3 = -5 110 = -2
+ 101 = -3

= 1011 = 3 (error)

- 3+2 = 5 011 = 3
010 = 2

= 101 = -3 (error)

- -2-3 = -5 110 = -2
- Overflow rule: If two numbers with the same sign bit (both positive or both negative) are added, the overflow occurs if and only if the result has the opposite sign.

000

111

0

001

1

-1

2

010

110

-2

3

-3

011

-4

101

100

ELEC 5200-001/6200-001 Lecture 7

Test 1 Result

Average = 19.9

Max = 23.5

Min = 14.5

Number of students

Score range

ELEC 5200-001/6200-001 Lecture 7

half_sum

XOR

b

carry_out

AND

Design Hardware Bit by Bit- Adding two bits:
a b half_sum carry_out

0 0 0 0

0 1 1 0

1 0 1 0

1 1 0 1

- Half-adder circuit

ELEC 5200-001/6200-001 Lecture 7

One-bit Full-Adder

- One-bit full-adder truth table
a b ci half_sum carry_out sum co

0 0 0 0 0 0 0

0 0 1 0 0 1 0

0 1 0 1 0 1 0

0 1 1 1 0 0 1

1 0 0 1 0 1 0

1 0 1 1 0 0 1

1 1 0 0 1 0 1

1 1 1 0 1 1 1

ELEC 5200-001/6200-001 Lecture 7

One-bit Full-Adder Circuit

ci

FAi

XOR

sumi

ai

XOR

AND

bi

AND

OR

Ci+1

ELEC 5200-001/6200-001 Lecture 7

32-bit Ripple-Carry Adder

c0

a0

b0

sum0

FA0

sum1

a1

b1

FA1

sum2

a2

b2

FA2

sum31

FA31

a31

b31

ELEC 5200-001/6200-001 Lecture 7

How Fast is Ripple-Carry Adder?

- Longest delay path (critical path) runs from cin to sum31.
- Suppose delay of full-adder is 100ps.
- Critical path delay = 3,200ps
- Clock rate cannot be higher than 1012/3,200 = 312MHz.
- Must use more efficient ways to handle carry.

ELEC 5200-001/6200-001 Lecture 7

16-bit

ripple

carry

adder

b0-b15

cin

Speeding Up the Addersum0-sum15

a16-a31

16-bit

ripple

carry

adder

0

b16-b31

0

sum16-sum31

Multiplexer

a16-a31

16-bit

ripple

carry

adder

1

b16-b31

This is known as

carry-select adder

1

ELEC 5200-001/6200-001 Lecture 7

Fast Adders

- In general, any output of a 32-bit adder can be evaluated as a logic expression in terms of all 65 inputs.
- Levels of logic in the circuit can be reduced to log2N for N-bit adder. Ripple-carry has N levels.
- More gates are needed, about log2N times that of ripple-carry design.
- Fastest design is known as carry lookahead adder.

ELEC 5200-001/6200-001 Lecture 7

N-bit Adder Design Options

Reference: J. L. Hennessy and D. A. Patterson, Computer Architecture:

A Quantitative Approach, Second Edition, San Francisco, California,

1990.

ELEC 5200-001/6200-001 Lecture 7

MIPS Instructions (see p. 175)

- Arithmetic: add, sub, addi, addu, subu, addiu, mfc0
- Data transfer: lw, sw, lhu, sh, lbu, sb, lui
- Logical: and, or, nor, andi, ori, sll, srl
- Conditional branch: beq, bne, slt, slti, sltu, sltiu
- Unconditional jump: j, jr, jal

ELEC 5200-001/6200-001 Lecture 7

Exception or Interrupt

- If an overflow is detected while executing add, addi or sub, then the address of that instruction is placed in a register called exception program counter (EPC).
- Instruction mfc0 can copy $epc to any other register, e.g., mfc0 $s1, $epc
- Unsigned operations, addu, addiu and subu do not cause an exception or interrupt.

ELEC 5200-001/6200-001 Lecture 7

Multifunction ALU

operation

c0

a0

b0

result0

ALU0

result1

a1

b1

ALU1

result2

operation

a2

b2

ALU2

ci

ai

bi

FAi

3

NOR

2

resulti

mux

OR

1

result31

ALU31

a31

b31

AND

0

ELEC 5200-001/6200-001 Lecture 7

Binary Multiplication (Unsigned)

1 0 0 0 two = 8ten multiplicand

1 0 0 1 two = 9ten multiplier

____________

1 0 0 0

0 0 0 0 partial products

0 0 0 0

1 0 0 0

____________

1 0 0 1 0 0 0two = 72ten

ELEC 5200-001/6200-001 Lecture 7

Multiplication Flowchart

Start

Initialize product register to 0

Partial product number, i = 1

LSB

of multiplier

?

Add multiplicand to

product and place result

in product register

1

0

Left shift multiplicand register 1 bit

Right shift multiplier register 1 bit

i = 32

i < 32

i = ?

Done

i = i + 1

ELEC 5200-001/6200-001 Lecture 7

Serial Multiplication

shift left

shift right

Multiplicand

32-bit multiplier

64

LSB

32 times

64

Control

test

64-bit ALU

64

64-bit product register

write

3 operations per bit

ELEC 5200-001/6200-001 Lecture 7

Serial Multiplication (Improved)

Multiplicand

2 operations per bit

32

32

Control

test

32-bit ALU

LSB

32 times

32

write

64-bit product register

shift right

00000 . . . 00000 32-bit multiplier

Initialized prod. Reg.

ELEC 5200-001/6200-001 Lecture 7

Example: 0010two× 0011two

ELEC 5200-001/6200-001 Lecture 7

Signed Multiplication

- Convert numbers into magnitudes.
- Multiply the two magnitudes through 32 iterations.
- Negate the result if the signs of the multiplicand and multiplier differed.
- A better method: Booth Algorithm.
- Time of serial multiplication: O(N) clock cycles for N-bit integers.

ELEC 5200-001/6200-001 Lecture 7

Booth Multiplier Algorithm

- A. D. Booth, “A Signed Binary Multiplication Technique,” Quarterly Journal of Mechanics and Applied Math., vol. 4, pt. 2, pp. 236-240, 1951.
- Direct multiplication of positive and negative integers using two’s complement addition.

ELEC 5200-001/6200-001 Lecture 7

Booth Algorithm: Basic Idea

- Consider a multiplier, 00011110 (30)
- We can write, 30 = 32 – 2, or
00100000 (32) = 25

+11111110 (-2) = -21

00011110 30

- We can write, 30 = 32 – 2, or
- Interpret multiplier (scan right to left), check bit-pairs:
- kth bit is 1, (k-1)th bit is 0, multiplier contains -2k term
- kth bit is 0, (k-1)th bit is 1, multiplier contains 2kterm
- kth bit is 0, (k-1)th bit is 1, 2k is absent in multiplier
- kth bit is 1, (k-1)th bit is 0, 2k is absent in multiplier

- Product, M×30 = M×25 - M×21 M: multiplicand
- Multiplication by 2k means a k-bit left shift

ELEC 5200-001/6200-001 Lecture 7

Booth Algorithm: Example 1

- 7 × 3 = 21
0111 multiplicand = 7

×0011(0) multiplier = 3

11111001 bit-pair 10, add -7 in two’s compl.

bit-pair 11, do nothing

000111 bit-pair 01, add 7

bit-pair 00, do nothing

00010101 21

ELEC 5200-001/6200-001 Lecture 7

Booth Algorithm: Example 2

- 7 × (-3) = -21
0111 multiplicand = 7

×1101(0) multiplier = -3

11111001 bit-pair 10, add -7 in two’s compl.

0000111 bit-pair 01, add 7

111001 bit-pair 10, add -7 in two’s compl.

bit-pair 11, do nothing

11101011 -21

ELEC 5200-001/6200-001 Lecture 7

Booth Algorithm: Example 3

- -7 × 3 = -21
1001 multiplicand = -7 in two’s com.

×0011(0) multiplier = 3

00000111 bit-pair 10, add 7

bit-pair 11, do nothing

111001 bit-pair 01, add -7

bit-pair 00, do nothing

11101011 -21

ELEC 5200-001/6200-001 Lecture 7

Booth Algorithm: Example 4

- -7 × (-3) = 21
1001 multiplicand = -7 in two’s com.

×1101(0) multiplier = -3 in two’s com.

00000111 bit-pair 10, add 7

1111001 bit-pair 01, add -7 in two’s compl.

000111 bit-pair 10, add 7

bit-pair 11, do nothing

00010101 21

ELEC 5200-001/6200-001 Lecture 7

Booth Advantage

Serial addition

Booth algorithm

00010100 20

×00011110 30

00000000

00010100

00010100

00010100

00010100

00000000

00000000

00000000________

000001001011000 600

00010100 20

×00011110 30

111111111101100

00000010100

__________________

0000001001011000 600

Four partial product additions Two partial product additions

ELEC 5200-001/6200-001 Lecture 7

Array Multiplier (Fast)

y3 y2 y1 y0 Multiplicand

x3 x2 x1 x0 Multiplier

________________________

x0y3 x0y2 x0y1 x0y0

x1y3 x1y2 x1y1 x1y0 Partial

x2y3 x2y2 x2y1 x2y0 Products

x3y3 x3y2 x3y1 x3y0

__________________________________________________

p7 p6 p5 p4 p3 p2 p1 p0

Note: Carry is added to the next partial product. Adding the carry

from the final stage needs an extra stage.

ELEC 5200-001/6200-001 Lecture 7

Array Multiplier

x0

ppi

yj

0

xi

0

0

0

0

x1

ci

0

FA

x2

co

ppi+1

0

x3

Critical path

FA

FA

FA

FA

0

p2

p1

p0

p7

p6

p5

p4

p3

ELEC 5200-001/6200-001 Lecture 7

Types of Array Multipliers

- Baugh-Wooley Algorithm: Signed product by two’s complement addition or subtraction according to the MSB’s.
- Booth multiplier algorithm
- Tree multipliers
- Reference: N. H. E. Weste and D. Harris, CMOS VLSI Design, A Circuits and Systems Perspective, Third Edition, Boston: Addison-Wesley, 2005.

ELEC 5200-001/6200-001 Lecture 7

MIPS Multiply Instructions

- Separate 32-bit registers, Hi and Lo, to hold the 64-bit product.
- Multiply signed:
mult $s1, $s2 # $s1 × $s2 = Hi, Lo

- Multiply unsigned:
multu $s1, $s2 # $s1 × $s2 = Hi, Lo

- Product is copied into registers by
- mfhi $s1 # copies Hi into $s1
- mflo $s1 # copies Lo into $s1

ELEC 5200-001/6200-001 Lecture 7

Binary Division (Unsigned)

1 3 Quotient

1 1 / 1 4 7 Divisor / Dividend

1 1

3 7 Partial remainder

3 3

4 Remainder

0 0 0 0 1 1 0 1

1 0 1 1 / 1 0 0 1 0 0 1 0

1 0 1 1

0 0 1 1 1 0

1 0 1 1

0 0 1 1 1 1

1 0 1 1

1 0 0

ELEC 5200-001/6200-001 Lecture 7

Binary Division Flowchart

Start

$A=0, $M=Divisor, $Q=Dividend, count=n

$A and $M have

one extra sign bit

beyond 32 bits.

Shift 1-bit left $A, $Q

$A ← $A - $M

No

Yes

$A < 0?

$Q0=0

$R←$R+$M

$Q0=1

Restore $A

(remainder)

count = count - 1

Done

$Q=Quotient

$A= Remainder

No

count = 0?

Yes

ELEC 5200-001/6200-001 Lecture 7

Division

33-bit $M (Divisor)

Step 2: Subtract $A←$A-$M

33

33

32 times

33-bit ALU

Initialize

$A←0

Step 1: 1-bit left shift $A and $Q

32

33-bit $A (Remainder)

32-bit $Q (Dividend)

Step 3: If sign-bit ($A)=0, set Q0=1

If sign-bit ($A)=1, set Q0=0 and restore $A

V. C. Hamacher, Z. G. Vranesic and S. G. Zaky, Computer Organization, Fourth Edition,

New York: McGraw-Hill, 1996.

ELEC 5200-001/6200-001 Lecture 7

Example: 8/3 = 2 (Remainder=2)

Initialize $A = 0 0 0 0 0 $Q = 1 0 0 0 $M = 0 0 0 1 1

Step 1, L-shift $A = 0 0 0 0 1 $Q = 0 0 0 ?

Step 2, Add - $M 1 1 1 0 1

$A = 1 1 1 1 0

Step 3, Set Q0 $Q = 0 0 0 0

Restore + $M 0 0 0 1 1

$A = 0 0 0 0 1

Step 1, L-shift $A = 0 0 0 1 0 $Q = 0 0 0 ? $M = 0 0 0 1 1

Step 2, Add - $M 1 1 1 0 1

$A = 1 1 1 1 1

Step 3, Set Q0 $Q = 0 0 0 0

Restore + $M 0 0 0 1 1

$A = 0 0 0 1 0

Cycle 1

Cycle 2

ELEC 5200-001/6200-001 Lecture 7

Example: 8/3 = 2 (Remainder=2) (Continued)

$A = 0 0 0 1 0 $Q = 0 0 0 0 $M = 0 0 0 1 1

Step 1, L-shift $A = 0 0 1 0 0 $Q = 0 0 0 ? $M = 0 0 0 1 1

Step 2, Add - $M 1 1 1 0 1

$A = 0 0 0 0 1

Step 3, Set Q0 $Q = 0 0 0 1

Step 1, L-shift $A = 0 0 0 1 0 $Q = 0 0 1 ? $M = 0 0 0 1 1

Step 2, Add - $M 1 1 1 0 1

$A = 1 1 1 1 1

Step 3, Set Q0 $Q = 0 0 1 0

Restore + $M 0 0 0 1 1

$A = 0 0 0 1 0

Cycle 3

Cycle 4

Note “Restore $A” in Steps 1, 2 and 4. This method is known as

the RESTORING DIVISION.

ELEC 5200-001/6200-001 Lecture 7

Non-Restoring Division

- Avoids the addition in the restore operation – does exactly one add or subtract per cycle.
- Non-restoring division algorithm:
- Step 1: Repeat 32 times
- shift $A, $Q left one bit
- if sign of $A is 0,
Subtract, $A ← $A - $M

else (sign of $A is 1),

Add, $A ← $A + $M

- if sign of $A is 0,
Set Q0 = 1

else (sign of $A is 1),

Set Q0 = 0

- Step 2: (after 32 Step 1 iterations) if sign of $A is 1, add $A ← $A + $M

- Step 1: Repeat 32 times

ELEC 5200-001/6200-001 Lecture 7

Non-Restoring Division: 8/3 = 2 (Rem=2)

Initialize $A = 0 0 0 0 0 $Q = 1 0 0 0 $M = 0 0 0 1 1

Step 1, L-shift $A = 0 0 0 0 1 $Q = 0 0 0 ? $M = 0 0 0 1 1

Subtract - $M 1 1 1 0 1

$A = 1 1 1 1 0 $Q = 0 0 0 0

Step 1, L-shift $A = 1 1 1 0 0 $Q = 0 0 0? $M = 0 0 0 1 1

Add $M 0 0 0 1 1

$A = 1 1 1 1 1 $Q = 0 0 00

Step 1, L-shift $A = 1 1 1 1 0 $Q = 0 0 0 ? $M = 0 0 0 1 1

Add $M 0 0 0 1 1

$A = 0 0 0 0 1 $Q = 0 0 0 1

Step 1, L-shift $A = 0 0 0 1 0 $Q = 0 0 1 ? $M = 0 0 0 1 1

Subtract -$M 1 1 1 0 1

$A = 1 1 1 1 1 $Q = 0 0 1 0

Step 2, Add $A ← $A + $M = 11111+00011 = 00010 (Final remainder)

Cycle 1

Cycle 2

Cycle 3

Cycle 4

ELEC 5200-001/6200-001 Lecture 7

Signed Division

- Remember the signs and divide magnitudes.
- Negate the quotient if the signs of divisor and dividend disagree.
- There is no other direct division method for signed division.

ELEC 5200-001/6200-001 Lecture 7

MIPS Division

- div $s2, $s3 # Lo=quotient, Hi=remainder
- divu $s2, $s3 # Lo=quotient, Hi=remainder
- mflo and mfhi retrieve Lo and Hi.
- Hardware ignores overflow, so software should determine if the quotient is too large.
- Software should also check if divisor = 0.

ELEC 5200-001/6200-001 Lecture 7

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