Lecture 13 ? Today?s lecture: ? What about branches? ? Crystal ball ? Look at performance again 1 2 Adding hazard detection to the CPU IF/I D W rite Rs 0 1 Addr Instruction memory Instr Address Write data Data memory Read data 1 0 PC Extend ALUSrc Result Zero ALU Instr [15 - 0] RegDst Read register 1 Read register 2 Write register Write data Read data 2 Read data 1 Registers Rd Rt 0 1 IF/ID ID/EX EX/MEM MEM/WB EX M WB Control M WB WB Rs 0 1 2 0 1 2 Forwarding Unit EX/MEM.RegisterRd MEM/WB.RegisterRd Hazard Unit 0 1 0 ID/EX.MemRead PC W rite Rt ID/EX.RegisterRt 3 The hazard detection unit ? The hazard detection unit?s inputs are as follows. ? IF/ID.RegisterRs and IF/ID.RegisterRt, the source registers for the current instruction. ? ID/EX.MemRead and ID/EX.RegisterRt, to determine if the previous instruction is LW and, if so, which register it will write to. ? By inspecting these values, the detection unit generates three outputs. ? Two new control signals PCWrite and IF/ID Write, which determine whether the pipeline stalls or continues. ? A mux select for a new multiplexer, which forces control signals for the current EX and future MEM/WB stages to 0 in case of a stall. 4 Generalizing Forwarding/Stalling ? What if data memory access was so slow, we wanted to pipeline it over 2 cycles? ? How many bypass inputs would the muxes in EXE have? ? Which instructions in the following require stalling and/or bypassing? lw r13, 0(r11) add r7, r8, r9 add r15, r7, r13 Clock cycle 1 2 3 4 5 6 DMRegIM Reg 5 Branches in the original pipelined datapath Read address Instruction memory Instruction [31-0] Address Write data Data memory Read data MemWrite MemRead 1 0 MemToReg 4 Shift left 2 P C Add 1 0 PCSrc Sign extend ALUSrc Result ZeroALU ALUOp Instr [15 - 0] RegDst Read register 1 Read register 2 Write register Write data Read data 2 Read data 1 Registers RegWrite Add Instr [15 - 11] Instr [20 - 16] 0 1 0 1 IF/ID ID/EX EX/MEM MEM/WB EX M WB Control M WB WB When are they resolved?here 6 Branches ? Most of the work for a branch computation is done in the EX stage. ? The branch target address is computed. ? The source registers are compared by the ALU, and the Zero flag is set or cleared accordingly. ? Thus, the branch decision cannot be made until the end of the EX stage. ? But we need to know which instruction to fetch next, in order to keep the pipeline running! ? This leads to what?s called a control hazard. DMReg RegIMbeq $2, $3, Label ? ? ? IM Clock cycle 1 2 3 4 5 6 7 8 7 Stalling is one solution ? Again, stalling is always one possible solution. ? Here we just stall until cycle 4, after we do make the branch decision. DMReg RegIMbeq $2, $3, Label ? ? ? DMReg RegIM Clock cycle 1 2 3 4 5 6 7 8 IM 8 Branch prediction ? Another approach is to guess whether or not the branch is taken. ? In terms of hardware, it?s easier to assume the branch is not taken. ? This way we just increment the PC and continue execution, as for normal instructions. ? If we?re correct, then there is no problem and the pipeline keeps going at full speed. DMReg RegIMbeq $2, $3, Label next instruction 1 next instruction 2 DMReg RegIM Clock cycle 1 2 3 4 5 6 7 DMReg RegIM How would you guess? 9 Branch misprediction ? If our guess is wrong, then we would have already started executing two instructions incorrectly. We?ll have to discard, or flush, those instructions and begin executing the right ones from the branch target address, Label. DMReg RegIMbeq $2, $3, Label next instruction 1 next instruction 2 Label: . . . RegIM Clock cycle 1 2 3 4 5 6 7 8 IM DMReg RegIM flush flush 10 Performance gains and losses ? Overall, branch prediction is worth it. ? Mispredicting a branch means that two clock cycles are wasted. ? But if our predictions are even just occasionally correct, then this is preferable to stalling and wasting two cycles for every branch. ? All modern CPUs use branch prediction. ? Accurate predictions are important for optimal performance. ? Most CPUs predict branches dynamically?statistics are kept at run-time to determine the likelihood of a branch being taken. ? The pipeline structure also has a big impact on branch prediction. ? A longer pipeline may require more instructions to be flushed for a misprediction, resulting in more wasted time and lower performance. ? We must also be careful that instructions do not modify registers or memory before they get flushed. 11 Implementing branches ? We can actually decide the branch a little earlier, in ID instead of EX. ? Our sample instruction set has only a BEQ. ? We can add a small comparison circuit to the ID stage, after the source registers are read. ? Then we would only need to flush one instruction on a misprediction. DMReg RegIMbeq $2, $3, Label next instruction 1 Label: . . . IM Clock cycle 1 2 3 4 5 6 7 DMReg RegIM flush 12 Implementing flushes ? We must flush one instruction (in its IF stage) if the previous instruction is BEQ and its two source registers are equal. ? We can flush an instruction from the IF stage by replacing it in the IF/ID pipeline register with a harmless nop instruction. ? MIPS uses sll $0, $0, 0 as the nop instruction. ? This happens to have a binary encoding of all 0s: 0000 .... 0000. ? Flushing introduces a bubble into the pipeline, which represents the one- cycle delay in taking the branch. ? The IF.Flush control signal shown on the next page implements this idea, but no details are shown in the diagram. 13 Branching without forwarding and load stalls 0 1 Addr Instruction memory Instr Address Write data Data memory Read data 1 0 Extend ALUSrc Result Zero ALU RegDst Read register 1 Read register 2 Write register Write data Read data 2 Read data 1 Registers Rd Rt 0 1 IF/ID ID/EX EX/MEM MEM/WB EX M WB Control M WB WB = Add Shift left 2 4 P C 1 0 PCSrc IF.Flush 14 Timing ? If no prediction: IF ID EX MEM WB IF IF ID EX MEM WB --- lost 1 cycle ? If prediction: ? If Correct IF ID EX MEM WB IF ID EX MEM WB -- no cycle lost ? If Misprediction: IF ID EX MEM WB IF0 IF1 ID EX MEM WB --- 1 cycle lost Assuming extra comparison. 15 Summary of Pipeline Hazards ? Three kinds of hazards conspire to make pipelining difficult. ? Structural hazards result from not having enough hardware available to execute multiple instructions simultaneously. ? These are avoided by adding more functional units (e.g., more adders or memories) or by redesigning the pipeline stages. ? Data hazards can occur when instructions need to access registers that haven?t been updated yet. ? Hazards from R-type instructions can be avoided with forwarding. ? Loads can result in a ?true? hazard, which must stall the pipeline. ? Control hazards arise when the CPU cannot determine which instruction to fetch next. ? We can minimize delays by doing branch tests earlier in the pipeline. ? We can also take a chance and predict the branch direction, to make the most of a bad situation. 16 Performance ? Now we?ll discuss issues related to performance: ? Latency/Response Time/Execution Time vs. Throughput ? How do you make a reasonable performance comparison? ? The 3 components of CPU performance ? The 2 laws of performance 17 Why know about performance ? Purchasing Perspective: ? Given a collection of machines, which has the ? Best Performance? ? Lowest Price? ? Best Performance/Price? ? Design Perspective: ? Faced with design options, which has the ? Best Performance Improvement? ? Lowest Cost? ? Best Performance/Cost ? ? Both require ? Basis for comparison ? Metric for evaluation 18 Many possible definitions of performance ? Every computer vendor will select one that makes them look good. How do you make sense of conflicting claims? Q: Why do end users need a new performance metric? A: End users who rely only on megahertz as an indicator for performance do not have a complete picture of PC processor performance and may pay the price of missed expectations. 19 Two notions of performance ? Which has higher performance? ? Depends on the metric ? Time to do the task (Execution Time, Latency, Response Time) ? Tasks per unit time (Throughput, Bandwidth) ? Response time and throughput are often in opposition Plane DC to Paris Speed Passengers Throughput (pmph) 747 6.5 hours 610 mph 470 286,700 Concorde 3 hours 1350 mph 132 178,200 20 Some Definitions ? Performance is in units of things/unit time ? E.g., Hamburgers/hour ? Bigger is better ? If we are primarily concerned with response time ? Performance(x) = 1 execution_time(x) ? Relative performance: ?X is N times faster than Y? N =Performance(X) = execution_time(Y) Performance(Y) execution_time(X) 21 Some Examples ? Time of Concorde vs. 747? ? Throughput of Concorde vs. 747? Plane DC to Paris Speed Passengers Throughput (pmph) 747 6.5 hours 610 mph 470 286,700 Concorde 3 hours 1350 mph 132 178,200 22 ? When comparing systems, need to fix the workload ? Which workload? Basis of Comparison Workload Pros Cons Actual Target Workload Representative Very specific Non-portable Difficult to run/measure Full Application Benchmarks Portable Widely used Realistic Less representative Small ?Kernel? or ?Synthetic? Benchmarks Easy to run Useful early in design Easy to ?fool? Microbenchmarks Identify peak capability and potential bottlenecks Real application performance may be much below peak 23 Benchmarking ? Some common benchmarks include: ? Adobe Photoshop for image processing ? BAPCo SYSmark for office applications ? Unreal Tournament 2003 for 3D games ? SPEC2000 for CPU performance ? The best way to see how a system performs for a variety of programs is to just show the execution times of all of the programs. ? Here are execution times for several different Photoshop 5.5 tasks, from http://www.tech-report.com 24 Summarizing performance ? Summarizing performance with a single number can be misleading?just like summarizing four years of school with a single GPA! ? If you must have a single number, you could sum the execution times. This example graph displays the total execution time of the individual tests from the previous page. ? A similar option is to find the average of all the execution times. For example, the 800MHz Pentium III (in yellow) needed 227.3 seconds to run 21 programs, so its average execution time is 227.3/21 = 10.82 seconds. ? A weighted sum or average is also possible, and lets you emphasize some benchmarks more than others. 25 The components of execution time ? Execution time can be divided into two parts. ? User time is spent running the application program itself. ? System time is when the application calls operating system code. ? The distinction between user and system time is not always clear, especially under different operating systems. ? The Unix time command shows both. salary.125 > time distill 05-examples.ps Distilling 05-examples.ps (449,119 bytes) 10.8 seconds (0:11) 449,119 bytes PS => 94,999 bytes PDF (21%) 10.61u 0.98s 0:15.15 76.5% User time System time CPU usage = (User + System) / Total ?Wall clock? time (including other processes) 26 Three Components of CPU Performance Cycles Per Instruction CPU timeX,P = Instructions executedP * CPIX,P * Clock cycle timeX 27 ? Instructions executed: ? We are not interested in the static instruction count, or how many lines of code are in a program. ? Instead we care about the dynamic instruction count, or how many instructions are actually executed when the program runs. ? There are three lines of code below, but the number of instructions executed would be 2001. li $a0, 1000 Ostrich: sub $a0, $a0, 1 bne $a0, $0, Ostrich Instructions Executed 28 ? The average number of clock cycles per instruction, or CPI, is a function of the machine and program. ? The CPI depends on the actual instructions appearing in the program? a floating-point intensive application might have a higher CPI than an integer-based program. ? It also depends on the CPU implementation. For example, a Pentium can execute the same instructions as an older 80486, but faster. ? In CS231, we assumed each instruction took one cycle, so we had CPI = 1. ? The CPI can be >1 due to memory stalls and slow instructions. ? The CPI can be <1 on machines that execute more than 1 instruction per cycle (superscalar). CPI 29 ? One ?cycle? is the minimum time it takes the CPU to do any work. ? The clock cycle time or clock period is just the length of a cycle. ? The clock rate, or frequency, is the reciprocal of the cycle time. ? Generally, a higher frequency is better. ? Some examples illustrate some typical frequencies. ? A 500MHz processor has a cycle time of 2ns. ? A 2GHz (2000MHz) CPU has a cycle time of just 0.5ns (500ps). Clock cycle time 30 CPU timeX,P = Instructions executedP * CPIX,P * Clock cycle timeX ? The easiest way to remember this is match up the units: ? Make things faster by making any component smaller!! ? Often easy to reduce one component by increasing another Execution time, again Seconds = Instructions * Clock cycles * Seconds Program Program Instructions Clock cycle Program Compiler ISA Organization Technology Instruction Executed CPI Clock Cycle TIme 31 ? Let?s compare the performances two 8086-based processors. ? An 800MHz AMD Duron, with a CPI of 1.2 for an MP3 compressor. ? A 1GHz Pentium III with a CPI of 1.5 for the same program. ? Compatible processors implement identical instruction sets and will use the same executable files, with the same number of instructions. ? But they implement the ISA differently, which leads to different CPIs. CPU timeAMD,P = InstructionsP * CPIAMD,P * Cycle timeAMD = = CPU timeP3,P = InstructionsP * CPIP3,P * Cycle timeP3 = = Example 1: ISA-compatible processors 32 Example 2: Comparing across ISAs ? Intel?s Itanium (IA-64) ISA is designed facilitate executing multiple instructions per cycle. If an Itanium processor achieves an average CPI of .3 (3 instructions per cycle), how much faster is it than a Pentium4 (which uses the x86 ISA) with an average CPI of 1? a) Itanium is three times faster b) Itanium is one third as fast c) Not enough information 33 Improving CPI ? Many processor design techniques we?ll see improve CPI ? Often they only improve CPI for certain types of instructions ? Fi = Fraction of instructions of type i CPI = CPI F where F = I i = 1 n i i i i Instruction Count ? First Law of Performance: Make the common case fast 34 Example: CPI improvements ? Base Machine: ? How much faster would the machine be if: ? we added a cache to reduce average load time to 3 cycles? ? we added a branch predictor to reduce branch time by 1 cycle? ? we could do two ALU operations in parallel? Op Type Freq (fi) Cycles CPIi ALU 50% 3 Load 20% 5 Store 10% 3 Branch 20% 2 35 ? Amdahl?s Law states that optimizations are limited in their effectiveness. ? For example, doubling the speed of floating-point operations sounds like a great idea. But if only 10% of the program execution time T involves floating-point code, then the overall performance improves by just 5%. ? What is the maximum speedup from improving floating point? Amdahl?s Law Execution time after improvement = Time affected by improvement + Time unaffected by improvementAmount of improvement Execution time after improvement = 0.10 T + 0.90 T = 0.95 T2 ? Second Law of Performance: Make the fast case common 36 ? Performance is one of the most important criteria in judging systems. ? There are two main measurements of performance. ? Execution time is what we?ll focus on. ? Throughput is important for servers and operating systems. ? Our main performance equation explains how performance depends on several factors related to both hardware and software. CPU timeX,P = Instructions executedP * CPIX,P * Clock cycle timeX ? It can be hard to measure these factors in real life, but this is a useful guide for comparing systems and designs. ? Amdahl?s Law tell us how much improvement we can expect from specific enhancements. ? The best benchmarks are real programs, which are more likely to reflect common instruction mixes. Summary Howard Huang Stalls and flushes CS232 @ UIUC
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