CS152 Computer Architecture and Engineering Lecture 12. Introduction to Pipelining

Transcription

1 CS152 Comput chitctu a Egiig Lctu 12 Itouctio to Pipliig ctob 11, 1999 Joh Kubiatowicz (http.c.bkly.u/~kubito) lctu li: Rcap: Micopogammig Micopogammig i a covit mtho fo implmtig tuctu cotol tat iagam: Raom logic plac by mico quc a RM Each li of RM call a µituctio: cotai quc cotol + valu fo cotol poit limit tat taitio: bach to zo, xt qutial, bach to µituctio a fom iplatch RM Hoizotal µco: o cotol bit i µituctio fo vy cotol li i atapath Vtical µco: goup of cotol-li co togth i µituctio (.g. poibl t) Cotol ig uc to Micopogammig Pat of th ig poc i to vlop a laguag that cib cotol a i ay fo huma to uta Lc12.1 Lc12.2 Rcap: Micopogammig Excptio mico- pco quc cotol µ-quc: ftch,ipatch, qutial atapath cotol micoituctio (µ) ipatch RM co To atapath co Micopogammig i a fuamtal cocpt implmt a ituctio t by builig a vy impl poco a itptig th ituctio tial fo vy complx ituctio a wh fw git taf a poibl ovkill wh IS match atapath 1-1 µ-co RM co implmt ou µ-co laguag: Fo itac: t- - mm- Lc12.3 u pogam Excptio: omal cotol flow: qutial, jump, bach, call, tu Excptio = upogamm cotol taf ytm tak actio to hal th xcptio - mut co th a of th offig ituctio - co ay oth ifomatio cay to tu aftwa tu cotol to u mut av & to u tat llow cotuctio of a u vitual machi Sytm Excptio Hal tu fom xcptio Lc12.4

2 Two Typ of Excptio: Itupt a Tap Itupt cau by xtal vt: - Ntwok, Kyboa, ik I/, Tim aychoou to pogam xcutio - Mot itupt ca b iabl fo bif pio of tim - Som (lik Pow Failig ) a o-makabl (NMI) may b hal btw ituctio imply up a um u pogam Tap cau by ital vt - xcptioal coitio (ovflow) - o (paity) - fault (o-it pag) ychoou to pogam xcutio coitio mut b mi by th hal ituctio may b ti o imulat a pogam cotiu o pogam may b abot Lc12.5 MIPS covtio: xcptio ma ay uxpct chag i cotol flow, without itiguihig ital o xtal; u th tm itupt oly wh th vt i xtally cau. Typ of vt Fom wh? MIPS tmiology I/ vic qut Extal Itupt Ivok S fom u pogam Ital Excptio ithmtic ovflow Ital Excptio Uig a ufi ituctio Ital Excptio Hawa malfuctio Eith Excptio o Itupt Lc12.6 What happ to Ituctio with Excptio? MIPS achitctu fi th ituctio a havig o ffct if th ituctio cau a xcptio. Wh gt to vitual mmoy w will that ctai cla of xcptio mut pvt th ituctio fom chagig th machi tat. Thi apct of halig xcptio bcom complx a pottially limit pfomac => why it i ha Lc12.7 Pci Itupt Pci tat of th machi i pv a if pogam xcut up to th offig ituctio ll pviou ituctio complt ffig ituctio a all followig ituctio act a if thy hav ot v tat Sam ytm co will wok o ifft implmtatio Poitio claly tablih by IM ifficult i th pc of pipliig, out-ot-o xcutio,... MIPS tak thi poitio Impci ytm oftwa ha to figu out what i wh a put it all back togth Pfomac goal oft la ig to foak pci itupt ytm oftwa vlop, u, makt tc. uually wih thy ha ot o thi Mo tchiqu fo out-of-o xcutio a bach pictio hlp implmt pci itupt Lc12.8

3 ig Pictu: u / ytm mo y poviig two mo of xcutio (u/ytm) it i poibl fo th comput to maag itlf opatig ytm i a pcial pogam that u i th pivilg mo a ha acc to all of th ouc of th comput pt vitual ouc to ach u that a mo covit that th phyical ouc - fil v. ik cto - vitual mmoy v phyical mmoy potct ach u pogam fom oth potct ytm fom maliciou u. S i aum to kow bt, a i tut co, o t ytm mo o xcptio. Excptio allow th ytm to tak actio i po to vt that occu whil u pogam i xcutig: ig th Excptio Hal Taitioal ppoach: Itupt Vcto <- MEM[ IV_ba + cau 00] 370, 68000, Vax, 80x86,... iv_ba RISC Hal Tabl < IT_ba + cau 0000 av tat a jump Spac, P, M88K,... MIPS ppoach: fix ty < EXC_a ctually vy mall tabl - RESET ty Might povi upplmtal bhavio (alig with omal floatig-poit umb fo itac). - TL Uimplmt ituctio u to mulat ituctio that - oth 10/11/99 w ot iclu i hawa UC Fall (I MicoVax) Lc12.9 iv_ba cau cau hal co hal ty co Lc12.10 Savig Stat Puh it oto th tack Vax, 68k, 80x86 Sav it i pcial git MIPS E, ava, Statu, Cau Shaow it M88k Sav tat i a haow of th ital pipli git Lc12.11 itio to MIPS IS to uppot Excptio? Excptio tat i kpt i copoco 0. E a 32-bit git u to hol th a of th affct ituctio (git 14 of copoco 0). Cau a git u to co th cau of th xcptio. I th MIPS achitctu thi git i 32 bit, though om bit a cutly uu. um that bit 5 to 2 of thi git co th two poibl xcptio ouc mtio abov: ufi ituctio=0 a aithmtic ovflow=1 (git 13 of copoco 0). av - git cotai mmoy a at which mmoy fc occu (git 8 of copoco 0) Statu - itupt mak a abl bit (git 12 of copoco 0) Cotol igal to wit E, Cau, av, a Statu abl to wit xcptio a ito, ica mux to a a iput two ( hx ) May hav to uo = + 4, ic wat E to poit to offig ituctio (ot it ucco); = - 4 Lc12.12

4 Rcap: tail of Statu git Statu Mak = 1 bit fo ach of 5 hawa a 3 oftwa itupt lvl 1 => abl itupt 0 => iabl itupt k = kl/u 0 => wa i th kl wh itupt occu 1 => wa uig u mo = itupt abl 0 => itupt w iabl 1 => itupt w abl Wh itupt occu, 6 LS hift lft 2 bit, ttig 2 LS to 0 u i kl mo with itupt iabl Mak k k k ol pv cut Lc12.13 Rcap: tail of Cau git Statu Pig Pig itupt 5 hawa lvl: bit t if itupt occu but ot yt vic hal ca wh mo tha o itupt occu at am tim, o whil co itupt qut wh itupt iabl Excptio Co co ao fo itupt 0 (INT) => xtal itupt 4 (RL) => a o xcptio (loa o it ftch) 5 (RS) => a o xcptio (to) 6 (IUS) => bu o o ituctio ftch 7 (US) => bu o o ata ftch 8 (Sycall) => Sycall xcptio 9 (KPT) => akpoit xcptio 10 (RI) => Rv Ituctio xcptio 12 (VF) => ithmtic ovflow xcptio 5 2 Co Lc12.14 Exampl: How Cotol Hal Tap i ou FS Ufi Ituctio tct wh o xt tat i fi fom tat 1 fo th op valu. W hal thi xcptio by fiig th xt tat valu fo all op valu oth tha lw, w, 0 (R-typ), jmp, bq, a oi a w tat 12. Show ymbolically uig oth to iicat that th op fil o ot match ay of th opco that labl ac out of tat 1. ithmtic ovflow tct o op uch a ig a U to av a t xcptio hal Extal Itupt flagg by at itupt li gai, mut av a t xcptio hal Not: Challg i igig cotol of a al machi i to hal ifft itactio btw ituctio a oth xcptio-cauig vt uch that cotol logic mai mall a fat. Complx itactio mak th cotol uit th mot challgig apct of hawa ig Lc12.15 How a tap a itupt to tat iagam? R[] <= S 0101 ituctio ftch <= MEM[] <= R[t] <= S 0111 M <= MEM[S] 1001 R[t] <= M 1010 co MEM[S] <= 1100 Pig INT E <= - 4 S<= +SX <= xp_a E <= oth cau <= 12 (vf) <= xp_a cau <= 10 (RI) R-typ Ri LW EQ ovflow SW S <= If - = S <= fu S <= op ZX S <= + SX S <= + SX th <= S E <= - 4 <= xp_a cau <= 0(INT) Hal Itupt ufi ituctio Lc12.16

5 ut: What ha to chag i ou µ-quc? N cocpt of bach at mico-co lvl µ-offt 4? 1 o µ-bach Co Slct N? 1 Mux 0 pig itupt ovflow Mux mico Mux ipatch RM pco Sq Slct E <= - 4 <= xp_a cau <= 12 (vf) ovflow R-typ S <= fu 0100 µ Slct Logic Lc12.17 Exampl: Ca aily u with fo o-ial mmoy R-typ S <= fu R[] <= S <= + 4 Ri S <= o ZX R[t] <= S <= + 4 <= MEM[] ~wait <= R[] <= R[t] LW S <= + SX M <= MEM[S] ~wait wait R[t] <= M <= + 4 ituctio ftch wait co / opa ftch SW EQ <= S <= + SX Nxt() MEM[S] <= ~wait <= + 4 wait ut oy Wit-back Lc12.18 Summay: Micopogammig o ipiatio fo RISC If impl ituctio coul xcut at vy high clock at If you coul v wit compil to pouc micoituctio If mot pogam u impl ituctio a aig mo If micoco i kpt i RM ita of RM o a to fix bug If am mmoy u fo cotol mmoy coul b u ita a cach fo macoituctio Th why ot kip ituctio itptatio by a micopogam a imply compil ictly ito lowt laguag of machi? (micopogammig i ovkill wh IS match atapath 1-1) miitativ Iu: Rult of Mitm I Exam vag: 62, Staa v: 13.5 Popl ha toubl with qua-oot poblm Thi wa vy much lik ivi! - Lag hift git movig lft Som iu with micoco Mitm I itibutio Lc /11/ UC Fall Lc12.20

6 Squa oot xampl: coi mai hiftig LEFT Statig: M= R 0 = a S 0 = 0000 Ty: N 1 = (2 S ) 1000 R 1 = S 1 =S = 1000 Ty: N 2 = (2 S ) 0100 Rult < 0 S 2 =S 1 = 1000 R 2 = (uchag) Ty: N 3 = (2 S ) 0010 R 3 = Ty: N 4 = S 3 =S = 1010 (2 S ) 0001 Rult < 0 S 4 =S 3 = 1010 R 4 = (uchag) miitativ Iu (cotiu) Gt tat aig Chapt 6! Complt chapt o Pipliig... Nxt wk ctio => Coy 119 Comput i th Nw: Mc ilico ha fially light of ay Fial ult: o: = with mai = 10 with 18 mai! Lc12.21 Lc12.22 Th ig Pictu: Wh a W Now? Pipliig i Natual! Th Fiv Claic Compot of a Comput Poco Iput Cotol oy Lauy Exampl, ia, Cathy, av ach hav o loa of cloth to wah, y, a fol Wah tak 30 miut C Nxt Topic: atapath utput Pipliig by alogy miitivia; Cou oa map y tak 40 miut Fol tak 20 miut Lc12.23 Lc12.24

7 T a k Squtial Lauy C 6 PM Miight Tim Squtial lauy tak 6 hou fo 4 loa If thy la pipliig, how log woul lauy tak? Lc12.25 T a k Pipli Lauy: Stat wok SP C 6 PM Miight Tim Pipli lauy tak 3.5 hou fo 4 loa Lc12.26 T a k Pipliig Lo C 6 PM Tim Pipliig o t hlp latcy of igl tak, it hlp thoughput of ti wokloa Pipli at limit by lowt pipli tag Multipl tak opatig imultaouly uig ifft ouc Pottial pup = Numb pip tag Ubalac lgth of pip tag uc pup Tim to fill pipli a tim to ai it uc pup Stall fo pc Lc12.27 Th Fiv Stag of Loa Cycl 1 Cycl 2 Cycl 3 Cycl 4 Cycl 5 Loa Iftch /c W Iftch: Ituctio Ftch Ftch th ituctio fom th Ituctio oy /c: it Ftch a Ituctio co : Calculat th mmoy a : Ra th ata fom th ata oy W: Wit th ata back to th git fil Lc12.28

8 Not: Th 5 tag w th all alog! Ftch <= MEM[] <= Pipliig Impov pfomac by icaig thoughput co out <= +SX 0001 ut oy Wit-back R-typ out <= fu 0100 R[] <= out 0101 Ri out <= op ZX 0110 R[t] <= out 0111 LW out <= + SX 1000 M <= MEM[out] 1001 R[t] <= M 1010 SW out <= + SX 1011 MEM[out] <= 1100 EQ If = th <= out 0010 Lc12.29 Ial pup i umb of tag i th pipli. o w achiv thi? Lc12.30 aic Ia Gaphically Rptig Pipli Ca hlp with awig qutio lik: how may cycl o it tak to xcut thi co? what i th oig uig cycl 4? u thi ptatio to hlp uta atapath What o w to a to plit th atapath ito tag? Lc12.31 Lc12.32

9 Covtioal Pipli utio Rptatio Tim IFtch c W IFtch c W IFtch c W IFtch c W IFtch c W Pogam Flow IFtch c W Sigl Cycl, Multipl Cycl, v. Pipli Cycl 1 Cycl 2 Clk Sigl Cycl Implmtatio: Loa Sto Wat Cycl 1 Cycl 2 Cycl 3 Cycl 4 Cycl 5 Cycl 6 Cycl 7 Cycl 8 Cycl 9 Cycl 10 Clk Multipl Cycl Implmtatio: Loa Sto R-typ Iftch W Iftch Iftch Pipli Implmtatio: Loa Iftch W Sto Iftch W Lc12.33 R-typ Iftch W Lc12.34 Why Pipli? Why Pipli? cau th ouc a th! Suppo w xcut 100 ituctio Tim (clock cycl) Sigl Cycl Machi 45 /cycl x 1 CPI x 100 it = 4500 Multicycl Machi 10 /cycl x 4.6 CPI (u to it mix) x 100 it = 4600 Ial pipli machi 10 /cycl x (1 CPI x 100 it + 4 cycl ai) = 1040 I t. It 0 It 1 It 2 It 3 It 4 Im m Im m Im m Im m Im m Lc12.35 Lc12.36

10 Ca pipliig gt u ito toubl? Y: Pipli Haza tuctual haza: attmpt to u th am ouc two ifft way at th am tim - E.g., combi wah/y woul b a tuctual haza o fol buy oig omthig l (watchig TV) ata haza: attmpt to u itm bfo it i ay - E.g., o ock of pai i y a o i wah; ca t fol util gt ock fom wah though y - ituctio p o ult of pio ituctio till i th pipli cotol haza: attmpt to mak a ciio bfo coitio i vaulat - E.g., wahig football uifom a to gt pop tgt lvl; to aft y bfo xt loa i - bach ituctio Ca alway olv haza by waitig pipli cotol mut tct th haza tak actio (o lay actio) to olv haza Lc12.37 I t. Sigl oy i a Stuctual Haza Loa It 1 It 2 It 3 It 4 Tim (clock cycl) tctio i ay i thi ca! (ight half highlight ma a, lft half wit) Lc12.38 Stuctual Haza limit pfomac Cotol Haza Solutio #1: Stall Exampl: if 1.3 mmoy acc p ituctio a oly o mmoy acc p cycl th avag CPI 1.3 othwi ouc i mo tha 100% utiliz I t. q Loa Tim (clock cycl) Lot pottial Stall: wait util ciio i cla Impact: 2 lot cycl (i.. 3 clock cycl p bach ituctio) => low Mov ciio to of co av 1 cycl p bach Lc12.39 Lc12.40

11 I t. Cotol Haza Solutio #2: Pict q Loa Tim (clock cycl) Pict: gu o ictio th back up if wog Impact: 0 lot cycl p bach ituctio if ight, 1 if wog (ight - 50% of tim) N to Squah a tat followig ituctio if wog Pouc CPI o bach of (1 * *.5) = 1.5 Total CPI might th b: 1.5 * *.8 = 1.1 (20% bach) Mo yamic chm: hitoy of 1 bach (- 90%) Lc12.41 Cotol Haza Solutio #3: lay ach I t. q Mic Loa Tim (clock cycl) lay ach: Rfi bach bhavio (tak plac aft xt ituctio) Impact: 0 clock cycl p bach ituctio if ca fi ituctio to put i lot (- 50% of tim) lauch mo ituctio p clock cycl, l uful Lc12.42 ata Haza o 1 ata Haza o 1: pci backwa i tim a haza a 1,2,3 ub 4, 1,3 a 6, 1,7 o 8, 1,9 xo 10, 1,11 I t. Tim (clock cycl) IF I/RF EX MEM W a 1,2,3 ub 4,1,3 a 6,1,7 o 8,1,9 xo 10,1,11 Im m Im m Im m Im m Im m Lc12.43 Lc12.44

12 I t. ata Haza Solutio: Fowa ult fom o tag to aoth Tim (clock cycl) IF I/RF EX MEM W a 1,2,3 ub 4,1,3 a 6,1,7 o 8,1,9 xo 10,1,11 Im m Im m Im m Im m Im m Fowaig (o ypaig): What about Loa? pci backwa i tim a haza Tim (clock cycl) IF I/RF EX MEM W Im m lw 1,0(2) ub 4,1,3 Ca t olv with fowaig: Mut lay/tall ituctio pt o loa Im m o K if fi a/wit poply Lc12.45 Lc12.46 Fowaig (o ypaig): What about Loa pci backwa i tim a haza Tim (clock cycl) IF I/RF EX MEM W lw 1,0(2) ub 4,1,3 Im m Stall Im m igig a Pipli Poco Go back a xami you atapath a cotol iagam aociat ouc with tat u that flow o ot coflict, o figu out how to olv at cotol i appopiat tag Ca t olv with fowaig: Mut lay/tall ituctio pt o loa Lc12.47 Lc12.48

13 Cotol a atapath: Split tat iag ito 5 pic <- []; < +4; <- R[]; < R[t] Pipli Poco (almot) fo li What happ if w tat a w ituctio vy cycl? S < + ; S < o ZX; S < + SX; M < [S] S < + SX; [S] <- If Co < +SX; It. Vali c x Ex mm wb W R[] < S; R[t] < S; It. R[] < M; Equal S M ata. Lc12.49 S M ata. Equal Lc12.50 Pipli atapath (a i book); ha to a Pipliig th Loa Ituctio Cycl 1 Cycl 2 Cycl 3 Cycl 4 Cycl 5 Cycl 6 Cycl 7 Clock 1t lw Iftch /c W 2 lw Iftch /c W 3 lw Iftch /c W Th fiv ipt fuctioal uit i th pipli atapath a: Ituctio oy fo th Iftch tag it Ra pot (bu a bu) fo th /c tag fo th tag ata oy fo th tag it Wit pot (bu W) fo th W tag Lc12.51 Lc12.52

14 Th Fou Stag of R-typ Cycl 1 Cycl 2 Cycl 3 Cycl 4 Clock Pipliig th R-typ a Loa Ituctio Cycl 1 Cycl 2 Cycl 3 Cycl 4 Cycl 5 Cycl 6 Cycl 7 Cycl 8 Cycl 9 R-typ Iftch /c W R-typ Iftch /c W p! W hav a poblm! Iftch: Ituctio Ftch Ftch th ituctio fom th Ituctio oy /c: it Ftch a Ituctio co : opat o th two git opa Upat W: Wit th output back to th git fil R-typ Iftch /c W Loa Iftch /c W R-typ Iftch /c W R-typ Iftch /c W W hav pipli coflict o tuctual haza: Two ituctio ty to wit to th git fil at th am tim! ly o wit pot Lc12.53 Lc12.54 Impotat bvatio Solutio 1: It ubbl ito th Pipli Each fuctioal uit ca oly b u oc p ituctio Each fuctioal uit mut b u at th am tag fo all ituctio: Loa u it Wit Pot uig it 5th tag Loa Iftch /c W R-typ u it Wit Pot uig it 4th tag Clock Cycl 1 Cycl 2 Cycl 3 Cycl 4 Cycl 5 Cycl 6 Cycl 7 Cycl 8 Cycl 9 Iftch /c W Loa Iftch /c W R-typ Iftch /c W R-typ Iftch /c Pipli W R-typ Iftch ubbl /c W Iftch /c R-typ Iftch /c W 2 way to olv thi pipli haza. It a bubbl ito th pipli to pvt 2 wit at th am cycl Th cotol logic ca b complx. Lo ituctio ftch a iu oppotuity. No ituctio i tat i Cycl 6! Lc12.55 Lc12.56

15 Solutio 2: lay R-typ Wit by Cycl lay R-typ git wit by o cycl: Now R-typ ituctio alo u wit pot at Stag 5 tag i a NP tag: othig i big o R-typ Iftch /c W Moifi Cotol & atapath <- []; < +4; <- R[]; < R[t] S < + ; S < o ZX; S < + SX; S < + SX; if Co < +SX; Clock Cycl 1 Cycl 2 Cycl 3 Cycl 4 Cycl 5 Cycl 6 Cycl 7 Cycl 8 Cycl 9 M < S M < S M < [S] [S] <- R-typ Iftch /c W R-typ Iftch /c W R[] < M; R[t] < M; R[] < M; Equal Loa Iftch /c W R-typ Iftch /c W R-typ Iftch /c W Lc12.57 It. S M ata. Lc12.58 Th Fou Stag of Sto Th Th Stag of q Cycl 1 Cycl 2 Cycl 3 Cycl 4 Cycl 1 Cycl 2 Cycl 3 Cycl 4 Sto Iftch /c W q Iftch /c W Iftch: Ituctio Ftch Ftch th ituctio fom th Ituctio oy /c: it Ftch a Ituctio co : Calculat th mmoy a : Wit th ata ito th ata oy Iftch: Ituctio Ftch Ftch th ituctio fom th Ituctio oy /c: it Ftch a Ituctio co : compa th two git opa, lct coct bach tagt a latch ito Lc12.59 Lc12.60

16 Cotol iagam atapath + ata Statioay Cotol S < + ; S < o ZX; <- []; < +4; <- R[]; < R[t] S < + SX; S < + SX; If Co < +SX; It. fu op t co t v w wb m x im v w wb m v w wb W M < S R[] < S; M < S R[t] < S; M < [S] R[] < M; Equal [S] <- S M ata. It. S M ata. Lc12.61 Lc12.62 Lt Ty it ut Stat: Ftch lw 1, 2(35) 14 ai 2, 2, 3 20 ub 3, 4, 5 24 bq 6, 7, oi 8, 9, a 10, 11, a 13, 14, 15 th a a octal It. co t im S 10 M ata W. 10 lw 1, 2(35) 14 ai 2, 2, 3 20 ub 3, 4, 5 24 bq 6, 7, oi 8, 9, a 10, 11, 12 Lc a 13, 14, 15 Lc12.64

17 Ftch 14, co 10 Ftch 20, co 14, 10 It. lw 1, 2(35) co 2 t im W It. ai 2, 2, 3 co 2 t lw 1 35 W S M. 2 S M. ata 10 lw 1, 2(35) 14 ai 2, 2, 3 ata 10 lw 1, 2(35) 14 ai 2, 2, ub 3, 4, 5 24 bq 6, 7, oi 8, 9, ub 3, 4, 5 24 bq 6, 7, oi 8, 9, a 10, 11, a 10, 11, a 13, 14, 15 Lc a 13, 14, 15 Lc12.66 Ftch 24, co 20, 14, 10 Ftch 30, c 24, Ex 20, 14, W 10 It. ub 3, 4, 5 co ai 2, 2, 3 lw 1 W It. bq 6, co 6 7 ub 3 ai 2 lw 1 W M M[2+35]. ata 10 lw 1, 2(35) 14 ai 2, 2, 3 ata 10 lw 1, 2(35) 14 ai 2, 2, ub 3, 4, 5 24 bq 6, 7, oi 8, 9, ub 3, 4, 5 24 bq 6, 7, oi 8, 9, a 10, 11, a 10, 11, a 13, 14, 15 Lc a 13, 14, 15 Lc12.68

18 Ftch 34, c 30, Ex 24, 20, W 14 Ftch 100, c 34, Ex 30, 24, W 20 It. oi 8, 9 17 co 9 xx bq ub ai W. 1=M[2+35] It. a 10, 11, 12 co oi 8 9 x bq xxx ub W. 1=M[2+35] 2 = 2+3 ata 10 lw 1, 2(35) 14 ai 2, 2, 3 ata 10 lw 1, 2(35) 14 ai 2, 2, ub 3, 4, 5 24 bq 6, 7, oi 8, 9, ub 3, 4, 5 24 bq 6, 7, oi 8, 9, a 10, 11, a 10, 11, a 13, 14, 15 Lc12.69 ooop, w houl hav oly o lay ituctio 100 a 13, 14, 15 Lc12.70 Ftch 104, c 100, Ex 34, 30, W 24 It. a 13, 14, 15 co a 10 xx oi bq xxx ata Squah th xta ituctio i th bach haow! W. 1=M[2+35] 2 = = lw 1, 2(35) 14 ai 2, 2, 3 20 ub 3, 4, 5 24 bq 6, 7, oi 8, 9, a 10, 11, a 13, 14, 15 Lc12.71 Ftch 108, c 104, Ex 100, 34, W 30 It. co a 13 xx a oi ata Squah th xta ituctio i th bach haow! W. 1=M[2+35] 2 = = lw 1, 2(35) 14 ai 2, 2, 3 20 ub 3, 4, 5 24 bq 6, 7, oi 8, 9, a 10, 11, a 13, 14, 15 Lc12.72

19 Ftch 114, c 110, Ex 104, 100, W 34 It. co a & R a ata Squah th xta ituctio i th bach haow! W. N W N vflow 1=M[2+35] 2 = = = lw 1, 2(35) 14 ai 2, 2, 3 20 ub 3, 4, 5 24 bq 6, 7, oi 8, 9, a 10, 11, a 13, 14, 15 Lc12.73 Summay: Pipliig What mak it ay all ituctio a th am lgth jut a fw ituctio fomat mmoy opa appa oly i loa a to What mak it ha? tuctual haza: uppo w ha oly o mmoy cotol haza: to woy about bach ituctio ata haza: a ituctio p o a pviou ituctio W ll buil a impl pipli a look at th iu W ll talk about mo poco a what ally mak it ha: xcptio halig tyig to impov pfomac with out-of-o xcutio, tc. Lc12.74 Summay Pipliig i a fuamtal cocpt multipl tp uig itict ouc Utiliz capabiliti of th atapath by pipli ituctio pocig tat xt ituctio whil wokig o th cut o limit by lgth of logt tag (plu fill/fluh) tct a olv haza Lc12.75