CS152 Computer Architecture and Engineering Lecture 12. Introduction to Pipelining
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1 CS152 Compu Achicu a Egiig Lcu 12 Ioucio o Pipliig Fbuay 27, 2001 Joh Kubiaowicz (hp.c.bkly.u/~kubio) lcu li: hp://www-i.c.bkly.u/~c152/ Rcap: Micopogammig Micopogammig i a covi mho fo implmig ucu cool a iagam: Raom logic plac by micopc quc a RM Each li of RM call a Piucio: coai quc cool + valu fo cool poi limi a aiio: bach o zo, x quial, bach o Piucio a fom iplach RM Hoizoal PCo: o cool bi i PIucio fo vy cool li i aapah Vical PCo: goup of cool-li co ogh i PIucio (.g. poibl ) Cool ig uc o Micopogammig Pa of h ig poc i o vlop a laguag ha cib cool a i ay fo huma o ua Lc12.1 Lc12.2 Rcap: Micopogammig Rcap: Excpio mico-pc pco quc cool µ-quc: fch,ipach, quial aapah cool micoiucio (µ) Dipach RM Dco To DaaPah Dco Micopogammig i a fuamal cocp implm a iucio by builig a vy impl poco a ipig h iucio ial fo vy complx iucio a wh fw gi af a poibl ovkill wh ISA mach aapah 1-1 µ-co RM Dco implm ou P-co laguag: Fo iac: - - mm- Lc12.3 u pogam Excpio: omal cool flow: quial, jump, bach, call, u Excpio = upogamm cool af ym ak acio o hal h xcpio - mu co h a of h offig iucio - co ay oh ifomaio cay o u afwa u cool o u mu av & o u a Allow coucio of a u viual machi Sym Excpio Hal u fom xcpio Lc12.4
2 Rcap: Two Typ of Excpio: Iup a Tap Iup cau by xal v: - Nwok, Kyboa, Dik I/, Tim aychoou o pogam xcuio - Mo iup ca b iabl fo bif pio of im - Som (lik Pow Failig ) a o-makabl (NMI) may b hal bw iucio imply up a um u pogam Tap cau by ial v - xcpioal coiio (ovflow) - o (paiy) - faul (o-i pag) ychoou o pogam xcuio coiio mu b mi by h hal iucio may b i o imula a pogam coiu o pogam may b abo Lc12.5 Rcap: Pci Iup Pci Ÿ a of h machi i pv a if pogam xcu up o h offig iucio All pviou iucio compl ffig iucio a all followig iucio ac a if hy hav o v a Sam ym co will wok o iff implmaio Poiio claly ablih by IBM Difficul i h pc of pipliig, ou-o-o xcuio,... MIPS ak hi poiio Impci Ÿ ym ofwa ha o figu ou wha i wh a pu i all back ogh Pfomac goal of la ig o foak pci iup ym ofwa vlop, u, mak c. uually wih hy ha o o hi Mo chiqu fo ou-of-o xcuio a bach picio hlp implm pci iup Lc12.6 Aig h Excpio Hal Savig Sa Taiioal Appoach: Iup Vco PC <- MEM[ IV_ba + cau 00] 370, 68000, Vax, 80x86,... iv_ba RISC Hal Tabl PC < IT_ba + cau 0000 av a a jump Spac, PA, M88K,... MIPS Appoach: fix y PC < EXC_a Acually vy mall abl - RESET y - TLB - oh iv_ba cau cau hal co hal y co Lc12.7 Puh i oo h ack Vax, 68k, 80x86 Sav i i pcial gi MIPS EPC, BaVa, Sau, Cau Shaow Rgi M88k Sav a i a haow of h ial pipli gi Lc12.8
3 Aiio o MIPS ISA o uppo Excpio? Excpio a i kp i copoco 0. U mfc0 a co of h gi Evy gi i bi, bu may b oly paially fi BaVA (gi 8) gi coai mmoy a a which mmoy fc occu Sau (gi 12) iup mak a abl bi Cau (gi 13) h cau of h xcpio Bi 5 o 2 of hi gi co h xcpio yp (.g ufi iucio=10 a aihmic ovflow=12) EPC (gi 14) a of h affc iucio (gi 14 of copoco 0). Cool igal o wi BaVA, Sau, Cau, a EPC o wi xcpio a io PC ( hx ) May hav o uo PC = PC + 4, ic wa EPC o poi o offig iucio (o i ucco): PC = PC - 4 Lc12.9 Rcap: Dail of Sau gi Sau Mak = 1 bi fo ach of 5 hawa a 3 ofwa iup lvl 1 => abl iup 0 => iabl iup k = kl/u 0 => wa i h kl wh iup occu 1 => wa uig u mo = iup abl 0 => iup w iabl 1 => iup w abl Wh iup occu, 6 LSB hif lf 2 bi, ig 2 LSB o 0 u i kl mo wih iup iabl Mak k k k ol pv cu Lc12.10 Rcap: Dail of Cau gi Sau Pig Pig iup 5 hawa lvl: bi if iup occu bu o y vic hal ca wh mo ha o iup occu a am im, o whil co iup qu wh iup iabl Excpio Co co ao fo iup 0 (INT) => xal iup 4 (ADDRL) => a o xcpio (loa o i fch) 5 (ADDRS) => a o xcpio (o) 6 (IBUS) => bu o o iucio fch 7 (DBUS) => bu o o aa fch 8 (Sycall) => Sycall xcpio 9 (BKPT) => Bakpoi xcpio 10 (RI) => Rv Iucio xcpio 12 (VF) => Aihmic ovflow xcpio 5 2 Co Lc12.11 Exampl: How Cool Hal Tap i ou FSD Ufi Iucio c wh o x a i fi fom a 1 fo h op valu. W hal hi xcpio by fiig h x a valu fo all op valu oh ha lw, w, 0 (R-yp), jmp, bq, a oi a w a 12. Show ymbolically uig oh o iica ha h op fil o o mach ay of h opco ha labl ac ou of a 1. Aihmic ovflow c o op uch a ig a U o av PC a xcpio hal Exal Iup flagg by a iup li Agai, mu av PC a xcpio hal No: Challg i al machi i o hal: Iacio bw iucio a oh xcpio-cauig v uch ha cool logic mai mall a fa. Complx iacio mak h cool ui h mo challgig apc of hawa ig Lc12.12
4 How a ap a iup o a iagam? R[] <= S 0101 iucio fch IR <= MEM[PC] PC <= PC R[] <= S 0111 M <= MEM[S] 1001 R[] <= M 1010 co EPC <= PC - 4 S<= PC +SX PC <= xp_a EPC <= PC oh cau <= 12 (vf) PC <= xp_a cau <= 10 (RI) R-yp Ri LW BEQ ovflow SW S <= A If -A B= B S <= A fu B S <= A op ZX S <= A + SX S <= A + SX h PC <= S MEM[S] <= B 1100 Pig INT EPC <= PC - 4 PC <= xp_a cau <= 0(INT) Hal Iup ufi iucio Lc12.13 Bu: Wha ha o chag i ou P-quc? N cocp of bach a mico-co lvl µ-off 4? 1 Do µ-bach Co Slc N? 1 0 pig iup ovflow A micopc Dipach RM pco Sq Slc EPC <= PC - 4 PC <= xp_a cau <= 12 (vf) ovflow R-yp S <= A fu B 0100 µa Slc Logic Lc12.14 Exampl: Ca aily u wih fo o-ial mmoy R-yp S <= A fu B R[] <= S PC <= PC + 4 Ri S <= A o ZX R[] <= S PC <= PC + 4 IR <= MEM[PC] ~wai A <= R[] B <= R[] LW S <= A + SX M <= MEM[S] ~wai wai R[] <= M PC <= PC + 4 iucio fch wai co / opa fch SW BEQ PC <= S <= A + SX Nx(PC) MEM[S] <= B ~wai PC <= PC + 4 wai Excu Mmoy Wi-back Lc12.15 Amiiaiv Iu Mim I: Thuay 3/1 (ay af omoow) 5:30 8:30 i 277 Coy Clo book, bu ca hav o 8 ½ u 11 h of hawi o Cov: Chap 1 5, Appic A C Mak u o chck ou h ampl quizz o h Wb! G a aig Chap 6! Compl chap o Pipliig... Scio: Thi wk cio Ÿ 433 Laim (a uual) Nx wk cio Ÿ Coy You will moa you poco o a myy pogam - Rpo ill u a miigh Lab Rpo: Up o you o o a goo job of ummaizig you wok Pa of ga will b o qualiy of you wiig - Pu co a chmaic i appic appopialy fc - U acual wopoco (Micoof Wo oli) Lc12.16
5 Quio #1: Why o micocoig? If impl iucio coul xcu a vy high clock a If you coul v wi compil o pouc micoiucio If mo pogam u impl iucio a aig mo If micoco i kp i RAM ia of RM o a o fix bug If am mmoy u fo cool mmoy coul b u ia a cach fo macoiucio Th why o kip iucio ipaio by a micopogam a imply compil icly io low laguag of machi? (micopogammig i ovkill wh ISA mach aapah 1-1) Rcall: Pfomac Evaluaio Wha i h avag CPI? a iagam giv CPI fo ach iucio yp wokloa giv fqucy of ach yp Typ CPI i fo yp Fqucy CPI i x fqi i Aih/Logic 4 40% 1.6 Loa 5 30% 1.5 So 4 10% 0.4 bach 3 20% 0.6 Avag CPI:4.1 Lc12.17 Lc12.18 Quio #2: Ca w g CPI < 4.1? Sm o b lo of il hawa Why o ovlap iucio??? PCW PCWCo PCSc Zo IoD MmW IRW RgD RgW SlA 1 PC 0 0 R Zo 0 Ra RA 5 R Rb bua A 1 Ial 0 5 Rg Fil 1 Mmoy R 4 0 WA Rw B Di Dou R 1 buw bub << 2 Cool Iucio Rg Mm Daa Rg u Th Big Picu: Wh a W Now? Th Fiv Claic Compo of a Compu Nx Topic: Poco Ipu Cool Mmoy Daapah upu Pipliig by Aalogy Pipli haza Imm Ex 16 Exp MmoRg p SlB Lc12.19 Lc12.20
6 Pipliig i Naual! Lauy Exampl A, Bia, Cahy, Dav ach hav o loa of cloh o wah, y, a fol Wah ak 30 miu Dy ak 40 miu Fol ak 20 miu A B C D Lc12.21 T a k Squial Lauy A B C D 6 PM Miigh Tim Squial lauy ak 6 hou fo 4 loa If hy la pipliig, how log woul lauy ak? Lc12.22 T a k Pipli Lauy: Sa wok ASAP A B C D 6 PM Miigh Tim Pipli lauy ak 3.5 hou fo 4 loa Lc12.23 T a k Pipliig Lo A B C D 6 PM Tim Pipliig o hlp lacy of igl ak, i hlp houghpu of i wokloa Pipli a limi by low pipli ag Mulipl ak opaig imulaouly uig iff ouc Poial pup = Numb pip ag Ubalac lgh of pip ag uc pup Tim o fill pipli a im o ai i uc pup Sall fo Dpc Lc12.24
7 Th Fiv Sag of Loa Loa Cycl 1 Cycl 2 Cycl 3 Cycl 4 Cycl 5 Ifch Rg/Dc Exc Mm W Ifch: Iucio Fch Fch h iucio fom h Iucio Mmoy Rg/Dc: Rgi Fch a Iucio Dco Exc: Calcula h mmoy a Mm: Ra h aa fom h Daa Mmoy W: Wi h aa back o h gi fil Lc12.25 Dco Fch No: Th 5 ag w h all alog! Excu Mmoy Wi-back R-yp ou <= A fu B 0100 R[] <= ou 0101 Ri ou <= A op ZX 0110 R[] <= ou 0111 IR <= MEM[PC] PC <= PC ou <= PC +SX 0001 LW ou <= A + SX 1000 M <= MEM[ou] 1001 R[] <= M 1010 SW ou <= A + SX 1011 MEM[ou] <= B 1100 BEQ If A = B h PC <= ou 0010 Lc12.26 Pipliig Impov pfomac by icaig houghpu Baic Ia Ial pup i umb of ag i h pipli. Do w achiv hi? Lc12.27 Wha o w o a o pli h aapah io ag? Lc12.28
8 Gaphically Rpig Pipli Covioal Pipli Excuio Rpaio Tim IFch Dc Exc Mm WB IFch Dc Exc Mm WB IFch Dc Exc Mm WB IFch Dc Exc Mm WB Ca hlp wih awig quio lik: how may cycl o i ak o xcu hi co? wha i h oig uig cycl 4? u hi paio o hlp ua aapah Pogam Flow IFch Dc Exc Mm WB IFch Dc Exc Mm WB Lc12.29 Lc12.30 Sigl Cycl, Mulipl Cycl, v. Pipli Cycl 1 Cycl 2 Clk Sigl Cycl Implmaio: Loa So Wa Cycl 1 Cycl 2 Cycl 3 Cycl 4 Cycl 5 Cycl 6 Cycl 7 Cycl 8 Cycl 9 Cycl 10 Clk Mulipl Cycl Implmaio: Loa So R-yp Ifch Rg Exc Mm W Ifch Rg Exc Mm Ifch Why Pipli? Suppo w xcu 100 iucio Sigl Cycl Machi 45 /cycl x 1 CPI x 100 i = 4500 Mulicycl Machi 10 /cycl x 4.6 CPI (u o i mix) x 100 i = 4600 Ial pipli machi 10 /cycl x (1 CPI x 100 i + 4 cycl ai) = 1040 Pipli Implmaio: Loa Ifch Rg Exc Mm W So Ifch Rg Exc Mm W R-yp Ifch Rg Exc Mm W Lc12.31 Lc12.
9 I. Why Pipli? Bcau w ca! I 0 I 1 I 2 I 3 I 4 Tim (clock cycl) Lc12.33 Ca pipliig g u io oubl? Y: Pipli Haza ucual haza: amp o u h am ouc wo iff way a h am im - E.g., combi wah/y woul b a ucual haza o fol buy oig omhig l (wachig TV) cool haza: amp o mak a ciio bfo coiio i valua - E.g., wahig fooball uifom a o g pop g lvl; o af y bfo x loa i - bach iucio aa haza: amp o u im bfo i i ay - E.g., o ock of pai i y a o i wah; ca fol uil g ock fom wah hough y - iucio p o ul of pio iucio ill i h pipli Ca alway olv haza by waiig pipli cool mu c h haza ak acio (o lay acio) o olv haza Lc12.34 Sigl Mmoy i a Sucual Haza Sucual Haza limi pfomac I. Loa I 1 I 2 I 3 I 4 Tim (clock cycl) Mm Rg Mm Rg Exampl: if 1.3 mmoy acc p iucio a oly o mmoy acc p cycl h avag CPI 1.3 ohwi ouc i mo ha 100% uiliz Dcio i ay i hi ca! (igh half highligh ma a, lf half wi) Lc12.35 Lc12.36
10 I. Cool Haza Soluio #1: Sall A Bq Loa Tim (clock cycl) Sall: wai uil ciio i cla Impac: 2 lo cycl (i.. 3 clock cycl p bach iucio) => low Mov ciio o of co av 1 cycl p bach Lo poial Mm Rg Mm Rg Lc12.37 I. Cool Haza Soluio #2: Pic A Bq Loa Tim (clock cycl) Mm Pic: gu o icio h back up if wog Impac: 0 lo cycl p bach iucio if igh, 1 if wog (igh - 50% of im) N o Squah a a followig iucio if wog Pouc CPI o bach of (1 * *.5) = 1.5 Toal CPI migh h b: 1.5 * *.8 = 1.1 (20% bach) Mo yamic chm: hioy of 1 bach (- 90%) Rg Mm Rg Lc12.38 Cool Haza Soluio #3: Dlay Bach Daa Haza o 1 I. A Bq Mic Loa Tim (clock cycl) Mm Dlay Bach: Rfi bach bhavio (ak plac af x iucio) Impac: 0 clock cycl p bach iucio if ca fi iucio o pu i lo (- 50% of im) A lauch mo iucio p clock cycl, l uful Rg Mm Rg Mm Rg Mm Rg Lc12.39 a 1,2,3 ub 4,1,3 a 6,1,7 o 8,1,9 xo 10,1,11 Lc12.40
11 Daa Haza o 1: Daa Haza Soluio: Dpci backwa i im a haza Fowa ul fom o ag o aoh I. Tim (clock cycl) IF ID/RF EX MEM WB a 1,2,3 ub 4,1,3 a 6,1,7 o 8,1,9 xo 10,1,11 Im Rg Dm Rg I. Tim (clock cycl) IF ID/RF EX MEM WB a 1,2,3 ub 4,1,3 a 6,1,7 o 8,1,9 xo 10,1,11 Im Rg Dm Rg Lc12.41 o K if fi a/wi poply Lc12.42 Fowaig (o Bypaig): Wha abou Loa? Dpci backwa i im a haza Tim (clock cycl) IF ID/RF EX MEM WB lw 1,0(2) Im Rg Dm Rg Fowaig (o Bypaig): Wha abou Loa Dpci backwa i im a haza Tim (clock cycl) IF ID/RF EX MEM WB lw 1,0(2) Im Rg Dm Rg ub 4,1,3 ub 4,1,3 Sall Ca olv wih fowaig: Mu lay/all iucio p o loa Ca olv wih fowaig: Mu lay/all iucio p o loa Lc12.43 Lc12.44
12 Digig a Pipli Poco Cool a Daapah: Spli a iag io 5 pic Go back a xami you aapah a cool iagam aocia ouc wih a IR <- Mm[PC]; PC < PC+4; A <- R[]; B< R[] u ha flow o o coflic, o figu ou how o olv S < A + B; S < A o ZX; S < A + SX; S < A + SX; If Co PC < PC+SX; a cool i appopia ag M < Mm[S] Mm[S] <- B R[] < S; R[] < S; R[] < M; Equal Lc12.45 Nx PC PC I. Mm IR Rg Fil A B Exc S D Mm Acc M Daa Mm Rg. Fil Lc12.46 Summay: Pipliig Ruc CPI by ovlappig may iucio Avag houghpu of appoximaly 1 CPI wih fa clock Uiliz capabilii of h Daapah a x iucio whil wokig o h cu o limi by lgh of log ag (plu fill/fluh) c a olv haza Wha mak i ay all iucio a h am lgh ju a fw iucio foma mmoy opa appa oly i loa a o Summay: Wh hi cla i goig W ll buil a impl pipli a look a h iu Lab 5 Ÿ Pipli Poco Lab 6 Ÿ Wih cach W ll alk abou mo poco a wha ally ha: xcpio halig yig o impov pfomac wih ou-of-o xcuio, c. Tyig o g CPI < 1 (Supcala xcuio) Wha mak i ha? ucual haza: uppo w ha oly o mmoy cool haza: o woy abou bach iucio aa haza: a iucio p o a pviou iucio Lc12.47 Lc12.48
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