In Review: A Single Cycle Datapath We have everything! Now we just need to know how to BUILD CONTROL

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Neil Arnold
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S6 L2 PU ign: ontol II n Piplining I () int.c.bly.u/~c6c S6 : Mchin Stuctu Lctu 2 PU ign: ontol II & Piplining I Noh Johnon 2-7-26 In Rviw: Singl ycl tpth W hv vything!

MmtoRg http://xc.com/676/ S6 L2 PU ign: ontol II n Piplining I (2) Extp Sc Summy: Singl-cycl Poco 5 tp to ign poco. nlyz intuction t tpth quimnt 2. Slct t of tpth componnt & tblih cloc mthoology 3.

1 S6 L2 PU ign: ontol II n Piplining I () int.c.bly.u/~c6c S6 : Mchin Stuctu Lctu 2 PU ign: ontol II & Piplining I Noh Johnon In Rviw: Singl ycl tpth W hv vything! Now w jut n to now how to UIL NRL Rgt R RgW R Rt buw cl imm6 np_l Rt Rw R Rb RgFil 6 cl Extn bu bu int ftch unit zo Intuction<3:> <2:25> R ct Z t In cl <6:2> Rt MmW R <:5> WEn <:5> t Mmoy Imm6 MmtoRg S6 L2 PU ign: ontol II n Piplining I (2) Extp Sc Summy: Singl-cycl Poco 5 tp to ign poco. nlyz intuction t tpth quimnt 2. Slct t of tpth componnt & tblih cloc mthoology 3. mbl tpth mting th quimnt 4. nlyz implmnttion of ch intuction to tmin tting of contol point tht ffct th git tnf. 5. mbl th contol logic Fomult Logic Eqution ign icuit Poco ontol Mmoy Input Stp 4: Givn tpth: RL ontol Int Mmoy Intuction<3:> <26:3> <:5> p Fun Rt <2:25> R <6:2> np_l RgW Rgt Extp Sc ct MmW MmtoRg R <:5> ontol <:5> Imm6 tpth utput PH S6 L2 PU ign: ontol II n Piplining I (3) S6 L2 PU ign: ontol II n Piplining I (4) Summy of th ontol Signl (/2) int Rgit nf R[] R[] + R[t]; P P + 4 c = Rg, ct =, Rgt =, RgW, np_l = +4 ub R[] R[] R[t]; P P + 4 c = Rg, ct = SU, Rgt =, RgW, np_l = +4 oi R[t] R[] + zo_xt(imm6); P P + 4 c = Im, Extop = Z,ct = R, Rgt = t,rgw, np_l = +4 lw R[t] MEM[ R[] + ign_xt(imm6)]; P P + 4 c = Im, Extop = n, ct =, MmtoRg, Rgt = t, RgW, np_l = +4 w MEM[ R[] + ign_xt(imm6)] R[]; P P + 4 c = Im, Extop = n, ct =, MmW, np_l = +4 bq if ( R[] == R[t] ) thn P P + ign_xt(imm6)] l P P + 4 np_l = b, ct = SU S6 L2 PU ign: ontol II n Piplining I (5) Summy of th ontol Signl (2/2) S func W on t :-) ppnix op ub oi lw w bq Rgt x x x Sc x MmtoRg x x x RgWit MmWit npl? Extp x x x x ct<2:> Subtct Subtct x R-typ op t hmt funct, ub I-typ op t immit oi, lw, w, bq J-typ op tgt S6 L2 PU ign: ontol II n Piplining I (6)

2 S6 L2 PU ign: ontol II n Piplining I (7) ooln Expion fo ontoll ontoll Implmnttion Rgt = + ub Sc = oi + lw + w MmtoRg = lw RgWit = + ub + oi + lw MmWit = w npl = bq = Extp = lw + w ct[] = ub + bq (um ct i, : SU, : R) ct[] = o wh, typ = ~op 5 ~op 4 ~op 3 ~op 2 ~op ~op, oi = ~op 5 ~op 4 op 3 op 2 ~op op lw = op 5 ~op 4 ~op 3 ~op 2 op op w = op 5 ~op 4 op 3 ~op 2 op op bq = ~op 5 ~op 4 ~op 3 op 2 ~op ~op = ~op 5 ~op 4 ~op 3 ~op 2 op ~op How o w implmnt thi in gt? opco func N logic ub oi lw w bq R logic Rgt Sc MmtoRg RgWit MmWit npl Extp ct[] ct[] = typ func 5 ~func 4 ~func 3 ~func 2 ~func ~func ub = typ func 5 ~func 4 ~func 3 ~func 2 func ~func S6 L2 PU ign: ontol II n Piplining I (8) P Intuction np_l R Rt Rgt RgW 5 R Rt 5 5 ct buw Rw R Rb bu -bit Rgit imm6 6 bu Extn ) MmoRg= x & ct= ub. SU o EQ? Extp 2) ct=. Which ignl i iffnt fo ll 3 of:, LW, & SW? Rgt o Extp? Sc S6 L2 PU ign: ontol II n Piplining I (9) Intuction Ftch Unit t In Intuction<3:> <2:25> <6:2> WEn t Mmoy <:5> <:5> Rt R R Imm6 Zo MmW MmtoRg 2 ) SR b) SE c) R ) E Summy: Singl-cycl Poco 5 tp to ign poco. nlyz intuction t tpth quimnt 2. Slct t of tpth componnt & tblih cloc mthoology 3. mbl tpth mting th quimnt 4. nlyz implmnttion of ch intuction to tmin tting of contol point tht ffct th git tnf. 5. mbl th contol logic Fomult Logic Eqution ign icuit S6 L2 PU ign: ontol II n Piplining I () Poco ontol tpth Mmoy Input utput Rviw: Singl cycl tpth 5 tp to ign poco. nlyz intuction t tpth quimnt 2. Slct t of tpth componnt & tblih cloc mthoology 3. mbl tpth mting th quimnt 4. nlyz implmnttion of ch intuction to tmin tting of contol point tht ffct th git tnf. 5. mbl th contol logic Poco ontol i th h pt MIPS m tht i ontol Intuction m iz tpth Souc git lwy in m plc Immit m iz, loction ption lwy on git/immit Mmoy Input utput How W uil h ontoll Rgt = + ub Sc = oi + lw + w ub MmtoRg = lw RgWit = + ub + oi + lw MmWit = w N logic oi lw w R logic bq npl = bq = Extp = lw + w ct[] = ub + bq (um ct i, : SU, : R) ct[] = o wh, typ = ~op 5 ~op 4 ~op 3 ~op 2 ~op ~op, oi = ~op 5 ~op 4 op 3 op 2 ~op op lw = op 5 ~op 4 ~op 3 ~op 2 op op w = op 5 ~op 4 op 3 ~op 2 op op bq = ~op 5 ~op 4 ~op 3 op 2 ~op ~op = ~op 5 ~op 4 ~op 3 ~op 2 op ~op opco func = typ func 5 ~func 4 ~func 3 ~func 2 ~func ~func ub = typ func 5 ~func 4 ~func 3 ~func 2 func ~func Rgt Sc MmtoRg RgWit MmWit npl Extp ct[] ct[] migoh omigoh, o you now wht thi mn? S6 L2 PU ign: ontol II n Piplining I () S6 L2 PU ign: ontol II n Piplining I (2)

3 S6 L2 PU ign: ontol II n Piplining I (3) Poco Pfomnc Gott o Luny n w timt th cloc t (fquncy) of ou ingl-cycl poco? W now: cycl p intuction lw i th mot mning intuction. um th ly fo mjo pic of th tpth: Int. Mm,, t Mm : 2n ch, gfil n Intuction xcution qui: = 8n 25 MHz Wht cn w o to impov cloc t? Will thi impov pfomnc wll? W wnt inc in cloc t to ult in pogm xcuting quic. nn, in, thy, v ch hv on lo of cloth to wh, y, fol, n put wy Wh t 3 minut y t 3 minut Fol t 3 minut Sth t 3 minut to put cloth into w S6 L2 PU ign: ontol II n Piplining I (4) Squntil Luny Piplin Luny 6 PM M 6 PM M im Squntil luny t 8 hou fo 4 lo S6 L2 PU ign: ontol II n Piplining I (5) Piplin luny t 3.5 hou fo 4 lo! S6 L2 PU ign: ontol II n Piplining I (6) im Gnl finition Piplining Lon (/2) Ltncy: tim to compltly xcut ctin t fo xmpl, tim to cto fom i i i cc tim o i ltncy houghput: mount of wo tht cn b on ov pio of tim 6 PM im Piplining on t hlp ltncy of ingl t, it hlp thoughput of nti wolo Multipl t opting imultnouly uing iffnt ouc Potntil pup = Numb pip tg im to fill piplin n tim to in it uc pup: 2.3X v. 4X in thi xmpl S6 L2 PU ign: ontol II n Piplining I (7) S6 L2 PU ign: ontol II n Piplining I (8)

4 S6 L2 PU ign: ontol II n Piplining I (9) Piplining Lon (2/2) Stp in Excuting MIPS 6 PM im Suppo nw Wh t 2 minut, nw Sth t 2 minut. How much ft i piplin? Piplin t limit by lowt piplin tg Unblnc lngth of pip tg uc pup ) IFtch: Intuction Ftch, Incmnt P 2) c: Intuction co, R Rgit 3) Exc: Mm-f: lcult ith-log: Pfom ption 4) Mm: Lo: R t fom Mmoy Sto: Wit t to Mmoy 5) W: Wit t c to Rgit S6 L2 PU ign: ontol II n Piplining I (2) Piplin Hz: Mtching oc in lt lo minitivi 6 PM M E F bubbl pn on ; tll inc fol ti up S6 L2 PU ign: ontol II n Piplining I (2) im HW8 u tomoow Pojct 2 u nxt Mony Nwgoup poblm Rmin: Mitm g u toy S6 L2 PU ign: ontol II n Piplining I (22) Poblm fo Piplining PU Stuctul Hz #: Singl Mmoy (/2) Limit to piplining: Hz pvnt nxt intuction fom xcuting uing it ignt cloc cycl Stuctul hz: HW cnnot uppot om combintion of intuction (ingl pon to fol n put cloth wy) ontol hz: Piplining of bnch cu lt intuction ftch to wit fo th ult of th bnch t hz: Intuction pn on ult of pio intuction till in th piplin (miing oc) h might ult in piplin tll o bubbl in th piplin. S6 L2 PU ign: ontol II n Piplining I (23) I n Lo t Int. Int 2 Int 3 Int 4 R m mmoy twic in m cloc cycl S6 L2 PU ign: ontol II n Piplining I (24) im (cloc cycl)

S6 L2 PU ign: ontol II n Piplining I (25) Stuctul Hz #: Singl Mmoy (2/2) Solution: infibl n infficint to ct con mmoy (W ll ln bout thi mo nxt w) o imult thi by hving two Lvl ch ( tmpoy mll [of uully

S6 L2 PU ign: ontol II n Piplining I (26) im (cloc cycl) I$ Rg $ Rg Stuctul Hz #2: Rgit (2/2) P Intuction wo iffnt olution hv bn u: ) RgFil cc i VERY ft: t l thn hlf th tim of tg Wit to Rgit uing fit

5 S6 L2 PU ign: ontol II n Piplining I (25) Stuctul Hz #: Singl Mmoy (2/2) Solution: infibl n infficint to ct con mmoy (W ll ln bout thi mo nxt w) o imult thi by hving two Lvl ch ( tmpoy mll [of uully mot cntly u] copy of mmoy) hv both n L Intuction ch n n L t ch n mo complx hw to contol whn both cch mi I n t. Stuctul Hz #2: Rgit (/2) w Int Int 2 Int 3 Int 4 n w n wit to git imultnouly? S6 L2 PU ign: ontol II n Piplining I (26) im (cloc cycl) I$ Rg $ Rg Stuctul Hz #2: Rgit (2/2) P Intuction wo iffnt olution hv bn u: ) RgFil cc i VERY ft: t l thn hlf th tim of tg Wit to Rgit uing fit hlf of ch cloc cycl R fom Rgit uing con hlf of ch cloc cycl 2) uil RgFil with inpnnt n wit pot Rult: cn pfom R n Wit uing m cloc cycl ) hn to piplining, I hv uc th tim it too m to wh my on hit. 2) Long piplin lwy win (inc l wo p tg & ft cloc). 2 ) FF b) F c) F ) S6 L2 PU ign: ontol II n Piplining I (27) S6 L2 PU ign: ontol II n Piplining I (28) hing to Rmmb onu li ptiml Piplin Ech tg i xcuting pt of n intuction ch cloc cycl. n intuction finih uing ch cloc cycl. n vg, xcut f mo quicly. Wht m thi wo? Similiti btwn intuction llow u to u m tg fo ll intuction (gnlly). Ech tg t bout th m mount of tim ll oth: littl wt tim. S6 L2 PU ign: ontol II n Piplining I (3) h xt li tht u to b inclu in lctu not, but hv bn mov to thi, th bonu to v upplmnt. h li will pp in th o thy woul hv in th noml pnttion S6 L2 PU ign: ontol II n Piplining I (3)

6 imm6 S6 L2 PU ign: ontol II n Piplining I (33) imm6 h Singl ycl tpth uing 3 J-typ op Nw P = { P[3..28], tgt, } = Intuction<3:> np_l=? Intuction R Rt Ftch Unit Rgt = x RgW = R Rt ct =x Rt R R Imm MmtoRg = x bu Zo MmW = Rw R Rb buw -bit Rgit bu WEn imm6 6 Extn Extp = x tgt Sc = x t In <2:25> <6:2> t Mmoy <:5> <:5> <:25> Intuction Ftch Unit t th En of 3 J-typ Nw P = { P[3..28], tgt, } np_l Zo 4 op Int Mmoy np_mux_l P S6 L2 PU ign: ontol II n Piplining I (34) tgt Intuction<3:> How o w moify thi to ccount fo? Intuction Ftch Unit t th En of 3 J-typ Nw P = { P[3..28], tgt, } np_l Zo 4 op np_mux_l 26 4 (MS) tgt Int Mmoy P Intuction<3:> Quy n Zo till gt t? o np_l n to b? If not, wht? S6 L2 PU ign: ontol II n Piplining I (35)

Instruction Execution

Instruction Execution MIPS Piplining Cpt280 D Cuti Nlon Intuction Excution C intuction: x = a + b; Ambly intuction: a a,b,x Stp 1: Stp 2: Stp 3: Stp : Stp 5: Stp 6: Ftch th intuction Dtmin it i an a intuction Ftch th ata a