1 Lecture 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 notes2 The Assembly Line A Start and finish a job before moving to the next Time Jobs Break the job into smaller stages B C A B C A B C A B C Unpipelined Pipelined3 Quantitative 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 speed4 A 5-Stage Pipeline5 A 5-Stage Pipeline Use the PC to access the I-cache and increment PC by 46 A 5-Stage Pipeline Read registers, compare registers, compute branch target; for now, assume branches take 2 cyc (there is enough work that branches can easily take more)7 A 5-Stage Pipeline ALU computation, effective address computation for load/store8 A 5-Stage Pipeline Memory access to/from data cache, stores finish in 4 cycles9 A 5-Stage Pipeline Write result of ALU computation or load into register file10 Conflicts/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)11 Hazards • 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 ways12 Structural 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)13 Data Hazards14 Bypassing • Some data hazard stalls can be eliminated: bypassing15 Example add R1, R2, R3 lw R4, 8(R1)16 Example lw R1, 8(R2) lw R4, 8(R1)17 Example lw R1, 8(R2) sw R1, 8(R3)18 Summary • 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, F419 Title •
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