KEY CENG 334. Midterm. Question 1. Question 2. Question 3. Question 4. Question 5. Question 6. Total. Name, SURNAME and ID

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1 Name, SURNAME and ID Middle East Technical University Department of Computer Engineering KEY CENG 334 Section 2 and 3 Spring Midterm Duration: 120 minutes. Exam: This is a closed book, closed notes exam. No attempts of cheating will be tolerated. In case such attempts are observed, the students who took part in the act will be prosecuted. The legal code states that students who are found guilty of cheating shall be expelled from the university for a minimum of one semester! About the exam questions: The points assigned for each question are shown in parenthesis next to the question. Whereever available, use the boxes to write down your answers. This booklet consists of 8 pages including this page. Check that you have them all! GOOD LUCK! Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Total

2 1 (15 pts) (a) (b) (c) (d) (e) Write down one advantage and one disadvantage of using user-level threads to kernel-level threads. Answer: Pros: Faster context-switch, Cons: inability to make use of multiple processors. the blocking of a single thread would block the process (hence all the other threads). What s a trap? What generates it and what does the OS do when it gets one? Answer: Trap is software interrupt. System calls generate it and the OS responds by calling the proper interrupt handler, that implements the system call itself. What s a process and what constitutes its essential components of a process? Answer: A process is a running program. It has three essential components: CPU state, Address Space (or memory state) and OS resources (such as open files, sockets etc.). What s preemption? What can a preemptive scheduler do that non-preemptive schedulers can t do? Asnwer: Preemption is the act of taking a resource from a process without its will. A preemptive scheduler can take away the CPU from a process without waiting for it to yield() or to go for I/O. What s priority inversion? Give a scheduling policy in which priority inversion can happen? How can it be solved? Answer: Priority inversion is the phenomenon of a higher-priority process/thread being blocked by a lower-priority process/thread (since it holds a resource, such as a mutex, that cannot be preempted). It can be solved by priority inheritance, that is by giving the lower-priority process/thread the same level of high-priority until it releases the resource.

3 2 (15 pts) Consider the C program below. (For space reasons, we are not checking error return codes, so assume that all functions return normally.) main() { if (fork() == 0) { if (fork() == 0) { printf("3"); else { pid_t pid; int status; if ((pid = wait(&status)) > 0) { printf("4"); else { if (fork() == 0) { printf("1"); exit(0); printf("2"); printf("0"); return 0; Out of the 5 outputs listed below, circle only the valid outputs of this program. Assume that all processes run to normal completion. Note that circling wrong outputs will be punished such that random guesses will receive zero point on average. A B C D E Answer: A,C,E Grading: If none of the outputs are circled then 0 points, else check the status (circled/uncircled) of each output. Each correct status would get +3 pts. Each wrong status would get -3 pts. Negative points are capped at -3 pts.

4 3 (8 pts) Consider the following synchronization code for two threads: int turn = 0; // shared void thread0(){ while (1){ while (turn!=0) ; critical_section(); turn = 1; uncritical_section(); void thread1(){ while(1){ while (turn!= 1) ; critical_section(); turn = 0; uncritical_section(); Is this a good solution to the critical-section problem? Consider the four requirements for synchronization, and discuss which ones it satisfy and which ones it doesn t? Answer: Recall the four requirements : (1 pts) Mutual exclusion: Satisfied. (4 pts) Progress: Not satisfied, since the threads must strictly alternate entering their critical sections, a thread wanting to enter its critical section twice in a row will be blocked until the other thread decides to enter (and leave) its critical section. (1 pts) Bounded waiting: Satisfied. (2 pts) Performance: Not very efficient, since it does spinlocking/

5 4 (14 pts) Consider 4 threads, called as A, B, C and D. Using semaphores, ensure that A completes its job() before any other process (B, C, or D) starts, and B completes its job() before C or D, but C and D may execute their job()concurrently. You can declare and initialize semaphores such as sem = semaphore(1); by calling two methods as sem.up(); or sem.down();. // declarations a= semaphore(0); b=semaphore(0); void A(){ a.up(); void B(){ a.down(); b.up(); b.up(); void C(){ b.down(); void D(){ b.down(); Grading: A->B synchronization: 6 pts. B->C,D synchronization: 8 pts. wrong initialization: -4 pts. Unnecessary complexity: -2 pts. and use them

6 5 (30 pts) We are interested in cooking a pan of menemen which requires two eggs and one tomato. Create a Mesa-type monitor with methods eggcomes() and tomatocomes(), which wait until a pan of menemen can be cooked. Don t worry about explicitly cooking menemen; just wait until two egg threads and one tomato thread can be grouped together. For example, if two egg threads call eggcomes(), and then a third tomato thread calls tomatocomes(), the third thread should wake up the first two threads and they should then all return. Use the following skeleton to write your code. Note that although a number of declaration are provided for you, you are free to declare and use more int s or condition variables as necessary. Assume that the functions wait(cv), notify(cv), and notifyall(cv), are available where cv is a declared condition variable. monitor menemen{ int we= 0; // Number of waiting eggs int wt= 0; // Number of waiting tomatoes condition waitinge; // used by tomatoes condition waitingt; // used by eggs void eggcomes(){ we++; if (we>=2 && wt >=1){ we -=2; wt--; notify(waitinge); notify(waitingt); else{ wait(waitingt); void tomatocomes(){ wt++; if (we>=2 && wt >= 1){ we -=2; wt--; notify(waitingt); notify(waitingt); else{ wait(waitinge); Grading: Trace for three specific cases, and evaluate whether all three threads are released after the arrival of the third one: - Egg-Egg-Tomato (10 pts) - Egg-Tomato-Egg (10 pts) - Tomato-Egg-Egg (10 pts) Unnecessary complexity: -5 pts Not keeping track of we and wt properly: -5

7 6 (18 pts) Three processes are at the ready queue of the scheduler in the order A, B and E at time 0. If you use the FCFS policy the scheduling executed is as shown in the Gantt chart below. In the chart, A, B, and E represents the CPU use of these processes whereas a, b and e denote the I/O s of processes respectively. Fill in the scheduling executed by Round robin (RR) with a timeslice of 1 unit, Shortest-Job- First (SJF), and Shortest-Remaining-Time-First (SRTF). For SJF and SRTF, assume that A, B and E uses CPU bursts of 5, 1 and 6 time units. If at a given time slice the CPU is being used by, say A, whereas the I/O requests of B and E are active, then fill in that time slice with Abe vertically. Also assume that the I/O device can handle multiple I/O requests concurrently. In SRTF, if the remaining CPU burst of the running process equals to the remaining CPU burst of another process in the ready list, then no context switch should occur. Grading: RR: 4pts; SJF: 6 pts; SRTF: 8 pts. 0 1 Time FCFS A a A a B b B b RR A B E A E A E B A E A E E A A A A A b b b b b b b b e e a a a a SJF B A A A A A B E E E E E E A A A A A b b b a a e e a a b b b b b SRTF B A A A B A A E E E E E E A A A A A b b b b b b b b a a e e a a 2 3 Time FCFS b E e RR SJF SRTF Grading: Each correct 5 time slot period gets 1/4 of the total points assigned to the policy