Computer Arithmetic Chapter 3

1 / 54

# Computer Arithmetic Chapter 3 - PowerPoint PPT Presentation

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

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

## PowerPoint Slideshow about 'Computer Arithmetic Chapter 3' - ostinmannual

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

### Computer ArithmeticChapter 3

Vishwani D. Agrawal

James J. Danaher Professor

Department of Electrical and Computer Engineering

Auburn University

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

[email protected]

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

• 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

000

001

0

111

1

-1

2

010

110

-2

-3

3

011

-4

101

100

Two’s Complement Numbers

ELEC 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

• 0 + 0 = 0
• 0 + 1 = 1
• 1 + 0 = 1
• 1 + 1 = (1)0

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)

• 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

a

half_sum

XOR

b

carry_out

AND

Design Hardware Bit by Bit

a b half_sum carry_out

0 0 0 0

0 1 1 0

1 0 1 0

1 1 0 1

ELEC 5200-001/6200-001 Lecture 7

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

ci

FAi

XOR

sumi

ai

XOR

AND

bi

AND

OR

Ci+1

ELEC 5200-001/6200-001 Lecture 7

c0

a0

b0

sum0

FA0

sum1

a1

b1

FA1

sum2

a2

b2

FA2

sum31

FA31

a31

b31

ELEC 5200-001/6200-001 Lecture 7

• 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

a0-a15

16-bit

ripple

carry

b0-b15

cin

sum0-sum15

a16-a31

16-bit

ripple

carry

0

b16-b31

0

sum16-sum31

Multiplexer

a16-a31

16-bit

ripple

carry

1

b16-b31

This is known as

1

ELEC 5200-001/6200-001 Lecture 7

• 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.

ELEC 5200-001/6200-001 Lecture 7

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)
• 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

?

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

• 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

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.

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

bit-pair 11, do nothing

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.

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

bit-pair 11, do nothing

00010101 21

ELEC 5200-001/6200-001 Lecture 7

Booth algorithm

00010100 20

×00011110 30

00000000

00010100

00010100

00010100

00010100

00000000

00000000

00000000________

000001001011000 600

00010100 20

×00011110 30

111111111101100

00000010100

__________________

0000001001011000 600

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

y3 y2 y1 y0

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

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