PowerPoint PresentationSlide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 13Slide 14Slide 15Slide 16Slide 17Slide 18Slide 191Lecture 3: Pipelining Basics • Biggest contributors to performance: clock speed, parallelism• Today: basic pipelining implementation (Sections A.1-A.3)• Reminders: Sign up for the class mailing list First assignment is on-line, due next Thursday TA office hours: Jim King, Wed 1-2pm, MEB 3423; Bharathan Rajaram, TBA Class notes2The Assembly LineAStart and finish a job before moving to the nextTimeJobsBreak the job into smaller stagesB CA B CA B CA B CUnpipelinedPipelined3Quantitative Effects• As a result of pipelining: Time in ns per instruction goes up Number of cycles per instruction goes up (note the increase in clock speed) Total execution time goes down, resulting in lower time per instruction Average cycles per instruction increases slightly Under ideal conditions, speedup = ratio of elapsed times between successive instruction completions = number of pipeline stages = increase in clock speed4A 5-Stage Pipeline5A 5-Stage Pipeline Use the PC to access the I-cache and increment PC by 46A 5-Stage PipelineRead registers, compare registers, compute branch target; for now, assumebranches take 2 cyc (there is enough work that branches can easily take more)7A 5-Stage PipelineALU computation, effective address computation for load/store8A 5-Stage PipelineMemory access to/from data cache, stores finish in 4 cycles9A 5-Stage PipelineWrite result of ALU computation or load into register file10Conflicts/Problems• I-cache and D-cache are accessed in the same cycle – it helps to implement them separately• Registers are read and written in the same cycle – easy to deal with if register read/write time equals cycle time/2 (else, use bypassing)• Branch target changes only at the end of the second stage -- what do you do in the meantime?• Data between stages get latched into registers (overhead that increases latency per instruction)11Hazards• Structural hazards: different instructions in different stages (or the same stage) conflicting for the same resource• Data hazards: an instruction cannot continue because it needs a value that has not yet been generated by an earlier instruction• Control hazard: fetch cannot continue because it does not know the outcome of an earlier branch – special case of a data hazard – separate category because they are treated in different ways12Structural Hazards• Example: a unified instruction and data cache stage 4 (MEM) and stage 1 (IF) can never coincide• The later instruction and all its successors are delayed until a cycle is found when the resource is free these are pipeline bubbles• Structural hazards are easy to eliminate – increase the number of resources (for example, implement a separate instruction and data cache)13Data Hazards14Bypassing• Some data hazard stalls can be eliminated: bypassing15Exampleadd R1, R2, R3lw R4, 8(R1)16Example lw R1, 8(R2) lw R4, 8(R1)17Example lw R1, 8(R2) sw R1, 8(R3)18Summary• For the 5-stage pipeline, bypassing can eliminate delays between the following example pairs of instructions: add/sub R1, R2, R3 add/sub/lw/sw R4, R1, R5 lw R1, 8(R2) sw R1, 4(R3)• The following pairs of instructions will have intermediate stalls: lw R1, 8(R2) add/sub/lw R3, R1, R4 or sw R3, 8(R1) fmul F1, F2, F3 fadd F5, F1, F419Title•
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