Comp 411: Computer Organization Fall 2007 - Course Introduction - Prof. Leona Mcmillan, Study notes of Computer Architecture and Organization

Download Comp 411: Computer Organization Fall 2007 - Course Introduction - Prof. Leona Mcmillan and more Study notes Computer Architecture and Organization in PDF only on Docsity! L0 1 - In tr o d u ct io n 1 C o m p 4 11 – F a ll 2 0 0 7 8 /2 2 /2 0 0 6 W e lc om e t o C om p 4 1 1 ! I t h o u g h t th is c o u rs e w a s c a lle d “C o m p u te r O rg a n iz a ti o n ” D a vi d M a ca ul a y 1) C o u rs e M ec h a n ic s 2 ) C o u rs e O bj ec ti ve s 3 ) In fo rm a ti o n L0 1 - In tr o d u ct io n 2 C o m p 4 11 – F a ll 2 0 0 7 8 /2 2 /2 0 0 6 M e e t th e C re w … Le ct u re s : Le o n a rd M cM ill a n ( S N -2 5 8 ) O ff ic e H o u rs T 2 -3 TA : TB A ( W e’ ll fi n d o u t n ex t M o n d a y) B o o k: P a tt er s o n & H en n es s y C o m p u te r O rg a n iz a ti o n & D es ig n 3 rd E d it io n , I S B N : 1 -5 5 8 6 0 -6 0 4 -1 (H o w ev er , y o u w o n ’t n ee d it fo r th e n ex t co u p le o f w ee ks ) L0 1 - In tr o d u ct io n 5 C o m p 4 11 – F a ll 2 0 0 7 8 /2 2 /2 0 0 6 G oa l 1 : D e m y st if y C om pu te rs S tr a n g el y, m o s t p eo p le ( ev en s o m e co m p u te r s ci en ti s ts ) a re a fr a id o f co m p u te rs . W e a re o n ly a fr a id o f th in g s w e d o n o t u n d er s ta n d ! I d o n o t fe a r co m p u te rs . I f ea r th e la ck o f th em . - Is a a c A s im o v (1 9 2 0 - 19 9 2 ) F ea r is t h e m a in s o u rc e o f s u p er s ti ti o n , a n d o n e o f th e m a in s o u rc es o f cr u el ty . T o c o n q u er f ea r is t h e be g in n in g o f w is d o m . - B er tr a n d R u s s el l ( 18 72 – 19 70 ) L0 1 - In tr o d u ct io n 6 C o m p 4 11 – F a ll 2 0 0 7 8 /2 2 /2 0 0 6 G oa l 2 : Po w e r of A b st ra ct io n D ef in e a f u n ct io n , d ev el o p a r o bu s t im p le m en ta ti o n , a n d th en p u t a b o x a ro u n d it . A bs tr a ct io n e n a bl es u s t o c re a te u n fa th o m a bl e s ys te m s (i n cl u d in g c o m p u te rs ). W h y d o w e n ee d A B S TR A C TI O N … Im a g in e a b ill io n - -- 1, 0 0 0 ,0 0 0 ,0 0 0 L0 1 - In tr o d u ct io n 7 C o m p 4 11 – F a ll 2 0 0 7 8 /2 2 /2 0 0 6 T h e k e y t o b ui ld in g sy st e m s w it h > 1 G c om po ne nt s P er s o n a l C o m p u te r: H a rd w a re & S o ft w a re C ir cu it B o a rd : ≈ ≈≈≈ 8 / s ys te m 1- 2 G d ev ic es In te g ra te d C ir cu it : ≈ ≈≈≈ 8 -1 6 / P C B .2 5 M -1 6 M d ev ic es M o d u le : ≈ ≈≈≈ 8 -1 6 / IC 10 0 K d ev ic es C el l: ≈ ≈≈≈ 1K -1 0 K / M o d u le 16 -6 4 d ev ic es G a te : ≈ ≈≈≈ 2 - 1 6 / C e ll 8 d e vi ce s S ch em e fo r re p re s en ti n g in fo rm a ti o n M O S F E T L0 1 - In tr o d u ct io n 1 0 C o m p 4 11 – F a ll 2 0 0 7 8 /2 2 /2 0 0 6 O ur P la n of A tt a ck …



U n d er s ta n d h o w t h in g s w o rk , b y a lt er n a ti n g b et w ee n lo w -l ev el ( bo tt o m -u p ) a n d h ig h le ve l( to p -d o w n ) co n ce p ts



E n ca p s u la te o u r u n d er s ta n d in g u s in g a p p ro p ri a te a bs tr a ct io n s
S tu d y o rg a n iz a ti o n a l p ri n ci p le s : h ie ra rc h y, in te rf a ce s , A P Is .



R o ll u p o u r s le ev es a n d d es ig n a t ea ch le ve l o f h ie ra rc h y



Le a rn e n g in ee ri n g t ri ck s - fr o m a h is to ri ca l p er s p ec ti ve - u s in g s ys te m a ti c d es ig n a p p ro a ch es - d ia g n o s e, f ix , a n d a vo id b u g s L0 1 - In tr o d u ct io n 1 1 C o m p 4 11 – F a ll 2 0 0 7 8 /2 2 /2 0 0 6 W h a t is “ C om pu ta ti on ”? C om pu ta ti on i s ab ou t “p ro ce ss in g in fo rm at io n” - Tr a n s fo rm in g in fo rm a ti o n fr o m o n e fo rm to a n o th er - D er iv in g n ew in fo rm a ti o n fr o m o ld - F in d in g in fo rm a ti o n a s s o ci a te d w it h a g iv en in p u t - “C om pu ta ti on ” d es cr ib es t h e m o ti o n o f in fo rm a ti o n th ro u g h t im e - “C om m un ic at io n” d es cr ib es t h e m o ti o n o f in fo rm a ti o n th ro u g h s p a ce L0 1 - In tr o d u ct io n 1 2 C o m p 4 11 – F a ll 2 0 0 7 8 /2 2 /2 0 0 6 W h a t is “ I nf or m a ti on ”? in fo rm a ti o n , n . K n o w le d g e co m m u n ic a te d o r re ce iv ed co n ce rn in g a p a rt ic u la r fa ct o r ci rc u m s ta n ce . In fo rm at io n re so lv es u nc er ta in ty . In fo rm a ti o n is s im p ly t h a t w h ic h ca n n o t be p re d ic te d . T h e le s s p re d ic ta bl e a m es s a g e is , t h e m o re in fo rm a ti o n it c o n ve ys ! T ar h ee ls w on ! A re y ou s ur e? I t is f oo tb al l se as on … “ 1 0 P ro b le m s et s, 2 qu iz ze s, a nd a f in al !” A C om pu te r Sc ie nt is t’ s D ef in it io n: L0 1 - In tr o d u ct io n 1 5 C o m p 4 11 – F a ll 2 0 0 7 8 /2 2 /2 0 0 6 Q ua nt if y in g I nf or m a ti on (C la ud e S h a nn on , 1 9 4 8 ) S u p p o s e yo u ’r e fa ce d w it h N e q u a lly p ro ba bl e ch o ic es , a n d I g iv e yo u a f a ct t h a t n a rr o w s it d o w n t o M c h o ic es . T h en yo u ’v e be en g iv en : lo g 2 (N /M ) b it s of i nf or m a ti on E xa m p le s :



in fo rm a ti o n in o n e co in f lip : l o g 2 (2 /1 ) = 1 b it



ro ll o f a s in g le d ie : l o g 2 (6 /1 ) = ~ 2 .6 b it s



o u tc o m e o f a F o o tb a ll g a m e: 1 b it (w el l, a ct u a lly , “ th ey w o n ” m a y co n ve y m o re in fo rm a ti o n t h a n “ th ey lo s t” … ) In fo rm at io n is m ea su re d in b it s (b in ar y d ig it s) = nu m b er o f 0 /1 ’s r eq ui re d to e nc od e ch oi ce (s ) L0 1 - In tr o d u ct io n 1 6 C o m p 4 11 – F a ll 2 0 0 7 8 /2 2 /2 0 0 6 E x a m pl e : S um o f 2 d ic e 2 3 4 5 6 7 8 9 1 0 1 1 1 2 i 2 = l og 2 (3 6 /1 ) = 5 .1 7 0 b it s i 3 = l og 2 (3 6 /2 ) = 4 .1 7 0 b it s i 4 = l og 2 (3 6 /3 ) = 3 .5 8 5 b it s i 5 = l og 2 (3 6 /4 ) = 3 .1 7 0 b it s i 6 = l og 2 (3 6 /5 ) = 2 .8 4 8 b it s i 7 = l og 2 (3 6 /6 ) = 2 .5 8 5 b it s i 8 = l og 2 (3 6 /5 ) = 2 .8 4 8 b it s i 9 = l og 2 (3 6 /4 ) = 3 .1 7 0 b it s i 1 0 = l og 2 (3 6 /3 ) = 3 .5 8 5 b it s i 1 1 = l og 2 (3 6 /2 ) = 4 .1 7 0 b it s i 1 2 = l og 2 (3 6 /1 ) = 5 .1 7 0 b it s T h e a ve ra ge in fo rm a ti on p ro vi d e d b y t h e s um o f 2 d ic e : b it s 2 7 4 3 p p i i i 2 i 1 2 2 i MN 2 NM i i . ) ( lo g ) ( lo g av e = − = = ∑ ∑ = E nt ro py L0 1 - In tr o d u ct io n 1 7 C o m p 4 11 – F a ll 2 0 0 7 8 /2 2 /2 0 0 6 S h ow M e t h e B it s! C a n t h e s u m o f tw o d ic e R E A LL Y b e re p re s en te d u s in g 3 .2 74 b it s ? If s o , h o w ? Th e fa ct is , t h e a ve ra g e in fo rm a ti o n c o n te n t is a s tr ic t *l o w er -b o u n d * o n h o w s m a ll o f a r ep re s en ta ti o n th a t w e ca n a ch ie ve . In p ra ct ic e, it is d if fi cu lt to r ea ch t h is b o u n d . B u t, w e ca n c o m e ve ry c lo s e. L0 1 - In tr o d u ct io n 2 0 C o m p 4 11 – F a ll 2 0 0 7 8 /2 2 /2 0 0 6 H uf fm a n C od in g • A s im p le * g re ed y* a lg o ri th m f o r a p p ro xi m a ti n g a n en tr o p y ef fi ci en t en co d in g 1. F in d t h e 2 it em s w it h t h e s m a lle s t p ro ba bi lit ie s 2 . J o in t h em in to a n ew * m et a * it em w h o s e p ro ba bi lit y is t h e s u m 3 . R em o ve t h e tw o it em s a n d in s er t th e n ew m et a it em 4 . R ep ea t fr o m s te p 1 u n ti l t h er e is o n ly o n e it em 3 6 /3 6 1 1 2 /3 6 3 2/3 6 4 /3 6 4 3/3 6 7 /3 6 9 4/3 6 5 4/3 6 8 /3 6 1 5 /3 6 1 2 1 /3 6 2 1/3 6 2 /3 6 7 6/3 6 1 1 /3 6 8 5/3 6 6 5/3 6 1 0 /3 6 2 1 /3 6 1 0 3 /3 6 5 /3 6 H uf fm a n d e co d in g tr e e L0 1 - In tr o d u ct io n 2 1 C o m p 4 11 – F a ll 2 0 0 7 8 /2 2 /2 0 0 6 C on ve rt in g T re e t o E nc od in g 3 6 /3 6 4 3/3 6 1 1 2 /3 6 3 2/3 6 4 /3 6 7 /3 6 9 4/3 6 5 4/3 6 8 /3 6 1 5 /3 6 7 6/3 6 1 0 3 /3 6 1 2 1 /3 6 2 1/3 6 2 /3 6 5 /3 6 1 1 /3 6 8 5/3 6 6 5/3 6 1 0 /3 6 2 1 /3 6 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 H uf fm a n d e co d in g tr e e O n ce t h e *t re e* is c o n s tr u ct ed , l a be l i ts e d g es c o n s is te n tl y a n d f o llo w t h e p a th s f ro m t h e la rg es t *m et a * it em t o ea ch o f th e re a l i te m t o f in d t h e en co d in g . 2 - 1 0 0 1 1 3 - 0 1 0 1 4 - 0 1 1 5 - 0 0 1 6 - 1 1 1 7 - 1 0 1 8 - 1 1 0 9 - 0 0 0 1 0 - 1 0 0 0 1 1 - 0 1 0 0 1 2 - 1 0 0 1 0 L0 1 - In tr o d u ct io n 2 2 C o m p 4 11 – F a ll 2 0 0 7 8 /2 2 /2 0 0 6 E nc od in g E ff ic ie nc y H o w d o es t h is e n co d in g s tr a te g y co m p a re t o t h e in fo rm a ti o n c o n te n t o f th e ro ll? P re tt y cl o s e. R ec a ll th a t th e lo w er b o u n d w a s 3 .2 74 b it s . H o w ev er , a n e ff ic ie n t en co d in g ( a s d ef in ed b y h a vi n g a n a ve ra g e co d e s iz e cl o s e to t h e in fo rm a ti o n c o n te n t) is n o t a lw a ys w h a t w e w a n t! 3 0 6 3 b 5 4 4 3 3 3 3 3 3 4 5 b av e 3 61 3 62 3 63 3 64 3 65 3 66 3 65 3 64 3 63 3 62 3 61 av e . = + + + + + + + + + + = L0 1 - In tr o d u ct io n 2 5 C o m p 4 11 – F a ll 2 0 0 7 8 /2 2 /2 0 0 6 Pr op e rt y 1 : Pa ri ty Th e s u m o f th e bi ts in e a ch s ym bo l i s e ve n . (t h is is h o w e rr o rs a re d et ec te d ) 2 -1 11 10 0 0 = 1 + 1 + 1 + 1 + 0 + 0 + 0 = 4 3 -1 11 11 0 1 = 1 + 1 + 1 + 1 + 1 + 0 + 1 = 6 4 -0 0 11 = 0 + 0 + 1 + 1 = 2 5 -0 10 1 = 0 + 1 + 0 + 1 = 2 6 -0 11 0 = 0 + 1 + 1 + 0 = 2 7- 0 0 0 0 = 0 + 0 + 0 + 0 = 0 8 -1 0 0 1 = 1 + 0 + 0 + 1 = 2 9 -1 0 10 = 1 + 0 + 1 + 0 = 2 10 -1 10 0 = 1 + 1 + 0 + 0 = 2 11 -1 11 11 10 = 1 + 1 + 1 + 1 + 1 + 1 + 0 = 6 12 -1 11 10 11 = 1 + 1 + 1 + 1 + 0 + 1 + 1 = 6 H ow m uc h in fo rm a ti on is i n th e la st b it ? L0 1 - In tr o d u ct io n 2 6 C o m p 4 11 – F a ll 2 0 0 7 8 /2 2 /2 0 0 6 Pr op e rt y 2 : S e pa ra ti on E a ch e n co d in g d if fe rs f ro m a ll o th er s b y a t le a s t tw o b it s in t h ei r o ve rl a p p in g p a rt s 3 4 5 6 7 8 9 1 0 1 1 1 2 1 1 1 1 1 0 1 0 0 1 1 0 1 0 1 0 1 1 0 0 0 0 0 1 0 0 1 1 0 1 0 1 1 0 0 1 1 1 1 1 1 0 1 1 1 1 0 1 1 2 1 1 1 1 0 0 0 1 1 1 1 x 0 x x x 1 1 x 1 x 1 x 1 1 x x x x x 1 x x 1 1 x 1 x 1 1 x x 1 1 1 1 x x 0 1 1 1 1 0 x x 3 1 1 1 1 1 0 1 x x 1 1 x 1 x 1 x 1 1 x x x x x 1 x x 1 1 x 1 x 1 1 x x 1 1 1 1 1 x x 1 1 1 1 x x 1 4 0 0 1 1 0 x x 1 0 x 1 x 0 0 x x x 0 x 1 x 0 1 x x x x x x x 1 1 x x 1 1 5 0 1 0 1 0 1 x x 0 x 0 x x x 0 1 x x x x x 1 0 x x 1 x 1 x 1 x 1 6 0 1 1 0 0 x x 0 x x x x x x 1 0 x 1 x 0 x 1 1 x x 1 1 x 7 0 0 0 0 x 0 0 x x 0 x 0 x x 0 0 x x x x x x x x 8 1 0 0 1 1 0 x x 1 x 0 x 1 x x 1 1 x x 1 9 1 0 1 0 1 x x 0 1 x 1 x 1 x 1 x 1 0 1 1 0 0 1 1 x x 1 1 x x 1 1 1 1 1 1 1 1 0 1 1 1 1 x 1 x T h is d if fe re nc e i s ca ll e d th e “ H a m m in g d is ta nc e ” “A H am m in g d is ta nc e o f on e -b it is n e e d e d t o un iq ue ly i d e nt if y an e nc od in g” L0 1 - In tr o d u ct io n 2 7 C o m p 4 11 – F a ll 2 0 0 7 8 /2 2 /2 0 0 6 E rr or c or re ct in g co d e s W e ca n a ct ua lly c o rr ec t 1- bi t er ro rs in e nc od in gs s ep a ra te d b y a H a m m in g d is ta nc e of t hr ee . T hi s is p os si bl e be ca us e th e se ts o f bi t pa tt er ns lo ca te d a H a m m in g d is ta nc e of 1 f ro m o ur e nc od in gs a re d is ti nc t. H ow ev er , a tt em p ti n g e rr o r co rr ec ti o n w it h s u ch a s m a ll s ep a ra ti o n is d a n g er o u s . S up po se , w e ha ve a 2 -b it e rr or . O ur e rr or co rr ec ti on s ch em e w ill t he n m is in te rp re t th e en co d in g. M is in te rp re ta ti on s a ls o oc cu rr ed w he n w e ha d 2 -b it e rr or s in o ur 1- bi t- er ro r- d et ec ti on ( pa ri ty ) sc he m es . A s a fe 1 -b it e rr or c or re ct io n sc he m e w ou ld c or re ct a ll 1- bi t er ro rs a nd d et ec t a ll 2 -b it e rr or s. W ha t H a m m in g d is ta nc e is n ee d ed b et w ee n en co d in gs t o a cc om pl is h th is ? 0 0 0 0 0 1 0 1 0 1 0 0 0 1 1 1 1 1 1 0 1 1 1 0 T h is i s ju st a vo ti ng sc h e m e