CPE 731 Advanced Computer Architecture Pipelining Review Dr. Gheith Abandah Adapted from the slides of Prof. David Patterson, University of California, Berkeley

10/19/2015CPE 731, Pipelining 2 Outline 5 stage pipelining Structural and Data Hazards Forwarding Branch Schemes Exceptions and Interrupts Multi-cycle Operations Conclusion

10/19/2015CPE 731, Pipelining 3 Five-Stage Pipelining Figure A.2, Page A-8 I n s t r. O r d e r Time (clock cycles) Reg ALU DMemIfetch Reg ALU DMemIfetch Reg ALU DMemIfetch Reg ALU DMemIfetch Reg Cycle 1Cycle 2Cycle 3Cycle 4Cycle 6Cycle 7Cycle 5

10/19/2015CPE 731, Pipelining 4 Pipelining is not quite that easy! Limits to pipelining: Hazards prevent next instruction from executing during its designated clock cycle –Structural hazards: HW cannot support this combination of instructions (single person to fold and put clothes away) –Data hazards: Instruction depends on result of prior instruction still in the pipeline (missing sock) –Control hazards: Caused by delay between the fetching of instructions and decisions about changes in control flow (branches and jumps).

10/19/2015CPE 731, Pipelining 5 One Memory Port/Structural Hazards Figure A.4, Page A-14 I n s t r. O r d e r Time (clock cycles) Load Instr 1 Instr 2 Instr 3 Instr 4 Reg ALU DMem Ifetch Reg ALU DMemIfetch Reg ALU DMemIfetch Reg ALU DMem Ifetch Reg Cycle 1Cycle 2Cycle 3Cycle 4Cycle 6Cycle 7Cycle 5 Reg ALU DMemIfetch Reg

10/19/2015CPE 731, Pipelining 6 One Memory Port/Structural Hazards (Similar to Figure A.5, Page A-15) I n s t r. O r d e r Time (clock cycles) Load Instr 1 Instr 2 Stall Instr 3 Reg ALU DMem Ifetch Reg ALU DMemIfetch Reg ALU DMemIfetch Reg Cycle 1Cycle 2Cycle 3Cycle 4Cycle 6Cycle 7Cycle 5 Reg ALU DMemIfetch Reg Bubble How do you “bubble” the pipe?

10/19/2015CPE 731, Pipelining 7 Speed Up Equation for Pipelining For simple RISC pipeline, CPI = 1:

10/19/2015CPE 731, Pipelining 8 Example: Dual-port vs. Single-port Machine A: Dual ported memory (“Harvard Architecture”) Machine B: Single ported memory, but its pipelined implementation has a 1.05 times faster clock rate Ideal CPI = 1 for both Loads are 40% of instructions executed SpeedUp A = Pipeline Depth/(1 + 0) x (clock unpipe /clock pipe ) = Pipeline Depth SpeedUp B = Pipeline Depth/( x 1) x (clock unpipe /(clock unpipe / 1.05) = (Pipeline Depth/1.4) x 1.05 = 0.75 x Pipeline Depth SpeedUp A / SpeedUp B = Pipeline Depth/(0.75 x Pipeline Depth) = 1.33 Machine A is 1.33 times faster

10/19/2015CPE 731, Pipelining 9 I n s t r. O r d e r add r1,r2,r3 sub r4,r1,r3 and r6,r1,r7 or r8,r1,r9 xor r10,r1,r11 Reg ALU DMemIfetch Reg ALU DMemIfetch Reg ALU DMemIfetch Reg ALU DMemIfetch Reg ALU DMemIfetch Reg Data Hazard on R1 Figure A.6, Page A-17 Time (clock cycles) IFID/RF EX MEM WB

10/19/2015CPE 731, Pipelining 10 Read After Write (RAW) Instr J tries to read operand before Instr I writes it Caused by a “Dependence” (in compiler nomenclature). This hazard results from an actual need for communication. Three Generic Data Hazards I: add r1,r2,r3 J: sub r4,r1,r3

10/19/2015CPE 731, Pipelining 11 Write After Read (WAR) Instr J writes operand before Instr I reads it Called an “anti-dependence” by compiler writers. This results from reuse of the name “r1”. Can’t happen in MIPS 5 stage pipeline because: – All instructions take 5 stages, and – Reads are always in stage 2, and – Writes are always in stage 5 I: sub r4,r1,r3 J: add r1,r2,r3 K: mul r6,r1,r7 Three Generic Data Hazards

10/19/2015CPE 731, Pipelining 12 Three Generic Data Hazards Write After Write (WAW) Instr J writes operand before Instr I writes it. Called an “output dependence” by compiler writers This also results from the reuse of name “r1”. Can’t happen in MIPS 5 stage pipeline because: – All instructions take 5 stages, and – Writes are always in stage 5 Will see WAR and WAW in more complicated pipes I: sub r1,r4,r3 J: add r1,r2,r3 K: mul r6,r1,r7

10/19/2015CPE 731, Pipelining 13 Outline 5 stage pipelining Structural and Data Hazards Forwarding Branch Schemes Exceptions and Interrupts Multi-cycle Operations Conclusion

10/19/2015CPE 731, Pipelining 14 Time (clock cycles) Forwarding to Avoid Data Hazard Figure A.7, Page A-19 I n s t r. O r d e r add r1,r2,r3 sub r4,r1,r3 and r6,r1,r7 or r8,r1,r9 xor r10,r1,r11 Reg ALU DMemIfetch Reg ALU DMemIfetch Reg ALU DMemIfetch Reg ALU DMemIfetch Reg ALU DMemIfetch Reg

10/19/2015CPE 731, Pipelining 15 HW Change for Forwarding Figure A.23, Page A-37 MEM/WR ID/EX EX/MEM Data Memory ALU mux Registers NextPC Immediate mux What circuit detects and resolves this hazard?

10/19/2015CPE 731, Pipelining 16 Time (clock cycles) Forwarding to Avoid LW-SW Data Hazard Figure A.8, Page A-20 I n s t r. O r d e r add r1,r2,r3 lw r4, 0(r1) sw r4,12(r1) or r8,r6,r9 xor r10,r9,r11 Reg ALU DMemIfetch Reg ALU DMemIfetch Reg ALU DMemIfetch Reg ALU DMemIfetch Reg ALU DMemIfetch Reg

10/19/2015CPE 731, Pipelining 17 Time (clock cycles) I n s t r. O r d e r lw r1, 0(r2) sub r4,r1,r6 and r6,r1,r7 or r8,r1,r9 Data Hazard Even with Forwarding Figure A.9, Page A-21 Reg ALU DMemIfetch Reg ALU DMemIfetch Reg ALU DMemIfetch Reg ALU DMemIfetch Reg

10/19/2015CPE 731, Pipelining 18 Data Hazard Even with Forwarding (Similar to Figure A.10, Page A-21) Time (clock cycles) or r8,r1,r9 I n s t r. O r d e r lw r1, 0(r2) sub r4,r1,r6 and r6,r1,r7 Reg ALU DMemIfetch Reg Ifetch ALU DMem Reg Bubble Ifetch ALU DMem Reg Bubble Reg Ifetch ALU DMem Bubble Reg H ow is this detected?

10/19/2015CPE 731, Pipelining 19 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 Ra,a LW Re,e LW Rf,f SUB Rd,Re,Rf SWRd,d Software Scheduling to Avoid Load Hazards Fast code: LW Rb,b LW Rc,c LW Re,e ADD Ra,Rb,Rc LW Rf,f SW Ra,a SUB Rd,Re,Rf SWRd,d Compiler optimizes for performance. Hardware checks for safety.

10/19/2015CPE 731, Pipelining 20 Outline 5 stage pipelining Structural and Data Hazards Forwarding Branch Schemes Exceptions and Interrupts Multi-cycle Operations Conclusion

10/19/2015CPE 731, Pipelining 21 Control Hazard on Branches Three Stage Stall 10: beq r1,r3,36 14: and r2,r3,r5 18: or r6,r1,r7 22: add r8,r1,r9 36: xor r10,r1,r11 Reg ALU DMemIfetch Reg ALU DMemIfetch Reg ALU DMemIfetch Reg ALU DMemIfetch Reg ALU DMemIfetch Reg What do you do with the 3 instructions in between? How do you do it? Where is the “commit”?

10/19/2015CPE 731, Pipelining 22 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 MIPS branch tests if register = 0 or  0 MIPS 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

10/19/2015CPE 731, Pipelining 23 Adder IF/ID Pipelined MIPS Datapath Figure A.24, page A-38 Memory Access Write Back Instruction Fetch Instr. Decode Reg. Fetch Execute Addr. Calc ALU Memory Reg File MUX Data Memory MUX Sign Extend Zero? MEM/WB EX/MEM 4 Adder Next SEQ PC RD WB Data Interplay of instruction set design and cycle time. Next PC Address RS1 RS2 Imm MUX ID/EX

10/19/2015CPE 731, Pipelining 24 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% MIPS branches not taken on average –PC+4 already calculated, so use it to get next instruction #3: Predict Branch Taken –53% MIPS branches taken on average –But haven’t calculated branch target address in MIPS »MIPS still incurs 1 cycle branch penalty »Other machines: branch target known before outcome

10/19/2015CPE 731, Pipelining 25 Four Branch Hazard Alternatives #4: Delayed Branch –Define branch to take place AFTER a following instruction branch instruction sequential successor 1 sequential successor sequential successor n branch target if taken –1 slot delay allows proper decision and branch target address in 5 stage pipeline –MIPS uses this Branch delay of length n

10/19/2015CPE 731, Pipelining 26 Scheduling Branch Delay Slots (Fig A.14) A is the best choice, fills delay slot & reduces instruction count (IC) In B, the sub instruction may need to be copied, increasing IC In B and C, must be okay to execute sub when branch fails add $1,$2,$3 if $2=0 then delay slot A. From before branchB. From branch targetC. From fall through add $1,$2,$3 if $1=0 then delay slot add $1,$2,$3 if $1=0 then delay slot sub $4,$5,$6 becomes if $2=0 then add $1,$2,$3 if $1=0 then sub $4,$5,$6 add $1,$2,$3 if $1=0 then sub $4,$5,$6

10/19/2015CPE 731, Pipelining 27 Delayed Branch 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: As processor go to deeper pipelines and multiple issue, the branch delay grows and need more than one delay slot –Delayed branching has lost popularity compared to more expensive but more flexible dynamic approaches –Growth in available transistors has made dynamic approaches relatively cheaper

10/19/2015CPE 731, Pipelining 28 Evaluating Branch Alternatives Assume 4% unconditional branch, 6% conditional branch- untaken, 10% conditional branch-taken SchedulingBranchCPIspeedup v.speedup v. scheme penaltyunpipelinedstall Stall pipeline Predict taken Predict not taken Delayed branch

10/19/2015CPE 731, Pipelining 29 Outline 5 stage pipelining Structural and Data Hazards Forwarding Branch Schemes Exceptions and Interrupts Multi-cycle Operations Conclusion

10/19/2015CPE 731, Pipelining 30 Problems with Pipelining Exception: An unusual event happens to an instruction during its execution –Examples: divide by zero, undefined opcode Interrupt: Hardware signal to switch the processor to a new instruction stream –Example: a sound card interrupts when it needs more audio output samples (an audio “click” happens if it is left waiting) Problem: It must appear that the exception or interrupt must appear between 2 instructions (I i and I i+1 ) –The effect of all instructions up to and including I i is totalling complete – No effect of any instruction after I i can take place The interrupt (exception) handler either aborts program or restarts at instruction I i+1

10/19/2015CPE 731, Pipelining 31 Precise Exceptions in Static Pipelines Key observation: architected state only change in memory and register write stages.

10/19/2015CPE 731, Pipelining 32 Outline 5 stage pipelining Structural and Data Hazards Forwarding Branch Schemes Exceptions and Interrupts Multi-cycle Operations Conclusion

10/19/2015CPE 731, Pipelining 33 Multi-cycle Operations

10/19/2015CPE 731, Pipelining 34 Multi-cycle Operations - Example

10/19/2015CPE 731, Pipelining 35 Multi-cycle Operations - Example UnitLatencyInitiation Interval Integer ALU01 Memory11 FP Adder31 FP Multiplier61 FP Divider2425

10/19/2015CPE 731, Pipelining 36 Problems1/5 1.Structural hazard on divide Solution: Detect at ID stage and stall 2.May have multiple register writes Example: mul.d f0, f4, f6 F D M1 M2 M3 M4 M5 M6 M7 M W... F D E M W add.d f2, f4, f6 F D A1 A2 A3 A4 M W... F D E M W ld.d f2, 0(r2) F D E M W

10/19/2015CPE 731, Pipelining 37 Problems2/5 2.May have multiple register writes Solution: Use shift register to reserve when need to write and stall on conflict Before|0|0|0|0| add.d|0|0|0|1| ld.d|1|0|1|0| ld.d|1|1|0|0| ld.d|1|0|0|0| Need to stall this

10/19/2015CPE 731, Pipelining 38 Problems3/5 3.There are long stalls due to RAW dependencies Example: ld.d f4, 0(r2) mul.d f0, f4, f6 1-cycle stall add.d f2, f0, f8 long stall s.D f2, 0(r2) long stall Solution: Mark registers pending as destinations, then in ID stage stall any instruction that uses the marked registers.

10/19/2015CPE 731, Pipelining 39 Problems4/5 4.WAW hazards are possible Example: mul.d f0, f4, f6 add.d f0, f6, f8 Solution: Stall the second instruction if its destination register is pending as a destination.

10/19/2015CPE 731, Pipelining 40 Problems5/5 5.Out-of-order completion creates problems in restarting after exceptions Example: div.d f0, f4, f6 Assume a div by 0 exception ld.d f2, 0(r2) Finishes before div Solution: 1) Accept imprecise exceptions 2) Buffer results and write-back in order.

10/19/2015CPE 731, Pipelining 41 Outline 5 stage pipelining Structural and Data Hazards Forwarding Branch Schemes Exceptions and Interrupts Multi-cycle Operations Conclusion

10/19/2015CPE 731, Pipelining 42 And In Conclusion: Control and Pipelining Just overlap tasks; easy if tasks are independent Speed Up  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 Exceptions, Interrupts add complexity Multi-cycle operations introduce several problems