Pipelined Implementation CSCi 2021: Computer Architecture and Organization

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# Pipelined Implementation CSCi 2021: Computer Architecture and Organization

## Pipelined Implementation CSCi 2021: Computer Architecture and Organization

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##### Presentation Transcript

1. Pipelined ImplementationCSCi 2021: Computer Architecture and Organization Chapter 4.4, 4.5 (through 4.5.7)

2. Today • Pipelining • Performance • Hazards • Deviate a bit from the slides (defer branch prediction)

3. Last Time • Sequential processor implementation • Instruction execution

4. Idea Divide process into independent stages Move objects through stages in sequence At any given times, multiple objects being processed Parallel Sequential Pipelined Real-World Pipelines: Car Washes

5. 300 ps 20 ps Combinational logic R e g Delay = 320 ps Throughput = 3.12 GOPS gigainstructions/ops per sec Clock Computational Example • System • Computation requires total of 300 picoseconds • Additional 20 picoseconds to save result in register

6. 100 ps 20 ps 100 ps 20 ps 100 ps 20 ps Comb. logic A R e g Comb. logic B R e g Comb. logic C R e g Delay = 360 ps Throughput = 8.33 GOPS Clock 3-Way Pipelined Version • System • Divide combinational logic into 3 blocks of 100 ps each • Can begin new operation as soon as previous one passes through stage A • Begin new operation every 120 ps • Overall latency increases • 360 ps from start to finish

7. 50 ps 20 ps 150 ps 20 ps 100 ps 20 ps Comb. logic A R e g Comb. logic B R e g Comb. logic C R e g Delay = 510 ps Throughput = 5.88 GOPS Clock OP1 A A A B B B C C C OP2 OP3 Time Limitations: Nonuniform Delays • Throughput limited by slowest stage: clock ~ this value (worst case) • Other stages sit idle for much of the time • Challenging to partition system into balanced stages

8. Delay = 420 ps, Throughput = 14.29 GOPS 50 ps 20 ps 50 ps 20 ps 50 ps 20 ps 50 ps 20 ps 50 ps 20 ps 50 ps 20 ps Comb. logic R e g Comb. logic R e g Comb. logic R e g Comb. logic R e g Comb. logic R e g Comb. logic R e g Clock Limitations: Register Overhead • As try to deepen pipeline, overhead of loading registers becomes more significant • Percentage of clock cycle spent loading register: • 3-stage pipeline: 16.67% (60/360) • 6-stage pipeline: 28.57% (?)

9. Example

10. R e g Combinational logic Clock OP1 OP2 OP3 Time Data Dependencies • System • Each operation may depend on result from preceding one • Not a car wash!

11. Data Dependencies in Processors • Result from one instruction used as operand for another • Read-after-write (RAW) dependency • Very common in actual programs • Must make sure our pipeline handles these properly • Get correct results • Minimize performance impact 1irmovl \$50, %eax 2addl %eax , %ebx 3mrmovl 100( %ebx ), %edx

12. A A A A B B B B C C C C Comb. logic A R e g Comb. logic B R e g Comb. logic C R e g OP1 OP2 OP3 OP4 Time Clock Data Hazards • Result does not feed back around in time for next operation • Pipelining has changed behavior of system

13. Control Dependency loop: sub1 %edx, %ebx jnetarg irmovl \$10, %edx jmp loop targ: Problem? take wrong path => control hazard

14. Pipeline Stages • Fetch • Select current PC • Read instruction • Compute incremented PC • Decode • Read program registers • Execute • Operate ALU • Memory • Read or write data memory • Write Back • Update register file

15. PIPE- Hardware • Pipeline registers hold intermediate values from instruction execution • Forward (Upward) Paths • Values passed from one stage to next

16. 1 2 3 4 5 6 7 8 9 irmovl \$1,%eax #I1 F D E M W irmovl \$2,%ecx #I2 F D E M W F E W M D irmovl \$3,%edx #I3 I4 I3 I2 I1 I5 irmovl \$4,%ebx #I4 F D E M W halt #I5 F D E M W F D E M W Cycle 5 Pipeline Example

17. 1 2 3 4 5 6 7 8 F D E M W 0x000: irmovl \$10,% edx F D E M W 0x006: irmovl \$3,% eax F D E M W 0x00c: addl % edx ,% eax F D E M W 0x00e: halt Cycle 4 M M_ valE = 10 M_ dstE = % edx E f e_ valE 0 + 3 = 3 E_ dstE = % eax D D Error f f valA valA R[ R[ ] ] = = 0 0 % % edx edx f f valB valB R[ R[ ] ] = = 0 0 % % eax eax Data Dependencies

18. 1 2 3 4 5 6 7 8 9 F D E M W 0x000: irmovl \$10,% edx F D E M W 0x006: irmovl \$3,% eax F F D D E E M M W W 0x00c: nop F F D D E E M M W W 0x00d: addl % edx ,% eax F F D D E E M M W W 0x00f: halt Cycle 5 W W f f R[ R[ ] ] 10 10 % % edx edx M M_ valE = 3 M_ dstE = % eax • • • D D Error f f valA valA R[ R[ ] ] = = 0 0 % % edx edx f f valB valB R[ R[ ] ] = = 0 0 % % eax eax Data Dependencies: 1 Nop

19. 1 2 3 4 5 6 7 8 9 10 F F D D E E M M W W 0x000: irmovl \$10,% edx F F D D E E M M W W 0x006: irmovl \$3,% eax F F D D E E M M W W 0x00c: nop F F D D E E M M W W 0x00d: nop F F D D E E M M W W 0x00e: addl % edx ,% eax F F D D E E M M W W 0x010: halt Cycle 6 W W W f f f R[ R[ R[ ] ] ] 3 3 3 % % % eax eax eax • • • • • • D D D f f f valA valA valA R[ R[ R[ ] ] ] = = = 10 10 10 Error % % % edx edx edx f f f valB valB valB R[ R[ R[ ] ] ] = = = 0 0 0 % % % eax eax eax Data Dependencies: 2 Nop’s

20. 1 2 3 4 5 6 7 8 9 10 11 F F D D E E M M W W 0x000: irmovl \$10,% edx F F D D E E M M W W 0x006: irmovl \$3,% eax F F D D E E M M W W 0x00c: nop F F D D E E M M W W 0x00d: nop F F D D E E M M W W 0x00e: nop F F D D E E M M W W 0x00f: addl % edx ,% eax F F D D E E M M W W 0x011: halt Cycle 6 W W f f R[ R[ ] ] 3 3 % % eax eax Cycle 7 D D f f valA valA R[ R[ ] ] = = 10 10 % % edx edx f f valB valB R[ R[ ] ] = = 3 3 % % eax eax Data Dependencies: 3 Nop’s

21. E M W F F D E M W Stalling for Data Dependencies 1 2 3 4 5 6 7 8 9 10 11 • If instruction follows too closely after one that writes register, slow it down • Hold instruction in decode • Dynamically inject nop(“bubble”)into execute stage 0x000: irmovl \$10,%edx 0x006: irmovl \$3,%eax F D E M W 0x00c: nop F D E M W 0x00d: nop F D E M W bubble 0x00e: addl %edx,%eax D D E M W 0x010: halt F F D E M W

22. E M W F F D E M W Cycle 6 W W_dstE = %eax W_valE = 3 • • • D srcA = %edx srcB = %eax Detecting Stall Condition 1 2 3 4 5 6 7 8 9 10 11 0x000: irmovl \$10,%edx 0x006: irmovl \$3,%eax F D E M W 0x00c: nop F D E M W 0x00d: nop F D E M W bubble 0x00e: addl %edx,%eax D D E M W 0x010: halt F F D E M W

23. Data Forwarding • Simple Pipeline • Register isn’t written until completion of write-back stage • Source operands read from register file in decode stage • Needs to be in register file at start of stage • Trick • Pass value directly from generating instruction to decode stage • Why is this preferred over bubbles?

24. E M W F F D E M W Cycle 6 W W_dstE = %eax W_valE = 3 • • • D srcA = %edx srcB = %eax Forwarding 1 2 3 4 5 6 7 8 9 10 11 0x000: irmovl \$10,%edx 0x006: irmovl \$3,%eax F D E M W 0x00c: nop F D E M W 0x00d: nop F D E M W bubble 0x00e: addl %edx,%eax D D E M W 0x010: halt F F D E M W add control logic and lines

25. Pipeline Summary • Concept • Break instruction execution into 5 stages • Run instructions through in pipelined mode • Limitations • Data dependencies/hazards • One instruction writes register, later one reads it • Control dependency/hazards • Instruction sets PC in way that pipeline did not predict correctly • Mispredictedbranch, return, etc • Fixing the Pipeline • bubbles, forwarding

26. Next Time • Optimizing Performance • Chapter 5 • Have a great weekend!

27. Predicting the PC • Start fetch of new instruction after current one has completed fetch stage • Not enough time to reliably determine next instruction • Guess which instruction will follow • Recover if prediction was incorrect

28. Our Prediction Strategy • Instructions that Don’t Transfer Control • Predict next PC to be valP • Always reliable • Call and Unconditional Jumps • Predict next PC to be valC (destination) • Always reliable • Conditional Jumps • Predict next PC to be valC (destination) • Only correct if branch is taken • Typically right 60% of time • Return Instruction • Don’t try to predict

29. 1 2 3 4 5 6 7 8 9 F D E M W 0x000: xorl % eax ,% eax F D E M W 0x002: jne t # Not taken … F D E M W 0x011: t: irmovl \$3, % edx # Target F D E M W 0x017: irmovl \$4, % ecx # Target+1 F F D D E E M M W W 0x007: irmovl \$1, % eax # Fall Through Cycle 5 M M_Bch = 0 M_ valA = 0x007 E E f f valE valE 3 3 dstE dstE = = % % edx edx D D valC valC = = 4 4 dstE dstE = = % % ecx ecx F F f f valC valC 1 1 f f rB rB % % eax eax Branch Misprediction Trace …… • Incorrectly executes instructions at branch target • Solution: instruction cancellation or squashing before E stage • generate bubbles