CS252 Graduate Computer Architecture Lecture 2  0 Review of Instruction Sets, Pipelines, and Caches

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# CS252 Graduate Computer Architecture Lecture 2  0 Review of Instruction Sets, Pipelines, and Caches

## CS252 Graduate Computer Architecture Lecture 2  0 Review of Instruction Sets, Pipelines, and Caches

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1. CS252Graduate Computer ArchitectureLecture 20 Review of Instruction Sets, Pipelines, and Caches September 1, 2000 Prof. John Kubiatowicz

2. Review, #1 • Technology is changing rapidly: • Capacity Speed • Logic 2x in 3 years 2x in 3 years • DRAM 4x in 3 years 2x in 10 years • Disk 4x in 3 years 2x in 10 years • Processor ( n.a.) 2x in 1.5 years • What was true five years ago is not necessarily true now. • Execution time is the REAL measure of computer performance! • Not clock rate, not CPI • “X is n times faster than Y” means:

3. Review, #2 • Amdahl’s Law: (or Law of Diminishing Returns) • CPI Law: • The “End to End Argument” is what RISC was ultimately about -- it is the performance of the complete system that matters, not individual components! CPU time = Seconds = Instructions x Cycles x Seconds Program Program InstructionCycle

4. Today: Quick review of everything you should have learned0( A countably-infinite set of computer architecture concepts )

5. CPU time = Seconds = Instructions x Cycles x Seconds Program Program Instruction Cycle Aspects of CPU Performance Inst Count CPI Clock Rate Program X Compiler X (X) Inst. Set. X X Organization X X Technology X

6. Cycles Per Instruction(Throughput) “Average Cycles per Instruction” Invest Resources where time is Spent! • CPI = (CPU Time * Clock Rate) / Instruction Count • = Cycles / Instruction Count “Instruction Frequency”

7. Example: Calculating CPI Base Machine (Reg / Reg) Op Freq Cycles CPI(i) (% Time) ALU 50% 1 .5 (33%) Load 20% 2 .4 (27%) Store 10% 2 .2 (13%) Branch 20% 2 .4 (27%) 1.5 Typical Mix

8. Integrated Circuits Costs Die Cost goes roughly with die area4

9. Real World Examples Chip Metal Line Wafer Defect Area Dies/ Yield Die Cost layers width cost /cm2 mm2 wafer 386DX 2 0.90 \$900 1.0 43 360 71% \$4 486DX2 3 0.80 \$1200 1.0 81 181 54% \$12 PowerPC 601 4 0.80 \$1700 1.3 121 115 28% \$53 HP PA 7100 3 0.80 \$1300 1.0 196 66 27% \$73 DEC Alpha 3 0.70 \$1500 1.2 234 53 19% \$149 SuperSPARC 3 0.70 \$1700 1.6 256 48 13% \$272 Pentium 3 0.80 \$1500 1.5 296 40 9% \$417 • From "Estimating IC Manufacturing Costs,” by Linley Gwennap, Microprocessor Report, August 2, 1993, p. 15

10. 1 Alpha/ 0 Beta/ 1 1 0 0 0 106 0 1 0 1 0 Delta/ 2 0 1 2 2 1 Mod 3 1 1 Finite State Machines: • System state is explicit in representation • Transitions between states represented as arrows with inputs on arcs. • Output may be either part of state or on arcs “Mod 3 Machine” Input (MSB first) 1 1 1 0

11. 1/0 “Moore Machine” Latch Alpha/ 0 Beta/ 1 1/1 Combinational Logic 0/0 0/1 0/0 Delta/ 2 “Mealey Machine” 1/1 Implementation as Combinational logic + Latch

12. State w/ Address Instruction Branch Combinational Logic/ Controlled Machine 0: forw 35 xxx 1: b_no_obstacles 000 2: back 10 xxx 3: rotate 90 xxx 4: goto 001 Control ROM (Instructions) Addr Branch PC + 1 Next Address MUX Microprogrammed Controllers • State machine in which part of state is a “micro-pc”. • Explicit circuitry for incrementing or changing PC • Includes a ROM with “microinstructions”. • Controlled logic implements at least branches and jumps

13. A B C D Pipelining: Its Natural! • Laundry Example • Ann, Brian, Cathy, Dave each have one load of clothes to wash, dry, and fold • Washer takes 30 minutes • Dryer takes 40 minutes • “Folder” takes 20 minutes

14. A B C D Sequential Laundry 6 PM Midnight 7 8 9 11 10 Time • Sequential laundry takes 6 hours for 4 loads • If they learned pipelining, how long would laundry take? 30 40 20 30 40 20 30 40 20 30 40 20 T a s k O r d e r

15. 30 40 40 40 40 20 A B C D Pipelined LaundryStart work ASAP 6 PM Midnight 7 8 9 11 10 • Pipelined laundry takes 3.5 hours for 4 loads Time T a s k O r d e r

16. 30 40 40 40 40 20 A B C D Pipelining Lessons 6 PM 7 8 9 • Pipelining doesn’t help latency of single task, it helps throughput of entire workload • Pipeline rate limited by slowest pipeline stage • Multiple tasks operating simultaneously • Potential speedup = Number pipe stages • Unbalanced lengths of pipe stages reduces speedup • Time to “fill” pipeline and time to “drain” it reduces speedup Time T a s k O r d e r

17. Computer Pipelines • Execute billions of instructions, so throughput is what matters • DLX desirable features: all instructions same length, registers located in same place in instruction format, memory operands only in loads or stores

18. A "Typical" RISC • 32-bit fixed format instruction (3 formats) • 32 32-bit GPR (R0 contains zero, DP take pair) • 3-address, reg-reg arithmetic instruction • Single address mode for load/store: base + displacement • no indirection • Simple branch conditions • Delayed branch see: SPARC, MIPS, HP PA-Risc, DEC Alpha, IBM PowerPC, CDC 6600, CDC 7600, Cray-1, Cray-2, Cray-3

19. Example: MIPS (­ DLX) Register-Register 6 5 11 10 31 26 25 21 20 16 15 0 Op Rs1 Rs2 Rd Opx Register-Immediate 31 26 25 21 20 16 15 0 immediate Op Rs1 Rd Branch 31 26 25 21 20 16 15 0 immediate Op Rs1 Rs2/Opx Jump / Call 31 26 25 0 target Op

20. Adder 4 Address Inst ALU 5 Steps of DLX DatapathFigure 3.1, Page 130 Instruction Fetch Instr. Decode Reg. Fetch Execute Addr. Calc Memory Access Write Back Next PC MUX Next SEQ PC Zero? RS1 Reg File MUX RS2 Memory Data Memory L M D RD MUX MUX Sign Extend Imm WB Data

21. MEM/WB ID/EX EX/MEM IF/ID Adder 4 Address ALU 5 Steps of DLX DatapathFigure 3.4, Page 134 Instruction Fetch Execute Addr. Calc Memory Access Instr. Decode Reg. Fetch Write Back Next PC MUX Next SEQ PC Next SEQ PC Zero? RS1 Reg File MUX Memory RS2 Data Memory MUX MUX Sign Extend WB Data Imm RD RD RD • Data stationary control • local decode for each instruction phase / pipeline stage

22. Reg Reg Reg Reg Reg Reg Reg Reg Ifetch Ifetch Ifetch Ifetch DMem DMem DMem DMem ALU ALU ALU ALU Cycle 1 Cycle 2 Cycle 3 Cycle 4 Cycle 5 Cycle 6 Cycle 7 Visualizing PipeliningFigure 3.3, Page 133 Time (clock cycles) I n s t r. O r d e r

23. Its Not That Easy for Computers • 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).

24. Reg Reg Reg Reg Reg Reg Reg Reg Reg Reg Ifetch Ifetch Ifetch Ifetch DMem DMem DMem DMem ALU ALU ALU ALU ALU One Memory Port/Structural HazardsFigure 3.6, Page 142 Time (clock cycles) Cycle 1 Cycle 2 Cycle 3 Cycle 4 Cycle 5 Cycle 6 Cycle 7 I n s t r. O r d e r Load DMem Instr 1 Instr 2 Instr 3 Ifetch Instr 4

25. Reg Reg Reg Reg Reg Reg Reg Reg Ifetch Ifetch Ifetch Ifetch DMem DMem DMem ALU ALU ALU ALU Bubble Bubble Bubble Bubble Bubble One Memory Port/Structural HazardsFigure 3.7, Page 143 Time (clock cycles) Cycle 1 Cycle 2 Cycle 3 Cycle 4 Cycle 5 Cycle 6 Cycle 7 I n s t r. O r d e r Load DMem Instr 1 Instr 2 Stall Instr 3

26. Speed Up Equation for Pipelining For simple RISC pipeline, CPI = 1:

27. 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 SpeedUpA = Pipeline Depth/(1 + 0) x (clockunpipe/clockpipe) = Pipeline Depth SpeedUpB = Pipeline Depth/(1 + 0.4 x 1) x (clockunpipe/(clockunpipe / 1.05) = (Pipeline Depth/1.4) x 1.05 = 0.75 x Pipeline Depth SpeedUpA / SpeedUpB = Pipeline Depth/(0.75 x Pipeline Depth) = 1.33 • Machine A is 1.33 times faster

28. Reg Reg Reg Reg Reg Reg Reg Reg Reg Reg ALU ALU ALU ALU ALU Ifetch Ifetch Ifetch Ifetch Ifetch DMem DMem DMem DMem DMem EX WB MEM IF ID/RF 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 Data Hazard on R1Figure 3.9, page 147 Time (clock cycles)

29. Three Generic Data Hazards • Read After Write (RAW)InstrJ tries to read operand before InstrI writes it • Caused by a “Dependence” (in compiler nomenclature). This hazard results from an actual need for communication. I: add r1,r2,r3 J: sub r4,r1,r3

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

31. I: sub r1,r4,r3 J: add r1,r2,r3 K: mul r6,r1,r7 Three Generic Data Hazards • Write After Write (WAW)InstrJ writes operand before InstrI writes it. • Called an “output dependence” by compiler writersThis also results from the reuse of name “r1”. • Can’t happen in DLX 5 stage pipeline because: • All instructions take 5 stages, and • Writes are always in stage 5 • Will see WAR and WAW in later more complicated pipes

32. CS 252 Administrivia • Sign up today! Web site is: http://www.cs.berkeley.edu/~kubitron/cs252-F00 • In class exam on Wednesday Sept 6th • Improve 252 experience if recapture common background • Bring 1 sheet of paper with notes on both sides • Doesn’t affect grade, only admission into class • 2 grades: Admitted or audit/take CS 152 1st (before class Friday) • Review: Chapters 1- 3, CS 152 home page, maybe “Computer Organization and Design (COD)2/e” • If did take a class, be sure COD Chapters 2, 5, 6, 7 are familiar • Copies in Bechtel Library on 2-hour reserve • Mark Whitney will be holding a review session this Sunday night at 6:00 in 310 Soda.

33. CS 252 Administrivia • Resources for course on web site: • Check out the ISCA (International Symposium on Computer Architecture) 25th year retrospective on web site.Look for “Additional reading” below text-book description • Pointers to previous CS152 exams and resources • Lots of old CS252 material • Interesting pointers at bottom. Check out the:WWWComputer Architecture Home Page • To give proper attention to projects (as well as homeworks and quizes), I can handle ~36 students • First priority is students taking ARCH prelims in next year • Second priority is students taking this for breadth • Third priority is EECS students • Fourth priority College of Engineering grad students

34. Reg Reg Reg Reg Reg Reg Reg Reg Reg Reg ALU ALU ALU ALU ALU Ifetch Ifetch Ifetch Ifetch Ifetch DMem DMem DMem DMem DMem 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 Forwarding to Avoid Data HazardFigure 3.10, Page 149 Time (clock cycles)

35. ALU HW Change for ForwardingFigure 3.20, Page 161 ID/EX EX/MEM MEM/WR NextPC mux Registers Data Memory mux mux Immediate

36. Reg Reg Reg Reg Reg Reg Reg Reg ALU Ifetch Ifetch Ifetch Ifetch DMem DMem DMem DMem ALU ALU ALU lwr1, 0(r2) I n s t r. O r d e r sub r4,r1,r6 and r6,r1,r7 or r8,r1,r9 Data Hazard Even with ForwardingFigure 3.12, Page 153 Time (clock cycles)

37. Reg Reg Reg Ifetch Ifetch Ifetch Ifetch DMem ALU Bubble ALU ALU Reg Reg DMem DMem Bubble Reg Reg Data Hazard Even with ForwardingFigure 3.13, Page 154 Time (clock cycles) I n s t r. O r d e r lwr1, 0(r2) sub r4,r1,r6 and r6,r1,r7 Bubble ALU DMem or r8,r1,r9

38. Software Scheduling to Avoid Load Hazards 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 a,Ra LW Re,e LW Rf,f SUB Rd,Re,Rf SW d,Rd Fast code: LW Rb,b LW Rc,c LW Re,e ADD Ra,Rb,Rc LW Rf,f SW a,Ra SUB Rd,Re,Rf SW d,Rd

39. Reg Reg Reg Reg Reg Reg Reg Reg Reg Reg ALU ALU ALU ALU ALU Ifetch Ifetch Ifetch Ifetch Ifetch DMem DMem DMem DMem DMem 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 Control Hazard on BranchesThree Stage Stall

40. 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 • DLX branch tests if register = 0 or  0 • DLX 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

41. MEM/WB ID/EX EX/MEM IF/ID Adder 4 Address ALU Pipelined DLX DatapathFigure 3.22, page 163 Instruction Fetch Execute Addr. Calc Memory Access Instr. Decode Reg. Fetch Write Back Next SEQ PC Next PC MUX Adder Zero? RS1 Reg File Memory RS2 Data Memory MUX MUX Sign Extend WB Data Imm RD RD RD • Data stationary control • local decode for each instruction phase / pipeline stage

42. 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% DLX branches not taken on average • PC+4 already calculated, so use it to get next instruction #3: Predict Branch Taken • 53% DLX branches taken on average • But haven’t calculated branch target address in DLX • DLX still incurs 1 cycle branch penalty • Other machines: branch target known before outcome

43. Four Branch Hazard Alternatives #4: Delayed Branch • Define branch to take place AFTER a following instruction branch instruction sequential successor1 sequential successor2 ........ sequential successorn branch target if taken • 1 slot delay allows proper decision and branch target address in 5 stage pipeline • DLX uses this Branch delay of length n

44. Delayed Branch • Where to get instructions to fill branch delay slot? • Before branch instruction • From the target address: only valuable when branch taken • From fall through: only valuable when branch not taken • Canceling branches allow more slots to be filled • 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: 7-8 stage pipelines, multiple instructions issued per clock (superscalar)

45. Evaluating Branch Alternatives Scheduling Branch CPI speedup v. speedup v. scheme penalty unpipelined stall Stall pipeline 3 1.42 3.5 1.0 Predict taken 1 1.14 4.4 1.26 Predict not taken 1 1.09 4.5 1.29 Delayed branch 0.5 1.07 4.6 1.31 Conditional & Unconditional = 14%, 65% change PC

46. Now, Review of Memory Hierarchy

47. Recap: Who Cares About the Memory Hierarchy? Processor-DRAM Memory Gap (latency) µProc 60%/yr. (2X/1.5yr) 1000 CPU “Moore’s Law” 100 Processor-Memory Performance Gap:(grows 50% / year) Performance 10 DRAM 9%/yr. (2X/10 yrs) DRAM 1 1980 1981 1982 1983 1984 1985 1986 1987 1988 1989 1990 1991 1992 1993 1994 1995 1996 1997 1998 1999 2000 Time

48. Levels of the Memory Hierarchy Upper Level Capacity Access Time Cost Staging Xfer Unit faster CPU Registers 100s Bytes <10s ns Registers prog./compiler 1-8 bytes Instr. Operands Cache K Bytes 10-100 ns 1-0.1 cents/bit Cache cache cntl 8-128 bytes Blocks Main Memory M Bytes 200ns- 500ns \$.0001-.00001 cents /bit Memory OS 512-4K bytes Pages Disk G Bytes, 10 ms (10,000,000 ns) 10 - 10 cents/bit Disk -6 -5 user/operator Mbytes Files Larger Tape infinite sec-min 10 Tape Lower Level -8

49. The Principle of Locality • The Principle of Locality: • Program access a relatively small portion of the address space at any instant of time. • Two Different Types of Locality: • Temporal Locality (Locality in Time): If an item is referenced, it will tend to be referenced again soon (e.g., loops, reuse) • Spatial Locality (Locality in Space): If an item is referenced, items whose addresses are close by tend to be referenced soon (e.g., straightline code, array access) • Last 15 years, HW relied on locality for speed

50. Lower Level Memory Upper Level Memory To Processor Blk X From Processor Blk Y Memory Hierarchy: Terminology • Hit: data appears in some block in the upper level (example: Block X) • Hit Rate: the fraction of memory access found in the upper level • Hit Time: Time to access the upper level which consists of RAM access time + Time to determine hit/miss • Miss: data needs to be retrieve from a block in the lower level (Block Y) • Miss Rate = 1 - (Hit Rate) • Miss Penalty: Time to replace a block in the upper level + Time to deliver the block the processor • Hit Time << Miss Penalty (500 instructions on 21264!)