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FTC.W99 1

Lecture 1:Cost/Performance, DLX, Pipelining,

Caches, Branch Prediction

Prof. Fred ChongECS 250A Computer Architecture

Winter 1999

(Slides based upon Patterson UCB CS252 Spring 1998)

FTC.W99 2

Computer Architecture Is …

the attributes of a [computing] system as seenby the programmer, i.e., the conceptualstructure and functional behavior, as distinctfrom the organization of the data flows andcontrols the logic design, and the physicalimplementation.

Amdahl, Blaaw, and Brooks, 1964

SOFTWARESOFTWARE

FTC.W99 3

Computer Architecture’sChanging Definition

• 1950s to 1960s: Computer Architecture CourseComputer Arithmetic

• 1970s to mid 1980s: Computer Architecture CourseInstruction Set Design, especially ISA appropriatefor compilers

• 1990s: Computer Architecture CourseDesign of CPU, memory system, I/O system,Multiprocessors

FTC.W99 4

Computer Architecture Topics

Instruction Set Architecture

Pipelining, Hazard Resolution,Superscalar, Reordering, Prediction, Speculation,Vector, DSP

Addressing,Protection,Exception Handling

L1 Cache

L2 Cache

DRAM

Disks, WORM, Tape

Coherence,Bandwidth,Latency

Emerging TechnologiesInterleavingBus protocols

RAID

VLSI

Input/Output and Storage

MemoryHierarchy

Pipelining and Instruction Level Parallelism

FTC.W99 5

Computer Architecture Topics

M

Interconnection NetworkS

PMPMPMP° ° °

Topologies,Routing,Bandwidth,Latency,Reliability

Network Interfaces

Shared Memory,Message Passing,Data Parallelism

Processor-Memory-Switch

MultiprocessorsNetworks and Interconnections

FTC.W99 6

ECS 250A Course Focus

Understanding the design techniques, machinestructures, technology factors, evaluationmethods that will determine the form ofcomputers in 21st Century

Technology ProgrammingLanguages

OperatingSystems History

ApplicationsInterface Design

(ISA)

Measurement &Evaluation

Parallelism

Computer Architecture:• Instruction Set Design• Organization• Hardware

FTC.W99 7

Topic CoverageTextbook: Hennessy and Patterson, ComputerArchitecture: A Quantitative Approach, 2nd Ed., 1996.

• Performance/Cost, DLX, Pipelining, Caches, Branch Prediction• ILP, Loop Unrolling, Scoreboarding, Tomasulo, Dynamic Branch

Prediction

• Trace Scheduling, Speculation• Vector Processors, DSPs• Memory Hierarchy

• I/O• Interconnection Networks• Multiprocessors

FTC.W99 8

ECS250A: Staff

Instructor: Fred Chong

Office: EUII-3031 chong@cs

Office Hours: Mon 4-6pm or by appt.

T. A: Diana Keen

Office: EUII-2239 keend@cs

TA Office Hours: Fri 1-3pm

Class: Mon 6:10-9pm

Text: Computer Architecture: A Quantitative Approach,Second Edition (1996)

Web page: http://arch.cs.ucdavis.edu/~chong/250A/

Lectures available online before 1PM day of lecture

Newsgroup: ucd.class.cs250a{.d}

FTC.W99 9

Grading• Problem Sets 35%

• 1 In-class exam (prelim simulation) 20%• Project Proposals and Drafts 10%• Project Final Report 25%

• Project Poster Session (CS colloquium) 10%

FTC.W99 10

VLSI Transistors

G

A

B

B

A G

FTC.W99 11

CMOS Inverter

In OutOut In

FTC.W99 12

CMOS NAND Gate

A B

C

A

BC

FTC.W99 13

IC cost = Die cost + Testing cost + Packaging cost

Final test yieldDie cost = Wafer cost Dies per Wafer * Die yield

Dies per wafer = š * ( Wafer_diam / 2)2 – š * Wafer_diam – Test dies Die Area ¦ 2 * Die Area

Die Yield = Wafer yield * 1 +Defects_per_unit_area * Die_Area

αα

Integrated Circuits Costs

Die Cost goes roughly with die area4

{− α− α

}

FTC.W99 14

Real World Examples

Chip Metal Line Wafer Defect Area Dies/ Yield Die Cost layers width cost /cm2 mm2 wafer386DX 2 0.90 $900 1.0 43 360 71% $4486DX2 3 0.80 $1200 1.0 81 181 54% $12

PowerPC 601 4 0.80 $1700 1.3 121 115 28% $53HP PA 7100 3 0.80 $1300 1.0 196 66 27% $73DEC Alpha 3 0.70 $1500 1.2 234 53 19% $149SuperSPARC 3 0.70 $1700 1.6 256 48 13% $272

Pentium 3 0.80 $1500 1.5 296 40 9% $417

– From "Estimating IC Manufacturing Costs,” by Linley Gwennap,Microprocessor Report, August 2, 1993, p. 15

FTC.W99 15

Cost/PerformanceWhat is Relationship of Cost to Price?

• Component Costs• Direct Costs (add 25% to 40%) recurring costs: labor,

purchasing, scrap, warranty

• Gross Margin (add 82% to 186%) nonrecurring costs:R&D, marketing, sales, equipment maintenance, rental, financingcost, pretax profits, taxes

• Average Discount to get List Price (add 33% to 66%): volumediscounts and/or retailer markup

ComponentCost

Direct Cost

GrossMargin

AverageDiscount

Avg. Selling Price

List Price

15% to 33% 6% to 8%

34% to 39%

25% to 40%

FTC.W99 16

• Assume purchase 10,000 units

Chip Prices (August 1993)

Chip Area Mfg. Price Multi- Commentmm2 cost plier

386DX 43 $9 $31 3.4 Intense CompetitionIntense Competition

486DX2 81 $35 $245 7.0 No CompetitionNo CompetitionPowerPC 601 121 $77 $280 3.6

DEC Alpha 234 $202 $1231 6.1 Recoup R&D?

Pentium 296 $473 $965 2.0 Early in shipments

FTC.W99 17

Summary: Price vs. Cost

0%

20%

40%

60%

80%

100%

Mini W/S PC

Average Discount

Gross Margin

Direct Costs

Component Costs

0

1

2

3

4

5

Mini W/S PC

Average Discount

Gross Margin

Direct Costs

Component Costs

4.73.8

1.8

3.5

2.5

1.5

FTC.W99 18

Year

1000

10000

100000

1000000

10000000

100000000

1970 1975 1980 1985 1990 1995 2000

i80386

i4004

i8080

Pentium

i80486

i80286

i8086

Technology Trends:Microprocessor Capacity

CMOS improvements:• Die size: 2X every 3 yrs• Line width: halve / 7 yrs

Alpha 21264: 15 millionPentium Pro: 5.5 millionPowerPC 620: 6.9 millionAlpha 21164: 9.3 millionSparc Ultra: 5.2 million

Moore’s Law

FTC.W99 19

Memory Capacity(Single Chip DRAM)

size

Year

1000

10000

100000

1000000

10000000

100000000

1000000000

1970 1975 1980 1985 1990 1995 2000

year size(Mb) cyc time1980 0.0625 250 ns1983 0.25 220 ns1986 1 190 ns1989 4 165 ns

1992 16 145 ns1996 64 120 ns2000 256 100 ns

FTC.W99 20

Technology Trends(Summary)

Capacity Speed (latency)

Logic 2x in 3 years 2x in 3 years

DRAM 4x in 3 years 2x in 10 years

Disk 4x in 3 years 2x in 10 years

FTC.W99 21

Processor PerformanceTrends

Microprocessors

Minicomputers

Mainframes

Supercomputers

Year

0.1

1

10

100

1000

1965 1970 1975 1980 1985 1990 1995 2000

FTC.W99 22

Processor Performance(1.35X before, 1.55X now)

0

200

400

600

800

1000

1200

87 88 89 90 91 92 93 94 95 96 97

DEC Alpha 21264/600

DEC Alpha 5/500

DEC Alpha 5/300

DEC Alpha 4/266IBM POWER 100

DEC AXP/500

HP 9000/750

Sun-4/

260

IBMRS/

6000

MIPS M/

120

MIPS M

2000

1.54X/yr

FTC.W99 23

Performance Trends(Summary)

• Workstation performance (measured in SpecMarks) improves roughly 50% per year(2X every 18 months)

• Improvement in cost performance estimatedat 70% per year

FTC.W99 24

Computer EngineeringMethodology

TechnologyTrends

FTC.W99 25


Evaluate ExistingEvaluate ExistingSystems for Systems for BottlenecksBottlenecks

TechnologyTrends

Benchmarks

FTC.W99 26



Simulate NewSimulate NewDesigns andDesigns and

OrganizationsOrganizations

TechnologyTrends

Benchmarks

WorkloadsFTC.W99 27



Simulate NewSimulate NewDesigns andDesigns and

OrganizationsOrganizations

Implement NextImplement NextGeneration SystemGeneration System

TechnologyTrends

Benchmarks

Workloads

ImplementationComplexity

FTC.W99 28

Measurement Tools

• Benchmarks, Traces, Mixes• Hardware: Cost, delay, area, power estimation

• Simulation (many levels)– ISA, RT, Gate, Circuit

• Queuing Theory

• Rules of Thumb• Fundamental “Laws”/Principles

FTC.W99 29

The Bottom Line:Performance (and Cost)

• Time to run the task (ExTime)– Execution time, response time, latency

• Tasks per day, hour, week, sec, ns … (Performance)– Throughput, bandwidth

Plane

Boeing 747

BAD/SudConcodre

Speed

610 mph

1350 mph

DC to Paris

6.5 hours

3 hours

Passengers

470

132

Throughput(pmph)

286,700

178,200

FTC.W99 30

The Bottom Line:Performance (and Cost)

"X is n times faster than Y" means

ExTime(Y) Performance(X)

--------- = ---------------

ExTime(X) Performance(Y)

• Speed of Concorde vs. Boeing 747

• Throughput of Boeing 747 vs. Concorde

FTC.W99 31

Amdahl's LawSpeedup due to enhancement E: ExTime w/o E Performance w/ ESpeedup(E) = ------------- = -------------------

ExTime w/ E Performance w/o E

Suppose that enhancement E accelerates a fraction Fof the task by a factor S, and the remainder of thetask is unaffected

FTC.W99 32

Amdahl’s Law

ExTimenew = ExTimeold x (1 - Fractionenhanced) + Fractionenhanced

Speedupoverall =ExTimeold

ExTimenew

Speedupenhanced

=

1

(1 - Fractionenhanced) + Fractionenhanced

Speedupenhanced

FTC.W99 33

Amdahl’s Law

• Floating point instructions improved to run 2X;but only 10% of actual instructions are FP

Speedupoverall =

ExTimenew =

FTC.W99 34

Amdahl’s Law

• Floating point instructions improved to run 2X;but only 10% of actual instructions are FP

Speedupoverall = 1

0.95= 1.053

ExTimenew = ExTimeold x (0.9 + .1/2) = 0.95 x ExTimeold

FTC.W99 35

Metrics of Performance

Compiler

Programming Language

Application

DatapathControl

Transistors Wires Pins

ISA

Function Units

(millions) of Instructions per second: MIPS(millions) of (FP) operations per second: MFLOP/s

Cycles per second (clock rate)

Megabytes per second

Answers per monthOperations per second

FTC.W99 36

Aspects of CPU PerformanceCPU time = Seconds = Instructions x Cycles x Seconds

Program Program Instruction Cycle

CPU time = Seconds = Instructions x Cycles x Seconds


Inst Count CPI Clock RateProgram X

Compiler X (X)

Inst. Set. X X

Organization X X

Technology X

FTC.W99 37

Cycles Per Instruction

CPU time = CycleTime * Σ CPI * Ii = 1

n

i i

CPI = Σ CPI * F where F = I i = 1

n

i i i i

Instruction Count

“Instruction Frequency”

Invest Resources where time is Spent!

CPI = (CPU Time * Clock Rate) / Instruction Count = Cycles / Instruction Count

“Average Cycles per Instruction”

FTC.W99 38

Example: Calculating CPI

Typical Mix

Base Machine (Reg / Reg)Op Freq Cycles CPI(i) (% Time)ALU 50% 1 .5 (33%)

Load 20% 2 .4 (27%)Store 10% 2 .2 (13%)Branch 20% 2 .4 (27%) 1.5

FTC.W99 39

SPEC: System PerformanceEvaluation Cooperative

• First Round 1989– 10 programs yielding a single number (“SPECmarks”)

• Second Round 1992– SPECInt92 (6 integer programs) and SPECfp92 (14 floating point

programs)» Compiler Flags unlimited. March 93 of DEC 4000 Model 610:spice: unix.c:/def=(sysv,has_bcopy,”bcopy(a,b,c)=

memcpy(b,a,c)”

wave5: /ali=(all,dcom=nat)/ag=a/ur=4/ur=200nasa7: /norecu/ag=a/ur=4/ur2=200/lc=blas

• Third Round 1995– new set of programs: SPECint95 (8 integer programs) and

SPECfp95 (10 floating point)– “benchmarks useful for 3 years”– Single flag setting for all programs: SPECint_base95,

SPECfp_base95

FTC.W99 40

How to Summarize Performance• Arithmetic mean (weighted arithmetic mean)

tracks execution time: ΣΣ(Ti)/n or ΣΣ(Wi*Ti)

• Harmonic mean (weighted harmonic mean) ofrates (e.g., MFLOPS) tracks execution time:

n/ ΣΣ(1/Ri) or n/ ΣΣ(Wi/Ri)

• Normalized execution time is handy for scalingperformance (e.g., X times faster thanSPARCstation 10)

• But do not take the arithmetic mean ofnormalized execution time,

use the geometric: (Π Π xi)^1/n)

FTC.W99 41

SPEC First Round• One program: 99% of time in single line of code• New front-end compiler could improve dramatically

Benchmark

0

100

200

300

400

500

600

700

800

gcc

epre

sso

spic

e

dodu

c

nasa

7 li

eqnt

ott

mat

rix3

00 fppp

p

tom

catv

FTC.W99 42

Impact of Means onSPECmark89 for IBM 550

Ratio to VAX: Time: Weighted Time:Program Before After Before After Before Aftergcc 30 29 49 51 8.91 9.22espresso 35 34 65 67 7.64 7.86spice 47 47 510 510 5.69 5.69doduc 46 49 41 38 5.81 5.45nasa7 78 144 258 140 3.43 1.86li 34 34 183 183 7.86 7.86eqntott 40 40 28 28 6.68 6.68matrix300 78 730 58 6 3.43 0.37fpppp 90 87 34 35 2.97 3.07tomcatv 33 138 20 19 2.01 1.94Mean 54 72 124 108 54.42 49.99

Geometric Arithmetic Weighted Arith.Ratio 1.33 Ratio 1.16 Ratio 1.09

FTC.W99 43

Performance Evaluation

• “For better or worse, benchmarks shape a field”• Good products created when have:

– Good benchmarks– Good ways to summarize performance

• Given sales is a function in part of performancerelative to competition, investment in improvingproduct as reported by performance summary

• If benchmarks/summary inadequate, then choosebetween improving product for real programs vs.improving product to get more sales;Sales almost always wins!

• Execution time is the measure of computerperformance!

FTC.W99 44

Instruction Set Architecture (ISA)

instruction set

software

hardware

FTC.W99 45

Interface Design

A good interface:

• Lasts through many implementations (portability,compatibility)

• Is used in many differeny ways (generality)

• Provides convenient functionality to higher levels

• Permits an efficient implementation at lower levels

Interfaceimp 1

imp 2

imp 3

use

use

use

time

FTC.W99 46

Evolution of Instruction SetsSingle Accumulator (EDSAC 1950)

Accumulator + Index Registers(Manchester Mark I, IBM 700 series 1953)

Separation of Programming Model from Implementation

High-level Language Based Concept of a Family(B5000 1963) (IBM 360 1964)

General Purpose Register Machines

Complex Instruction Sets Load/Store Architecture

RISC

(Vax, Intel 432 1977-80) (CDC 6600, Cray 1 1963-76)

(Mips,Sparc,HP-PA,IBM RS6000, . . .1987)FTC.W99 47

Evolution of Instruction Sets

• Major advances in computer architecture aretypically associated with landmark instructionset designs

– Ex: Stack vs GPR (System 360)

• Design decisions must take into account:– technology

– machine organization– programming languages– compiler technology– operating systems

• And they in turn influence these

FTC.W99 48

A "Typical" RISC

• 32-bit fixed format instruction (3 formats)• 32 32-bit GPR (R0 contains zero, DP take pair)• 3-address, reg-reg arithmetic instruction

• Single address mode for load/store:base + displacement

– no indirection

• Simple branch conditions

• Delayed branch

see: SPARC, MIPS, HP PA-Risc, DEC Alpha, IBM PowerPC, CDC 6600, CDC 7600, Cray-1, Cray-2, Cray-3

FTC.W99 49

Example: MIPS

Op

31 26 01516202125

Rs1 Rd immediate

Op

31 26 025

Op

31 26 01516202125

Rs1 Rs2

target

Rd Opx

Register-Register

561011

Register-Immediate

Op

31 26 01516202125

Rs1 Rs2/Opx immediate

Branch

Jump / Call

FTC.W99 50

Summary, #1• Designing to Last through Trends

Capacity Speed

Logic 2x in 3 years 2x in 3 years

DRAM 4x in 3 years 2x in 10 years

Disk 4x in 3 years 2x in 10 years

• 6yrs to graduate => 16X CPU speed, DRAM/Disk size

• Time to run the task– Execution time, response time, latency

• Tasks per day, hour, week, sec, ns, …– Throughput, bandwidth

• “X is n times faster than Y” means ExTime(Y) Performance(X)

--------- = --------------

ExTime(X) Performance(Y)FTC.W99 51

Summary, #2

• Amdahl’s Law:

• CPI Law:

• Execution time is the REAL measure of computerperformance!

• Good products created when have:– Good benchmarks, good ways to summarize performance

• Die Cost goes roughly with die area4

• Can PC industry support engineering/researchinvestment?

Speedupoverall =ExTimeold

ExTimenew

=

1

(1 - Fractionenhanced) + Fractionenhanced

Speedupenhanced





FTC.W99 52

Pipelining: Its Natural!

• Laundry Example• Ann, Brian, Cathy, Dave

each have one load of clothesto wash, dry, and fold

• Washer takes 30 minutes

• Dryer takes 40 minutes

• “Folder” takes 20 minutes

A B C D

FTC.W99 53

Sequential Laundry

• Sequential laundry takes 6 hours for 4 loads

• If they learned pipelining, how long would laundry take?

A

B

C

D

30 40 20 30 40 20 30 40 20 30 40 20

6 PM 7 8 9 10 11 Midnight

Task

Order

Time

FTC.W99 54

Pipelined LaundryStart work ASAP

• Pipelined laundry takes 3.5 hours for 4 loads

A

B

C

D

6 PM 7 8 9 10 11 Midnight

Task

Order

Time

30 40 40 40 40 20

FTC.W99 55

Pipelining Lessons• Pipelining doesn’t help

latency of single task, ithelps throughput ofentire workload

• Pipeline rate limited byslowest pipeline stage

• Multiple tasks operatingsimultaneously

• Potential speedup =Number pipe stages

• Unbalanced lengths ofpipe stages reducesspeedup

• Time to “fill” pipeline andtime to “drain” it reducesspeedup

A

B

C

D

6 PM 7 8 9

Task

Order

Time

30 40 40 40 40 20

FTC.W99 56

Computer Pipelines

• Execute billions of instructions, sothroughput is what matters

• DLX desirable features: all instructions samelength, registers located in same place ininstruction format, memory operands only inloads or stores

FTC.W99 57

5 Steps of DLX DatapathFigure 3.1, Page 130

MemoryAccess

WriteBack

InstructionFetch

Instr. DecodeReg. Fetch

ExecuteAddr. Calc

IRLMD

FTC.W99 58

Pipelined DLX DatapathFigure 3.4, page 137

MemoryAccess

WriteBack

InstructionFetch Instr. Decode

Reg. FetchExecute

Addr. Calc.

• Data stationary control– local decode for each instruction phase / pipeline stage FTC.W99 59

Visualizing PipeliningFigure 3.3, Page 133

Instr.

Order

Time (clock cycles)

FTC.W99 60

Its Not That Easy forComputers

• Limits to pipelining: Hazards prevent nextinstruction from executing during its designatedclock cycle

– Structural hazards: HW cannot support this combination ofinstructions (single person to fold and put clothes away)

– Data hazards: Instruction depends on result of priorinstruction still in the pipeline (missing sock)

– Control hazards: Pipelining of branches & otherinstructionsstall the pipeline until the hazardbubbles” in thepipeline

FTC.W99 61

One Memory Port/Structural HazardsFigure 3.6, Page 142

Instr.

Order

Time (clock cycles)

Load

Instr 1

Instr 2

Instr 3

Instr 4FTC.W99 62

One Memory Port/Structural HazardsFigure 3.7, Page 143

Instr.

Order

Time (clock cycles)

Load

Instr 1

Instr 2

stall

Instr 3FTC.W99 63

Speed Up Equation forPipelining

CPIpipelined = Ideal CPI+ Pipeline stall clock cycles per instr

Speedup = Ideal CPI x Pipeline depth Clock Cycleunpipelined Ideal CPI + Pipeline stall CPI Clock Cyclepipelined

Speedup = Pipeline depth Clock Cycleunpipelined 1 + Pipeline stall CPI Clock Cyclepipelined

x

x

FTC.W99 64

Example: Dual-port vs. Single-port

• Machine A: Dual ported memory

• Machine B: Single ported memory, but its pipelinedimplementation has a 1.05 times faster clock rate

• Ideal CPI = 1 for both• Loads are 40% of instructions executed SpeedUpA = Pipeline Depth/(1 + 0) x (clockunpipe/clockpipe) = Pipeline Depth SpeedUpB = Pipeline Depth/(1 + 0.4 x 1)

x (clockunpipe/(clockunpipe / 1.05) = (Pipeline Depth/1.4) x 1.05 = 0.75 x Pipeline Depth

SpeedUpA / SpeedUpB = Pipeline Depth/(0.75 x Pipeline Depth) = 1.33

• Machine A is 1.33 times fasterFTC.W99 65

Data Hazard on R1Figure 3.9, page 147

Instr.

Order

Time (clock cycles)

add r1,r2,r3

sub r4,r1,r3

and r6,r1,r7

or r8,r1,r9

xor r10,r1,r11

IF ID/RF EX MEM WB

FTC.W99 66

Three Generic Data HazardsInstrI followed by InstrJ

• Read After Write (RAW)InstrJ tries to read operand before InstrI writes it

FTC.W99 67


• Write After Read (WAR)InstrJ tries to write operand before InstrI reads i

– Gets wrong operand

• Can’t happen in DLX 5 stage pipeline because:– All instructions take 5 stages, and

– Reads are always in stage 2, and– Writes are always in stage 5

FTC.W99 68


• Write After Write (WAW)InstrJ tries to write operand before InstrI writes it

– Leaves wrong result ( InstrI not InstrJ )

• Can’t happen in DLX 5 stage pipeline because:– All instructions take 5 stages, and

– Writes are always in stage 5

• Will see WAR and WAW in later more complicatedpipes

FTC.W99 69

Forwarding to Avoid Data HazardFigure 3.10, Page 149

Instr.

Order

Time (clock cycles)

add r1,r2,r3

sub r4,r1,r3

and r6,r1,r7

or r8,r1,r9

xor r10,r1,r11

FTC.W99 70

HW Change for ForwardingFigure 3.20, Page 161

FTC.W99 71

Instr.

Order

Time (clock cycles)

lw r1, 0(r2)

sub r4,r1,r6

and r6,r1,r7

or r8,r1,r9

Data Hazard Even with ForwardingFigure 3.12, Page 153

FTC.W99 72

Data Hazard Even with ForwardingFigure 3.13, Page 154

Instr.

Order

Time (clock cycles)

lw r1, 0(r2)

sub r4,r1,r6

and r6,r1,r7

or r8,r1,r9

FTC.W99 73

Try producing fast code for

a = b + c;d = e – f;

assuming a, b, c, d ,e, and f in memory.Slow code:

LW Rb,bLW Rc,cADD Ra,Rb,RcSW a,RaLW Re,e

LW Rf,fSUB Rd,Re,RfSW d,Rd

Software Scheduling to AvoidLoad Hazards

Fast code:LW Rb,bLW Rc,cLW Re,eADD Ra,Rb,Rc

LW Rf,fSW a,RaSUB Rd,Re,RfSW d,Rd FTC.W99 74

Control Hazard on BranchesThree Stage Stall

FTC.W99 75

Branch Stall Impact

• If CPI = 1, 30% branch, Stall 3 cycles => new CPI = 1.9!

• Two part solution:– Determine branch taken or not sooner, AND– Compute taken branch address earlier

• DLX branch tests if register = 0 or ° 0• DLX Solution:

– Move Zero test to ID/RF stage– Adder to calculate new PC in ID/RF stage

– 1 clock cycle penalty for branch versus 3

FTC.W99 76

Pipelined DLX DatapathFigure 3.22, page 163

MemoryAccess

WriteBack

InstructionFetch

Instr. DecodeReg. Fetch

ExecuteAddr. Calc.

This is the correct 1 cyclelatency implementation!

FTC.W99 77

Four Branch Hazard Alternatives

#1: Stall until branch direction is clear

#2: Predict Branch Not Taken– Execute successor instructions in sequence– “Squash” instructions in pipeline if branch actually taken

– Advantage of late pipeline state update– 47% DLX branches not taken on average– PC+4 already calculated, so use it to get next instruction

#3: Predict Branch Taken– 53% DLX branches taken on average

– But haven’t calculated branch target address in DLX» DLX still incurs 1 cycle branch penalty» Other machines: branch target known before outcome

FTC.W99 78

Four Branch Hazard Alternatives

#4: Delayed Branch– Define branch to take place AFTER a following instruction

branch instructionsequential successor1sequential successor2........sequential successorn

branch target if taken

– 1 slot delay allows proper decision and branch targetaddress in 5 stage pipeline

– DLX uses this

Branch delay of length n

FTC.W99 79

Delayed Branch

• Where to get instructions to fill branch delay slot?– Before branch instruction– From the target address: only valuable when branch taken– From fall through: only valuable when branch not taken– Cancelling branches allow more slots to be filled

• Compiler effectiveness for single branch delay slot:– Fills about 60% of branch delay slots– About 80% of instructions executed in branch delay slots useful

in computation– About 50% (60% x 80%) of slots usefully filled

• Delayed Branch downside: 7-8 stage pipelines,multiple instructions issued per clock (superscalar)

FTC.W99 80

Evaluating Branch Alternatives

Scheduling Branch CPI speedup v. speedup v.scheme penalty unpipelined stall

Stall pipeline 3 1.42 3.5 1.0Predict taken 1 1.14 4.4 1.26Predict not taken 1 1.09 4.5 1.29Delayed branch 0.5 1.07 4.6 1.31

Conditional & Unconditional = 14%, 65% change PC

Pipeline speedup = Pipeline depth1 +Branch frequency ×Branch penalty

FTC.W99 81

Pipelining Summary

• Just overlap tasks, and easy if tasks are independent• Speed Up Š Pipeline Depth; if ideal CPI is 1, then:

• Hazards limit performance on computers:– Structural: need more HW resources– Data (RAW,WAR,WAW): need forwarding, compiler scheduling– Control: delayed branch, prediction

Speedup =Pipeline Depth

1 + Pipeline stall CPIX

Clock Cycle Unpipelined

Clock Cycle Pipelined

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