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1 Lecture 2: Measuring Performance/Cost/Power Today’s topics: (Sections 1.6, 1.4, 1.7, 1.8)  Quantitative principles of computer design  Measuring cost.

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Presentation on theme: "1 Lecture 2: Measuring Performance/Cost/Power Today’s topics: (Sections 1.6, 1.4, 1.7, 1.8)  Quantitative principles of computer design  Measuring cost."— Presentation transcript:

1 1 Lecture 2: Measuring Performance/Cost/Power Today’s topics: (Sections 1.6, 1.4, 1.7, 1.8)  Quantitative principles of computer design  Measuring cost  Real industrial examples

2 2 Summarizing Performance Recall discussion on AM versus GM GM: does not require a reference machine, but does not predict performance very well  So we multiplied execution times and determined that sys-A is 1.2x faster…but on what workload? AM: does predict performance for a specific workload, but that workload was determined by executing programs on a reference machine  Every year or so, the reference machine will have to be updated

3 3 Example We fixed a reference machine X and ran 4 programs A, B, C, D on it such that each program ran for 1 second The exact same workload (the four programs execute the same number of instructions that they did on machine X) is run on a new machine Y and the execution times for each program are 0.8, 1.1, 0.5, 2 With AM of normalized execution times, we can conclude that Y is 1.1 times slower than X – perhaps, not for all workloads, but definitely for one specific workload (where all programs run on the ref-machine for an equal #cycles) With GM, you may find inconsistencies

4 4 GM Example Computer-A Computer-B Computer-C P1 1 sec 10 secs 20 secs P2 1000 secs 100 secs 20 secs Conclusion with GMs: (i) A=B (ii) C is ~1.6 times faster For (i) to be true, P1 must occur 100 times for every occurrence of P2 With the above assumption, (ii) is no longer true Hence, GM can lead to inconsistencies

5 5 CPU Performance Equation CPU time = clock cycle time x cycles per instruction x number of instructions Influencing factors for each:  clock cycle time: technology and organization  CPI: organization and instruction set design  instruction count: instruction set design and compiler CPI (cycles per instruction) or IPC (instructions per cycle) can not be accurately estimated analytically

6 6 Measuring System CPI Assume that an architectural innovation only affects CPI For 3 programs, base CPIs: 1.2, 1.8, 2.5 CPIs for proposed model: 1.4, 1.9, 2.3 What is the best way to summarize performance with a single number? AM, HM, or GM of CPIs?

7 7 Example AM of CPI for base case = 1.2 cyc + 1.8 cyc + 2.5 cyc instr instr instr 5.5 cycles is execution time if each program ran for one instruction – therefore, AM of CPI defines a workload where every program runs for an equal #instrs HM of CPI = 1 / AM of IPC ; defines a workload where every program runs for an equal number of cycles GM of CPI: warm fuzzy number, not necessarily representing any workload

8 8 Amdahl’s Law Architecture design is very bottleneck-driven – make the common case fast, do not waste resources on a component that has little impact on overall performance/power Amdahl’s Law: performance improvements through an enhancement is limited by the fraction of time the enhancement comes into play Example: a web server spends 40% of time in the CPU and 60% of time doing I/O – a new processor that is ten times faster results in a 36% reduction in execution time (speedup of 1.56) – Amdahl’s Law states that maximum execution time reduction is 40% (max speedup of 1.66)

9 9 Principle of Locality Most programs are predictable in terms of instructions executed and data accessed The 90-10 Rule: a program spends 90% of its execution time in only 10% of the code Temporal locality: a program will shortly re-visit X Spatial locality: a program will shortly visit X+1

10 10 Exploit Parallelism Most operations do not depend on each other – hence, execute them in parallel At the circuit level, simultaneously access multiple ways of a set-associative cache At the organization level, execute multiple instructions at the same time At the system level, execute a different program while one is waiting on I/O

11 11 Factors Determining Cost Cost: amount spent by manufacturer to produce a finished good High volume  faster learning curve, increased manufacturing efficiency (10% lower cost if volume doubles), lower R&D cost per produced item Commodities: identical products sold by many vendors in large volumes (keyboards, DRAMs) – low cost because of high volume and competition among suppliers

12 12 Wafers and Dies An entire wafer is produced and chopped into dies that undergo testing and packaging

13 13 Integrated Circuit Cost Cost of an integrated circuit = (cost of die + cost of packaging and testing) / final test yield Cost of die = cost of wafer / (dies per wafer x die yield) Dies/wafer = wafer area / die area -  wafer diam / die diag Die yield = wafer yield x (1 + (defect rate x die area) /  ) -  Thus, die yield depends on die area and complexity arising from multiple manufacturing steps (  ~ 4.0)

14 14 Integrated Circuit Cost Examples A 30 cm diameter wafer cost $5-6K in 2001 Such a wafer yields about 366 good 1 cm 2 dies and 1014 good 0.49 cm 2 dies (note the effect of area and yield) Die sizes: Alpha 21264 1.15 cm 2, Itanium 3.0 cm 2, embedded processors are between 0.1 – 0.25 cm 2

15 15 Contribution of IC Costs to Total System Cost SubsystemFraction of total cost Cabinet: sheet metal, plastic, power supply, fans, cables, nuts, bolts, manuals, shipping box 6% Processor22% DRAM (128 MB)5% Video card5% Motherboard5% Processor board subtotal37% Keyboard and mouse3% Monitor19% Hard disk (20 GB)9% DVD drive6% I/O devices subtotal37% Software (OS + Office)20%

16 16 Cost and Price A $1000 increase in cost may result in a $3000 increase in price – hence, important to understand the relationship The relationship is complex – for example, a company may underprice a product that has heavy competition and overprice a product that has no competition

17 17 Computing Price Component costs: developing wafers, testing, packaging Direct costs: 10-30% of component costs: labor, warranty Gross margin (indirect costs): 10-45% of sum of these three (average selling price): R&D, marketing, sales, building rental, profits Retail mark-up: sum of all the above gives list price Low-end PCs may have low gross margins – low R&D, low cost for sales, low profits R&D costs are only 4-12%

18 18 Desktop Prices All systems have similar configurations – price variations due to expandability, expensive disks/memory/processor/OS, commoditization VendorModelProcessorClock speed MHz Price CompaqPresario 7000AMD Athlon1,400$2,091 DellPrecision 420Intel Pentium III1,000$3,834 DellPrecision 530Intel Pentium 41,700$4,175 HPWorkstation c3600PA 8600552$12,631 IBMRS6000 44P/170IBM III-2450$13,889 SunSunblade 100UltraSPARC II-e500$2,950 SunSunblade 1000UltraSPARC III750$9,950

19 19 Desktop Price-Performance (SPEC CPU2k)

20 20 Server Prices SystemCPUsPrice IBM xSeries 370 c/s280 Pentium III$15.5 M Compaq AlphaServer GS 32032 Alpha 21264$10.3 M Fujitsu PRIMEPOWER 2000048 SPARC64 GP$9.7 M IBM pSeries 680 7017-S8524 IBM RS64-IV$7.5 M HP 9000 Enterprise Server48 HP PA-RISC 8600$8.5 M IBM iSeries 400 840-242024 iSeries400 Model 840$8.4 M Dell PowerEdge 64003 Pentium III$131 K IBM xSeries 250 c/s4 Pentium III$297 K Compaq Proliant ML5704 Pentium III$375 K HP NetServer LH 60006 Pentium III$373 K NEC Express 5800/1808 Pentium III$683 K HP 9000/L20004 PA-RISC 8500$368 K

21 21 Server Performance Using the OLTP benchmark TPC-C

22 22 Server Price-Performance

23 23 Embedded Prices Does not include the prices and power of support chips High variance in functionality, price, performance, power The IBM and AMD processors are used in network switches and laptops, the NEC VR 5432 is used in laser printers, the NEC VR 4122 is used in PDAs ProcessorIssue rateTypical power (mW)Price AMD Elan SC52011600$38 AMD K6-2E+3+9600$78 IBM PowerPC 750CX46000$94 NEC VR 543222088$25 NEC VR 41221700$33

24 24 Embedded Price-Performance-Power

25 25 Title Bullet


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