Do not start the test until instructed to do so!
|
|
- Bathsheba Flowers
- 6 years ago
- Views:
Transcription
1 CS 3204 Operating Systems Midterm (Abrams) Spring 2004 VIRG INIA POLYTECHNIC INSTITUTE AND STATE U T PRO SI M UNI VERSI TY Instructions: Do not start the test until instructed to do so! Print your name in the space provided below. This examination is closed book and closed notes. No calculators or other computing devices may be used. Answer each question in the space provided. If you need to continue an answer onto the back of a page, clearly indicate that and label the continuation with the question number. If you want partial credit, justify your answers, even when justification is not explicitly required. When you have completed the test, sign the pledge below and turn in the test. Note that either failing to return this test, or discussing its content with a student who has not taken it is a violation of the Honor Code. Print Your Name: Pledge: On my honor, I have neither given nor received unauthorized aid on this examination. Sign your name: Problem Long1 Long2 Long3 Long4 Short1 Short2 Short3 Short4 Short5 Short6 Total Grade /10 /20 /25 /24 /3 /5 /2 /1 /5 /5 /100 1 of 1
2 CS 3204 Operating Systems Midterm (Abrams) Spring 2004 Long Answer L1. [10 points] Consider the simple run/ready/block state transition scheme for process scheduling. Draw the state diagram, showing all the possible transitions. Label each transition with a brief, precise description of an event that might cause a process to make that transition. 2 of 2
3 CS 3204 Operating Systems Midterm (Abrams) Spring 2004 L2. [20 points] Consider executing the two threads below, with the shared global variables that are shown. //Shared variables 1 boolean t1 = ; 2 boolean t2 = ; //Thread T1 T1.4 void main() { T1.5 while ( true ) { T1.6 t1 = ; T1.7 while ( t2 ); T1.8 //critical section code T1.9 t1 = ; T1.10 //code outside critical section T1.11 } //end while loop T1.12 } //end thread T1 // Thread T2 T2.4 void main() { T2.5 while ( true ) { T2.6 t2 = ; T2.7 while ( t1 ); T2.8 //critical section code T2.9 t2 = ; T2.10 //code outside critical section T2.11 } //end while loop T2.12 } //end thread T2 (a) [10 points] Fill in the blanks in the code above (with TRUE or FALSE) to guarantee that mutual exclusion will be achieved. (b) [10 points] Does program with your modifications in (a) guarantee that the program is free from deadlock? YES NO If you answer YES, justify your answer. If you answer NO, clearly describe an execution sequence (using line numbers) that would lead to a deadlock of the two threads. 3 of 3
4 CS 3204 Operating Systems Midterm (Abrams) Spring 2004 L3. [25 points] Consider executing the two threads below, with the shared global variables that are shown. //Shared variables 1 Semaphore valueproduced = new Semaphore(1); 2 Semaphore valueconsumed = new Semaphore(1); 3 int sharedvalue=2; // variable shared by Producer and Consumer //Producer thread 4 void main() { 5 int v=0; 6 while ( true ) { 7 v++; 8 P(valueConsumed); 9 sharedvalue=v; 10 V(valueProduced); 11 } // end while 12 } // end producer thread //Consumer thread 13 void main() { 14 int v; 15 while ( true ) { 16 P(valueProduced); 17 v = sharedvalue; 18 V(valueConsumed); 19 printf( %d, v); 21 } // end consumer thread (a) [5 points] List all line numbers at which a thread can potentially block. (b) [10 points] If the code as shown above is executed, would either thread block forever? YES NO If your answer is YES clearly describe an execution sequence (using line numb ers) that would lead to a thread blocking forever. If your answer is NO, justify your answer by giving a logical argument why neither thread could block forever. 4 of 4
5 (c) [10 points] The specification for the program above says that all executions of this program must write to standard output the sequence of integers 1,2,3,4,5,. Does the program meet the specification? YES NO If your answer is YES, justify your answer by giving a logical argument as to why this is the only possible output for any execution order. If your answer is NO, clearly describe an execution sequence (using line numbers) that generates a different output, and state what the program would write to standard output. 5
6 L4. [24 points, 4 points per box] After graduation from Virginia Tech, you work at a software company that is writing a new operating system. The designers are trying to decide whether to use user-level threads, kernel-level threads, or a hybrid threading model. Because you are fresh out of CS3204, you are in a great position to advise them. In the boxes below, make a list of the pros and cons of each option. User level pros: User level cons: Kernel level pros: Kernel level cons: Hybrid pros: Hybrid cons: Short Answer S1. [3 points] If three threads want to achieve mutual exclusion, which of the following algorithms could be used? a. Peterson s YES NO b. Lamport s bakery YES NO c. Dekker s YES NO S2. [5 points] Why would an operating system that uses a microkernel run faster on a multiprocessor computer than on using a monolithic design? 6
7 S3. [2 points] What is the difference between a system call and a normal call of a function or method in an API? S4. [1 point] Give one reason why a DVD allows storage of more data than a CD. S5. [5 points] Which of the following is a true statement: a. RISC and CISC typically use the same number of registers b. CISC microprocessors generally contain more registers c. RISC microprocessors generally uses more registers Justify your answer: S6. [5 points] Why is a kernel call often implemented by a software interrupt? Have a good break! 7
8 8
9 Solution: L1: See Figure 3.1 L2: This is the algorithm in Figure 5.8 in the book. (a) Blanks are filled in as follows: boolean t1=false; boolean t2=false; t1=true t1=false t2=true t2=false (b) NO (In the following, a correct answer can omit the items in [] or <>.) There are several correct sequences. But ANY correct answer must have T1.6 AND T2.6 executed BEFORE either of T1.7 and T2.7, and NO statements executed after T1.7 and T2.7. Here are two correct answers: [T1.4,] T1.5, T1.6, [<context switch>,] [T2.4,] T2.5, T2.6, T2.7, <T2 stays in T2.7 forever>, T1.7 <T1 stays in T2.7 forever DEADLOCK> or [T2.4,] T2.5, T2.6, T2.7, [<context switch>,] [T1.4,] T1.5, T1.6, T1.7, <DEADLOCK> L3: Comment: A key difference of the code in problem L3 from Figure 5.16 in the book is listed in item 2 in Self Review on page 231 valueproduced is initialized to 1 and not zero. (a) 8 and 16. That is because only the P() operation blocks. The V() operation and all other statements used in the program never block.. NOTE: It is optional to list statement 19. The process blocks to execute 19, unless the compiler/os do programmed I/O (which an old computer and OS or an embedded OS would do). (b) NO This is the hardest problem in the exam. To be rigorous, the answer requires a proof (or a proof outline, similar to the arguments given in Chapter 5 for why the shared memory mutual exclusion algorithms do not block a thread forever). However, full credit will be given to an answer that has the seed of the ideas needed in a proof at least the fact that the value of the semaphore named in each P operation is always larger than 1 when the P is executed (either because the thread is in its first loop iteration, or, by inductive reasoning, the thread did not block at P in the last loop iteration, and the thread only got past the P in the last iteration if the semaphore value was incremented to a value larger than 0 (by the other thread)). For the record, here is a proof outline Proof by contradiction. Assume that one thread does block forever. Let s look at the 3 possible blocking statements from part (a). [It is OPTIONAL to discuss statement 19] Statement 19 cannot block forever, because I/O to standard out will eventually complete. Stdout is either the console so nothing would block display of letters on the console or a file 9
10 (if stdio is redirected to a file), but even if the file system becomes full an error message would be generated and printf() would return in error. Hence statement 19 cannot block forever. Exactly one of statement 8 or 16 (or both) are blocked forever. It is impossible for exactly one of the two statements to block forever, because the other thread would eventually execute a V() operation on the semaphore whose P operation is causing blocking forever. That V operation would unblock the blocked thread. Both statement 8 AND 16 are forever blocked. This can only happen if both semaphores are 0. However, it is impossible for this case to arise. The proof is by induction. o On first iteration or each thread s loop, the semaphore is 1, so the P decrements the semaphore to 0 and does not block. Furthermore, the thread does a V on the other semaphore, so that other semaphore now has value 1 or 2. o Now assume that a thread has not blocked on its first i-1 iterations (for i>0). On the ith iteration of a thread s loop, the semaphore has value of greater than 0 (or else the iteration i-1 would have blocked; since i-1 did not block we know that the thread completed execution of the P operation, which would only have happened if the semaphore value was incremented by a V operation. Therefore, there is no statement at which either thread could block forever. (c) NO. 1,2,13,14,15,16,17,18,19 would generate 2 as the first output number. 10
11 L4: User level pros: Doesn t require kernel changes Facilitates portability User level cons: All threads in a process are context switched as one unit by OS. Thus if one thread did I/O, all threads would block. A process containing many threads cannot use more than one processor of a a multiprocessor. Threads get shorter timeslices because all threads share timeslice of one process. Kernel level pros: Allows every thread to be scheduled separately by OS, yielding maximum potential for parallelism, and allowing one thread to block while others continue. Can utilize multiple processors. Harder to handle signals and terminating threads. Kernel level cons: Harder to implement than user-level threading. Also, hard to make programs portable. Can create more overhead in OS needs to be one scheduler activation for every thread, which means schedule is controlling more activations (and hence larger kernel data structures), which could reduce OS performance. Hybrid pros: Allows pros of kernel-level. Allows multiple user threads to be mapped to one kernel scheduler activation ( m-to-n thread mapping ) negating last con listed for kernel level. Hybrid cons: Can be harder to create a portable program that works on any thread model. Programmer must be careful about choosing which threads to map to a single kernel activation record. If the programmer makes a bad choice, they get the disadvantages of user or kernel level threads. (For example, if the programmer maps two threads that both do I/O to the same kernel scheduler activation, the if one thread blocks for I/O the other thread must wait also.) 11
12 S1: a. NO (only does 2-way) b. YES (does N-way) c. NO (only does 2-way) S2: All calls to the OS kernel are serialized. Hence the kernel can run on only one processor in a multiprocessor. Therefore, to allow maximum concurrency, the kernel should do as little work as possible to minimize its execution time. A microkernel moves as much OS functionality as possible outside the kernel, thereby minimizing the kernel s execution time. All the functionality that a microkernel OS puts outside the kernel can potentially be run concurrently on multiple processors, thereby improving OS performance. S3: System call invokes operation inside the OS and ultimately the kernel, whereas normal API call invokes user space code. S4: A correct answer is either of the following: DVDs use 2 layer. DVDs use thinner tracks. S5: Correct answer is c, for either of these reasons: With a simpler instruction set, RISC does not typically have ALU instructions that access memory [just load and store instructions to access memory]. Hence RISC requires more quantities to be in registers than CISC. RISC processors have multiple copies of processor registers to permit fast context switching among a limited number of processes without requiring registers to be copied to a process control block in main memory. S6: Either of these answers is correct: An interrupt is a safe way to enter the kernel when a software interrupt occurs, the PC is changed by the processor to an address in the interrupt vector table (set by the kernel during boot time), so that the next instruction executed is inside the kernel. The processor must switch from user to privileged mode when the program counter changes to an instruction inside the kernel code. For security, a means is needed for this to happen only when a kernel call occurs a user must not be able to switch to privileged mode without entering the kernel. A software interrupt permits the mode change in a safe way. (In the above answers, safe means that a nasty user cannot enter the kernel or change to privileged mode without raising a software interrupt.) 12
Solution printed. Do not start the test until instructed to do so! CS 2504 Intro Computer Organization Test 2 Spring 2006.
VIRG INIA POLYTECHNIC INSTITUTE AND STATE U T PROSI M UNI VERSI TY Instructions: Print your name in the space provided below. This examination is closed book and closed notes, aside from the permitted
Solution printed. Do not start the test until instructed to do so! CS 2604 Data Structures Midterm Spring Instructions:
VIRG INIA POLYTECHNIC INSTITUTE AND STATE U T PROSI M UNI VERSI TY Instructions: Print your name in the space provided below. This examination is closed book and closed notes, aside from the permitted
Solution printed. Do not start the test until instructed to do so! CS 2604 Data Structures Midterm Summer I Instructions:
VIRG INIA POLYTECHNIC INSTITUTE AND STATE U T PROSI M UNI VERSI TY Instructions: Print your name in the space provided below. This examination is closed book and closed notes, aside from the permitted
CS4411 Intro. to Operating Systems Exam 1 Fall points 10 pages
CS4411 Intro. to Operating Systems Exam 1 Fall 2005 (October 6, 2005) 1 CS4411 Intro. to Operating Systems Exam 1 Fall 2005 150 points 10 pages Name: Most of the following questions only require very short
Chapter 5 Asynchronous Concurrent Execution
Chapter 5 Asynchronous Concurrent Execution Outline 5.1 Introduction 5.2 Mutual Exclusion 5.2.1 Java Multithreading Case Study 5.2.2 Critical Sections 5.2.3 Mutual Exclusion Primitives 5.3 Implementing
CS 2604 Data Structures Midterm Summer 2000 U T. Do not start the test until instructed to do so!
VIRG INIA POLYTECHNIC INSTITUTE AND STATE U T PROSI M UNI VERSI TY Instructions: Print your name in the space provided below. This examination is closed book and closed notes. No calculators or other computing
CS140 Operating Systems and Systems Programming Midterm Exam
CS140 Operating Systems and Systems Programming Midterm Exam October 28 th, 2005 (Total time = 50 minutes, Total Points = 50) Name: (please print) In recognition of and in the spirit of the Stanford University
Midterm on next week Tuesday May 4. CS 361 Concurrent programming Drexel University Fall 2004 Lecture 9
CS 361 Concurrent programming Drexel University Fall 2004 Lecture 9 Bruce Char and Vera Zaychik. All rights reserved by the author. Permission is given to students enrolled in CS361 Fall 2004 to reproduce
CS 361 Concurrent programming Drexel University Fall 2004 Lecture 8. Proof by contradiction. Proof of correctness. Proof of mutual exclusion property
CS 361 Concurrent programming Drexel University Fall 2004 Lecture 8 Bruce Char and Vera Zaychik. All rights reserved by the author. Permission is given to students enrolled in CS361 Fall 2004 to reproduce
CS140 Operating Systems and Systems Programming Midterm Exam
CS140 Operating Systems and Systems Programming Midterm Exam October 31 st, 2003 (Total time = 50 minutes, Total Points = 50) Name: (please print) In recognition of and in the spirit of the Stanford University
CROWDMARK. Examination Midterm. Spring 2017 CS 350. Closed Book. Page 1 of 30. University of Waterloo CS350 Midterm Examination.
Times: Thursday 2017-06-22 at 19:00 to 20:50 (7 to 8:50PM) Duration: 1 hour 50 minutes (110 minutes) Exam ID: 3520593 Please print in pen: Waterloo Student ID Number: WatIAM/Quest Login Userid: Sections:
CS140 Operating Systems and Systems Programming Midterm Exam
CS140 Operating Systems and Systems Programming Midterm Exam October 28 th, 2002 (Total time = 50 minutes, Total Points = 50) Name: (please print) In recognition of and in the spirit of the Stanford University
THE AUSTRALIAN NATIONAL UNIVERSITY Mid-semester Quiz, Second Semester COMP2310 (Concurrent and Distributed Systems)
THE AUSTRALIAN NATIONAL UNIVERSITY Mid-semester Quiz, Second Semester 2001 COMP2310 (Concurrent and Distributed Systems) Writing Period: 65 minutes duration Study Period: 0 minutes duration Permitted Materials:
CS 571 Operating Systems. Midterm Review. Angelos Stavrou, George Mason University
CS 571 Operating Systems Midterm Review Angelos Stavrou, George Mason University Class Midterm: Grading 2 Grading Midterm: 25% Theory Part 60% (1h 30m) Programming Part 40% (1h) Theory Part (Closed Books):
CS 2505 Computer Organization I
Instructions: Print your name in the space provided below. This examination is closed book and closed notes, aside from the permitted one-page formula sheet. No calculators or other computing devices may
CIS Operating Systems Synchronization based on Busy Waiting. Professor Qiang Zeng Spring 2018
CIS 3207 - Operating Systems Synchronization based on Busy Waiting Professor Qiang Zeng Spring 2018 Previous class IPC for passing data Pipe FIFO Message Queue Shared Memory Compare these IPCs for data
Operating Systems. Lecture 4 - Concurrency and Synchronization. Master of Computer Science PUF - Hồ Chí Minh 2016/2017
Operating Systems Lecture 4 - Concurrency and Synchronization Adrien Krähenbühl Master of Computer Science PUF - Hồ Chí Minh 2016/2017 Mutual exclusion Hardware solutions Semaphores IPC: Message passing
Chapter 5: Process Synchronization. Operating System Concepts 9 th Edition
Chapter 5: Process Synchronization Silberschatz, Galvin and Gagne 2013 Chapter 5: Process Synchronization Background The Critical-Section Problem Peterson s Solution Synchronization Hardware Mutex Locks
Synchronization for Concurrent Tasks
Synchronization for Concurrent Tasks Minsoo Ryu Department of Computer Science and Engineering 2 1 Race Condition and Critical Section Page X 2 Algorithmic Approaches Page X 3 Hardware Support Page X 4
Pre- and post- CS protocols. CS 361 Concurrent programming Drexel University Fall 2004 Lecture 7. Other requirements for a mutual exclusion algorithm
CS 361 Concurrent programming Drexel University Fall 2004 Lecture 7 Bruce Char and Vera Zaychik. All rights reserved by the author. Permission is given to students enrolled in CS361 Fall 2004 to reproduce
Synchronization. Before We Begin. Synchronization. Credit/Debit Problem: Race Condition. CSE 120: Principles of Operating Systems.
CSE 120: Principles of Operating Systems Lecture 4 Synchronization January 23, 2006 Prof. Joe Pasquale Department of Computer Science and Engineering University of California, San Diego Before We Begin
CMPS 111 Spring 2013 Prof. Scott A. Brandt Midterm Examination May 6, Name: ID:
CMPS 111 Spring 2013 Prof. Scott A. Brandt Midterm Examination May 6, 2013 Name: ID: This is a closed note, closed book exam. There are 23 multiple choice questions, 6 short answer questions. Plan your
CS153: Midterm (Winter 19)
CS153: Midterm (Winter 19) Name: Student ID: Answer all questions. State any assumptions clearly. Problem 1: (24 points; 5 minutes) Indicate whether each of the following statements is true or false: (T)
KEY CENG 334. Midterm. Question 1. Question 2. Question 3. Question 4. Question 5. Question 6. Total. Name, SURNAME and ID
00 11 00 11 01 01 01 01 01 01 00 11 00 11 Name, SURNAME and ID Middle East Technical University Department of Computer Engineering KEY CENG 334 Section 2 and 3 Spring 2010-2011 Midterm Duration: 120 minutes.
University of Illinois at Chicago Department of Computer Science. Final Examination. CS 151 Mathematical Foundations of Computer Science Fall 2012
University of Illinois at Chicago Department of Computer Science Final Examination CS 151 Mathematical Foundations of Computer Science Fall 2012 Thursday, October 18, 2012 Name: Email: Print your name
CS140 Operating Systems and Systems Programming
CS140 Operating Systems and Systems Programming Midterm Exam July 25th, 2006 Total time = 60 minutes, Total Points = 100 Name: (please print) In recognition of and in the spirit of the Stanford University
Fall 2004 CS414 Prelim 1
Fall 2004 CS414 Prelim 1 1. The Sim City Smoking Ban problem: In the Sim-City community Woobish most people smoke, but the laws of Sim City require that non-smokers be protected from passive smoke. So
ECE 612: Embedded and Real-Time Systems
ECE 612: Embedded and Real-Time Systems Instructor: A. Cyrus Sabzevari Last updated: 1/16/2013 Email: asabzeva@gmu.edu Spring 2013 Monday 7:20-10 PM Office Hours: Monday 6:10 7:10 in the adjunct faculty
(b) External fragmentation can happen in a virtual memory paging system.
Alexandria University Faculty of Engineering Electrical Engineering - Communications Spring 2015 Final Exam CS333: Operating Systems Wednesday, June 17, 2015 Allowed Time: 3 Hours Maximum: 75 points Note:
COS 318: Midterm Exam (October 23, 2012) (80 Minutes)
COS 318: Midterm Exam (October 23, 2012) (80 Minutes) Name: This exam is closed-book, closed-notes. 1 single-sided 8.5x11 sheet of notes is permitted. No calculators, laptop, palmtop computers are allowed
University of Waterloo Midterm Examination Model Solution CS350 Operating Systems
University of Waterloo Midterm Examination Model Solution CS350 Operating Systems Fall, 2003 1. (10 total marks) Suppose that two processes, a and b, are running in a uniprocessor system. a has three threads.
COS 126 General Computer Science Spring Written Exam 1
COS 126 General Computer Science Spring 2017 Written Exam 1 This exam has 9 questions (including question 0) worth a total of 70 points. You have 50 minutes. Write all answers inside the designated spaces.
CS 2505 Computer Organization I Test 1. Do not start the test until instructed to do so! printed
Instructions: Print your name in the space provided below. This examination is closed book and closed notes, aside from the permitted one-page formula sheet. This examination is closed book and closed
CS 153 Design of Operating Systems Winter 2016
CS 153 Design of Operating Systems Winter 2016 Lecture 7: Synchronization Administrivia Homework 1 Due today by the end of day Hopefully you have started on project 1 by now? Kernel-level threads (preemptable
CSE 451 Midterm 1. Name:
CSE 451 Midterm 1 Name: 1. [2 points] Imagine that a new CPU were built that contained multiple, complete sets of registers each set contains a PC plus all the other registers available to user programs.
Synchronization. Before We Begin. Synchronization. Example of a Race Condition. CSE 120: Principles of Operating Systems. Lecture 4.
CSE 120: Principles of Operating Systems Lecture 4 Synchronization October 7, 2003 Before We Begin Read Chapter 7 (Process Synchronization) Programming Assignment #1 Due Sunday, October 19, midnight Prof.
CS 3114 Data Structures and Algorithms Test 1 READ THIS NOW!
READ THIS NOW! Print your name in the space provided below. There are 7 short-answer questions, priced as marked. The maximum score is 100. This examination is closed book and closed notes, aside from
CS4411 Intro. to Operating Systems Exam 2 Fall 2009
CS4411 Intro. to Operating Systems Exam 2 { Fall 2009 1 CS4411 Intro. to Operating Systems Exam 2 Fall 2009 150 points { 8 pages Name: Most of the following questions only require short answers. Usually
Locks. Dongkun Shin, SKKU
Locks 1 Locks: The Basic Idea To implement a critical section A lock variable must be declared A lock variable holds the state of the lock Available (unlocked, free) Acquired (locked, held) Exactly one
Operating Systems EDA092, DIT 400 Exam
Chalmers University of Technology and Gothenburg University Operating Systems EDA092, DIT 400 Exam 2015-04-14 Date, Time, Place: Tuesday 2015/04/14, 14:00 18:00, Väg och vatten -salar Course Responsible:
CS 2504 Intro Computer Organization Test 1
Instructions: Print your name in the space provided below. This examination is closed book and closed notes, aside from the permitted one-page formula sheet and the MIPS reference card. No calculators
CSE 120: Principles of Operating Systems. Lecture 4. Synchronization. October 7, Prof. Joe Pasquale
CSE 120: Principles of Operating Systems Lecture 4 Synchronization October 7, 2003 Prof. Joe Pasquale Department of Computer Science and Engineering University of California, San Diego 2003 by Joseph Pasquale
CS 2506 Computer Organization II Test 2
Instructions: Print your name in the space provided below. This examination is closed book and closed notes, aside from the permitted one-page formula sheet. No calculators or other computing devices may
Operating System Architecture. CS3026 Operating Systems Lecture 03
Operating System Architecture CS3026 Operating Systems Lecture 03 The Role of an Operating System Service provider Provide a set of services to system users Resource allocator Exploit the hardware resources
Eastern Mediterranean University School of Computing and Technology Department of Information Technology. ITEC202 Operating Systems
Page 1 of 8 ITEC202 Operating Systems, Midterm Exam Eastern Mediterranean University School of Computing and Technology Department of Information Technology ITEC202 Operating Systems ITEC 202 Midterm Examination
CPSC/ECE 3220 Fall 2017 Exam Give the definition (note: not the roles) for an operating system as stated in the textbook. (2 pts.
CPSC/ECE 3220 Fall 2017 Exam 1 Name: 1. Give the definition (note: not the roles) for an operating system as stated in the textbook. (2 pts.) Referee / Illusionist / Glue. Circle only one of R, I, or G.
Operating Systems Comprehensive Exam. Spring Student ID # 3/16/2006
Operating Systems Comprehensive Exam Spring 2006 Student ID # 3/16/2006 You must complete all of part I (60%) You must complete two of the three sections in part II (20% each) In Part I, circle or select
Semaphores. May 10, Mutual exclusion with shared variables is difficult (e.g. Dekker s solution).
Semaphores May 10, 2000 1 Introduction Mutual exclusion with shared variables is difficult (e.g. Dekker s solution). Generalising to an arbitrary number of processes is also nontrivial (e.g. Lamport s
UNIT:2. Process Management
1 UNIT:2 Process Management SYLLABUS 2.1 Process and Process management i. Process model overview ii. Programmers view of process iii. Process states 2.2 Process and Processor Scheduling i Scheduling Criteria
CSE 131 Introduction to Computer Science Fall Exam I
CSE 131 Introduction to Computer Science Fall 2015 Given: 24 September 2015 Exam I Due: End of session This exam is closed-book, closed-notes, no electronic devices allowed. The exception is the sage page
Process Management And Synchronization
Process Management And Synchronization In a single processor multiprogramming system the processor switches between the various jobs until to finish the execution of all jobs. These jobs will share the
Concurrent & Distributed Systems Supervision Exercises
Concurrent & Distributed Systems Supervision Exercises Stephen Kell Stephen.Kell@cl.cam.ac.uk November 9, 2009 These exercises are intended to cover all the main points of understanding in the lecture
CS 2505 Computer Organization I Test 1. Do not start the test until instructed to do so! printed
Instructions: Print your name in the space provided below. This examination is closed book and closed notes, aside from the permitted one-page formula sheet. This examination is closed book and closed
CHAPTER 6: PROCESS SYNCHRONIZATION
CHAPTER 6: PROCESS SYNCHRONIZATION The slides do not contain all the information and cannot be treated as a study material for Operating System. Please refer the text book for exams. TOPICS Background
Midterm Exam Amy Murphy 6 March 2002
University of Rochester Midterm Exam Amy Murphy 6 March 2002 Computer Systems (CSC2/456) Read before beginning: Please write clearly. Illegible answers cannot be graded. Be sure to identify all of your
Subject: Operating System (BTCOC403) Class: S.Y.B.Tech. (Computer Engineering)
A. Multiple Choice Questions (60 questions) Subject: Operating System (BTCOC403) Class: S.Y.B.Tech. (Computer Engineering) Unit-I 1. What is operating system? a) collection of programs that manages hardware
Sistemi in tempo reale Anno accademico
Sistemi in tempo reale Anno accademico 2006-2007 Concorrenza - II Giuseppe Lipari http://feanor.sssup.it/~lipari Scuola Superiore Sant Anna Outline 1 Introduction to concurrency 2 Models of concurrency:
Remaining Contemplation Questions
Process Synchronisation Remaining Contemplation Questions 1. The first known correct software solution to the critical-section problem for two processes was developed by Dekker. The two processes, P0 and
CSC369 Lecture 2. Larry Zhang
CSC369 Lecture 2 Larry Zhang 1 Announcements Lecture slides Midterm timing issue Assignment 1 will be out soon! Start early, and ask questions. We will have bonus for groups that finish early. 2 Assignment
CS4411 Intro. to Operating Systems Exam 1 Fall points 9 pages
CS4411 Intro. to Operating Systems Exam 1 Fall 2009 1 CS4411 Intro. to Operating Systems Exam 1 Fall 2009 150 points 9 pages Name: Most of the following questions only require very short answers. Usually
CSE 131 Introduction to Computer Science Fall 2016 Exam I. Print clearly the following information:
CSE 131 Introduction to Computer Science Fall 2016 Given: 29 September 2016 Exam I Due: End of Exam Session This exam is closed-book, closed-notes, no electronic devices allowed The exception is the "sage
Mutual Exclusion. 1 Formal problem definitions. Time notion CSE /17/2015. Outline of this lecture:
CSE 539 03/17/2015 Mutual Exclusion Lecture 15 Scribe: Son Dinh Outline of this lecture: 1. Formal problem definitions 2. Solution for 2 threads 3. Solution for n threads 4. Inherent costs of mutual exclusion
Lecture Topics. Announcements. Today: Concurrency: Mutual Exclusion (Stallings, chapter , 5.7)
Lecture Topics Today: Concurrency: Mutual Exclusion (Stallings, chapter 5.1-5.4, 5.7) Next: Concurrency: Deadlock and Starvation (Stallings, chapter 6.1, 6.6-6.8) 1 Announcements Self-Study Exercise #5
CSE 247 Data Structures and Algorithms Fall Exam I
CSE 247 Data Structures and Algorithms Fall 2016 Given: 29 February 2016 Exam I Due: End of session This exam is closed-book, closed-notes. No electronic devices or resources of any kind are allowed. The
Do not start the test until instructed to do so!
Instructions: Print your name in the space provided below. This examination is closed book and closed notes, aside from the permitted one-page formula sheet and the MIPS reference card. No calculators
Operating Systems Comprehensive Exam. Spring Student ID # 2/17/2011
Operating Systems Comprehensive Exam Spring 2011 Student ID # 2/17/2011 You must complete all of Section I You must complete two of the problems in Section II If you need more space to answer a question,
CS143 Final Fall 2009
CS143 Final Fall 2009 Please read all instructions (including these) carefully. There are 4 questions on the exam, all with multiple parts. You have 2 hours to work on the exam. The exam is closed book,
CSC Operating Systems Spring Lecture - XII Midterm Review. Tevfik Ko!ar. Louisiana State University. March 4 th, 2008.
CSC 4103 - Operating Systems Spring 2008 Lecture - XII Midterm Review Tevfik Ko!ar Louisiana State University March 4 th, 2008 1 I/O Structure After I/O starts, control returns to user program only upon
Chapter 6: Process [& Thread] Synchronization. CSCI [4 6] 730 Operating Systems. Why does cooperation require synchronization?
Chapter 6: Process [& Thread] Synchronization CSCI [4 6] 730 Operating Systems Synchronization Part 1 : The Basics Why is synchronization needed? Synchronization Language/Definitions:» What are race conditions?»
CS420: Operating Systems. Process Synchronization
Process Synchronization James Moscola Department of Engineering & Computer Science York College of Pennsylvania Based on Operating System Concepts, 9th Edition by Silberschatz, Galvin, Gagne Background
Q1. State True/false with jusification if the answer is false:
Paper Title: Operating System (IOPS332C) Quiz 1 Time : 1 hr Q1. State True/false with jusification if the answer is false: a. Multiprogramming (having more programs in RAM simultaneously) decreases total
Spring 2017 CS 1110/1111 Exam 1
CS 1110/1111 Spring 2017 Exam 1 page 1 of 6 Spring 2017 CS 1110/1111 Exam 1 Bubble in your computing ID in the footer of this page. We use an optical scanner to read it, so fill in the bubbles darkly.
Chapters 5 and 6 Concurrency
Operating Systems: Internals and Design Principles, 6/E William Stallings Chapters 5 and 6 Concurrency Patricia Roy Manatee Community College, Venice, FL 2008, Prentice Hall Concurrency When several processes/threads
CS 2505 Computer Organization I
Instructions: Print your name in the space provided below. This examination is closed book and closed notes, aside from the permitted one-page formula sheet. No calculators or other computing devices may
COS 126 Written Exam 2 (Spring 2015)
COS 126 Written Exam 2 (Spring 2015) There are 8 questions on this exam, weighted as indicated below. This exam is closed book. You may use a single-page two-sided hand-written cheatsheet. There is a blank
University of California at Berkeley College of Engineering Department of Electrical Engineering and Computer Science
University of California at Berkeley College of Engineering Department of Electrical Engineering and Computer Science CS 162 Spring 2010 I. Stoica FIRST MIDTERM EXAMINATION Tuesday, March 9, 2010 INSTRUCTIONS
Student Name: University of California at Berkeley College of Engineering Department of Electrical Engineering and Computer Science
University of California at Berkeley College of Engineering Department of Electrical Engineering and Computer Science CS 162 Spring 2011 I. Stoica FIRST MIDTERM EXAMINATION Wednesday, March 9, 2011 INSTRUCTIONS
CS3331 Concurrent Computing Exam 1 Solutions Fall 2018
CS3331 Concurrent Computing Exam 1 Solutions Fall 2018 1 CS3331 Concurrent Computing Exam 1 Solutions Fall 2018 1. Basic Concepts (a) [10 points] Explain interrupts and traps, and provide a detailed account
THE UNIVERSITY OF WESTERN AUSTRALIA SAMPLE EXAM QUESTIONS 2007 WITH SOLUTIONS SCHOOL OF COMPUTER SCIENCE CITS3213 CONCURRENT PROGRAMMING (PART II)
THE UNIVERSITY OF WESTERN AUSTRALIA SAMPLE EXAM QUESTIONS 2007 WITH SOLUTIONS SCHOOL OF COMPUTER SCIENCE CITS3213 CONCURRENT PROGRAMMING (PART II) The exam will contain: 6 questions (3 for each part) Time
Midterm Exam #2 Solutions October 25, 2016 CS162 Operating Systems
University of California, Berkeley College of Engineering Computer Science Division EECS all 2016 Anthony D. Joseph Midterm Exam #2 Solutions October 25, 2016 CS162 Operating Systems Your Name: SID AND
CS 2506 Computer Organization II Test 1
Instructions: Print your name in the space provided below. This examination is closed book and closed notes, aside from the permitted one-page formula sheet. No calculators or other computing devices may
CS 111X - Fall Test 1
CS 111X - Fall 2016 - Test 1 1/9 Computing ID: CS 111X - Fall 2016 - Test 1 Name: Computing ID: On my honor as a student, I have neither given nor received unauthorized assistance on this exam. Signature:
Synchronization. CSE 2431: Introduction to Operating Systems Reading: Chapter 5, [OSC] (except Section 5.10)
Synchronization CSE 2431: Introduction to Operating Systems Reading: Chapter 5, [OSC] (except Section 5.10) 1 Outline Critical region and mutual exclusion Mutual exclusion using busy waiting Sleep and
Problem Score Max Score 1 Syntax directed translation & type
CMSC430 Spring 2014 Midterm 2 Name Instructions You have 75 minutes for to take this exam. This exam has a total of 100 points. An average of 45 seconds per point. This is a closed book exam. No notes
Dr. D. M. Akbar Hussain DE5 Department of Electronic Systems
Concurrency 1 Concurrency Execution of multiple processes. Multi-programming: Management of multiple processes within a uni- processor system, every system has this support, whether big, small or complex.
CS162, Spring/1992 Midterm #2 Professor Thomas Anderson
CS162, Spring/1992 Midterm #2 Professor Thomas Anderson General Information: This is a closed book examination. You have 60 minutes to answer as many question as possible. The number in parentheses at
CS 2506 Computer Organization II Test 1. Do not start the test until instructed to do so! printed
Instructions: Print your name in the space provided below. This examination is closed book and closed notes, aside from the permitted one-page formula sheet. No calculators or other computing devices may
CS370 Operating Systems
CS370 Operating Systems Colorado State University Yashwant K Malaiya Spring 1018 L14 Deadlocks Slides based on Text by Silberschatz, Galvin, Gagne Various sources 1 1 FAQ Mutex vs Semaphore 2 2 FAQ Does
Dept. of CSE, York Univ. 1
EECS 3221.3 Operating System Fundamentals No.5 Process Synchronization(1) Prof. Hui Jiang Dept of Electrical Engineering and Computer Science, York University Background: cooperating processes with shared
EECS 452 Midterm Closed book part Fall 2010
EECS 452 Midterm Closed book part Fall 2010 Name: unique name: Sign the honor code: I have neither given nor received aid on this exam nor observed anyone else doing so. Scores: # Points Closed book Page
! Why is synchronization needed? ! Synchronization Language/Definitions: ! How are locks implemented? Maria Hybinette, UGA
Chapter 6: Process [& Thread] Synchronization CSCI [4 6] 730 Operating Systems Synchronization Part 1 : The Basics! Why is synchronization needed?! Synchronization Language/Definitions:» What are race
Operating Systems. Designed and Presented by Dr. Ayman Elshenawy Elsefy
Operating Systems Designed and Presented by Dr. Ayman Elshenawy Elsefy Dept. of Systems & Computer Eng.. AL-AZHAR University Website : eaymanelshenawy.wordpress.com Email : eaymanelshenawy@yahoo.com Reference
COS 126 General Computer Science Fall Written Exam 1
COS 26 General Computer Science Fall 27 Written Exam This exam has 7 questions (including question ) worth a total of 7 points. You have 5 minutes. Write all answers inside the designated spaces. Policies.
Introduction to OS Synchronization MOS 2.3
Introduction to OS Synchronization MOS 2.3 Mahmoud El-Gayyar elgayyar@ci.suez.edu.eg Mahmoud El-Gayyar / Introduction to OS 1 Challenge How can we help processes synchronize with each other? E.g., how
Midterm Exam Amy Murphy 19 March 2003
University of Rochester Midterm Exam Amy Murphy 19 March 2003 Computer Systems (CSC2/456) Read before beginning: Please write clearly. Illegible answers cannot be graded. Be sure to identify all of your
CSE 131S Introduction to Computer Science Summer SON Exam I
CSE 131S Introduction to Computer Science Summer SON 2014 Exam I Given: 1 July 2014 Due: End of live session This exam is closed-book, closed-notes, no electronic devices allowed except for downloading
Do not start the test until instructed to do so!
Instructions: Print your name in the space provided below. This examination is closed book and closed notes, aside from the permitted one-page formula sheet. No calculators or other electronic devices
Introduction to Operating Systems
Introduction to Operating Systems Lecture 4: Process Synchronization MING GAO SE@ecnu (for course related communications) mgao@sei.ecnu.edu.cn Mar. 18, 2015 Outline 1 The synchronization problem 2 A roadmap
Computer Science CS221 Test 2 Name. 1. Give a definition of the following terms, and include a brief example. a) Big Oh
Computer Science CS221 Test 2 Name 1. Give a definition of the following terms, and include a brief example. a) Big Oh b) abstract class c) overriding d) implementing an interface 10/21/1999 Page 1 of
R13 SET - 1 2. Answering the question in Part-A is compulsory 1 a) Define Operating System. List out the objectives of an operating system. [3M] b) Describe different attributes of the process. [4M] c)