CSE 520 Computer Architecture Lec 4 – Quantifying Cost, Energy- Consumption, Performance, and Dependability (Chapter 1) Sandeep K. S. Gupta School of Computing and Informatics Arizona State University Based on Slides by David Patterson and M. Younis

CSE 520 Fall Moore’s Law gets life-term extension “Intel, IBM unveil new chip technology - Breakthrough, using new material, will allow processors to become smaller and more powerful” – CNN Money –“Intel Corp. and IBM have announced one of the biggest advances in transistors in four decades, overcoming a frustrating obstacle by ensuring microchips can get even smaller and more powerful.” –“The latest breakthrough means Intel, IBM and others can proceed with technology roadmaps that call for the next generation of chips to be made with circuitry as small as 45 nanometers, about 1/2000th the width of a human hair.” –“Researchers are optimistic the new technology can be used at least through two more technology generations out, when circuitry will be just 22 nanometers. “ “This gives the entire chip industry a new life in terms of Moore’s Law, in all three of the big metrics – performance, power consumption, and transistor density,” – David Lammers, director, WeSRCH.com – social networking site for semiconductor enthusiasts (part of VLSI Research Inc.)

CSE 520 Fall What is the Breakthrough? “How to reduce energy loss in microchips transistors as the technology shrinks as the transistor shrinks to the atomic scale?” “The problem is that the silicon dioxide used for mote than 40 years as an insulator inside transistors has been shaved so thin that an increasing amount of current is seeping through, wasting electricity and generating unnecessary heat.” “[Intel and IBM] have discovered a way to replace [SiO2] with various metals [e.g Intel is using silvery metal called hafnium] in parts called the gate, which turns the transistor on and off, and the gate dielectric, and insulating layer, which helps improve transistor performance and retain more energy.”

CSE 520 Fall What does it mean for Intel and Arizona? “The chip to be used in [Intel’s] new “Penryn” microprocessor, will be produced in Intel facilities throughout the world. But the new [$3B] plant, called Fab 32, will allow Chandler to remain a key site for the company’s manufacturing operation.” – The Arizona Republic, “New Intel chip is fab news for Chandler”, M. Jarman, Jan. 28, “Intel says the new chip is a result of the biggest breakthrough in transistor technology in 40 years.” “It also ratchets up the competition between Intel and rival chipmaker Advanced Micro devices Inc., which helped IBM develop the technology along with electronics maker Sony Corp. and Toshiba Corp.” “Intel will be the first to have this in production, but IBM could potentially have a density advantage compared with Intel’s scheme. But both should get the gold medal.” Source: The Arizona Republic, Jan 28, 2007.

CSE 520 Fall Recap Execution (CPU) time is the only true measure of performance. One must be careful when using other measures such as MIPS. Computer architects (Industry) need to be aware of Technology trends to design computer architectures which address the various “walls”. Increasing proportion of Static (or leakage) current (in comparison to Dynamic current) is a cause of concern One of the motivation for multicore design is to reduce Thermal dissipation

CSE 520 Fall q A common theme in Hardware design is to make the common case fast q Increasing the clock rate would not affect memory access time q Using a floating point processing unit does not speed integer ALU operations Example: Floating point instructions improved to run 2 times faster; but only 10% of the actual instructions are floating point Exec-Time new = Exec-Time old x ( /2) = 0.95 x Exec-Time old Speedup overall = Exec-Time new / Exec-Time old = 1/0.95 = The performance enhancement possible with a given improvement is limited by the amount that the improved feature is used Amdahl’s Law Slide by M. Younis

CSE 520 Fall Processor Performance Equation Where: C i is the count of number of instructions of class i executed CPI i is the average number of cycles per instruction for that instruction class n is the number of different instruction classes Slide by M. Younis

CSE 520 Fall Performance Metrics - Summary è Maximizing performance means minimizing response (execution) time Compiler Programming Language Application Datapath Control TransistorsWiresPins IS A Function Units (millions) of Instructions per second: MIPS (millions) of (FP) operations per second: MFLOP/s Cycles per second (clock rate) Megabytes per second Operations per second Designer User * Figure is courtesy of Dave Patterson

CSE 520 Fall Chapter 1: Fundamentals of Computer Design Technology Trends: Culture of tracking, anticipating and exploiting advances in technology Understanding “Cost” Careful, quantitative comparisons: 1.Define, quantity, and summarize relative performance 2.Define and quantity relative cost 3.Define and quantity dependability 4.Define and quantity power

CSE 520 Fall Moore’s Law: 2X transistors / “year” “Cramming More Components onto Integrated Circuits” –Gordon Moore, Electronics, 1965 # on transistors / cost-effective integrated circuit double every N months (12 ≤ N ≤ 24)

CSE 520 Fall Latency Lags Bandwidth (last ~20 years) Performance Milestones Processor: ‘286, ‘386, ‘486, Pentium, Pentium Pro, Pentium 4 (21x,2250x) Ethernet: 10Mb, 100Mb, 1000Mb, Mb/s (16x,1000x) Memory Module: 16bit plain DRAM, Page Mode DRAM, 32b, 64b, SDRAM, DDR SDRAM (4x,120x) Disk : 3600, 5400, 7200, 10000, RPM (8x, 143x) CPU high, Memory low (“Memory Wall”)

CSE 520 Fall Rule of Thumb for Latency Lagging BW In the time that bandwidth doubles, latency improves by no more than a factor of 1.2 to 1.4 (and capacity improves faster than bandwidth) Stated alternatively: Bandwidth improves by more than the square of the improvement in Latency

CSE 520 Fall Reasons Latency Lags Bandwidth 1.Moore’s Law helps BW more than latency 2.Distance limits latency 3.Bandwidth easier to sell (“bigger=better”) 4.Latency helps BW, but not vice versa 5.Bandwidth hurts latency 6.Operating System overhead hurts Latency more than Bandwidth

CSE 520 Fall Summary of Technology Trends For disk, LAN, memory, and microprocessor, bandwidth improves by square of latency improvement –In the time that bandwidth doubles, latency improves by no more than 1.2X to 1.4X Lag probably even larger in real systems, as bandwidth gains multiplied by replicated components –Multiple processors in a cluster or even in a chip –Multiple disks in a disk array –Multiple memory modules in a large memory –Simultaneous communication in switched LAN HW and SW developers should innovate assuming Latency Lags Bandwidth –If everything improves at the same rate, then nothing really changes –When rates vary, require real innovation

CSE 520 Fall Chapter 1: Fundamentals of Computer Design Technology Trends: Culture of tracking, anticipating and exploiting advances in technology Understanding “Cost” Careful, quantitative comparisons: 1.Define, quantity, and summarize relative performance 2.Define and quantity relative cost 3.Define and quantity dependability 4.Define and quantity power

CSE 520 Fall Trends in Cost Textbooks usually ignore cost half of cost- performance because costs change. Yet understanding cost and its factors is essential for designer’s to make intelligent decisions about what features to include in designs when costs is an issue Agenda: Study impact of time, volume and commodification Underlying principle: learning curve – manufacturing costs decrease over time –Measured by change in yield – the percentage of manufactured devices that survives the testing procedure

CSE 520 Fall Integrated Circuits: Fueling Innovation Chip manufacturing begins with silicon, a substance found in sand Silicon does not conduct electricity well and thus called semiconductor A special chemical process can transform tiny areas of silicon to either: 1. Excellent conductors of electricity (like copper) 2. Excellent insulator from electricity (like glass) 3. Areas that can conduct or insulate under a special condition (a switch) A transistor is simply an on/off switch controlled by electricity Integrated circuits combines dozens of hundreds of transistors in a chip Advances of the IC technology affect H/W and S/W design philosophy

CSE 520 Fall Microelectronics Process Silicon ingot are 6-12 inches in diameter and about inches long The manufacturing process of integrated circuits is critical to the cost of a chip Impurities in the wafer can lead to defective devices and reduces the yield

CSE 520 Fall Integrated Circuits Costs Die cost roughly goes with die area 4 * Slide is courtesy of Dave Patterson

CSE 520 Fall Example: Dies per Wafer Find the number of dies per 300 mm (30 cm) wafer for a die that is 1.5 cm on a side’

CSE 520 Fall Example: Dies per Wafer Find the number of dies per 300 mm (30 cm) wafer for a die that is 1.5 cm on a side’ Die Area = 2.25 cm 2 Dies per wafer = = (  x (30/2) 2 )/2.25 – (  x 30)/  (2 x 2.25) = (706.9/2.25) – (94.2/2.12) = 270

CSE 520 Fall Example: Die Yield Find the die yield for dies that are 1.5 cm on a side and 1.0 cm on a side, assuming a defect density of 0.4 per cm 2 and α is 4.

CSE 520 Fall Example: Die Yield Find the die yield for dies that are 1.5 cm on a side and 1.0 cm on a side, assuming a defect density of 0.4 per cm 2 and α is 4. Dies areas: 2.25 cm 2 and 1.00 cm 2, respectively. For larger die, yield = (1 + (0.4x 2.25)/4.0) -4 = 0.44 For smaller die, yield = (1 + (0.4x 1)/4.0) -4 = 0.68 i.e. less than half of all the large dies are good, but more than two-thirds of the small dies are good.

CSE 520 Fall Real World Examples From "Estimating IC Manufacturing Costs,” by Linley Gwennap, Microprocessor Report, August 2, 1993, p. 15 * Slide is courtesy of Dave Patterson

CSE 520 Fall Costs and Trends in Cost  Understanding trends in component costs (how they will change over time) is an important issue for designers  Component prices drop over time without major improvements in manufacturing technology What affect cost Œ Learning curve:  The more experience in manufacturing a component, the better the yield (the number of good devices/ total number of devices)  In general, a chip, board or system with twice the yield will have half the cost  The learning curve is different for different components, thus complicating new system design decisions  Volume  Larger volume increases rate of learning curve and manufacturing efficiency  Doubling the volume typically reduce cost by 10% Ž Commodities  Are essentially identical products sold by multiple vendors in large volumes  Aid the competition and drive the efficiency higher and thus the cost down

CSE 520 Fall $/DRAM chip Cost Trends for DRAM One dollar in 1977  $2.95 in 2001 Cost/MB = $500 in 1997 = $0.35 in 2000 = $0.08 in 2001 Demand exceeded supply  price slow drop Each generation drops in dollar price by a factor of 10 to 30 over its lifetime

CSE 520 Fall Cost Trends for Processors Intel List price for 1000 units of the Pentium III Price drop due yield enhancements

CSE 520 Fall Cost vs. Price  Component Costs: raw material cost for the system’s building blocks  Direct Costs (add 25% to 40%) recurring costs: labor, purchasing, scrap, warranty  Gross Margin (add 82% to 186%) nonrecurring costs: R&D, marketing, sales, equipment maintenance, rental, financing cost, pretax profits, taxes  Average Discount to get List Price (add 33% to 66%): volume discounts and/or retailer markup Component Cost Direct Cost Gross Margin Average Discount Avg. Selling Price List Price 15% to 33% 6% to 8% 34% to 39% 25% to 40% * Slide is courtesy of Dave Patterson

CSE 520 Fall Example: Price vs. Cost * Slide is courtesy of Dave Patterson Chip Prices (August 1993) for a volume of 10,000 units

CSE 520 Fall Outline Technology Trends: Culture of tracking, anticipating and exploiting advances in technology Understanding Cost Careful, quantitative comparisons: 1.Define and quantity power 2.Define and quantity dependability 3.Define, quantity, and summarize relative performance 4.Define and quantity relative cost

CSE 520 Fall Define and quantity power ( 1 / 2) For CMOS chips, traditional dominant energy consumption has been in switching transistors, called dynamic power For mobile devices, energy better metric For a fixed task, slowing clock rate (frequency switched) reduces power, but not energy Capacitive load a function of number of transistors connected to output and technology, which determines capacitance of wires and transistors Dropping voltage helps both, so went from 5V to 1V To save energy & dynamic power, most CPUs now turn off clock of inactive modules (e.g. Fl. Pt. Unit)

CSE 520 Fall Example of quantifying power Suppose 15% reduction in voltage results in a 15% reduction in frequency. What is impact on dynamic power?

CSE 520 Fall Define and quantity power (2 / 2) Because leakage current flows even when a transistor is off, now static power important too Leakage current increases in processors with smaller transistor sizes Increasing the number of transistors increases power even if they are turned off In 2006, goal for leakage is 25% of total power consumption; high performance designs at 40% Very low power systems even gate voltage to inactive modules to control loss due to leakage

CSE 520 Fall Outline Review Technology Trends: Culture of tracking, anticipating and exploiting advances in technology Careful, quantitative comparisons: 1.Define and quantity power 2.Define and quantity dependability 3.Define, quantity, and summarize relative performance 4.Define and quantity relative cost

CSE 520 Fall Define and quantity dependability (1/3) How decide when a system is operating properly? Infrastructure providers now offer Service Level Agreements (SLA) to guarantee that their networking or power service would be dependable Systems alternate between 2 states of service with respect to an SLA: 1.Service accomplishment, where the service is delivered as specified in SLA 2.Service interruption, where the delivered service is different from the SLA Failure = transition from state 1 to state 2 Restoration = transition from state 2 to state 1

CSE 520 Fall Define and quantity dependability (2/3) Module reliability = measure of continuous service accomplishment (or time to failure). 2 metrics 1.Mean Time To Failure (MTTF) measures Reliability 2.Failures In Time (FIT) = 1/MTTF, the rate of failures Traditionally reported as failures per billion hours of operation Mean Time To Repair (MTTR) measures Service Interruption –Mean Time Between Failures (MTBF) = MTTF+MTTR Module availability measures service as alternate between the 2 states of accomplishment and interruption (number between 0 and 1, e.g. 0.9) Module availability = MTTF / ( MTTF + MTTR)

CSE 520 Fall Example calculating reliability If modules have exponentially distributed lifetimes (age of module does not affect probability of failure), overall failure rate is the sum of failure rates of the modules Calculate FIT and MTTF for 10 disks (1M hour MTTF per disk), 1 disk controller (0.5M hour MTTF), and 1 power supply (0.2M hour MTTF):

CSE 520 Fall Example calculating reliability If modules have exponentially distributed lifetimes (age of module does not affect probability of failure), overall failure rate is the sum of failure rates of the modules Calculate FIT and MTTF for 10 disks (1M hour MTTF per disk), 1 disk controller (0.5M hour MTTF), and 1 power supply (0.2M hour MTTF):

CSE 520 Fall Outline Review Technology Trends: Culture of tracking, anticipating and exploiting advances in technology Careful, quantitative comparisons: 1.Define and quantity power 2.Define and quantity dependability 3.Define, quantity, and summarize relative performance 4.Define and quantity relative cost

CSE 520 Fall Performance(X) Execution_time(Y) n == Performance(Y) Execution_time(X) Definition: Performance Performance is in units of things per sec –bigger is better If we are primarily concerned with response time performance(x) = 1 execution_time(x) " X is n times faster than Y" means

CSE 520 Fall Performance: What to measure Usually rely on benchmarks vs. real workloads To increase predictability, collections of benchmark applications, called benchmark suites, are popular SPECCPU: popular desktop benchmark suite –CPU only, split between integer and floating point programs –SPECint2000 has 12 integer, SPECfp2000 has 14 integer pgms –SPECCPU announced in Spring 2006 –SPECSFS (NFS file server) and SPECWeb (WebServer) added as server benchmarks Transaction Processing Council measures server performance and cost-performance for databases –TPC-C Complex query for Online Transaction Processing –TPC-H models ad hoc decision support –TPC-W a transactional web benchmark –TPC-App application server and web services benchmark

CSE 520 Fall Performance Tuning Cycle Workload Benchmarks Independent Software Vendors Evaluation: Simulation/Silicon Satisfactory? OK H/W or S/W changes Product No Based on talk with Jim Abele, Intel Chandler (8/30/07)

CSE 520 Fall Some Comments Usually the industry teams look far in future –Currently Intel Chandler team is looking at workloads for 2012 The Workstation workload of today are PC workloads of tomorrow Independent S/W vendors (such as Microsoft/Adobe) may or may not work with chip manufacturers to make changes in their products. Modern chips provide many performance counters and ‘event” tracing – can be used in conjunction with performance enhancement tools such as VTune from Intel.

CSE 520 Fall How Summarize Suite Performance (1/5) Arithmetic average of execution time of all pgms? –But they vary by 4X in speed, so some would be more important than others in arithmetic average Could add a weights per program, but how pick weight? –Different companies want different weights for their products SPECRatio: Normalize execution times to reference computer, yielding a ratio proportional to performance = time on reference computer time on computer being rated

CSE 520 Fall How Summarize Suite Performance (2/5) If program SPECRatio on Computer A is 1.25 times bigger than Computer B, then Note that when comparing 2 computers as a ratio, execution times on the reference computer drop out, so choice of reference computer is irrelevant

CSE 520 Fall How Summarize Suite Performance (3/5) Since ratios, proper mean is geometric mean (SPECRatio unitless, so arithmetic mean meaningless) 1.Geometric mean of the ratios is the same as the ratio of the geometric means 2.Ratio of geometric means = Geometric mean of performance ratios  choice of reference computer is irrelevant! These two points make geometric mean of ratios attractive to summarize performance

CSE 520 Fall How Summarize Suite Performance (4/5) Does a single mean well summarize performance of programs in benchmark suite? Can decide if mean a good predictor by characterizing variability of distribution using standard deviation Like geometric mean, geometric standard deviation is multiplicative rather than arithmetic Can simply take the logarithm of SPECRatios, compute the standard mean and standard deviation, and then take the exponent to convert back:

CSE 520 Fall How Summarize Suite Performance (5/5) Standard deviation is more informative if know distribution has a standard form –bell-shaped normal distribution, whose data are symmetric around mean –lognormal distribution, where logarithms of data--not data itself--are normally distributed (symmetric) on a logarithmic scale For a lognormal distribution, we expect that 68% of samples fall in range 95% of samples fall in range Note: Excel provides functions EXP(), LN(), and STDEV() that make calculating geometric mean and multiplicative standard deviation easy

CSE 520 Fall Example Standard Deviation (1/2) GM and multiplicative StDev of SPECfp2000 for Itanium 2

CSE 520 Fall Example Standard Deviation (2/2) GM and multiplicative StDev of SPECfp2000 for AMD Athlon

CSE 520 Fall Comments on Itanium 2 and Athlon Standard deviation of 1.98 for Itanium 2 is much higher-- vs so results will differ more widely from the mean, and therefore are likely less predictable Falling within one standard deviation: –10 of 14 benchmarks (71%) for Itanium 2 –11 of 14 benchmarks (78%) for Athlon Thus, the results are quite compatible with a lognormal distribution (expect 68%)

CSE 520 Fall Comparing & Summarizing Performance q Wrong summary can present a confusing picture è A is 10 times faster than B for program 1 è B is 10 times faster than A for program 2 q Total execution time is a consistent summary measure q The relative execution times for the same workload is an informative performance summary è Assuming that programs 1 and 2 are executing for the same number of times on computers A and B Execution time is the only valid and unimpeachable measure of performance

CSE 520 Fall Performance Reports Guiding principle is reproducibility (report environment & experiments setup)

CSE 520 Fall And in conclusion … Computer Architecture >> ISA Tracking and extrapolating technology part of architect’s responsibility Expect Bandwidth in disks, DRAM, network, and processors to improve by at least as much as the square of the improvement in Latency Quantify dynamic and static power –Capacitance x Voltage 2 x frequency, Energy vs. power Quantify dependability –Reliability (MTTF, FIT), Availability (99.9…) Quantify and summarize performance –Ratios, Geometric Mean, Multiplicative Standard Deviation Next Week: Quiz Chapter 1, ILP – Ch2 – Assumes Appendix A