short cuts for multiply and divide for positive numbers 1. multiply by 2 k is the same as shift k to...

Computer Architecture CSE 3322Lecture 13

HW Due 3/24/09 3.4.2a, 3.7.2a, 3.8.1a

Short Cuts for Multiply and Divide

For Positive Numbers 1. Multiply by 2k is the same as shift k to the left, 0 fill 2. Divide by 2k is the same as shift k to the right, 0 fill

0

11

22

31

1

1

2222

2

ccc...cc

cc

nn

in

ii



1

1

2

in

iicc

For 2’s Complement Numbers It does not always work!



1

1

2

in

iicc

For 2’s Complement Numbers It does not always work!Ex: -5 = -0101 , 2’s Complement = 1011Divide by 4 by shift right 2, fill with ?



1

1

2

in

iicc

For 2’s Complement Numbers It does not always work!Ex: -5 = -0101 , 2’s Complement = 1011Divide by 4 by shift right 2, fill with ?1110 is 0001+1=0010 = 2 WRONG!

Floating Point NumbersNormalized Mantissa or Significand and Exponent

Add 0.713 x 10-2 and 0.964 x 10-1

Floating Point NumbersNormalized Mantissa or Significand and Exponent

Add 0.713 x 10-2 and 0.964 x 10-1 1. Align the exponents by shifting the smaller

0.0713 x 10-1

0.9640 x 10-1

Floating Point Numbers Normalized Mantissa or Significand and Exponent


0.0713 x 10-1

0.9640 x 10-1

2. Add 1.0353 x 10-1

Floating Point Numbers


0.0713 x 10-1

0.9640 x 10-1

2. Add 1.0353 x 10-1

3. Normalize 0.10353 x 100



0.0713 x 10-1

0.9640 x 10-1

2. Add 1.0353 x 10-1

3. Normalize 0.10353 x 100

4. Round off 0.104 x 100

Normalized REAL Binary Number: ±1.yyyyyyyyy x 2eeee


Leading 1 Binary Point Significand = 1.yyyyyyyyyExponent(signed)Arithmetic


Leading 1 Binary Point Significand = 1.yyyyyyyyyExponent(signed)Arithmetic

16

181

41

212 43210

1

0

ccccccc in

ii

143210 nccccc.c


Add 1.0101 x 23 and 1.1011 x 21


Add 1.0101 x 23 and 1.1011 x 21 1. Align the exponents by shifting the smaller

1.0101 x 23

0.011011 x 23



1.0101 x 23

0.011011 x 23

2. Add 1.101111 x 23



1.0101 x 23

0.011011 x 23

2. Add 1.101111 x 23

3. Normalize – Check for Overflow and Underflow

1.101111 x 23

Round - Off

4.56 3 significant digits + .0357 4.5957

Round - Off


4.60 Rounded Off

Round - Off


4.60 Rounded Off

What if 4.5950 ?

Round - Off


4.60 Rounded Off

What if 4.5950 ? 4.60 Round to even

Round - Off

Binary Decimal.00 .00.01 .25.10 .50.11 .75

Round - Off

Binary Decimal.00 .00.01 .25.10 .50 Are there any trailing 1’s ?

If not, round to even.11 .75



1.0101 x 23

0.011011 x 23

2. Add 1.101111 x 23


1.101111 x 23

4. Round to 4 bits Guard bit = 1 and Round bit = 1



1.0101 x 23

0.011011 x 23

2. Add 1.101111 x 23


1.101111 x 23

4. Round to 4 bits Guard bit = 1 and Round bit = 1

1.1100 x 23

IEEE 754 Floating Point StandardNormalized REAL Binary Number:

±1.yyyyyyyyy x 2eeee

IEEE 754(-1)S x ( 1 + F) x 2E-127

s exponent+127 significand - 1

1 bit E (8 bits) F (23 bits)



IEEE 754(-1)S x ( 1 + F) x 2E-127



Bias Exponent E>0



IEEE 754(-1)S x ( 1 + F) x 2E-127



• Only Zero is F = 0 and E = 0• Simplifies data exchange• Compare using integer processes• Accuracy and round-off & Overflow and Underflow

IEEE 754 Floating Point Standard

1 10000010 01010000000000000000000

161

412

23

1

i

iixcF = .25+.0625 = .3125


1 10000010 01010000000000000000000

161

412

23

1

i

iixcF = .25+.0625 = .3125

E = 27 + 2 = 128 + 2 = 130


1 10000010 01010000000000000000000

161

412

23

1

i

iixcF = .25+.0625 = .3125

E = 27 + 2 = 128 + 2 = 130

Number = (-1)S x ( 1 + F) x 2E-127


1 10000010 01010000000000000000000

161

412

23

1

i

iixcF = .25+.0625 = .3125

E = 27 + 2 = 128 + 2 = 130

Number = (-1)S x ( 1 + F) x 2E-127

= – ( 1 + .3125) x 2130 – 127


1 10000010 01010000000000000000000

161

412

23

1

i

iixcF = .25+.0625 = .3125

E = 27 + 2 = 128 + 2 = 130

Number = (-1)S x ( 1 + F) x 2E-127

= – ( 1 + .3125) x 2130 – 127

= – 1.3125 x 23 = – 1.3125 x 8 = – 10.5


Consider representing –5 in IEEE 754 Floating Point Format

IEEE 754 Floating Point StandardConsider representing –5 in IEEE 754 Floating Point Format

5 = 101.0 x 20 = 1.01 x 22

IEEE 754 Floating Point StandardConsider representing –5 in IEEE 754 Floating Point Format

5 = 101.0 x 20 = 1.01 x 22

IEEE 754(-1)S x ( 1 + F) x 2E-127


1bit E (8 bits) F (23 bits)


Consider representing –5 in IEEE 754 Floating Point Format5 = 101.0 x 20 = 1.01 x 22

IEEE 754(-1)S x ( 1 + F) x 2E-127



E –127 = 2E = 2 + 127 = 129 = 1000 0001



IEEE 754(-1)S x ( 1 + F) x 2E-127



E –127 = 2E = 2 + 127 = 129 = 1000 0001

1 +F = 1.01000 ...



IEEE 754(-1)S x ( 1 + F) x 2E-127



1 1000 0001 0100 0000 .... 0000

E = 2 + 127 = 129 = 1000 0001

1 +F = 1.01000 ...



Double PrecisionIEEE 754(-1)S x ( 1 + F) x 2E-1023



significand – 1 (continued)

F (32 bits)

Sequential Network Structures - Review

X1X2

Xn

Y1Y2

Ym

Q1Q2

Qm

Z1 Z2 Zk

Combinational Logic

Register Q

Clock

Stability Conditionclki Y Q

Input

Output

Flip - Flop with NOR Gates

Q = R+Q

Q = S+Q

R

S

PresentState RS Q 00 01 10 11 0 1 Next State Q


Q = R+Q

Q = S+Q

R

S

PresentState RS Q 00 01 10 11 0 0 1 1 Next State Q


Q = R+Q

Q = S+Q

R

S

PresentState RS Q 00 01 10 11 0 0 1 1 1 1 Next State Q


Q = R+Q

Q = S+Q

R

S

PresentState RS Q 00 01 10 11 0 0 1 0 1 1 1 0 Next State Q


Q = R+Q

Q = S+Q

R

S

PresentState RS Q 00 01 10 11 0 0 1 0 ? 1 1 1 0 ? Next State Q

D-latch

C

D

Q

D

C

Q

QS

R

D flip-flop

Output changes only on the trailing clock edge

QQ

_Q

Dl a t c h

D

C

Dl a t c h

D

C

D

C

Q

D

C

Q

Sequential Network Structures - Review

X1X2

Xn

Y1Y2

Ym

Q1Q2

Qm

Z1 Z2 Zk

Combinational Logic

Register Q

Clock

Stability Conditionclki Y Q

Input

Output

Five Components of Computers

Input

Output

MemoryControl

Datapath

Processor

Start by Building the Datapath

1. Access the Instruction from Memory2. Access the Data from Registers3. Perform the Instruction4. Write the Result

PC

InstructionMemory

Next PC Logic

Instruction

Address

Simplified Overview

Access the Instruction from Memory

PC

InstructionMemory

Next PC Logic

Instruction

Address

RegisterFile

Simplified Overview

Access the Data from Registers

PC

InstructionMemory

Next PC Logic

Instruction

Address

RegisterFile

ALU

Simplified Overview

Perform the Instruction

PC

InstructionMemory

Next PC Logic

Instruction

Address

RegisterFile

ALU DataMemory

Addr

Data In

DataOut

Simplified Overview

Write the Result

PC

InstructionMemory

Next PC Logic

Instruction

Address

RegisterFile

ALU DataMemory

Addr

Data In

DataOut

Simplified Overview

Timing Assumption

short cuts for multiply and divide for positive numbers 1. multiply by 2 k is the same as shift k to...

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