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### cs320-2008-t2-sample-final

Course: CPSC 344, Fall 2010
School: UBC
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Word Count: 1666

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320 CPSC Sample Final Examination April 2009 Name: Signature: Student ID: You have 2.5 hours to write the 8 questions on this examination. A total of 110 marks are available. Justify all of your answers, except if the question says not to. Question No notes or electronic equipment are allowed, except for one 8.5 11 sheet of paper, handwritten. 2 Keep your answers short. If you run out of space for a...

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UBC - CPSC - 344
CPSC 320 Sample Final ExaminationApril 2009[10] 1. Answer each of the following questions with true or false. Give a short justicationfor each of your answers.[5] a. 6n O(5n )This is false:66nlim n = limn 5n + 5= +Solution :nwhich means that
UBC - CPSC - 344
CPSC 320 Midterm 1Monday, October 18th, 2010[13] 1. Short answers[4] a. We frequently prove an O(n2 ) bound on the worst-case running time T (n) of analgorithm, without proving any bound on T (n). However, we would never provean (n2 ) bound on T (n)
UBC - CPSC - 344
CPSC 320 Midterm 2Monday, November 15th, 2010[16] 1. Short Answers[4] a. Your boss asks you to design a divide-and-conquer algorithm to solve a problemwhose input is an array of measurements that contain temperature data. What are thetwo main issues
UBC - CPSC - 344
CPSC 320 Sample Final ExaminationDecember 2010Name:Signature:Student ID: You have 2.5 hours to write the 8 questions on this examination. A total of 100 marks are available. Justify all of your answers, except if the questionsays not to.Question
UBC - CPSC - 344
CPSC 320 Sample Final ExaminationDecember 2010[10] 1. Answer each of the following questions with true or false. Give a short justicationfor each of your answers.[5] a. 6n O(5n )This is false:6n6lim n = limn 5n + 5= +Solution :nwhich means t
UBC - CPSC - 344
CPSC 320 Sample Midterm 1October 2010Name:Signature:Student ID: You have 50 minutes to write the 4 questions on this examination.A total of 40 marks are available. Justify all of your answers. You are allowed to bring in one hand-written, double-s
UBC - CPSC - 344
CPSC 320 Sample Midterm 1October 2010[12] 1. Answer each of the questions with either true or false. You must justify each of youranswers; an answer without a justication will be worth at most 1.5 out of 4.[4] a. 3n+2 + 5 O(3n1 ).Solution : This is t
UBC - CPSC - 344
CPSC 320 Sample Midterm 2November 2010Name:Signature:Student ID: You have 50 minutes to write the 5 questions on this examination.A total of 40 marks are available. Justify all of your answers. You are allowed to bring in one hand-written, double-
UBC - CPSC - 344
CPSC 320 Sample Midterm 2November 2010[6] 1. Short Answers[3] a. If we can not use a recursion tree to prove a tight bound on the value of a function T (n) dened by a recurrence relation, can we use the Master theorem instead?Explain why or why not.S
UBC - CPSC - 344
CPSC 313, 05w Term 1 Final ExamDate: December 12, 2005; Instructor: Mike FeeleyThis is a closed book exam; no notes; you may use calculators to perform simple arithmetic calculations.Pages of potentially useful notes appear at the end of the exam. Answ
UBC - CPSC - 344
CPSC 313, 05w Term 1 Midterm 1Date: October 7, 2005; Instructor: Mike FeeleyThis is a closed book exam; no notes; you may use calculators to perform simple arithmetic calculations.Answer in the space provided; use the backs of pages if needed. There ar
UBC - CPSC - 344
CPSC 313, 05w Term 1 Midterm 1 SolutionsDate: October 7, 2005; Instructor: Mike Feeley1. (10 marks) Short answers.1a. What is the advantage of using two different registers (i.e., %ebp and %esp) to store virtualaddresses to the runtime stack?The stac
UBC - CPSC - 344
CPSC 313, 05w Term 1 Midterm 2Date: November 9, 2005; Instructor: Mike FeeleyThis is a closed book exam; no notes; you may use calculators to perform simple arithmetic calculations.NOTE: the last page shows the PIPE architecture diagram. Answer in the
UBC - CPSC - 344
CPSC 313, 05w Term 1 Midterm 2 SolutionsDate: November 9, 2005; Instructor: Mike Feeley1. (10 marks) Short answers.1a. Briey explain how an n-channel MOSFET transistor closes the circuit between its source anddrain when its gate voltage is high.The h
UBC - CPSC - 344
CPSC 313, 06w Term 1 Midterm 1Date: October 4, 2006; Instructor: Norm HutchinsonThis is a closed book exam; no notes; you may use calculators only if you dont trust your own brain toperform simple arithmetic calculations. Answer in the space provided;
UBC - CPSC - 344
CPSC 313, 06w Term 1 Midterm 1 SolutionsDate: October 4, 2006; Instructor: Norm Hutchinson1. (8 marks)1a.Short answers.Is the address of a local variable in a C function determined statically or dynamically?Briey explain.(2 marks)It is determined
UBC - CPSC - 344
CPSC 313, 06w Term 1 Midterm 2Date: November 17, 2006; Instructor: Norm HutchinsonThis is a closed book exam; no notes; you may use a calculator if you wish. Answer in the space provided;use the backs of pages if needed. There are 6 questions on 8 page
UBC - CPSC - 344
CPSC 313, 06w Term 1 Midterm 2 SolutionsDate: November 17, 2006; Instructor: Norm Hutchinson1. (10 marks)Short answers.1a.(2 marks) In the standard pipeline (F, D, E, M, W), does stalling stage D one cycle cause stalls orbubbles for any other cycles
UBC - CPSC - 344
CPSC 313, 06w Term 2 Midterm 1Date: February 9, 2007; Instructor: Norm HutchinsonThis is a closed book exam; no notes; no calculators. Answer in the space provided; use the backs of pagesif needed.There are 6 questions on 4 pages, totaling 42 marks.Y
UBC - CPSC - 344
CPSC 313, 06w Term 2 Midterm 1 SolutionsDate: February 9, 2007; Instructor: Norm Hutchinson1. (8 marks)Short answers.1a.(2 marks) Is the address of a global variable in a C program determined statically (before theprogram runs) or dynamically (while
UBC - CPSC - 344
CPSC 313, 06w Term 2 Midterm 2Date: March 23, 2007; Instructor: Norm HutchinsonThis is a closed book exam; no notes; you may use a calculator if you wish. Answer in the space provided;use the backs of pages if needed.There are 6 questions on 8 pages,
UBC - CPSC - 344
CPSC 313, 06w Term 2 Midterm 2 SolutionsDate: March 23, 2007; Instructor: Norm Hutchinson1. (12 marks)Short answers.1a. (2 marks) Describe the difference between stalling and creating a bubble.Stalling means that an instruction remains in a particula
UBC - CPSC - 344
CPSC 31306W Term 2Problem Set #1Due: Sunday, January 21, 2006 at 11: 59 PM (thirteen-hour grace period)Instructions: Hand all of your solutions in on paper.1. On an IA32 machine (e.g., lin01.ugrad.cs.ubc.ca), using gcc, compile the following C progra
UBC - CPSC - 344
CPSC 31306W Term 2Problem Set #1 Solution1. I edited the output slightly to make it shorter..bss_counter:.space 4.text_swap:pushl%ebpmovl%esp, %ebpsubl\$8, %espmovl%ebx, (%esp)movl8(%ebp), %edxmovl16(%ebp), %ebxmovl%esi, 4(%esp)movl
UBC - CPSC - 344
CPSC 31306W Term 2Problem Set #2Due: Sunday, January 28, 2006 at 11:59 PM (thirteen-hour grace period)All of your solutions should be turned in on paper.1. Variables declared in a C program can be stored in either registers or memory, and if inmemor
UBC - CPSC - 344
CPSC 31306W Term 2Problem Set #2 Solution1. The only illegal operation is taking the address of i, since it is in a register. Note the similarity of allof the accesses to the variables that are stored in memory: a, l, and g. Despite the dierences in h
UBC - CPSC - 344
CPSC 31306W Term 2Problem Set #3Due: Sunday, February 4, 2007 at 11:59 PM (thirteen-hour grace period)1. Write in C a function int gcd(int x, int y); that returns the greatest common divisorof its two arguments. You may assume that they are both posi
UBC - CPSC - 344
CPSC 31306W Term 2Problem Set #3 - Solution1. (a) int gcd(int a, int b)cfw_if (a = b)return a;else if (a &gt; b)return gcd(a - b, b);elsereturn gcd(a, b - a);.file&quot;gcdrec.c&quot;.text.p2align 4,15.globl gcd.typegcd, @functiongcd:pushl%ebpmov
UBC - CPSC - 344
CS 313, Winter 2006 - Term 2Assignment 4: Defusing a Binary BombAssigned: February 5, Due: Sunday, February 18, 11:59:59 PM1IntroductionThe nefarious Dr. Evil has planted a slew of binary bombs on our machines. A binary bomb is a programthat consist
UBC - CPSC - 344
CS 313, Winter 2006 - Term 2Assignment 5: HCL and y86Assigned: February 26, Due: Sunday, March 4, 11:59PM (with theusual 13 hour grace period)Instructions: Hand in all solutions on paper.1. Write an HCL expression for a boolean signal implies, true w
UBC - CPSC - 344
CPSC 313 06W Term 2 Problem Set #5 - Solution 1. bool implies = !a | b; 2. The four signals divide into 2 classes, sort0 and sort3 choose the minimum or maximum elements, which look like this. sort3 is identical to sort0, with the comparisons changed from
UBC - CPSC - 344
CPSC 313, Winter 2006 - Term 2Understanding the Y86 Architectureand its Sequential ImplementationAssigned: March 2, Due: Sunday, March 11, 11:59PMFor this problem set, you may either work by yourself or in a group of two. If you choose to work in agr
UBC - CPSC - 344
CPSC 31306W Term 2Problem Set #6 - Solution1. iaddl:StageFetchDecodeExecuteMemoryWrite backPC updateiaddl V, rBicode:ifun M1 [PC]rA:rB M1 [PC + 1]valC M4 [PC + 2]valP PC + 6valB R[rB]valE valB + valCsetCCR[rB] valEPC valP2. leave:Sta
UBC - CPSC - 344
CPSC 313, Winter 2006 - Term 2PipeliningAssigned: March 9, Due: Sunday, March 18, 11:59PMAll of these questions are to be handed in on paper.1. Consider a ve-stage pipeline with stage gate delays of 150 ps, 75 ps, 100 ps, 100 ps and 175 ps anda memor
UBC - CPSC - 344
CPSC 31306W Term 2Problem Set #7 - Solution1.(a)1185ps= 5.41 109 cycles per second or 5.41 GHz(b) 5.41 Gops (instructions per second)(c) 5 * 185 = 925 ps per instruction(d)1i. 600ps = 1.66 109 cycles per second or 1.66 GHzii. 1.66 Gops (instr
UBC - CPSC - 344
CPSC 313, Winter 2006 - Term 2Problem Set #8Assigned: March 26, Due: Sunday, April 1, 11:59PMYou will hand in all of your solutions to this problem set on paper.1. What is SRAM? Where do you typically nd it? What is good about it? What is bad about it
UBC - CPSC - 344
CPSC 31306W Term 2Problem Set #8 - Solution1. SRAM stands for Static Random Access Memory. It can be typically found in caches, on or off thecore. It is fast and persistent, but it requires more transistors so is more expensive per unit of memorythat
UBC - CPSC - 344
CPSC 313, Winter 2006 - Term 2Problem Set #9Assigned: April 2Due Wednesday, April 11, 13:00 pm (no grace period)1. Textbook 10.112. Textbook 10.123. Textbook 10.134. Consider a virtual-memory system with the following parameters PTBR = physical ad
UBC - CPSC - 344
CPSC 31306W Term 2Problem Set #9 - Solution1.A. 00 0010 0111 1100B.VPNTLBITLBTTLB hit?page fault?PPN0x90x10x2NN0x17C. 0101 1111 1100D.COCICTcache hit?cache byte?2.0x00xf0x17N-A. 00 0011 1010 1001B.VPNTLBITLBTTLB hit?pa
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