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 an b Do th aition Sto th ult in x Goto tp 1 Intuction Ftch Intuction Dco pan Ftch Excut Rult Sto Nxt Intuction Cpt280, Autumn 2017 1
Typical Ftch-Excut Achitctu Pogam Count (PC) Pogam Mmoy Intuction Rgit A Contol R In Mmoy (Rgit Fil) Initializ PC to Point to Fit Intuction Pogam Count (PC) Pogam Mmoy Intuction Rgit A Contol R In Mmoy (Rgit Fil) Cpt280, Autumn 2017 2
Rout Intuction to Intuction Rgit Pogam Count (PC) Pogam Mmoy Intuction Rgit A Contol R In Mmoy (Rgit Fil) Slct Fom Rgit Fil Pogam Count (PC) Pogam Mmoy Intuction Rgit A Contol R In Mmoy (Rgit Fil) Cpt280, Autumn 2017 3
Rout to Aithmtic Unit () Pogam Count (PC) Pogam Mmoy Intuction Rgit A Contol R In Mmoy (Rgit Fil) Do th Computation Pogam Count (PC) Pogam Mmoy Intuction Rgit A Contol R In Mmoy (Rgit Fil) Cpt280, Autumn 2017
Sto th Rult Pogam Count (PC) Pogam Mmoy Intuction Rgit A Contol R In Mmoy (Rgit Fil) Incmnt th PC to Nxt Intuction Pogam Count (PC) Pogam Mmoy Intuction Rgit A Contol R In Mmoy (Rgit Fil) Cpt280, Autumn 2017 5
Piplining Piplining povi a mtho fo xcuting multipl intuction at th am tim. Exampl Ann, Bian, Cathy, Dav ach hav on loa of cloth to wah, y, an fol; Wah tak 30 minut; Dy tak 0 minut; Fol tak 20 minut. A B C D Squntial Launy 6 PM 7 8 9 10 11 Minight Tim T a k A B C D 30 0 20 30 0 20 30 0 20 30 0 20 Squntial launy tak 6 hou fo loa. Cpt280, Autumn 2017 6
Piplin Launy 6 PM 7 8 9 10 11 Minight Tim T a k A B C D 30 0 0 0 0 20 Piplin launy tak 3.5 hou fo loa. Piplining Lon T a k 6 PM 7 8 9 Tim 30 0 0 0 0 20 A B C D Piplining on t hlp latncy of ingl tak, it hlp thoughput of nti wokloa; Piplin at limit by lowt piplin tag; Piplining i multipl tak, opating imultanouly, uing iffnt ouc; Potntial pup = numb of pip tag; Unbalanc lngth of pip tag uc pup; Tim to fill piplin an tim to ain it uc pup; What low own th pip? Cpt280, Autumn 2017 7
Th Fiv Stag of th Loa Intuction Cycl 1 Cycl 2 Cycl 3 Cycl Cycl 5 Loa Iftch Rg/Dc Exc Mm W Iftch: Intuction ftch fom th intuction mmoy; Rg/Dc: Ftch ata fom git an co intuction; Exc: Calculat th mmoy a; Mm: Ra th ata fom ata mmoy; W: Wit th ata into th git fil. Piplin Excution Tim IFtch Dc Exc Mm WB IFtch Dc Exc Mm WB IFtch Dc Exc Mm WB IFtch Dc Exc Mm WB IFtch Dc Exc Mm WB Pogam Flow IFtch Dc Exc Mm WB Multipl intuction a in vaiou tag at th am tim. Aum: Each intuction tak fiv cycl. Cpt280, Autumn 2017 8
Gaphically Rpnting Piplin Tim (clock cycl) I n t. Int 0 Int 1 Sing th piplin can anw qution lik: How many cycl o it tak to xcut thi co? What i th oing uing cycl? A two intuction tying to u th am ouc at th am tim? Full Piplin Tim (clock cycl) I n t. Int 0 Int 1 Int 2 Int 3 Int Cpt280, Autumn 2017 9
Piplin Exampl CPI (Clock Cycl P Intuction) th mtic by which comput pfomanc i mau. Suppo: 100 intuction a xcut; Th ingl cycl machin (complt an intuction in on clock cycl) ha a cycl tim of 5 n; Th piplin machin ha a cycl tim of 10 n; Singl Cycl Machin 5 n/cycl x 1 CPI x 100 intuction = 500 n Ial Piplin Machin 10 n/cycl x (1 CPI x 100 intuction + cycl ain) = 100 n Ial piplin v. ingl cycl pup 500 n / 100 n =.33 Can Piplining gt u into Toubl? Y: Piplin Haza Stuctual haza - attmpt to u th am ouc two iffnt way at th am tim E.g., two intuction ty to a th am mmoy at th am tim. haza - attmpt to u an itm bfo it i ay Intuction pn on ult of pio intuction till in th piplin a 1, 2, 3 ub, 2, 1 Contol haza - attmpt to mak a ciion bfo conition i valuat Banch intuction a 1, 2, 3 bq 1, loop Can alway olv haza by waiting. Piplin contol mut tct an olv th haza. Cpt280, Autumn 2017 10
Singl Mmoy i a Stuctual Haza Tim (clock cycl) I n t. Loa Int 1 Int 2 Int 3 Int Mm Rg Mm Rg Mm Rg Mm Rg Mm Mm Rg Mm Rg Rg Mm Rg Mm Rg Mm Rg Stuctual Haza Limit Pfomanc Solution 1: U paat intuction an ata mmoi. Solution 2: Allow mmoy to a an wit mo than on wo p cycl. Solution 3: Stall. Cpt280, Autumn 2017 11
Haza Fo Rgit 1 Poblm: 1 cannot b a by oth intuction bfo it i wittn by th a intuction. a 1, 2, 3 ub, 1, 3 an 6, 1, 7 o 8, 1,9 xo 10, 1, 11 I n t. Haza on 1 Dpnnci backwa in tim a haza. Tim (clock cycl) IF ID/RF EX MEM WB a 1,2,3 ub,1,3 an 6,1,7 o 8,1,9 xo 10,1,11 Im Rg Dm Rg Cpt280, Autumn 2017 12
Haza Solution Th o i K if w fin a/wit poply. Fowa ult fom on tag to anoth. I n t. Tim (clock cycl) IF ID/RF EX MEM WB a 1,2,3 ub,1,3 an 6,1,7 o 8,1,9 xo 10,1,11 Im Rg Dm Rg Fowaing Loa Intuction Dpnnci backwa in tim a haza. Tim (clock cycl) IF ID/RF EX MEM WB lw 1,0(2) ub,1,3 Can t olv with fowaing - mut lay/tall intuction that a pnnt on loa. Cpt280, Autumn 2017 13
Contol Haza Solution 1 Stall, wait until ciion i cla. Solution 2 A mo hawa in 2 n tag to o th git compaion. I n t. A Bq Loa Tim (clock cycl) Mm Rg Mm Rg Mm Rg Mm Rg Mm Rg Mm Rg Contol Haza Solution Banch Piction Banch piction bit() Banch Hitoy Tabl I n t. A Bq Loa Tim (clock cycl) Mm Rg Mm Rg Mm Rg Mm Rg Mm Rg Mm Rg Cpt280, Autumn 2017 1
Summay Squntial xcution wa viw; Piplining wa intouc an illutat; Haza w not Stuctual; ; Contol. Solution to haza icu Stuctual mo hawa; - fowaing an bypaing; Contol banch piction. Cpt280, Autumn 2017 15