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Pipelining. By Pradondet Nilagupta Based on Lecture note on Advanced Computer Architecture Prof. Mike Schulte Prof. Yirng-An Chen. A. B. C. D. Introduction to Pipelining. Pipelining: An implementation technique that overlaps the execution of multiple instructions. Laundry Example

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pipelining

Pipelining

By Pradondet Nilagupta

Based on Lecture note on

Advanced Computer Architecture

Prof. Mike Schulte

Prof. Yirng-An Chen

introduction to pipelining

A

B

C

D

Introduction to Pipelining
  • Pipelining: An implementation technique that overlaps the execution of multiple instructions.
  • Laundry Example
  • Ann, Brian, Cathy, Dave each have one load of clothes to wash, dry, and fold
  • Washer takes 30 minutes
  • Dryer takes 40 minutes
  • “Folder” takes 20 minutes
sequential laundry

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B

C

D

Sequential Laundry

6 PM

Midnight

7

8

9

11

10

Time

  • Sequential laundry takes 6 hours for 4 loads
  • If they learned pipelining, how long would laundry take?

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pipelined laundry start work asap

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B

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Pipelined LaundryStart work ASAP

6 PM

Midnight

7

8

9

11

10

  • Pipelined laundry takes 3.5 hours for 4 loads
  • Speedup = 6/3.5 = 1.7

Time

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pipelining lessons

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Pipelining Lessons

6 PM

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  • Pipelining doesn’t help latency of single task, it helps throughput of entire workload
  • Pipeline rate limited by slowest pipeline stage
  • Multiple tasks operating simultaneously
  • Potential speedup = Number pipe stages
  • Unbalanced lengths of pipe stages reduces speedup
  • Time to “fill” pipeline and time to “drain” it reduces speedup

Time

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computer pipelines
Computer Pipelines
  • Execute billions of instructions, so throughput is what matters
  • RISC desirable features: all instructions same length, registers located in same place in instruction format, memory operands only in loads or stores
pipelining basics

30ns

3ns

Comb.

Logic

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E

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Clock

Pipelining Basics

Unpipelined

System

  • One operation must complete before next can begin
  • Operations spaced 33ns apart

Delay = 33ns

Throughput = 30MHz

Op1

Op2

Op3

??

Time

3 stage pipelining

10ns

3ns

10ns

3ns

10ns

3ns

Comb.

Logic

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Comb.

Logic

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Comb.

Logic

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Clock

3 Stage Pipelining
  • Space operations 13ns apart
  • 3 operations occur simultaneously

Delay = 39ns

Throughput = 77MHz

Op1

Op2

Op3

Op4

??

Time

limitation nonuniform pipelining

5ns

3ns

15ns

3ns

10ns

3ns

Com.

Log.

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Comb.

Logic

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Comb.

Logic

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Limitation: Nonuniform Pipelining
  • Throughput limited by slowest stage
    • Delay determined by clock period * number of stages
  • Must attempt to balance stages

Delay = 18 * 3 = 54 ns

Throughput = 55MHz

Clock

limitation deep pipelines

5ns

3ns

5ns

3ns

5ns

3ns

5ns

3ns

5ns

3ns

5ns

3ns

Com.

Log.

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Com.

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Com.

Log.

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Com.

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Com.

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Com.

Log.

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Clock

Limitation: Deep Pipelines
  • Diminishing returns as add more pipeline stages
  • Register delays become limiting factor
    • Increased latency
    • Small throughput gains

Delay = 48ns, Throughput = 128MHz

limitation sequential dependencies
Limitation: Sequential Dependencies

Comb.

Logic

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Comb.

Logic

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Comb.

Logic

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  • Op4 gets result from Op1 !
  • Pipeline Hazard

Clock

Op1

Op2

Op3

Op4

??

Time

speed up equation for pipelining

4

3

5

3

4

2

6

Pipeline registers

Speed Up Equation for Pipelining
  • Assumptions:
    • No delays except component’s latencies
    • A fixed pipeline overhead: 2ns.
  • What is the cycle time for the pipeline version of the circuit that maximizes performance without allocating multiple cycles to a stage?
  • What is the total execution time for the pipeline version?
  • What is the speedup versus a single-cycle unpipelined version?
multiple cycle dlx cycles 1 and 2
Multiple-Cycle DLX: Cycles 1 and 2
  • Most DLX instruction can be implemented in 5 clock cycles (see Figure 3.1 on page 130).
  • The first two clock cycles are the same for every instruction.

1. Instruction fectch cycle (IF)

IR <= Mem[PC] (load instruction)

NPC <= PC+4 (update program counter)

2. Instruction decode / register fetch cycle (ID)

A <= Regs[IR ] (fetch source reg1)

B <= Regs[IR ] (fetch source reg2)

Imm <= (IR ) ## IR (fetch and sign-ext imm.)

6...10

11…15

16

16

16…31

multiple cycle dlx cycle 3
Multiple-Cycle DLX: Cycle 3
  • The third cycle is known as the

Execution/ effective address cycle (EX)

  • The actions performed in this cycle depend on the type of operations.
    • Memory reference (e.g., LW R1, 30 (R2))

ALUOutput <= A + Imm (Calculate effective address)

    • Register-Register ALU op. (e.g., ADD R1, R2, R3)

ALUOutput <=A op B (Perform ALU operation)

    • Register-Immed. ALU op. (e.g., ADDI R1, R2, #3)

ALUOutput <=A op Imm (Perform ALU operation)

    • Branch (e.g., BEQZ R4, next)

ALUOutput <= NPC + Imm (Compute branch target)

Cond <= (A == 0) (Compare A to 0)

multiple cycle dlx cycle 4
Multiple-Cycle DLX: Cycle 4
  • The fourth cycle is known as the

Memory access / branch completion cycle (MEM)

  • The only DLX instructions active in this cycle are loads, stores, and branches
    • Loads (e.g., LW R1, 30 (R2))

LMD <= Mem[ALUOutput] (load memory onto processor)

    • Stores (e.g., 500(R4), R3)

Mem[ALUOutput] <= B (store data into memory)

    • Branch (e.g., BEQZ R4, next)

if (cond) PC <= ALUoutput (Set PC based on cond)

else PC <= NPC

multiple cycle dlx cycle 5
Multiple-Cycle DLX: Cycle 5
  • The fifth cycle is known as the

Write-back cycle (WB)

  • During this cycles, results are written to the register file
    • Register-Register ALU op. (e.g., ADD R1, R2, R3)

Regs[IR ] <= ALUOutput

    • Register-Immed. ALU op (e.g., ADD R1, R2, #3)

Regs[IR ] <= ALUOutput

    • Load Instruction (e.g., LW R1, 30 (R2))

Regs[IR ] <= LMD

16…20

11…15

11…15

5 steps of dlx datapath figure 3 1

Instruction

Fetch

Instr. Decode

Reg. Fetch

Execute

Addr. Calc

Memory

Access

Write

Back

M

U

X

Zero?

Cond.

+

NPC

4

M

U

X

A

PC

ALU

Output

Regs

ALU

LMD

Data

Mem.

M

U

X

M

U

X

B

Inst.

Mem.

IR

Sign

Ext.

Imm.

16

32

5 Steps of DLX DatapathFigure 3.1
cpi for the multiple cycle dlx
CPI for the Multiple-Cycle DLX
  • The multiple-cycle DLX requires 4 cycles for branches and stores and 5 cycles for the other operations.
  • Assuming 20% of the instructions are branches or loads, this gives a CPI of 4.80.
  • We could improve the CPI by allowing ALU operations to complete during memory cycle
  • Assuming 40% of the instructions are ALU operations, this would reduce the CPI to 4.40.
pipelining dlx
Pipelining DLX
  • To reduce the CPI, DLX can be implemented using a five stage pipeline.
  • In this example, it takes 10 cycles execute 5 instructions for a CPI of 2.
visualizing pipelining figure 3 3 page 133
Visualizing PipeliningFigure 3.3, Page 133

Time (clock cycles)

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pipelined dlx datapath figure 3 4 page 134
Pipelined DLX DatapathFigure 3.4 page 134

Instruction

Fetch

Instr. Decode

Reg. Fetch

Execute

Addr. Calc.

M

U

X

Zero?

+

Write

Back

Memory

Access

4

M

U

X

PC

ALU

Regs

Data

Mem.

M

U

X

M

U

X

Inst.

Mem.

16

32

Sign

Ext.

IF/ID

ID/EX

EX/MEM

MEM/WB

  • Pipeline registers are used to tranfer results from one pipeline stage to the next.
basic performance issues in pipelining
Basic Performance Issues in Pipelining
  • Pipelining increases the CPU instruction throughput - the number of instructions complete per unit of time - but it is not reduce the execution time of an individual instruction.
pipeline speedup example
Pipeline Speedup Example
  • Assume the multiple cycle DLX has a 10-ns clock cycle, loads take 5 clock cycles and account for 40% of the instructions, and all other instructions take 4 clock cycles.
  • If pipelining the machine add 1-ns to the clock cycle, how much speedup in instruction execution rate do we get from pipelining.

MC Ave Instr. Time = Clock cycle x Average CPI

= 10 ns x (0.6 x 4 + 0.4 x 5)

= 44 ns

PL Ave Instr. Time = 10 + 1 = 11 ns

Speedup = 44 / 11 = 4

  • This ignores time needed to fill & empty the pipeline and delays due to hazards.
its not that easy for computers
Its Not That Easy for Computers
  • Limits to pipelining: Hazards prevent next instruction from executing during its designated clock cycle
    • Structural hazards: Hardware cannot support this combination of instructions - two instructions need the same resource.
    • Data hazards: Instruction depends on result of prior instruction still in the pipeline
    • Control hazards: Pipelining of branches & other instructions that change the PC
  • Common solution is to stall the pipeline until the hazard is resolved, inserting one or more “bubbles” in the pipeline
speed up equations for pipelining
Speed Up Equations for Pipelining
  • Stalls reduce the speedup obtained from pipelining

Speedup from pipelining = Ave Instr Time unpipelined

Ave Instr Time pipelined

= CPIunpipelined x Clock Cycleunpipelined

CPIpipelined x Clock Cyclepipelined

CPIpipelined = Ideal CPI + Pipeline stall CPI

= 1 + Pipeline stall CPI

Speedup = CPIunpipelined Clock Cycleunpipelined

1 + Pipeline stall CPI Clock Cyclepipelined

Speedup < Pipeline depth

1 + Pipeline stall CPI

x

speed up equation for pipelining1
Speed Up Equation for Pipelining

CPIpipelined = Ideal CPI + Pipeline stall clock cycles per instr

Speedup = Ideal CPI x Pipeline depth Clock Cycleunpipelined

Ideal CPI + Pipeline stall CPI Clock Cyclepipelined

ASSUMING IDEAL CPI OF 1

Speedup = Pipeline depth Clock Cycleunpipelined

1 + Pipeline stall CPI Clock Cyclepipelined

x

x

structure hazards
Structure Hazards
  • Sometime called Resource Conflict.
  • Example.
    • Some pipelined machines have shared a single memory pipeline for a data and instruction. As a result, when an instruction contains a data memory reference, it will conflict with the instruction reference for a latter instruction.
one memory port structural hazards figure 3 6 page 142
One Memory Port/Structural HazardsFigure 3.6, Page 142

Load

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Instr 1

Instr 2

Instr 3

Instr 4

one memory port structural hazards figure 3 7 page 143
One Memory Port/Structural HazardsFigure 3.7, Page 143

Load

I

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O

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Instr 1

Instr 2

stall

Instr 3

example one or two memory ports
Example: One or Two Memory Ports?
  • Machine A has a two port memory - access instructions and data simultaneously.
  • Machine B has a one port memory, but its pipelined implementation has a 1.05 times faster clock rate
  • Ideal CPI = 1 for both
  • Loads are 40% of instructions executed

Ave Instr.Time A = Clock cycle A x CPI A

= Clock cycle A

Ave Instr.Time B = Clock cycle B x CPI B

= (Clock cycle A / 1.05) x (1 + 0.4)

= Clock cyle A x 1.33

Ave Instr.Time B = 1.33

Ave Instr.Time A

  • Machine A is 1.33 times faster
data hazard
Data Hazard
  • Data hazard occur when pipeline changes the order of read/write accesses to operands so that the order differs from the order seen by sequentially execution instructions on an unpipelined machine
three generic data hazards
Three Generic Data Hazards

InstrI followed be InstrJ

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

I: ADD R1, R2, R3 IF ID EX MEM WB

J: SUB R4, R1, R5 IF ID EX MEM WB

three generic data hazards1
Three Generic Data Hazards

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

I: LW R1, 0(R2) IF ID EX MEM1 MEM2 WB

J: ADD R1, R2, R3 IF ID EX WB

three generic data hazards2
Three Generic Data Hazards

InstrI followed be InstrJ

  • Write After Read (WAR)InstrJ tries to write operand before InstrI reads it
  • Can’t happen in the DLX 5 stage pipeline because:
    • All instructions take 5 stages,
    • Reads are always in stage 2, and
    • Writes are always in stage 5

I: SW 0(R1), R2 IF ID EX MEM1 MEM2 WB

J: ADD R2, R3, R4 IF ID EX WB

compiler scheduling for data hazards
Compiler Scheduling for Data Hazards
  • Rather than just allow the pipeline to stall, the compiler could try to schedule the pipeline to avoid these stalls by arranging the code sequence to eliminate the hazard. The technique, called pipeline scheduling or instruction scheduling
software scheduling to avoid load hazards
Software Scheduling to Avoid Load Hazards

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,b

LW Rc,c

ADD Ra,Rb,Rc

SW a,Ra

LW Re,e

LW Rf,f

SUB Rd,Re,Rf

SW d,Rd

Fast code:

LW Rb,b

LW Rc,c

LW Re,e

ADD Ra,Rb,Rc

LW Rf,f

SW a,Ra

SUB Rd,Re,Rf

SW d,Rd

pipelining summary
Pipelining Summary
  • Pipelining overlaps the execution of multiple instructions.
  • With an idea pipeline, the CPI is one, and the speedup is equal to the number of stages in the pipeline.
  • However, several factors prevent us from achieving the ideal speedup, including
    • Not being able to divide the pipeline evenly
    • The time needed to empty and flush the pipeline
    • Overhead needed for pipeling
    • Structural, data, and control harzards
pipelining summary1
Pipelining Summary
  • Just overlap tasks, and easy if tasks are independent
  • Speed Up VS. Pipeline Depth; if ideal CPI is 1, then:
  • Hazards limit performance on computers:
    • Structural: need more HW resources
    • Data: need forwarding, compiler scheduling
    • Control: discuss next time

Pipeline Depth

Clock Cycle Unpipelined

Speedup =

X

Clock Cycle Pipelined

1 + Pipeline stall CPI