1. Computer Architecture Lecture Notes Spring 2005 Dr. Michael P. Frank Competency Area 2: Performance Metrics Lecture 1

2. Performance Metrics Why is it necessary for us to study performance? — Performance is usually the key to the effectiveness of a system (hardware + software). — Performance is critical to customers (purchasers), thus, we as designers and architects must also make it a priority. — Performance must be assessed and understood in order for a system to communicate efficiently with peripheral devices.

3. Performance Metrics How can we determine performance? 0.017500 700 100605Car 0.04626,550 1,389 5313,20250Dash-8 0.145287,760 6,230 2,180119,501132Concorde 0.06354,516 2,406 70811,75077BAE-146-200 0.045100,440 4,442 83723,859120Airbus 340-300 0.039246,79612,493 869139,681284Airbus 340-300 0.032230,31010,548 85391,380270Boeing 767-300 0.048387,32010,734 920216,847421Boeing 747-400 CostThroughput Cruising Range Speed Fuel Capacity Passenger Capacity Aircraft Consider this example from the transportation industry:

4. Performance Example Fuel Capacity in liters Range in kilometers Speed in kilometers/hour Throughput is defined as (# of passengers) x (cruising speed) Cost is given as (fuel capacity) / (passengers x range) Which mode of transportation has the “best” performance?

5. Performance Example It depends on how we define performance. Consider raw speed: —Getting from one place to another quickly best worst

6. Performance Example What if we’re interested in the rate at which people are carried throughput: best worst

7. Performance Example Often times we relate performance and cost. Thus we can consider the amount of fuel used per passenger: Best plane Best overall

8. Performance Metrics Similar measures of performance are used for computers. —Number of computations done per unit of time —Cost of computations —Possibly several aspects of cost can be considered including initial purchase price, operating cost, cost of training users of system, etc. Common performance measures are 1.RESPONSE TIME – the amount of time it takes a program to complete (a.k.a execution time) 2.THROUGHPUT – the total amount of work done in a given amount of time

9. Performance Metrics Example: Given the following actions: 1. Replacing processor with a faster version 2. Adding additional processors to perform separate tasks in a multiprocessor system do they (a) increase throughput, (a) decrease response time or (c) both?

10. Defining Performance Our focus will be primarily on execution time. To maximize performance implies a minimization in execution time: For two machines: We say that machine Y is faster than machine X.

11. Performance Metrics Notes: (1) If X is n times faster than Y, then (2)To avoid confusion, we’ll use the following terminology: We sayWe mean “improve performance”  increase performance “improve execution time”  decrease execution time

12. Performance Example If machine A runs a program in 10 seconds and machine B runs the same program in 15 seconds, how much faster is A than B?

13. Performance Example If machine A runs a program in 10 seconds and machine B runs the same program in 15 seconds, how much faster is A than B?

14. Measuring Performance Quite simply, TIME is the measure of computer performance! The most straightforward definition of time in wall-clock time  elapsed time  response time. Total time to complete a task including system overhead activities such as Input/Output tasks, disk and memory accesses, etc.

15. Measuring Performance CPU Time is the time it takes to complete a task excluding the time it takes for I/O waits. USER CPU TIME The time CPU is busy executing the user’s code. CPU TIME SYSTEM CPU TIME The time CPU spends performing operating system tasks. Note:Sometimes system and user CPU times are difficult to distinguish since it is hard to assign responsibility for OS activities.

16. Measuring Performance Example, To understand the concept of CPUTime, consider the UNIX command ‘time’. Once typed, it may return a response similar to 90.7u 12.9s 2:39 65% What do these numbers mean?

17. Measuring Performance Example, To understand the concept of CPUTime, consider the UNIX command ‘time’. Once typed, it may return a response similar to 90.7u 12.9s 2:39 65% User CPU Time % of elapsed time that is CPU time System CPU Time Elapsed Time

18. Measuring Performance Example, To understand the concept of CPUTime, consider the UNIX command ‘time’. Once typed, it may return a response similar to 90.7u 12.9s 2:39 65% a.What is the total CPUTime? b.Percentage of time spent on I/O and other programs?

19. Measuring Performance Example, To understand the concept of CPUTime, consider the UNIX command ‘time’. Once typed, it may return a response similar to 90.7u 12.9s 2:39 65% a.What is the total CPUTime? b.Percentage of time spent on I/O and other programs?

20. Measuring Performance Other notes: 1.SYSTEM PERFORMANCE – reciprocal of elapsed time on an unloaded system (e.g. no user applications) 2.CPU PERFORMANCE – recip. of user CPU time 3.CLOCK CYCLES (CC) – discrete time intervals measured by the processor clock running at a constant rate. 4.CLOCK PERIOD – time it takes to complete a clock cycle 5.CLOCK RATE – inverse of clock period

21. Measuring Performance Consider CPU performance: Also,

22. Measuring Performance Since the execution time clearly depends on the number of instructions for a program, we must also define another performance metric: CPI = average number of clock cycles per instruction

23. Measuring Performance Now we have two more equations that we can define for CPUTime:

24. Measuring Performance In summary, performance metrics include: Components of Performance Units of Measure CPUTimeSeconds for program IC# of instructions for a program CPIAverage # of clock cycles per instructions t CC Seconds per clock cycle

25. Measuring Performance Example, Suppose Machine A implements the same ISA as Machine B. Given and for some program, and and for the same program, determine which machine is faster and by how much.

26. Breakdown by Instruction Category Recall CPI = Clock cycles (CC) per instruction But, CPI depends on many factors, including: —Memory system behavior —Processor structure —Availability special processor features –E.g., floating point, graphics, etc. To characterize the effect of changing specific aspects of the architecture, we find it helpful to break down CC into components due to different classes (categories) of instructions: —Where: –IC i = instruction count for class i –CPI i = avg. cycles for insts. in class i –n = the number of instruction classes

27. Example Suppose a processor has 3 categories of instructions A,B,C with the following CPIs: And, suppose a compiler designer is comparing two code sequences for a given program that have the following instruction counts: Determine: (i) Which code sequence executes the most instructions? (ii) Which will be faster? (iii) What is the average CPI for each code sequence? Instr. ClassCPI i A1 B2 C3 Code Seq. Inst. counts IC A IC B IC C 1212 2411

28. Solution to Example Part (i): — IC seq1 = 2 + 1 + 2 = 5 instructions — IC seq2 = 4 + 1 + 1 = 6 instructions —  Code sequence 2 executes more instructions Part (ii): — CC seq1 = ∑ i (CPI i xIC i ) = 1x2 + 2x1 + 3x2 = 10 cycles — CC seq2 = ∑ i (CPI i xIC i ) = 1x4 + 2x1 + 3x1 = 9 cycles —  Code sequence 2 takes fewer cycles  is faster! Part (iii): — CPI seq1 = {CC/IC} seq1 = 10 cyc./5 inst. = 2 — CPI seq2 = {CC/IC} seq2 = 9 cyc./6 inst. = 1.5 Which part should we consult to tell us which code sequence has better performance?

29. Importance of Benchmarks How do we evaluate and compare the performance of different architectures? —We use benchmarks  Programs that are specifically chosen to measure performance.  A workload is a set of programs.  Benchmarks consist of workloads that (user hopes) will predict the performance of the actual workload  It is important that benchmarks consist of realistic workloads  Not simple toy programs or code fragments  Manufacturers often try to fine-tune their machines to do well on popular benchmarks that were too simple  This does not always mean the machine will do well on real programs!

30. SPEC benchmark A popular source of benchmarks is SPEC —Standard Performance Evaluation Corporation General CPU benchmarks: CPU2000. —Includes programs such as: –gzip (compression), vpr (FPGA place & route), gcc (compiler), crafty (chess), vortex (database) SPEC also offers specialized benchmarks for: —Graphics, Parallel computing, Java, mail servers, network fileservers, web servers They publish reports on benchmark results for various systems. —Main metric: SPECRatio – Proportional to average inverse execution time. The bigger, the better! Reproducibility of results is very important!

31. Summarizing Performance How do we summarize performance in a way that accurately compares different machines? —One common approach: Total Execution Time (TET) –Based on: —Or, if the workload includes n different programs, we can calculate the average or Arithmetic Mean (AM): –Smaller AM  Improved performance —Other methods are also used: –Weighted arithmetic mean, geometric mean ratio.

32. Performance Improvement Recall the formula: CPUTime = IC × CPI / f cyc. —Thus, CPU performance is Perf = f / (IC ×CPI). Thus we can see 3 basic ways to improve CPU performance on a given task: —Increase clock frequency —Decrease CPI –by improved processor organization —Decrease instruction count –By compiler enhancement, –change in ISA design (new instructions), or –A more efficient application algorithm. However, we have to be careful! —Sometimes, improving one of these can hurt others!

33. Generalized Cost Measures In this course, we will often be focusing on ways to minimize execution time of programs. —Either CPU time, or number of clock cycles. Execution time is one example of what we may call a generalized cost measure (GCM). —A GCM is any property of a HW/SW design that tells us how much of some valued resource is used up when the system is manufactured or used. Other examples of important GCMs include: —Energy consumed by a computation —Silicon chip area used up by a circuit design —Dollar cost to manufacture a computer component We will study some general engineering principles that apply to the minimization of any GCM in any system.

34. Additive Cost Measures Let us suppose we have a GCM C for a system. Many times, the total cost C can be represented as a sum of independent cost components: —E.g., C = C 1 + C 2 + … + C n or. These could correspond to the resources used by individual subsystems of the whole system. —Or, used in doing particular categories of tasks. For example, execution time T can be broken down as the sum of time T fp taken by floating- point instructions and the time T oth for others. —That is, T = T fp + T oth.

35. Improving Part of a System Suppose a GCM is broken down as C = A + B. —The total cost is the sum of two components A & B. Now suppose you are considering making an improvement to the system design that affects only cost component B. —Suppose you reduce it by a factor f, to B′ = B/f. The new total cost is then C′ = A + B′. —The cost of component A is unaffected. Overall (total) cost has therefore been reduced by the factor:

36. Diminishing Returns Suppose we continue improving (reducing) a cost component by larger and larger factors. —Does this mean the system’s total cost will be reduced by correspondingly large factors?  NO! Even if we improved one cost component (B in our example) by a factor of f = ∞, note that: Even here, the overall cost reduction factor f overall would still be only the finite value 1+B/A! —The system can only be improved by at most this factor, if we improve just the one component B.

37. Diminishing Returns Example Suppose a particular chip contains B = 1 cm 2 of logic circuits, and A = 2 cm 2 of cache memory. —The total cost (in terms of area) is C = A+B = 3 cm 2. Now, let’s go crazy trying to simplify and shrink the design of just the logic circuit… —What is the maximum factor by which this tactic can reduce the area cost of the whole design (logic+memory)? Obviously, this can reduce the total area from 3 (cm 2 ) to no less than 2 (area of memory alone), —or, shrink it by a factor of f overall = 3/2 = 1.5. Note we could have obtained this same answer using the equation f overall,max = 1+B/A as well. Logic 1 cm 2 Memory 2 cm 2

38. Part/rest (initial) ( f ) (B/A) Graph Showing Diminishing Returns

39. Important Lessons to Take from This It’s probably not worth spending significant design time extensively improving just a single component of a system, —Unless that component accounts for a dominant part of the total cost (by some measure) to begin with. (B/A >> 1). It’s only worth improving a given component up to the point where it is no longer dominant. —Reducing it further won’t make a lot of difference. Therefore, all components with significant costs must be improved together in order to significantly improve an entire design. —Well-engineered systems will tend to have roughly comparable costs in all of their major components.

40. Other Ways to Calculate f overall Earlier, we saw this formula: —For the overall improvement factor f overall resulting from improving component B by the factor f. But, what if we don’t know the values of A and B? —What if we only know their relative sizes? –Fortunately, it turns out that we can still calculate f overall. Let us define frac enh = B/C = B/(A+B) to be the fraction of the original total system cost that is accounted for by the particular part B that is going to be enhanced. —Then, the fraction of cost accounted for by A (the rest of the system) is Our equation for f overall can then be reexpressed in terms of the quantities frac enh and 1−frac enh, as follows…

41. Calculating f overall in terms of frac enh Let’s re-express f overall in terms of frac enh : We will call this form for f overall the Generalized Amdahl’s Law. (We’ll see why in a moment.)

42. Amdahl’s Law Proper We saw that execution time is one valid cost measure. —In such a case, note that the factor by which a cost is reduced is the speedup, or the factor by which performance is improved. We thus rename the improvement factor f of B (the enhanced part) to speedup enh, and the overall improvement factor f overall becomes speedup overall, and we get: This is called Amdahl’s Law, and it is one of the most widely hyped quantitative principles of processor design. —But as we can see, it is not a special law of CPU architecture, but just an application of the universal engineering principle of diminishing returns which we discussed earlier.

43. Key Points from This Module Throughput vs. Response Time Performance as Inverse Execution Time Speedup Factors Averaging Benchmark Results CPU Performance Equation: —Execution time = IC × CPI × t cc —Performance = f cc / (IC × CPI) Amdahl’s Law: — C′ = A + B/f — Implies: C = Execution time after improvement B = Part of execution time affected by improvement f = Factor of improvement (speedup of enhanced part) A = Part of execution time unaffected by improvement

44. Example Performance Calculation Suppose program takes 10 secs. on computer A —And suppose computer A has a 4 GHz clock Want new computer B to run prg. in 6 seconds. —Suppose that increasing the clock speed is only possible with a substantial processor redesign, –which will result in 1.2 × as many clock cycles being needed to execute the program. What clock rate is needed? — Answer: 4 GHz × (10/6) × 1.2 = 8 GHz

45. Another Example Consider two different implementations of a given ISA, running a given benchmark: — Processor A has a cycle time of 250 ps –And a CPI of 2.0 — Processor B has a cycle time of 500 ps –And a CPI of 1.2 Which computer is faster on this benchmark, and by what factor? — Processor A takes 250 ps × 2.0 = 500 ps / instr. — Processor B takes 500 ps × 1.2 = 600 ps / instr. — Thus, A is faster by a factor of 6/5 = 1.2×.

46. Another example Suppose some Java application takes 15 seconds on a certain machine. A new Java compiler is released that requires only 0.6 as many dynamic instructions to run the application. — Unfortunately, it also increases the CPI by 1.1 × –Presumably, uses more multi-cycle instructions. How fast will the application run when compiled using the new compiler? —It will take 15 × 0.6 × 1.1 = 9.9 seconds to run —It will be 15/9.9 = 50/33 = 1.515…× faster –Only slightly more than 50% faster than before.

47. Another Example Consider the following measurements of execution time: Which of the following statements are true? — A is faster than B for program 1. — A is faster than B for program 2. — A is faster than B for a workload with equal numbers of executions of programs 1 and 2. — A is faster than B for a workload with twice as many executions of program 1 as of program 2. ProgramComputer AComputer B 12 sec.4 sec. 25 sec.2 sec.