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ELEC 5200-001/6200-001 Computer Architecture and Design Spring 2016 Pipelining (Chapter 6)

ELEC 5200-001/6200-001 Computer Architecture and Design Spring 2016 Pipelining (Chapter 6). Vishwani D. Agrawal James J. Danaher Professor Department of Electrical and Computer Engineering Auburn University, Auburn, AL 36849 http://www.eng.auburn.edu/~vagrawal vagrawal@eng.auburn.edu.

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ELEC 5200-001/6200-001 Computer Architecture and Design Spring 2016 Pipelining (Chapter 6)

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  1. ELEC 5200-001/6200-001Computer Architecture and DesignSpring 2016Pipelining (Chapter 6) Vishwani D. Agrawal James J. Danaher Professor Department of Electrical and Computer Engineering Auburn University, Auburn, AL 36849 http://www.eng.auburn.edu/~vagrawal vagrawal@eng.auburn.edu ELEC 5200-001/6200-001 Lecture 6

  2. ILP: Instruction Level Parallelism • Single-cycle and multi-cycle datapaths execute one instruction at a time. • How can we get better performance? • Answer: Execute multiple instructions at a time: • Pipelining – Enhance a multi-cycle datapath to fetch one instruction every cycle. • Parallelism – Fetch multiple instructions every cycle. ELEC 5200-001/6200-001 Lecture 6

  3. Automobile Team Assembly 1 hour 1 hour 1 hour 1 hour 1 car assembled every four hours 6 cars per day 180 cars per month 2,040 cars per year ELEC 5200-001/6200-001 Lecture 6

  4. Automobile Assembly Line Task 2 1 hour Task 3 1 hour Task 4 1 hour Task 1 1 hour Mecahnical Electrical Painting Testing First car assembled in 4 hours (pipeline latency) thereafter, 1 car completed per hour 21 cars on first day, thereafter 24 cars per day 717 cars per month 8,637 cars per year What gives 4X increase? ELEC 5200-001/6200-001 Lecture 6

  5. Throughput: Team Assembly Red car started Red car completed Mechanical Electrical Painting Testing Mechanical Electrical Painting Testing Blue car started Blue car completed Time Time of assembling one car = n hours where n is the number of nearly equal subtasks, each requiring 1 unit of time Throughput = 1/n cars per unit time ELEC 5200-001/6200-001 Lecture 6

  6. Throughput: Assembly Line Mechanical Electrical Painting Testing Car 1 Car 2 Car 3 Car 4 . . Mechanical Electrical Painting Testing Mechanical Electrical Painting Testing Mechanical Electrical Painting Testing Car 1 complete Car 2 complete time Time to complete first car = n time units (latency) Cars completed in time T = T – n + 1 Throughput = 1 – (n – 1)/ T cars per unit time Throughput (assembly line) 1 – (n – 1)/ Tn(n – 1) ─────────────────── = ──────── = n – ───── → n Throughput (team assembly) 1/nT as T→∞ ELEC 5200-001/6200-001 Lecture 6

  7. Some Features of Assembly Line Electrical parts delivered (JIT) Task 2 1 hour Task 3 1 hour Task 4 1 hour Task 1 1 hour Mechanical Electrical Painting Testing 3 cars in the assembly line are suspects, to be removed (flush pipeline) Defect found Stall assembly line to fix the cause of defect ELEC 5200-001/6200-001 Lecture 6

  8. Pros and Cons • Advantages: • Efficient use of labor. • Specialists can do better job. • Just in time (JIT) methodology eliminates warehouse cost. • Disadvantages: • Penalty of defect latency. • Lack of flexibility in production. • Assembly line work is monotonous and boring. • https://www.youtube.com/watch?v=IjarLbD9r30 • https://www.youtube.com/watch?v=ANXGJe6i3G8 • https://www.youtube.com/watch?v=5lp4EbfPAtI ELEC 5200-001/6200-001 Lecture 6

  9. A Single Cycle Example Delay of 1-bit full adder = 1ns Clock period ≥ 32ns a31 . . . a2 a1 a0 c32 1-b full adder s31 . . . s2 s1 s0 1-b full adder b31 . . . b2 b1 b0 1-b full adder 1-b full adder Time of adding words ~ 32ns Time of adding bytes ~ 32ns 0 Clock ELEC 5200-001/6200-001 Lecture 6

  10. A Multicycle Implementation a31 . . . a2 a1 a0 Delay of 1-bit full adder = 1ns Clock period ≥ 1ns Time of adding words ~ 32ns Time of adding bytes ~ 8ns Shift 1-b full adder b31 . . . b2 b1 b0 s31 . . . s2 s1 s0 c32 FF Shift Shift Initialize to 0 Clock ELEC 5200-001/6200-001 Lecture 6

  11. A Pipeline Implementation Delay of 1-bit full adder = 1ns Clock period ≥ 1ns a31 . . . a2 a1 a0 P1 P2 P3 .. P31 c32 1-b full adder s31 . . . s2 s1 s0 1-b full adder b31 . . . b2 b1 b0 1-b full adder 1-b full adder Time of adding words ~ 1ns Time of adding bytes ~ 1ns 0 Clock ELEC 5200-001/6200-001 Lecture 6

  12. Pipelining in a Computer • Divide datapath into nearly equal tasks, to be performed serially and requiring non-overlapping resources. • Insert registers at task boundaries in the datapath; registers pass the output data from one task as input data to the next task. • Synchronize tasks with a clock having a cycle time just enough to complete the longest task. • Break each instruction down into a set of tasks so that instructions can be executed in a staggered fashion. ELEC 5200-001/6200-001 Lecture 6

  13. Pipelining of RISC Instructions(From Lecture 3, Slide 6) Fetch Instruction Examine Opcode ALU Operation Memory Read/Write Store Result IF ID EX MEM WB Instruction Decode Execute Memory Write Fetch instruction and Operation Back Fetch operands to Reg file Although an instruction takes five clock cycles, one instruction is completed every cycle. ELEC 5200-001/6200-001 Lecture 6

  14. Pipelining a Single-Cycle Datapath No operation on data; idle times equalize instruction lengths. ELEC 5200-001/6200-001 Lecture 6

  15. Execution Time: Single-Cycle 0 2 4 6 8 10 12 14 16 . . Time (ns) lw $1, 100($0) lw $2, 200($0) lw $3, 300($0) IF ID EX MEM WB IF ID EX MEM WB IF ID EX MEM WB Clock cycle time = 8 ns Total time for executing three lw instructions = 24 ns ELEC 5200-001/6200-001 Lecture 6

  16. Pipelined Datapath No operation on data; idle time inserted to equalize instruction lengths. ELEC 5200-001/6200-001 Lecture 6

  17. Execution Time: Pipeline 0 2 4 6 8 10 12 14 16 . . Time (ns) lw $1, 100($0) lw $2, 200($0) lw $3, 300($0) IF ID EX MEM RW IF ID EX MEM RW IF ID EX MEM RW Clock cycle time = 2 ns, four times faster than single-cycle clock Total time for executing three lw instructions = 14 ns Single-cycle time 24 Performance ratio = ──────────── = ── = 1.7 Pipeline time 14 ELEC 5200-001/6200-001 Lecture 6

  18. Pipeline Performance Clock cycle time = 2 ns 1,003 lw instructions: Total time for executing 1,003 lw instructions = 2,014 ns Single-cycle time 8,024 Performance ratio = ──────────── = ──── = 3.98 Pipeline time 2,014 10,003 lw instructions: Performance ratio = 80,024 / 20,014 = 3.998 → Clock cycle ratio (4) Pipeline performance approaches clock-cycle ratio for long programs. ELEC 5200-001/6200-001 Lecture 6

  19. Single-Cycle Datapath WB: write back ID: Instr. decode, reg. file read EX: Execute, address calc. MEM: mem. access IF: Instr. fetch 4 Add 1 mux 0 ALU Branch opcode MemtoReg CONTROL 26-31 RegWrite ALUSrc 21-25 zero MemWrite MemRead ALU Instr. mem. PC Reg. File Data mem. 1 mux 0 16-20 0 mux 1 1 mux 0 11-15 RegDst ALUOp ALU Cont. Sign ext. Shift left 2 0-15 0-5 ELEC 5200-001/6200-001 Lecture 6

  20. Pipeline Registers IF/ID ID/EX EX/MEM 1 mux 0 4 Add ALU Branch opcode MemtoReg CONTROL 26-31 MEM/WB RegWrite ALUSrc 21-25 zero MemWrite MemRead ALU Instr. mem. PC Reg. File Data mem. 1 mux 0 16-20 0 mux 1 1 mux 0 11-15 This requires a CONTROL not too different from single-cycle RegDst ALUOp ALU Cont. Sign ext. Shift left 2 0-15 0-5 ELEC 5200-001/6200-001 Lecture 6

  21. Pipeline Register Functions • Four pipeline registers are added: ELEC 5200-001/6200-001 Lecture 6

  22. Pipelined Datapath IF/ID ID/EX EX/MEM MEM/WB 4 1 mux 0 Add ALU opcode Shift left 2 26-31 zero 21-25 Instr mem ALU 16-20 Data mem. Reg. File PC 1 mux 0 0 mux 1 Sign ext. 11-15 for R-type 16-20 for I-type lw 0-15 ELEC 5200-001/6200-001 Lecture 6

  23. Five-Cycle Pipeline CC1 CC2 CC3 CC4 CC5 DM IM ID, REG. FILE READ ALU REG. FILE WRITE MEM/WB PC IF/ID ID/EX EX/MEM ELEC 5200-001/6200-001 Lecture 6

  24. Add Instruction • add $t0, $s1, $s2 Machine instruction word 000000 10001 10010 01000 00000 100000 opcode $s1 $s2 $t0 function CC1 CC2 CC3 CC4 CC5 DM IM ID, REG. FILE READ ALU REG. FILE WRITE MEM/WB PC IF/ID ID/EX EX/MEM IF ID EX MEM WB read $s1 add write $t0 read $s2 $s1+$s2 ELEC 5200-001/6200-001 Lecture 6

  25. Pipelined Datapath Executing add IF/ID ID/EX EX/MEM MEM/WB 4 1 mux 0 Add ALU opcode Shift left 2 26-31 s1 zero $s1 21-25 Instr mem ALU addr Data mem data 16-20 Reg. File PC $s2 s2 1 mux 0 0 mux 1 Sign ext. 11-15 for R-type 16-20 for I-type lw t0 0-15 ELEC 5200-001/6200-001 Lecture 6

  26. DM IM ID, REG. FILE READ ALU REG. FILE WRITE MEM/WB IF/ID ID/EX EX/MEM Load Instruction • lw $t0, 1200 ($t1) 100011 01001 01000 0000 0100 1000 0000 opcode $t1 $t0 1200 CC1 CC2 CC3 CC4 CC5 PC IF ID EX MEM WB read $t1 add read write $t0 sign ext $t1+1200 M[addr] 1200 ELEC 5200-001/6200-001 Lecture 6

  27. Pipelined Datapath Executing lw IF/ID ID/EX EX/MEM MEM/WB 4 1 mux 0 Add ALU opcode Shift left 2 26-31 t1 zero $t1 21-25 Instr mem ALU addr Data mem data 16-20 Reg. File PC 1 mux 0 0 mux 1 Sign ext. 11-15 for R-type 16-20 for I-type lw t0 0-15 1200 ELEC 5200-001/6200-001 Lecture 6

  28. DM IM ID, REG. FILE READ ALU REG. FILE WRITE MEM/WB IF/ID ID/EX EX/MEM Store Instruction • sw $t0, 1200 ($t1) 101011 01001 01000 0000 0100 1000 0000 opcode $t1 $t0 1200 CC1 CC2 CC3 CC4 CC5 PC IF ID EX MEM WB read $t1 add write sign ext $t1+1200 M[addr] 1200 (addr) ← $t0 ELEC 5200-001/6200-001 Lecture 6

  29. Pipelined Datapath Executing sw IF/ID ID/EX EX/MEM MEM/WB 4 1 mux 0 Add ALU opcode Shift left 2 26-31 t1 zero $t1 21-25 Instr mem ALU addr Data mem data 16-20 Reg. File PC $t0 t0 1 mux 0 0 mux 1 Sign ext. 11-15 for R-type 16-20 for I-type lw 0-15 1200 ELEC 5200-001/6200-001 Lecture 6

  30. Executing a Program Consider a five-instruction segment: lw $10, 20($1) sub $11, $2, $3 add $12, $3, $4 lw $13, 24($1) add $14, $5, $6 ELEC 5200-001/6200-001 Lecture 6

  31. DM DM DM DM DM ID, REG. FILE READ IM ALU REG. FILE WRITE ID, REG. FILE READ ID, REG. FILE READ ALU ALU REG. FILE WRITE REG. FILE WRITE IM IM ID, REG. FILE READ ID, REG. FILE READ ALU ALU IM IM REG. FILE WRITE REG. FILE WRITE MEM/WB MEM/WB MEM/WB MEM/WB MEM/WB IF/ID EX/MEM ID/EX EX/MEM EX/MEM ID/EX ID/EX EX/MEM EX/MEM IF/ID IF/ID IF/ID IF/ID ID/EX ID/EX Program Execution time CC1 CC2 CC3 CC4 CC5 lw $10, 20($1) sub $11, $2, $3 PC Program instructions add $12, $3, $4 PC lw $13, 24($1) PC add $14, $5, $6 PC ELEC 5200-001/6200-001 Lecture 6

  32. CC5 MEM: sub $11, $2, $3 WB: lw $10, 20($1) IF: add $14, $5, $6 ID: lw $13, 24($1) EX: add $12, $3, $4 IF/ID ID/EX EX/MEM MEM/WB 4 1 mux 0 Add ALU opcode Shift left 2 26-31 zero 21-25 Instr mem ALU 16-20 Data mem. Reg. File PC 1 mux 0 0 mux 1 Sign ext. 11-15 for R-type 16-20 for I-type lw 0-15 ELEC 5200-001/6200-001 Lecture 6

  33. Advantages of Pipeline • After the fifth cycle (CC5), one instruction is completed each cycle; CPI ≈ 1, neglecting the initial pipeline latency of 5 cycles. • Pipeline latency is defined as the number of stages in the pipeline, or • The number of clock cycles after which the first instruction is completed. • The clock cycle time is about four times shorter than that of single-cycle datapath and about the same as that of multicycle datapath. • For multicycle datapath, CPI = 3. …. • So, pipelined execution is faster, but . . . ELEC 5200-001/6200-001 Lecture 6

  34. Science is always wrong. It never solves a problem without creating ten more. George Bernard Shaw ELEC 5200-001/6200-001 Lecture 6

  35. Pipeline Hazards • Definition: Hazard in a pipeline is a situation in which the next instruction cannot complete execution one clock cycle after completion of the presentinstruction. • Three types of hazards: • Structural hazard (resource conflict) • Data hazard • Control hazard ELEC 5200-001/6200-001 Lecture 6

  36. Structural Hazard • Two instructions cannot execute due to a resource conflict. • Example: Consider a computer with a common data and instruction memory. The fourth cycle of a lw instruction requires memory access (memory read) and at the same time the first cycle of the fourth instruction requires instruction fetch (memory read). This will cause a memory resource conflict. ELEC 5200-001/6200-001 Lecture 6

  37. Example of Structural Hazard time CC1 CC2 CC3 CC4 CC5 IM/DM ALU REG. FILE WRITE IM/DM ID, REG. FILE READ lw $10, 20($1) MEM/WB ID/EX EX/MEM IF/ID IM/DM ID, REG. FILE READ ALU REG. FILE WRITE IM/DM MEM/WB PC IF/ID ID/EX EX/MEM sub $11, $2, $3 Common data and instr. Mem. IM/DM ID, REG. FILE READ ALU IM/DM REG. FILE WRITE MEM/WB PC ID/EX IF/ID EX/MEM add $12, $3, $4 Program instructions IM/DM ID, REG. FILE READ ALU IM/DM REG. FILE WRITE MEM/WB PC IF/ID ID/EX EX/MEM lw $13, 24($1) Nedded by two instructions ELEC 5200-001/6200-001 Lecture 6

  38. Possible Remedies for Structural Hazards • Provide duplicate hardware resources in datapath. • Control unit or compiler can insert delays (no-op cycles) between instructions. This is known as pipeline stall or bubble. ELEC 5200-001/6200-001 Lecture 6

  39. Stall (Bubble) for Structural Hazard time CC1 CC2 CC3 CC4 CC5 IM/DM ID, REG. FILE READ IM/DM ALU REG. FILE WRITE lw $10, 20($1) MEM/WB ID/EX IF/ID EX/MEM IM/DM ALU IM/DM ID, REG. FILE READ REG. FILE WRITE sub $11, $2, $3 MEM/WB PC ID/EX IF/ID EX/MEM IM/DM IM/DM ID, REG. FILE READ ALU REG. FILE WRITE add $12, $3, $4 MEM/WB PC IF/ID ID/EX EX/MEM Program instructions Stall (bubble) IM/DM ID, REG. FILE READ ALU IM/DM REG. FILE WRITE lw $13, 24($1) MEM/WB PC EX/MEM IF/ID ID/EX ELEC 5200-001/6200-001 Lecture 6

  40. Data Hazard • Data hazard means that an instruction cannot be completed because the needed data, being generated by another instruction in the pipeline, is not available. • Example: consider two instructions: • add $s0, $t0, $t1 • sub $t2, $s0, $t3 # needs $s0 ELEC 5200-001/6200-001 Lecture 6

  41. Example of Data Hazard time CC1 CC2 CC3 CC4 CC5 Write s0 in CC5 DM ID, REG. FILE READ ALU REG. FILE WRITE IM add $s0, $t0, $t1 PC ID/EX EX/MEM MEM/WB IF/ID DM REG. FILE WRITE ID, REG. FILE READ IM ALU sub $t2, $s0, $t3 MEM/WB PC IF/ID ID/EX EX/MEM Read s0 and t3 in CC3 We need to read s0 from reg file in cycle 3 But s0 will not be written in reg file until cycle 5 However, s0 will only be used in cycle 4 And it is available at the end of cycle 3 Program instructions ELEC 5200-001/6200-001 Lecture 6

  42. Forwarding or Bypassing • Output of a resource used by an instruction is forwarded to the input of some resource being used by another instruction. • Forwarding can eliminate some, but not all, data hazards. ELEC 5200-001/6200-001 Lecture 6

  43. Forwarding for Data Hazard time CC1 CC2 CC3 CC4 CC5 Write s0 in CC5 DM ID, REG. FILE READ REG. FILE WRITE ALU IM add $s0, $t0, $t1 MEM/WB PC ID/EX EX/MEM IF/ID Forwarding DM ID, REG. FILE READ ALU IM REG. FILE WRITE sub $t2, $s0, $t3 IF/ID ID/EX EX/MEM PC MEM/WB Read s0 and t3 in CC3 Program instructions ELEC 5200-001/6200-001 Lecture 6

  44. Forwarding Unit Hardware ID/EX EX/MEM MEM/WB FORW. MUX ALU Data Mem. MUX Data to reg. file FORW. MUX Destination registers Source reg. IDs from opcode Forwarding Unit ELEC 5200-001/6200-001 Lecture 6

  45. Forwarding Alone May Not Work time CC1 CC2 CC3 CC4 CC5 Write s0 in CC5 DM ID, REG. FILE READ ALU REG. FILE WRITE IM lw $s0, 20($s1) PC MEM/WB EX/MEM ID/EX IF/ID DM ID, REG. FILE READ ALU REG. FILE WRITE IM sub $t2, $s0, $t3 MEM/WB PC IF/ID ID/EX EX/MEM Read s0 and t3 in CC3 Program instructions data needed by sub (data hazard) data available from memory only at the end of cycle 4 ELEC 5200-001/6200-001 Lecture 6

  46. Use Bubble and Forwarding time CC1 CC2 CC3 CC4 CC5 Write s0 in CC5 DM ALU REG. FILE WRITE ID, REG. FILE READ IM lw $s0, 20($s1) PC ID/EX MEM/WB EX/MEM IF/ID stall (bubble) Forwarding DM ALU IM REG. FILE WRITE ID, REG. FILE READ PC IF/ID ID/EX EX/MEM MEM/WB sub $t2, $s0, $t3 Program instructions ELEC 5200-001/6200-001 Lecture 6

  47. Hazard Detection Unit Hardware Disable write Hazard Detection Unit ID/EX EX/MEM MEM/WB Control FORW. MUX ALU NOP MUX Data Mem. 0 PC FORW. MUX Instruction to reg. file IF/ID Control signals Forwarding Unit register IDs from prev. instr. ELEC 5200-001/6200-001 Lecture 6

  48. Resolving Hazards • Hazards are resolved by Hazard detection and forwarding units. • Compiler’s understanding of how these units work can improve performance. ELEC 5200-001/6200-001 Lecture 6

  49. Avoiding Stall by Code Reorder C code: A = B + E; C = B + F; MIPS code: lw $t1, 0($t0) . $t1 written lw $t2, 4($t0) . . $t2 written add $t3, $t1, $t2 . . . $t1, $t2 needed sw $t3, 12($t0) . . . . lw $t4, 8($t0) . . . . . $t4 written add $t5, $t1, $t4 . . . . . $t4 needed sw $t5, 16($t0) . . . . . . . . . . . . . . . ELEC 5200-001/6200-001 Lecture 6

  50. Reordered Code C code: A = B + E; C = B + F; MIPS code: lw $t1, 0($t0) lw $t2, 4($t0) lw $t4, 8($t0) add $t3, $t1, $t2no hazard sw $t3, 12($t0) add $t5, $t1, $t4no hazard sw $t5, 16($t0) ELEC 5200-001/6200-001 Lecture 6

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