ECE/CS 552: Introduction To Computer Architecture 1

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1 ECE/CS 552: Perormace ad Cot Itructor:Mikko H Lipati Fall 200 Uiverity o Wicoi-Madio Lecture ote partially baed o et created by Mark Hill. Perormace ad Cot Which o the ollowig airplae ha the bet perormace? Airplae Paeger Rage (mi) Speed (mph) Boeig Boeig BAC/Sud Cocorde Dougla DC How much ater i the Cocorde v. the 747 How much bigger i the 747 v. DC-8? Perormace ad Cot Which computer i atet? Not o imple Scietiic imulatio FP perormace Program developmet Iteger perormace Databae workload Memory, I/O Perormace o Computer Wat to buy the atet computer or what you wat to do? Workload i all-importat Correct meauremet ad aalyi Wat to deig the atet computer or what the cutomer wat to pay? Cot i a importat criterio Forecat Time ad perormace Iro Law MIPS ad MFLOPS Which program ad how to average Amdahl law Deiig Perormace What i importat to whom? Computer ytem uer Miimize elaped time or program = time_ed time_tart Called repoe time Computer ceter maager Maximize completio rate = #job/ecod Called throughput Architecture

2 Repoe Time v. Throughput I throughput = /av. repoe time? Oly i NO overlap Otherwie, throughput > /av. repoe time E.g. a luch buet aume 5 etree Each pero take 2 miute/etrée Throughput i pero every 2 miute BUT time to ill up tray i 0 miute Why ad what would the throughput be otherwie? 5 people imultaeouly illig tray (overlap) Without overlap, throughput = /0 What i Perormace or u? For computer architect CPU time = time pet ruig a program Ituitively, bigger hould be ater, o: Perormace = /X time, where X i repoe, CPU executio, etc. Elaped time = CPU time + I/O wait We will cocetrate o CPU time Improve Perormace Improve (a) repoe time or (b) throughput? Fater CPU Help both (a) ad (b) Add more CPU Help (b) ad perhap (a) due to le queueig Perormace Compario Machie A i time ater tha machie B i per(a)/per(b) = time(b)/time(a) = Machie A i x% ater tha machie B i per(a)/per(b) = time(b)/time(a) = + x/00 E.g. time(a) = 0, time(b) = 5 5/0 =.5 => A i.5 time ater tha B 5/0 =.5 => A i 50% ater tha B Breakig Dow Perormace A program i broke ito itructio H/W i aware o itructio, ot program At lower level, H/W break itructio ito cycle Lower level tate machie chage tate every cycle For example: GHz Sapdrago ru 000M cycle/ec, cycle = 2.5GHz Core i7 ru 2.5G cycle/ec, cycle = 0.25 Iro Law Time Proceor Perormace = Program Itructio Cycle = Time X X Program Itructio Cycle (code ize) (CPI) (cycle time) Architecture --> Implemetatio --> Realizatio Compiler Deiger Proceor Deiger Chip Deiger Architecture 2

3 Iro Law Itructio/Program Itructio executed, ot tatic code ize Determied by algorithm, compiler, ISA Cycle/Itructio Determied by ISA ad CPU orgaizatio Overlap amog itructio reduce thi term Time/cycle Determied by techology, orgaizatio, clever circuit deig Our Goal Miimize time which i the product, NOT iolated term Commo error to mi term while deviig optimizatio E.g. ISA chage to decreae itructio cout BUT lead to CPU orgaizatio which make clock lower Bottom lie: term are iter-related Other Metric MIPS ad MFLOPS MIPS = itructio cout/(executio time x 0 6 ) = clock rate/(cpi x 0 6 ) But MIPS ha eriou hortcomig Problem with MIPS E.g. without FP hardware, a FP op may take 50 igle-cycle itructio With FP hardware, oly oe 2-cycle itructio Thu, addig FP hardware: CPI icreae (why?) Itructio/program decreae (why?) Total executio time decreae BUT, MIPS get wore! 50/50 => 2/ 50 => 50 => 2 50 MIPS => 2 MIPS Problem with MIPS Igore program Uually ued to quote peak perormace Ideal coditio => guarateed ot to exceed! Whe i MIPS ok? Same compiler, ame ISA E.g. ame biary ruig o AMD Pheom, Itel Core i7 Why? Itr/program i cotat ad ca be igored Other Metric MFLOPS = FP op i program/(executio time x 0 6 ) Aumig FP op idepedet o compiler ad ISA Ote ae or umeric code: matrix ize determie # o FP op/program However, ot alway ae: Miig itructio (e.g. FP divide) Optimizig compiler Relative MIPS ad ormalized MFLOPS Add to couio Architecture 3

4 Rule Ue ONLY Time Beware whe readig, epecially i detail are omitted Beware o Peak Guarateed ot to exceed Iro Law Example Machie A: clock, CPI 2.0, or program x Machie B: clock 2, CPI.2, or program x Which i ater ad how much? Time/Program = itr/program x cycle/itr x ec/cycle Time(A) = N x 2.0 x = 2N Time(B) = N x.2 x 2 = 2.4N Compare: Time(B)/Time(A) = 2.4N/2N =.2 So, Machie A i 20% ater tha Machie B or thi program Iro Law Example Keep ad For equal perormace, i CPI(B)=.2, what i CPI(A)? Time(B)/Time(A) = = (Nx2x.2)/(NxxCPI(A)) CPI(A) = 2.4 Iro Law Example Keep CPI(A)=2.0 ad CPI(B)=.2 For equal perormace, i clock(b)=2, what i clock(a)? Time(B)/Time(A) = = (N x 2.0 x clock(a))/(n x.2 x 2) clock(a) =.2 Which Program Executio time o what program? Bet cae your alway ru the ame et o program Port them ad time the whole workload I reality, ue bechmark Program choe to meaure perormace Predict perormace o actual workload Save eort ad moey Repreetative? Hoet? Bechmarketig How to Average Machie A Machie B Program 0 Program Total 00 0 Oe awer: or total executio time, how much ater i B? 9.x Architecture 4

5 How to Average Aother: arithmetic mea (ame reult) Arithmetic mea o time: time i AM(A) = 00/2 = ( ) i AM(B) = 0/2 = /55 = 9.x Valid oly i program ru equally ote, o ue weighted arithmetic mea: weight( i) time( i) i Other Average E.g., 30 mph or irt 0 mile, the 90 mph or ext 0 mile, what i average peed? Average peed = (30+90)/2 WRONG Average peed = total ditace / total time = (20 / (0/30 + 0/90)) = 45 mph Harmoic Mea Harmoic mea o rate = i rate( ) Ue HM i orced to tart ad ed with rate (e.g. reportig MIPS or MFLOPS) Why? Rate ha time i deomiator Mea hould be proportioal to ivere o um o time (ot um o ivere) See: J.E. Smith, Characterizig computer perormace with a igle umber, CACM Volume 3, Iue 0 (October 988), pp Dealig with Ratio Machie A Machie B Program 0 Program Total 00 0 I we take ratio with repect to machie A Machie A Machie B Program 0 Program 2 0. Dealig with Ratio Average or machie A i, average or machie B i 5.05 I we take ratio with repect to machie B Machie A Machie B Program 0. Program 2 0 Average 5.05 Ca t both be true!!! Do t ue arithmetic mea o ratio! Geometric Mea Ue geometric mea or ratio Geometric mea o ratio = ratio( i) i Idepedet o reerece machie I the example, GM or machie a i, or machie B i alo Normalized with repect to either machie Architecture 5

6 But GM o ratio i ot proportioal to total time AM i example ay machie B i 9. time ater GM ay they are equal I we took total executio time, A ad B are equal oly i Program i ru 00 time more ote tha program 2 Geerally, GM will mipredict or three or more machie Summary Ue AM or time Ue HM i orced to ue rate UeGMiorcedtoueratio orced ratio Bet o all, ue uormalized umber to compute time Bechmark: SPEC2000 Sytem Perormace Evaluatio Cooperative Formed i 80 to combat bechmarketig SPEC89, SPEC92, SPEC95, SPEC iteger ad 4 loatig-poit program Su Ultra-5 300MHz reerece machie ha core o 00 Report GM o ratio to reerece machie Bechmark: SPEC CINT2000 Bechmark Decriptio 64.gzip Compreio 75.vpr FPGA place ad route 76.gcc C compiler 8.mc Combiatorial optimizatio 86.craty Che 97.parer Word proceig, grammatical aalyi 252.eo Viualizatio (ray tracig) 253.perlbmk PERL cript executio 254.gap Group theory iterpreter 255.vortex Object-orieted databae 256.bzip2 Compreio 300.twol Place ad route imulator Bechmark: SPEC CFP2000 Bechmark Decriptio 68.wupwie Phyic/Quatum Chromodyamic 7.wim Shallow water modelig 72.mgrid Multi-grid olver: 3D potetial ield 73.applu Parabolic/elliptic PDE 77.mea 3-D graphic library 78.galgel Computatioal Fluid Dyamic 79.art Image Recogitio/Neural Network 83.equake Seimic Wave Propagatio Simulatio 87.acerec Image proceig: ace recogitio 88.ammp Computatioal chemitry 89.luca Number theory/primality tetig 9.ma3d Fiite-elemet Crah Simulatio 200.ixtrack High eergy uclear phyic accelerator deig 30.api Meteorology: Pollutat ditributio Bechmark Pitall Bechmark ot repreetative Your workload i I/O boud, SPEC i uele Bechmark i too old Bechmark age poorly; bechmarketig preure caue vedor to optimize compiler/hardware/otware to bechmark Need to be periodically rerehed Architecture 6

7 Amdahl Law Motivatio or optimizig commo cae Speedup = old time / ew time = ew rate / old rate Let a optimizatio peed ractio o time by a actor o oldtime Speedup oldtime oldtime Amdahl Law Example Your bo ak you to improve perormace by: Improve the ALU ued 95% o time by 0% Improve memory pipelie ued 5% o time by 0x Let =ractio ped up ad = peedup o that ractio New_time = (-) x old_time + (/) x old_time Speedup = old_time / ew_time Speedup = old_time / ((-) x old_time + (/) x old_time) Amdahl Law: Speedup Amdahl Law Example, cot d Speedup 95% % %.052 Amdahl Law: Limit Make commo cae at: Speedup lim Amdahl Law: Limit Coider ucommo cae! lim I (-) i otrivial Speedup i limited! Particularly true or exploitig parallelim i the large, where large i ot cheap GPU with e.g. 024 proceor (hader core) Parallel portio peed up by (024x) Serial portio o code (-) limit peedup E.g. 0% erial limit to 0x peedup! Summary Time ad perormace: Machie A time ater tha Machie B I Time(B)/Time(A) = Iro Law: Perormace = Time/program = Itructio Cycle = Time X X Program Itructio Cycle (code ize) (CPI) (cycle time) Architecture 7

8 Summary Cot d Other Metric: MIPS ad MFLOPS Beware o peak ad omitted detail Bechmark: SPEC2000 Summarize perormace: AM or time HM or rate GM or ratio Speedup Amdahl Law: Architecture 8

Performance of Computers. Performance of Computers. Defining Performance. Forecast

Performance of Computers. Performance of Computers. Defining Performance. Forecast Performance of Computers Which computer is fastest? Not so simple scientific simulation - FP performance program development - Integer performance commercial work - I/O Performance of Computers Want to