Instruction Execution
|
|
- Eric Robertson
- 6 years ago
- Views:
Transcription
1 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
2 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
3 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
4 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
5 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
6 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 Minight Tim T a k A B C D Squntial launy tak 6 hou fo loa. Cpt280, Autumn
7 Piplin Launy 6 PM Minight Tim T a k A B C D Piplin launy tak 3.5 hou fo loa. Piplining Lon T a k 6 PM Tim 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
8 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
9 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
10 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
11 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
12 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
13 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
14 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
15 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
Revision MIPS Pipelined Architecture
Rviion MIPS Piplin Achitctu D. Eng. Am T. Abl-Hami ELECT 1002 Sytm-n-a-Chip Dign Sping 2009 MIPS: A "Typical" RISC ISA 32-bit fix fomat intuction (3 fomat) 32 32-bit GPR (R0 contain zo, DP tak pai) 3-a,
More informationAgenda. Single Cycle Performance Assume >me for ac>ons are 100ps for register read or write; 200ps for other events. Review: Single- cycle Processor
Agna CS 61C: Gat Ia in Comput Achitctu (Machin Stuctu) Intuc>on Lvl Paalllim Intucto: Rany H. Katz Davi A. PaJon hjp://int.c.bkly.u/~c61c/fa1 Rviw Piplin Excu>on Piplin Datapath Aminitivia Piplin Haza
More informationIn Review: A Single Cycle Datapath We have everything! Now we just need to know how to BUILD CONTROL
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! Now w jut n to now how to UIL NRL
More informationCOMP303 Computer Architecture Lecture 11. An Overview of Pipelining
COMP303 Compute Achitectue Lectue 11 An Oveview of Pipelining Pipelining Pipelining povides a method fo executing multiple instuctions at the same time. Laundy Example: Ann, Bian, Cathy, Dave each have
More informationP&H 4.51 Pipelined Control. 3. Control Hazards. Hazards. Stall => 2 Bubbles/Clocks Time (clock cycles) Control Hazard: Branching 4/15/14
P&H.51 Piplind Contol CS 61C: Gat Ida in Comput Achitctu (Machin Stuctu) Lctu 2: Piplin Paalllim Intucto: Dan Gacia int.c.bkly.du/~c61c! Hazad SituaHon that pvnt tahng th nxt logical intuchon in th nxt
More informationCS152 Computer Architecture and Engineering Lecture 12. Introduction to Pipelining
CS152 Comput chitctu a Egiig Lctu 12 Itouctio to Pipliig a 10, 1999 Joh Kubiatowicz (http.c.bkly.u/~kubito) lctu li: http://www-it.c.bkly.u/~c152/ Rcap: icopogammig icopogammig i a covit mtho fo implmtig
More informationCS152 Computer Architecture and Engineering Lecture 12. Introduction to Pipelining
CS152 Comput chitctu a Egiig Lctu 12 Itouctio to Pipliig ctob 11, 1999 Joh Kubiatowicz (http.c.bkly.u/~kubito) lctu li: http://www-it.c.bkly.u/~c152/ Rcap: Micopogammig Micopogammig i a covit mtho fo implmtig
More informationRecap: Microprogramming. Specialize state-diagrams easily captured by microsequencer simple increment & branch fields datapath control fields
C 152 Lc13.1 Rcap: icopogammig C152: Compu chicu a Egiig Lcu 13 - oucio o Pipliig pcializ a-iagam aily capu by micoquc impl icm & bach fil aapah cool fil Cool ig uc o icopogammig icopogammig i a fuamal
More informationECE4680 Computer Organization and Architecture. Hazards in a Pipeline Processor. Pipeline is good but you need be careful.
ECE4680 Compu gaizaio a Achicu Haza i a Pipli Poco Pipli i goo bu you b caful. ECE4680 Haza.1 2002-4-3 Pipliig: Naual! Lauy Exampl A, Bia, Cahy, Dav ach hav o loa of cloh o wah, y, a fol A B C D Wah ak
More informationCENG 3420 Computer Organization and Design. Lecture 07: Pipeline Review. Bei Yu
CENG 3420 Compu gaizaio a Dig Lcu 07: Pipli Rviw Bi Yu CEG3420 L07.1 Spig 2016 Rviw: Sigl Cycl Diavaag & Avaag q U h clock cycl ifficily h clock cycl mu b im o accommoa h low i pcially poblmaic fo mo complx
More informationGreat Idea #4: Parallelism. CS 61C: Great Ideas in Computer Architecture. Pipelining Hazards. Agenda. Review of Last Lecture
CS 61C: Gat das i Comput Achitctu Pipliig Hazads Gu Lctu: Jui Hsia 4/12/2013 Spig 2013 Lctu #31 1 Gat da #4: Paalllism Softwa Paalll Rqus Assigd to comput.g. sach Gacia Paalll Thads Assigd to co.g. lookup,
More informationECE 361 Computer Architecture Lecture 13: Designing a Pipeline Processor
ECE 361 Compu Achicu Lcu 13: Digig a Pipli Poco 361 haza.1 Rviw: A Pipli Daapah Clk fch Rg/Dc Exc Mm W RgW Exp p Bach PC 1 0 PC+4 A Ui F/D Rgi PC+4 mm16 R Ra Rb R RFil R Rw Di R D/Ex Rgi 0 1 PC+4 mm16
More informationDigitalteknik och Datorarkitektur 5hp
Fch = + Digialkik och Daoakiku 5hp Sigl Cycl, Mulicycl & Pipliig 7 maj 8 kal.maklu@i.uu. Excu Dco Sigl Cycl pah wih Cool Ui På föa föläig a vi ihop all hä.. Hu u fö iukio av R-yp? p [3-6] Cool Ui Bach
More informationCS420/520 Computer Architecture I
CS420/520 Compu Achicu Haza i a Pipli Poco (CA4: Appix A) D. Xiaobo Zhou Dpam of Compu Scic CS420/520 pipli.1 Rviw: Pipliig Lo T a k 6 PM 7 8 9 Tim 30 40 40 40 40 20 A B C D Pipliig o hlp lacy of igl ak,
More informationCS 61C: Great Ideas in Computer Architecture Control and Pipelining, Part II. Anything can be represented as a number, i.e., data or instrucwons
CS 61C: Ga a i Compu Achicu Cool a Pipliig, Pa 10/29/12 uco: K Aaovic, Ray H. Kaz hdp://i.c.bkly.u/~c61c/fa12 Fall 2012 - - Lcu #28 1 Paalll Rqu Aig o compu.g., Sach Kaz Paalll Tha Aig o co.g., Lookup,
More informationCS 61C: Great Ideas in Computer Architecture (Machine Structures) Instruc(on Level Parallelism: Mul(ple Instruc(on Issue
CS 61C: Gat Ida in Comput Achitctu (Machin Stuctu) Intuc(on Lvl Paalllim: Mul(pl Intuc(on Iu Intucto: Randy H. Katz David A. PaGon hgp://int.c.bkly.du/~c61c/fa10 1 Paalll Rqut Aignd to comput.g., Sach
More informationLoad Instr 1. Instr 2 Instr 3. Instr 4. Outline & Announcements. EEL-4713C Computer Architecture Pipelined Processor - Hazards
uli & Aoucm EEL-4713C Compu Achicu Pipli Poco - Haza oucio o Haza Fowaig 1 cycl Loa Dlay 1 cycl Bach Dlay Wha mak pipliig ha EEL4713C A Goo-Ro.1 EEL4713C A Goo-Ro.2 Pipliig alig wih haza Sigl Mmoy i a
More informationSolutions to Supplementary Problems
Solution to Supplmntay Poblm Chapt Solution. Fomula (.4): g d G + g : E ping th void atio: G d 2.7 9.8 0.56 (56%) 7 mg Fomula (.6): S Fomula (.40): g d E ping at contnt: S m G 0.56 0.5 0. (%) 2.7 + m E
More informationCS61C Introduction to Pipelining. Lecture 25. April 28, 1999 Dave Patterson (http.cs.berkeley.edu/~patterson)
S61 Ioucio o Pipliig Lcu 25 pil 28, 1999 v Po (hp.c.bly.u/~po) www-i.c.bly.u/~c61c/chul.hml uli Rviw Pm Pig o Sc Pipliig logy Pipliig Iucio Excuio miiivi, Wh hi Suff fo? Hz o Pipliig Soluio o Hz vc Pipliig
More informationCOMPSCI 230 Discrete Math Trees March 21, / 22
COMPSCI 230 Dict Math Mach 21, 2017 COMPSCI 230 Dict Math Mach 21, 2017 1 / 22 Ovviw 1 A Simpl Splling Chck Nomnclatu 2 aval Od Dpth-it aval Od Badth-it aval Od COMPSCI 230 Dict Math Mach 21, 2017 2 /
More informationALU. Announcements. Lecture 9. Pipeline Hazards. Review: Single-cycle Datapath (load instruction) Review: Multi-cycle Datapath. R e g s.
Aoucm Lcu 9 Pipli Haza Chio Kozyaki Safo Uiviy hp://cla.afo.u/8b PA- i u oay Elcoic ubmiio Lab2 i u o Tuay 2/3 h Quiz ga will b availabl wk Soluio will b po o li omoow Tuay 2/3 h lcu will b a vio playback
More informationCDS 101: Lecture 7.1 Loop Analysis of Feedback Systems
CDS : Lct 7. Loop Analsis of Fback Sstms Richa M. Ma Goals: Show how to compt clos loop stabilit fom opn loop poptis Dscib th Nqist stabilit cition fo stabilit of fback sstms Dfin gain an phas magin an
More informationICS 233 Computer Architecture & Assembly Language
ICS 233 Computer Architecture & Assembly Language Assignment 6 Solution 1. Identify all of the RAW data dependencies in the following code. Which dependencies are data hazards that will be resolved by
More informationCS152 Computer Architecture and Engineering Lecture 12. Introduction to Pipelining
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
More informationWhat Makes Production System Design Hard?
What Maks Poduction Systm Dsign Had? 1. Things not always wh you want thm whn you want thm wh tanspot and location logistics whn invntoy schduling and poduction planning 2. Rsoucs a lumpy minimum ffctiv
More informationPeriod vs. Length of a Pendulum
Gaphcal Mtho n Phc Gaph Intptaton an Lnazaton Pat 1: Gaphng Tchnqu In Phc w u a vat of tool nclung wo, quaton, an gaph to mak mol of th moton of objct an th ntacton btwn objct n a tm. Gaph a on of th bt
More informationCMP N 301 Computer Architecture. Appendix C
CMP N 301 Computer Architecture Appendix C Outline Introduction Pipelining Hazards Pipelining Implementation Exception Handling Advanced Issues (Dynamic Scheduling, Out of order Issue, Superscalar, etc)
More informationSchool of Electrical Engineering. Lecture 2: Wire Antennas
School of lctical ngining Lctu : Wi Antnnas Wi antnna It is an antnna which mak us of mtallic wis to poduc a adiation. KT School of lctical ngining www..kth.s Dipol λ/ Th most common adiato: λ Dipol 3λ/
More informationKinetics. Central Force Motion & Space Mechanics
Kintics Cntal Foc Motion & Spac Mcanics Outlin Cntal Foc Motion Obital Mcanics Exampls Cntal-Foc Motion If a paticl tavls un t influnc of a foc tat as a lin of action ict towas a fix point, tn t motion
More informationMOS transistors (in subthreshold)
MOS tanito (in ubthhold) Hitoy o th Tanito Th tm tanito i a gnic nam o a olid-tat dvic with 3 o mo tminal. Th ild-ct tanito tuctu wa it dcibd in a patnt by J. Lilinld in th 193! t took about 4 ya bo MOS
More informationDifferential Kinematics
Lctu Diffntia Kinmatic Acknowgmnt : Pof. Ouama Khatib, Robotic Laboato, tanfo Univit, UA Pof. Ha Aaa, AI Laboato, MIT, UA Guiing Qution In obotic appication, not on th poition an ointation, but th vocit
More informationA Comparative Study and Analysis of an Optimized Control Strategy for the Toyota Hybrid System
Pag 563 Wol Elctic Vhicl Jounal Vol. 3 - ISSN 3-6653 - 9 AVERE EVS4 Stavang, Noway, May 13-16, 9 A Compaativ Stuy an Analysis of an Optimiz Contol Statgy fo th Toyota Hybi Systm Tho Hofman 1, Thijs Punot
More informationNeural Networks The ADALINE
Lat Lctu Summay Intouction to ua to Bioogica uon Atificia uon McCuoch an itt LU Ronbatt cton Aan Bnaino, a@i.it.ut.t Machin Laning, 9/ ua to h ADALI M A C H I L A R I G 9 / cton Limitation cton aning u
More informationCS152 Computer Architecture and Engineering Lecture 12. Exceptions (continued) Introduction to Pipelining
CS152 Compu Achicu a Egiig Lcu 12 Excpio (coiu) oucio o Pipliig cob 12 h, 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
More informationCS:APP Chapter 4 Computer Architecture Pipelined Implementation
CS:APP Chaptr 4 Computr Architctur Piplind Implmntation CS:APP2 Ovrviw Gnral Principls of Piplinin n Goal n Difficultis Cratin a Piplind Y86 Procssor n arranin SEQ n Insrtin piplin ristrs n Problms with
More informationThe angle between L and the z-axis is found from
Poblm 6 This is not a ifficult poblm but it is a al pain to tansf it fom pap into Mathca I won't giv it to you on th quiz, but know how to o it fo th xam Poblm 6 S Figu 6 Th magnitu of L is L an th z-componnt
More information(( ) ( ) ( ) ( ) ( 1 2 ( ) ( ) ( ) ( ) Two Stage Cluster Sampling and Random Effects Ed Stanek
Two ag ampling and andom ffct 8- Two Stag Clu Sampling and Random Effct Ed Stank FTE POPULATO Fam Labl Expctd Rpon Rpon otation and tminology Expctd Rpon: y = and fo ach ; t = Rpon: k = y + Wk k = indx
More informationSTRIPLINES. A stripline is a planar type transmission line which is well suited for microwave integrated circuitry and photolithographic fabrication.
STIPLINES A tiplin i a plana typ tanmiion lin hih i ll uitd fo mioav intgatd iuity and photolithogaphi faiation. It i uually ontutd y thing th nt onduto of idth, on a utat of thikn and thn oving ith anoth
More informationSOFTWARE. Computer Architecture Topics. Shared Memory, Message Passing, Data Parallelism. Network Interfaces. Interconnection Network
Lctu 1: Cot/Pfomac, DLX, Pipliig, Cach, Bach Pictio Pof. F Chog ECS 250A Comput Achitctu Wit 1999 Comput Achitctu I th attibut of a [computig] ytm a by th pogamm, i.., th cocptual tuctu a fuctioal bhavio,
More informationPhysics 111. Lecture 38 (Walker: ) Phase Change Latent Heat. May 6, The Three Basic Phases of Matter. Solid Liquid Gas
Physics 111 Lctu 38 (Walk: 17.4-5) Phas Chang May 6, 2009 Lctu 38 1/26 Th Th Basic Phass of Matt Solid Liquid Gas Squnc of incasing molcul motion (and ngy) Lctu 38 2/26 If a liquid is put into a sald contain
More informationEMPORIUM H O W I T W O R K S F I R S T T H I N G S F I R S T, Y O U N E E D T O R E G I S T E R.
H O W I T W O R K S F I R S T T H I N G S F I R S T, Y O U N E E D T O R E G I S T E R I n o r d e r t o b u y a n y i t e m s, y o u w i l l n e e d t o r e g i s t e r o n t h e s i t e. T h i s i s
More informationLecture 2: Frequency domain analysis, Phasors. Announcements
EECS 5 SPRING 24, ctu ctu 2: Fquncy domain analyi, Phao EECS 5 Fall 24, ctu 2 Announcmnt Th cou wb it i http://int.c.bkly.du/~5 Today dicuion ction will mt Th Wdnday dicuion ction will mo to Tuday, 5:-6:,
More informationy cos x = cos xdx = sin x + c y = tan x + c sec x But, y = 1 when x = 0 giving c = 1. y = tan x + sec x (A1) (C4) OR y cos x = sin x + 1 [8]
DIFF EQ - OPTION. Sol th iffrntial quation tan +, 0
More informationMid Year Examination F.4 Mathematics Module 1 (Calculus & Statistics) Suggested Solutions
Mid Ya Eamination 3 F. Matmatics Modul (Calculus & Statistics) Suggstd Solutions Ma pp-: 3 maks - Ma pp- fo ac qustion: mak. - Sam typ of pp- would not b countd twic fom wol pap. - In any cas, no pp maks
More informationSymmetry of Lagrangians of holonomic systems in terms of quasi-coordinates
Vol 18 No 8, Augut 009 c 009 Chin. Phy. Soc. 1674-1056/009/1808/3145-05 Chinee Phyic B an IOP Publihing Lt Symmety of Lagangian of holonomic ytem in tem of quai-cooinate Wu Hui-Bin an Mei Feng-Xiang School
More informationPartial Fraction Expansion
Paial Facion Expanion Whn ying o find h inv Laplac anfom o inv z anfom i i hlpfl o b abl o bak a complicad aio of wo polynomial ino fom ha a on h Laplac Tanfom o z anfom abl. W will illa h ing Laplac anfom.
More informationFinal Exam Solutions
CS 2 Advancd Data Structurs and Algorithms Final Exam Solutions Jonathan Turnr /8/20. (0 points) Suppos that r is a root of som tr in a Fionacci hap. Assum that just for a dltmin opration, r has no childrn
More informationBit-Alignment for Retargetable Code Generators
Bitlinmnt fo Rtatabl Co Gnatos Kon choofs Gt Goossns Huo Man y IMEC, Kaplf 75, B3 Luvn, Blium bstact Whn builin a bittu tatabl compil, vy sinal typ must b implmnt xactly as spcifi, vn whn th wolnth of
More informationStructural Hazard #1: Single Memory (1/2)! Structural Hazard #1: Single Memory (2/2)! Review! Pipelining is a BIG idea! Optimal Pipeline! !
S61 L21 PU ig: Pipliig (1)! i.c.bly.u/~c61c S61 : Mchi Sucu Lcu 21 PU ig: Pipliig 2010-07-27!!!uco Pul Pc! G GE FLL SESN KES NW! Foobll Su So ic ow o-l o icomig Fhm, f, Gu u (hʼ m!). h i o log xcu fo yo
More informationLaplace Transformation
Univerity of Technology Electromechanical Department Energy Branch Advance Mathematic Laplace Tranformation nd Cla Lecture 6 Page of 7 Laplace Tranformation Definition Suppoe that f(t) i a piecewie continuou
More informationCS 6353 Compiler Construction, Homework #1. 1. Write regular expressions for the following informally described languages:
CS 6353 Compilr Construction, Homwork #1 1. Writ rgular xprssions for th following informally dscribd languags: a. All strings of 0 s and 1 s with th substring 01*1. Answr: (0 1)*01*1(0 1)* b. All strings
More information6. KALMAN-BUCY FILTER
6. KALMAN-BUCY FILTER 6.1. Motivation and preliminary. A wa hown in Lecture 2, the optimal control i a function of all coordinate of controlled proce. Very often, it i not impoible to oberve a controlled
More informationsin sin 1 d r d Ae r 2
Diffction k f c f Th Huygn-Fnl Pincil tt: Evy unobtuct oint of vfont, t givn intnt, v ouc of hicl cony vlt (ith th m funcy tht of th imy v. Th mlitu of th oticl fil t ny oint byon i th uoition of ll th
More informationMichela Taufer CS:APP
Michla Taufr CS:APP Powrpoint Lctur Nots for Computr Systms: A Prorammr's Prspctiv,. Bryant and D. O'Hallaron, Prntic Hall, 2003 Ovrviw 2 CISC 360 Faʼ08 al-world Piplins: Car Washs Squntial Paralll Piplind
More informationFrom Elimination to Belief Propagation
School of omputr Scinc Th lif Propagation (Sum-Product lgorithm Probabilistic Graphical Modls (10-708 Lctur 5, Sp 31, 2007 Rcptor Kinas Rcptor Kinas Kinas X 5 ric Xing Gn G T X 6 X 7 Gn H X 8 Rading: J-hap
More informationPart II, Measures Other Than Conversion I. Apr/ Spring 1
Pt II, Msus Oth hn onvsion I p/7 11 Sping 1 Pt II, Msus Oth hn onvsion II p/7 11 Sping . pplictions/exmpls of th RE lgoithm I Gs Phs Elmnty Rction dditionl Infomtion Only fd P = 8. tm = 5 K =. mol/dm 3
More informationCDS 101/110: Lecture 7.1 Loop Analysis of Feedback Systems
CDS 11/11: Lctu 7.1 Loop Analysis of Fdback Systms Novmb 7 216 Goals: Intoduc concpt of loop analysis Show how to comput closd loop stability fom opn loop poptis Dscib th Nyquist stability cition fo stability
More informationHelping you learn to save. Pigby s tips and tricks
Hlpg yu lan t av Pigby tip and tick Hlpg vy littl av Pigby ha bn tachg hi find all abut ny and hw t av f what ty want. Tuffl i avg f a nw tappy bubbl d and Pi can t wait t b abl t buy nw il pat. Pigby
More informationESCI 341 Atmospheric Thermodynamics Lesson 16 Pseudoadiabatic Processes Dr. DeCaria
ESCI 34 Atmohi hmoynami on 6 Puoaiabati Po D DCaia fn: Man, A an FE obitaill, 97: A omaion of th uialnt otntial tmatu an th tati ngy, J Atmo Si, 7, 37-39 Btt, AK, 974: Futh ommnt on A omaion of th uialnt
More informationChemistry 222 DO NOT OPEN THE EXAM UNTIL YOU ARE READY TO TAKE IT! You may allocate a maximum of 80 continuous minutes for this exam.
Chmtry Sprg 09 Eam : Chaptr -5 Nam 80 Pot Complt fv (5) of th followg problm. CLEARLY mark th problm you o ot wat gra. You mut how your work to rcv crt for problm rqurg math. Rport your awr wth th approprat
More informationINTRODUCTION TO AUTOMATIC CONTROLS INDEX LAPLACE TRANSFORMS
adjoint...6 block diagram...4 clod loop ytm... 5, 0 E()...6 (t)...6 rror tady tat tracking...6 tracking...6...6 gloary... 0 impul function...3 input...5 invr Laplac tranform, INTRODUCTION TO AUTOMATIC
More informationY 0. Standing Wave Interference between the incident & reflected waves Standing wave. A string with one end fixed on a wall
Staning Wav Intrfrnc btwn th incint & rflct wavs Staning wav A string with on n fix on a wall Incint: y, t) Y cos( t ) 1( Y 1 ( ) Y (St th incint wav s phas to b, i.., Y + ral & positiv.) Rflct: y, t)
More informationORBITAL TO GEOCENTRIC EQUATORIAL COORDINATE SYSTEM TRANSFORMATION. x y z. x y z GEOCENTRIC EQUTORIAL TO ROTATING COORDINATE SYSTEM TRANSFORMATION
ORITL TO GEOCENTRIC EQUTORIL COORDINTE SYSTEM TRNSFORMTION z i i i = (coωcoω in Ωcoiinω) (in Ωcoω + coωcoiinω) iniinω ( coωinω in Ωcoi coω) ( in Ωinω + coωcoicoω) in icoω in Ωini coωini coi z o o o GEOCENTRIC
More informationThen the number of elements of S of weight n is exactly the number of compositions of n into k parts.
Geneating Function In a geneal combinatoial poblem, we have a univee S of object, and we want to count the numbe of object with a cetain popety. Fo example, if S i the et of all gaph, we might want to
More informationLecture 3.2: Cosets. Matthew Macauley. Department of Mathematical Sciences Clemson University
Lctu 3.2: Costs Matthw Macauly Dpatmnt o Mathmatical Scincs Clmson Univsity http://www.math.clmson.du/~macaul/ Math 4120, Modn Algba M. Macauly (Clmson) Lctu 3.2: Costs Math 4120, Modn Algba 1 / 11 Ovviw
More informationReview. Agenda. Dynamic Pipeline Scheduling. Specula>on. Why Do Dynamic Scheduling? 11/7/10
Rviw CS 61C: Ga a i Compu Achicu (Machi Sucu) uc>o Lvl Paalllim uco: Ray H. Kaz Davi A. PaJo hjp://i.c.bkly.u/~c61c/fa10 Pipli challg i haza Fowaig hlp w/may aa haza Dlay bach hlp wih cool haza i 5 ag
More informationPayroll Direct Deposit
Payroll Dirct Dposit Dirct Dposit for mploy paychcks allows cntrs to avoi printing an physically istributing papr chcks to mploys. Dirct posits ar ma through a systm known as Automat Claring Hous (ACH),
More informationEXAMPLES 4/12/2018. The MIPS Pipeline. Hazard Summary. Show the pipeline diagram. Show the pipeline diagram. Pipeline Datapath and Control
The MIPS Pipeline CSCI206 - Computer Organization & Programming Pipeline Datapath and Control zybook: 11.6 Developed and maintained by the Bucknell University Computer Science Department - 2017 Hazard
More informationChapter 7 Dynamic stability analysis I Equations of motion and estimation of stability derivatives - 4 Lecture 25 Topics
Chapt 7 Dynamic stability analysis I Equations of motion an stimation of stability ivativs - 4 ctu 5 opics 7.8 Expssions fo changs in aoynamic an populsiv focs an momnts 7.8.1 Simplifi xpssions fo changs
More informationWEEK 3 Effective Stress and Pore Water Pressure Changes
WEEK 3 Effctiv Str and Por Watr Prur Chang 5. Effctiv tr ath undr undraind condition 5-1. Dfinition of ffctiv tr: A rvi A you mut hav larnt that th ffctiv tr, σ, in oil i dfind a σ σ u Whr σ i th total
More informationNear-Optimal Relay Station Placement for Power Minimization in WiMAX Networks
Nea-Optimal elay Station lacement fo owe Minimization in WiMAX Netwok Dejun Yang, Xi Fang an Guoliang Xue Abtact In the IEEE 80.16j tana, the elay tation ha been intouce to inceae the coveage an the thoughput
More informationOverview. Real-World Pipelines: Car Washes. Computational Example. 3-Way Pipelined Version. Pipeline Diagrams
Ovrviw omputr rchitctur: Piplind Implmntation I Sci 2021: achin rchitctur and Oranization Lctur #20, arch 7th-9th, 2016 Your instructor: Stphn camant Gnral Principls of Piplinin Goal ifficultis ratin a
More informationOn the Hamiltonian of a Multi-Electron Atom
On th Hamiltonian of a Multi-Elctron Atom Austn Gronr Drxl Univrsity Philadlphia, PA Octobr 29, 2010 1 Introduction In this papr, w will xhibit th procss of achiving th Hamiltonian for an lctron gas. Making
More informationInvestment. Net Present Value. Stream of payments A 0, A 1, Consol: same payment forever Common interest rate r
Bfo going o Euo on buin, a man dov hi Roll-Royc o a downown NY Ciy bank and wn in o ak fo an immdia loan of $5,. Th loan offic, akn aback, qud collaal. "Wll, hn, h a h ky o my Roll-Royc", h man aid. Th
More informationT-72: Saratoga to Coyne ENVIRONMENTAL SCREENING ACCESS PLAN
on tl asting on tl asting st isconsin apids isconsin apids Bion oyn t -7: aatoga to oyn V G P VP F: B: Bak t Pot dwads V V P : PB ots: ata viwd and oucs: V U F GP PUP.. atways and tlands - isconsin, ydology
More informationEstimation and Confidence Intervals: Additional Topics
Chapte 8 Etimation and Confidence Inteval: Additional Topic Thi chapte imply follow the method in Chapte 7 fo foming confidence inteval The text i a bit dioganized hee o hopefully we can implify Etimation:
More informationACCURATE FLOATING-POINT SUMMATION IN CUB
ACCURATE FLOATING-POINT SUMMATION IN CUB URI VERNER Summe inten OUTLINE Who need accuate floating-point ummation?! Round-off eo: ouce and ecovey A new method fo accuate FP ummation on a GPU Added a a function
More informationCHAPTER 5 CIRCULAR MOTION
CHAPTER 5 CIRCULAR MOTION and GRAVITATION 5.1 CENTRIPETAL FORCE It is known that if a paticl mos with constant spd in a cicula path of adius, it acquis a cntiptal acclation du to th chang in th diction
More informationMinimum Spanning Trees
Yufi Tao ITEE Univrsity of Qunslan In tis lctur, w will stuy anotr classic prolm: finin a minimum spannin tr of an unirct wit rap. Intrstinly, vn tou t prolm appars ratr iffrnt from SSSP (sinl sourc sortst
More informationHomework #7 Solution. Solutions: ΔP L Δω. Fig. 1
Homework #7 Solution Aignment:. through.6 Bergen & Vittal. M Solution: Modified Equation.6 becaue gen. peed not fed back * M (.0rad / MW ec)(00mw) rad /ec peed ( ) (60) 9.55r. p. m. 3600 ( 9.55) 3590.45r.
More informationProblem Set 8 Solutions
Deign and Analyi of Algorithm April 29, 2015 Maachuett Intitute of Technology 6.046J/18.410J Prof. Erik Demaine, Srini Devada, and Nancy Lynch Problem Set 8 Solution Problem Set 8 Solution Thi problem
More informationEven/Odd Mode Analysis of the Wilkinson Divider
//9 Wilkinn Dividr Evn and Odd Md Analyi.dc / Evn/Odd Md Analyi f th Wilkinn Dividr Cnidr a matchd Wilkinn pwr dividr, with a urc at prt : Prt Prt Prt T implify thi chmatic, w rmv th grund plan, which
More informationFI 3103 Quantum Physics
7//7 FI 33 Quantum Physics Axan A. Iskana Physics of Magntism an Photonics sach oup Institut Tknoogi Banung Schoing Equation in 3D Th Cnta Potntia Hyognic Atom 7//7 Schöing quation in 3D Fo a 3D pobm,
More informationCBSE-XII-2013 EXAMINATION (MATHEMATICS) The value of determinant of skew symmetric matrix of odd order is always equal to zero.
CBSE-XII- EXAMINATION (MATHEMATICS) Cod : 6/ Gnal Instuctions : (i) All qustions a compulso. (ii) Th qustion pap consists of 9 qustions dividd into th sctions A, B and C. Sction A compiss of qustions of
More informationEE 361L Fall 2010 Pipelined MIPS L0 (PMIPS L0) and Pipelined MIPS L (PMIPS L)
EE 361L Fall 2010 iplind S L0 (S L0) and iplind S L (S L) Last updatd: Novmbr 8, 2010 1. ntroduction S L0 and S L ar piplind vrsions of SL (for S Lit). Appndix A has a dscription of th SL procssor. S L0
More informationy = 2xe x + x 2 e x at (0, 3). solution: Since y is implicitly related to x we have to use implicit differentiation: 3 6y = 0 y = 1 2 x ln(b) ln(b)
4. y = y = + 5. Find th quation of th tangnt lin for th function y = ( + ) 3 whn = 0. solution: First not that whn = 0, y = (1 + 1) 3 = 8, so th lin gos through (0, 8) and thrfor its y-intrcpt is 8. y
More informationThe Moúõ. ExplÉüers. Fun Facts. WÉüd Proèô. Parts oì Sp. Zoú Animal Roêks
onn C f o l b Ta 4 5 õ Inoåucio Pacic 8 L LoËíca c i c 3 a P L Uppca 35 k W h Day oì 38 a Y h Moõh oì WÉüld 44 o nd h a y a d h Bi 47 u g 3-D Fi 54 Zoú Animal 58 Éüm Landf 62 Roêk 68 Th Moúõ õ o 74 l k
More information/ / MET Day 000 NC1^ INRTL MNVR I E E PRE SLEEP K PRE SLEEP R E
05//0 5:26:04 09/6/0 (259) 6 7 8 9 20 2 22 2 09/7 0 02 0 000/00 0 02 0 04 05 06 07 08 09 0 2 ay 000 ^ 0 X Y / / / / ( %/ ) 2 /0 2 ( ) ^ 4 / Y/ 2 4 5 6 7 8 9 2 X ^ X % 2 // 09/7/0 (260) ay 000 02 05//0
More informationProject Two RISC Processor Implementation ECE 485
Project Two RISC Processor Implementation ECE 485 Chenqi Bao Peter Chinetti November 6, 2013 Instructor: Professor Borkar 1 Statement of Problem This project requires the design and test of a RISC processor
More informationExtinction Ratio and Power Penalty
Application Not: HFAN-.. Rv.; 4/8 Extinction Ratio and ow nalty AVALABLE Backgound Extinction atio is an impotant paamt includd in th spcifications of most fib-optic tanscivs. h pupos of this application
More information3 2x. 3x 2. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB
Math B Intgration Rviw (Solutions) Do ths intgrals. Solutions ar postd at th wbsit blow. If you hav troubl with thm, sk hlp immdiatly! () 8 d () 5 d () d () sin d (5) d (6) cos d (7) d www.clas.ucsb.du/staff/vinc
More information3. (2) What is the difference between fixed and hybrid instructions?
1. (2 pts) What is a "balanced" pipeline? 2. (2 pts) What are the two main ways to define performance? 3. (2) What is the difference between fixed and hybrid instructions? 4. (2 pts) Clock rates have grown
More informationMultiple Short Term Infusion Homework # 5 PHA 5127
Multipl Short rm Infusion Homwork # 5 PHA 527 A rug is aministr as a short trm infusion. h avrag pharmacokintic paramtrs for this rug ar: k 0.40 hr - V 28 L his rug follows a on-compartmnt boy mol. A 300
More information[ ] 1+ lim G( s) 1+ s + s G s s G s Kacc SYSTEM PERFORMANCE. Since. Lecture 10: Steady-state Errors. Steady-state Errors. Then
SYSTEM PERFORMANCE Lctur 0: Stady-tat Error Stady-tat Error Lctur 0: Stady-tat Error Dr.alyana Vluvolu Stady-tat rror can b found by applying th final valu thorm and i givn by lim ( t) lim E ( ) t 0 providd
More informationHistogram Processing
Hitogam Poceing Lectue 4 (Chapte 3) Hitogam Poceing The hitogam of a digital image with gay level fom to L- i a dicete function h( )=n, whee: i the th gay level n i the numbe of pixel in the image with
More informationAnouncements. Conjugate Gradients. Steepest Descent. Outline. Steepest Descent. Steepest Descent
oucms Couga Gas Mchal Kazha (6.657) Ifomao abou h Sma (6.757) hav b pos ol: hp://www.cs.hu.u/~msha Tch Spcs: o M o Tusay afoo. o Two paps scuss ach w. o Vos fo w s caa paps u by Thusay vg. Oul Rvw of Sps
More informationA s device signals an interrupt. time-> time T. A s device. starts device. starts device. A s ISR. WAIT/block. Process A. interrupt.
/1 /1 BAA F 67 :; - -, - % 67 :; = : J 3KJ AA L A s dvic signas an intrrut A s dvic tim T tim-> A s starts dvic starts dvic rocss A AT/bock rocss B AT/bock intrrut B s starts dvic B s dvic B s dvic signas
More information4. (3) What do we mean when we say something is an N-operand machine?
1. (2) What are the two main ways to define performance? 2. (2) When dealing with control hazards, a prediction is not enough - what else is necessary in order to eliminate stalls? 3. (3) What is an "unbalanced"
More informationGoals for Performance Lecture
Goals for Performance Lecture Understand performance, speedup, throughput, latency Relationship between cycle time, cycles/instruction (CPI), number of instructions (the performance equation) Amdahl s
More informationEEO 401 Digital Signal Processing Prof. Mark Fowler
EEO 401 Digital Signal Procssing Prof. Mark Fowlr Dtails of th ot St #19 Rading Assignmnt: Sct. 7.1.2, 7.1.3, & 7.2 of Proakis & Manolakis Dfinition of th So Givn signal data points x[n] for n = 0,, -1
More information