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Pipelined Datapath

Pipelined Datapath. Lecture notes from MKP, H. H. Lee and S. Yalamanchili. Reading. Sections 4.5 – 4.9. Pipeline Performance. Assume time for stages is 100ps for register read or write 200ps for other stages Compare pipelined datapath with single-cycle datapath. Pipeline Performance.

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Pipelined Datapath

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  1. Pipelined Datapath Lecture notes from MKP, H. H. Lee and S. Yalamanchili

  2. Reading • Sections 4.5 – 4.9

  3. Pipeline Performance • Assume time for stages is • 100ps for register read or write • 200ps for other stages • Compare pipelined datapath with single-cycle datapath

  4. Pipeline Performance Single-cycle (Tc= 800ps) Pipelined (Tc= 200ps)

  5. Pipeline Speedup • If all stages are balanced • i.e., all take the same time • If not balanced, speedup is less • Speedup due to increased throughput • Latency (time for each instruction) does not decrease

  6. Basic Idea All instructions are 32-bits Few & regular instruction formats Alignment of memory operands

  7. Pipelining • What makes it easy • All instructions are the same length • Simple instruction formats • Memory operands appear only in loads and stores • What makes it hard? • structural hazards: suppose we had only one memory • control hazards: need to worry about branch instructions • data hazards: an instruction depends on a previous instruction • What really makes it hard: • exception handling • trying to improve performance with out-of-order execution, etc.

  8. Pipeline registers • Need registers between stages • To hold information produced in previous cycle Pipeline stage execution time

  9. MEM IF ID EX WB add Graphically Representing Pipelines 2 4 6 8 10 Time lw MEM IF ID EX WB • Shading indicates the unit is being used by the instruction • Shading on the right half of the register file (ID or WB) or memory means the element is being read in that stage • Shading on the left half means the element is being written in that stage

  10. Graphically Representing Pipelines • Can help with answering questions like: • how many cycles does it take to execute this code? • what is the ALU doing during cycle 4? • use this representation to help understand datapaths T i m e ( i n c l o c k c y c l e s ) P r o g r a m C C 1 C C 2 C C 3 C C 4 C C 5 C C 6 e x e c u t i o n o r d e r ( i n i n s t r u c t i o n s ) l w $ 1 0 , 2 0 ( $ 1 ) I M R e g A L U D M R e g s u b $ 1 1 , $ 2 , $ 3 I M R e g D M R e g A L U

  11. MEM MEM MEM IF IF IF ID ID ID EX EX EX WB WB WB Structural Hazard 2 4 6 8 10 Time lw MEM IF ID EX WB add sub add Need to separate instruction and data memory

  12. IF for Load, Store, … Pipeline stage execution time

  13. ID for Load, Store, … Pipeline stage execution time

  14. EX for Load Pipeline stage execution time

  15. MEM for Load Pipeline stage execution time

  16. WB for Load Wrongregisternumber

  17. Corrected Datapath for Load Pipeline stage execution time

  18. EX for Store Pipeline stage execution time

  19. MEM for Store Pipeline stage execution time

  20. WB for Store Pipeline stage execution time

  21. Pipelining Example add $14, $5, $6 lw $13, 24($1) add $12, $3, $4 sub $11, $2, $3 lw $10, 20($1) Note what is happening in the register file 0 M u x 1 I F / I D I D / E X E X / M E M M E M / W B A d d A d d 4 A d d r e s u l t S h i f t l e f t 2 R e a d n o r e g i s t e r 1 i A d d r e s s P C t R e a d c u d a t a 1 r t R e a d s Z e r o n r e g i s t e r 2 I I n s t r u c t i o n R e g i s t e r s A L U R e a d A L U m e m o r y 0 R e a d W r i t e A d d r e s s d a t a 2 1 r e s u l t d a t a r e g i s t e r M M u D a t a u W r i t e x m e m o r y x d a t a 1 0 W r i t e d a t a 1 6 3 2 S i g n e x t e n d I n s t r u c t i o n 0 [ 2 0 – 1 6 ] M u I n s t r u c t i o n x [ 1 5 – 1 1 ] 1 Pipeline stage execution time R e g D s t

  22. Pipelined Control (Simplified)

  23. Pipelined Control • Control signals derived from instruction • As in single-cycle implementation • Pass control signals along like data

  24. Pipelined Control

  25. IF: lw $10, 9($1) P C S r c I D / E X 0 M W B u E X / M E M x 1 C o n t r o l M W B M E M / W B E X M W B I F / I D A d d A d d 4 A d d r e s u l t e t i r B r a n c h W S h i f t g e l e f t 2 e t i R r A L U S r c W m g R e a d n e e o i r e g i s t e r 1 M R A d d r e s s P C t R e a d c o t u d a t a 1 r m t R e a d s e Z e r o n r e g i s t e r 2 M I I n s t r u c t i o n R e g i s t e r s A L U R e a d A L U m e m o r y 0 R e a d W r i t e d a t a 2 A d d r e s s r e s u l t 1 d a t a r e g i s t e r M M D a t a u u m e m o r y W r i t e x x d a t a 1 0 W r i t e d a t a I n s t r u c t i o n 1 6 3 2 6 [ 1 5 – 0 ] S i g n A L U M e m R e a d e x t e n d c o n t r o l I n s t r u c t i o n [ 2 0 – 1 6 ] 0 A L U O p M u I n s t r u c t i o n x [ 1 5 – 1 1 ] 1 R e g D s t Datapath with Control

  26. IF: sub $11, $2, $3 ID: lw $10, 9($1) P C S r c I D / E X 0 11 M W B u E X / M E M “lw” x 010 1 C o n t r o l M W B M E M / W B 0001 E X M W B I F / I D A d d A d d 4 A d d r e s u l t e t i r B r a n c h W S h i f t g e l e f t 2 e t i R r A L U S r c W m g R e a d n e e o i r e g i s t e r 1 M R A d d r e s s P C t R e a d c o t u d a t a 1 r m t R e a d s e Z e r o n r e g i s t e r 2 M I I n s t r u c t i o n R e g i s t e r s A L U R e a d A L U m e m o r y 0 R e a d W r i t e d a t a 2 A d d r e s s r e s u l t 1 d a t a r e g i s t e r M M D a t a u u m e m o r y W r i t e x x d a t a 1 0 W r i t e d a t a I n s t r u c t i o n 1 6 3 2 6 [ 1 5 – 0 ] S i g n A L U M e m R e a d e x t e n d c o n t r o l I n s t r u c t i o n [ 2 0 – 1 6 ] 0 A L U O p M u I n s t r u c t i o n x [ 1 5 – 1 1 ] 1 R e g D s t Datapath with Control

  27. IF: and $12, $4, $5 ID: sub $11, $2, $3 EX: lw $10, 9($1) P C S r c I D / E X 0 11 10 M W B u E X / M E M “sub” x 010 000 1 C o n t r o l M W B 0 M E M / W B 1100 00 E X M W B I F / I D 1 A d d A d d 4 A d d r e s u l t e t i r B r a n c h W S h i f t g e l e f t 2 e t i R r A L U S r c W m g R e a d n e e o i r e g i s t e r 1 M R A d d r e s s P C t R e a d c o t u d a t a 1 r m t R e a d s e Z e r o n r e g i s t e r 2 M I I n s t r u c t i o n R e g i s t e r s A L U R e a d A L U m e m o r y 0 R e a d W r i t e d a t a 2 A d d r e s s r e s u l t 1 d a t a r e g i s t e r M M D a t a u u m e m o r y W r i t e x x d a t a 1 0 W r i t e d a t a I n s t r u c t i o n 1 6 3 2 6 [ 1 5 – 0 ] S i g n A L U M e m R e a d e x t e n d c o n t r o l I n s t r u c t i o n [ 2 0 – 1 6 ] 0 A L U O p M u I n s t r u c t i o n x [ 1 5 – 1 1 ] 1 R e g D s t Datapath with Control

  28. IF: or $13, $6, $7 ID: and $12, $4, $5 EX: sub $11, $2, $3 MEM: lw $10, 9($1) P C S r c I D / E X 0 10 10 M W B u E X / M E M “and” x 000 000 11 1 C o n t r o l M W B 0 1 M E M / W B 1100 1 10 E X M W B 0 I F / I D 0 A d d A d d 4 A d d r e s u l t e t i r B r a n c h W S h i f t g e l e f t 2 e t i R r A L U S r c W m g R e a d n e e o i r e g i s t e r 1 M R A d d r e s s P C t R e a d c o t u d a t a 1 r m t R e a d s e Z e r o n r e g i s t e r 2 M I I n s t r u c t i o n R e g i s t e r s A L U R e a d A L U m e m o r y 0 R e a d W r i t e d a t a 2 A d d r e s s r e s u l t 1 d a t a r e g i s t e r M M D a t a u u m e m o r y W r i t e x x d a t a 1 0 W r i t e d a t a I n s t r u c t i o n 1 6 3 2 6 [ 1 5 – 0 ] S i g n A L U M e m R e a d e x t e n d c o n t r o l I n s t r u c t i o n [ 2 0 – 1 6 ] 0 A L U O p M u I n s t r u c t i o n x [ 1 5 – 1 1 ] 1 R e g D s t Datapath with Control

  29. IF: add $14, $8, $9 ID: or $13, $6, $7 EX: and $12, $4, $5 MEM: sub $11, .. WB: lw $10, 9($1) P C S r c I D / E X 0 10 10 M W B u E X / M E M “or” x 000 000 10 1 C o n t r o l M W B 0 1 1 M E M / W B 1100 0 10 E X M W B 0 I F / I D 0 1 A d d A d d 4 A d d r e s u l t e t i r B r a n c h W S h i f t g e l e f t 2 e t i R r A L U S r c W m g R e a d n e e o i r e g i s t e r 1 M R A d d r e s s P C t R e a d c o t u d a t a 1 r m t R e a d s e Z e r o n r e g i s t e r 2 M I I n s t r u c t i o n R e g i s t e r s A L U R e a d A L U m e m o r y 0 R e a d W r i t e d a t a 2 A d d r e s s r e s u l t 1 d a t a r e g i s t e r M M D a t a u u m e m o r y W r i t e x x d a t a 1 0 W r i t e d a t a I n s t r u c t i o n 1 6 3 2 6 [ 1 5 – 0 ] S i g n A L U M e m R e a d e x t e n d c o n t r o l I n s t r u c t i o n [ 2 0 – 1 6 ] 0 A L U O p M u I n s t r u c t i o n x [ 1 5 – 1 1 ] 1 R e g D s t Datapath with Control

  30. IF: xxxx ID: add $14, $8, $9 EX: or $13, $6, $7 MEM: and $12… WB: sub $11, .. P C S r c I D / E X 0 10 10 M W B u E X / M E M “add” x 000 000 10 1 C o n t r o l M W B 0 1 1 M E M / W B 1100 0 10 E X M W B 0 I F / I D 0 0 A d d A d d 4 A d d r e s u l t e t i r B r a n c h W S h i f t g e l e f t 2 e t i R r A L U S r c W m g R e a d n e e o i r e g i s t e r 1 M R A d d r e s s P C t R e a d c o t u d a t a 1 r m t R e a d s e Z e r o n r e g i s t e r 2 M I I n s t r u c t i o n R e g i s t e r s A L U R e a d A L U m e m o r y 0 R e a d W r i t e d a t a 2 A d d r e s s r e s u l t 1 d a t a r e g i s t e r M M D a t a u u m e m o r y W r i t e x x d a t a 1 0 W r i t e d a t a I n s t r u c t i o n 1 6 3 2 6 [ 1 5 – 0 ] S i g n A L U M e m R e a d e x t e n d c o n t r o l I n s t r u c t i o n [ 2 0 – 1 6 ] 0 A L U O p M u I n s t r u c t i o n x [ 1 5 – 1 1 ] 1 R e g D s t Datapath with Control

  31. IF: xxxx ID: xxxx EX: add $14, $8, $9 MEM: or $13, .. WB: and $12… P C S r c 0 10 M I D / E X u E X / M E M x 000 10 W B 1 C o n t r o l W B 0 1 1 M E M / W B M 0 10 M W B 0 I F / I D 0 0 E X A d d A d d 4 A d d r e s u l t e t i r B r a n c h W S h i f t g e l e f t 2 e t i R r A L U S r c W m g R e a d n e e o i r e g i s t e r 1 M R A d d r e s s P C t R e a d c o t u d a t a 1 r m t R e a d s e Z e r o n r e g i s t e r 2 M I I n s t r u c t i o n R e g i s t e r s A L U R e a d A L U m e m o r y 0 R e a d W r i t e d a t a 2 A d d r e s s r e s u l t 1 d a t a r e g i s t e r M M D a t a u u m e m o r y W r i t e x x d a t a 1 0 W r i t e d a t a I n s t r u c t i o n 1 6 3 2 6 [ 1 5 – 0 ] S i g n A L U M e m R e a d e x t e n d c o n t r o l I n s t r u c t i o n [ 2 0 – 1 6 ] 0 A L U O p M u I n s t r u c t i o n x [ 1 5 – 1 1 ] 1 R e g D s t Datapath with Control

  32. IF: xxxx ID: xxxx EX: xxxx MEM: add $14, .. P C S r c I D / E X 0 M W B u E X / M E M x 10 1 M C o n t r o l W B 0 1 M E M / W B 0 E X M W B 0 I F / I D 0 A d d A d d 4 A d d r e s u l t e t i r B r a n c h W S h i f t g e l e f t 2 e t i R r A L U S r c W m g R e a d n e e o i r e g i s t e r 1 M R A d d r e s s P C t R e a d c o t u d a t a 1 r m t R e a d s e Z e r o n r e g i s t e r 2 M I I n s t r u c t i o n R e g i s t e r s A L U R e a d A L U m e m o r y 0 R e a d W r i t e d a t a 2 A d d r e s s r e s u l t 1 d a t a r e g i s t e r M M D a t a u u m e m o r y W r i t e x x d a t a 1 0 W r i t e d a t a I n s t r u c t i o n 1 6 3 2 6 [ 1 5 – 0 ] S i g n A L U M e m R e a d e x t e n d c o n t r o l I n s t r u c t i o n [ 2 0 – 1 6 ] 0 A L U O p M u I n s t r u c t i o n x [ 1 5 – 1 1 ] 1 R e g D s t Datapath with Control WB: or $13…

  33. P C S r c I D / E X 0 M W B u E X / M E M x 1 M C o n t r o l W B 1 M E M / W B E X M W B I F / I D 0 A d d A d d 4 A d d r e s u l t e t i r B r a n c h W S h i f t g e l e f t 2 e t i R r A L U S r c W m g R e a d n e e o i r e g i s t e r 1 M R A d d r e s s P C t R e a d c o t u d a t a 1 r m t R e a d s e Z e r o n r e g i s t e r 2 M I I n s t r u c t i o n R e g i s t e r s A L U R e a d A L U m e m o r y 0 R e a d W r i t e d a t a 2 A d d r e s s r e s u l t 1 d a t a r e g i s t e r M M D a t a u u m e m o r y W r i t e x x d a t a 1 0 W r i t e d a t a I n s t r u c t i o n 1 6 3 2 6 [ 1 5 – 0 ] S i g n A L U M e m R e a d e x t e n d c o n t r o l I n s t r u c t i o n [ 2 0 – 1 6 ] 0 A L U O p M u I n s t r u c t i o n x [ 1 5 – 1 1 ] 1 R e g D s t Datapath with Control IF: xxxx ID: xxxx EX: xxxx MEM: xxxx WB: add $14..

  34. Data Hazards • An instruction depends on completion of data access by a previous instruction • add $s0, $t0, $t1sub $t2, $s0, $t3

  35. Dependencies • Problem with starting next instruction before first is finished • dependencies that “go backward in time” are data hazards

  36. Software Solution • Have compiler guarantee no hazards • Where do we insert the “nops” ? sub $2, $1, $3 and $12, $2, $5 or $13, $6, $2 add $14, $2, $2 sw $15, 100($2) • Problem: this really slows us down!

  37. A Better Solution • Consider this sequence: sub $2, $1,$3and $12,$2,$5or $13,$6,$2add $14,$2,$2sw $15,100($2) • We can resolve hazards with forwarding • How do we detect when to forward?

  38. Dependencies & Forwarding Do not wait for results to be written to the register file – find them in the pipeline  forward to ALU

  39. IF: add $14, $8, $9 ID: or $13, $6, $7 EX: and$6, $4, $5 MEM: sub $11, .. WB: lw $10, 9($1) P C S r c I D / E X 0 10 10 M W B u E X / M E M “or” x 000 000 10 1 C o n t r o l M W B 0 1 1 M E M / W B 1100 0 10 E X M W B 0 I F / I D 0 1 A d d A d d 4 A d d r e s u l t e t i r B r a n c h W S h i f t g e l e f t 2 e t i R r A L U S r c W m g R e a d n e e o i r e g i s t e r 1 M R A d d r e s s P C t R e a d c o t u d a t a 1 r m t R e a d s e Z e r o n r e g i s t e r 2 M I I n s t r u c t i o n R e g i s t e r s A L U R e a d A L U m e m o r y 0 R e a d W r i t e d a t a 2 A d d r e s s r e s u l t 1 d a t a r e g i s t e r M M D a t a u u m e m o r y W r i t e x x d a t a 1 0 W r i t e d a t a I n s t r u c t i o n 1 6 3 2 6 [ 1 5 – 0 ] S i g n A L U M e m R e a d e x t e n d c o n t r o l I n s t r u c t i o n [ 2 0 – 1 6 ] 0 A L U O p M u I n s t r u c t i o n x [ 1 5 – 1 1 ] 1 R e g D s t Forwarding

  40. Forwarding (simplified) ID/EX EX/MEM MEM/WB Register File Data Memory ALU MUX

  41. MUX MUX Forwarding (from EX/MEM) ID/EX EX/MEM MEM/WB Register File Data Memory ALU MUX

  42. MUX MUX Forwarding (from MEM/WB) ID/EX EX/MEM MEM/WB Register File Data Memory ALU MUX

  43. MUX MUX Forwarding (operand selection) ID/EX EX/MEM MEM/WB Register File Data Memory ALU MUX Forwarding Unit

  44. MUX MUX MUX Forwarding (operand propagation) ID/EX EX/MEM MEM/WB Register File Data Memory ALU MUX Rd Rt EX/MEM Rd Forwarding Unit Rt Rs MEM/WB Rd Combinational Logic!

  45. Detecting the Need to Forward • Pass register numbers along pipeline • e.g., ID/EX.RegisterRs = register number for Rssitting in ID/EX pipeline register • ALU operand register numbers in EX stage are given by • ID/EX.RegisterRs, ID/EX.RegisterRt • Data hazards when 1a. EX/MEM.RegisterRd = ID/EX.RegisterRs 1b. EX/MEM.RegisterRd = ID/EX.RegisterRt 2a. MEM/WB.RegisterRd = ID/EX.RegisterRs 2b. MEM/WB.RegisterRd = ID/EX.RegisterRt Fwd fromEX/MEMpipeline reg Fwd fromMEM/WBpipeline reg

  46. Detecting the Need to Forward • But only if forwarding instruction will write to a register! • EX/MEM.RegWrite, MEM/WB.RegWrite • And only if Rd for that instruction is not $zero • EX/MEM.RegisterRd ≠ 0,MEM/WB.RegisterRd ≠ 0

  47. Forwarding Paths

  48. Forwarding Conditions • EX hazard • if (EX/MEM.RegWrite and (EX/MEM.RegisterRd ≠ 0) and (EX/MEM.RegisterRd = ID/EX.RegisterRs))ForwardA = 10 • if (EX/MEM.RegWrite and (EX/MEM.RegisterRd ≠ 0) and (EX/MEM.RegisterRd = ID/EX.RegisterRt))ForwardB = 10 • MEM hazard • if (MEM/WB.RegWrite and (MEM/WB.RegisterRd ≠ 0) and (MEM/WB.RegisterRd = ID/EX.RegisterRs))ForwardA = 01 • if (MEM/WB.RegWrite and (MEM/WB.RegisterRd ≠ 0) and (MEM/WB.RegisterRd = ID/EX.RegisterRt))ForwardB = 01

  49. Double Data Hazard • Consider the sequence: add $1,$1,$2add $1,$1,$3add $1,$1,$4 • Both hazards occur • Want to use the most recent • Revise MEM hazard condition • Only forward if EX hazard condition isn’t true

  50. Revised Forwarding Condition • MEM hazard • if (MEM/WB.RegWrite and (MEM/WB.RegisterRd ≠ 0)and not (EX/MEM.RegWrite and (EX/MEM.RegisterRd ≠ 0) and (EX/MEM.RegisterRd = ID/EX.RegisterRs)) and (MEM/WB.RegisterRd = ID/EX.RegisterRs))ForwardA = 01 • if (MEM/WB.RegWrite and (MEM/WB.RegisterRd ≠ 0)and not (EX/MEM.RegWrite and (EX/MEM.RegisterRd ≠ 0) and (EX/MEM.RegisterRd = ID/EX.RegisterRt)) and (MEM/WB.RegisterRd = ID/EX.RegisterRt))ForwardB = 01 • Checking precedence of EX hazard

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