Computer Architecture and Operating Systems Course: International University Bremen Date: Final Examination

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1 Computer Architecture and Operating Systems Course: International University Bremen Date: Dr. Jürgen Schönwälder Type: open book Final Examination Problem F.1: operating systems ( =10 points) Indicate which of the following statements are correct or incorrect by marking the appropriate boxes. For every correctly marked box, you will earn two points. For every incorrectly marked box, you will loose one point. Statements which are not marked or which are marked as and will be ignored. The minimum number of points you can achieve is zero. Caches usually work well for applications that exhibit spatial and temporal locality. A hard link can span multiple file systems. Hard real-time operating systems cannot use virtual memory systems. Context switches require to reload or switch the page table in the memory management unit on paged virtual memory systems. Virtualization of the physical hardware allows to execute multiple operating systems concurrently where every concurrent operating system schedules its own set of potentially multi-threaded processes. Caches usually work well for applications that exhibit spatial and temporal locality. A hard link can span multiple file systems. Hard real-time operating systems cannot use virtual memory systems. Context switches require to reload or switch the page table in the memory management unit on paged virtual memory systems. Virtualization of the physical hardware allows to execute multiple operating systems concurrently where every concurrent operating system schedules its own set of potentially multi-threaded processes. Problem F.2: processes and threads (10+5+5=20 points) a) Consider the C program shown below. How many processes will be created when the program is executed? (Assume that all system calls succeed.) Draw a graph that shows the parent/child relationships among the processes. Hint: It usually helps to write down values for a and b for each process. /* fork.c */ #include <sys/types.h> #include <unistd.h> int main(int argc, char **argv) { pid_t a = 0, b = 0;

2 (void) fork(); a = fork(); if (a == 0) { b = fork(); if (a b) { (void) fork(); return 0; b) In a operating system with threads, is there one stack per thread or one stack per process? Explain. c) What is the main difference between context switch between processes and context switches between threads of the same process on a modern desktop computer? a) A total of 10 processes will be created (including the process entering the main() function). The resulting process tree looks likes this: +-fork-+-fork-+-fork-+-fork +-fork +-fork +-fork-+-fork +-fork +-fork b) Every thread must have its own stack to maintain the function call stack. A single shared stack would make it very difficult to return from functions. c) The main difference between threads and processes is that processes have their own logical address space while threads share the address space of the process. As a consequence, context switches between processes requires to potentially reload the memory management unit while a context switch between threads of the same process does not require to switch to another logical memory space. Problem F.3: process synchronization (20 points) The following synchronization problem is known as the unisex bathroom problem : A single bathroom is used as a unisex bathroom. The following two constraints must be maintained: 1. There cannot be men and women in the bathroom at the same time. 2. There should never be more than three people in the bathroom. Your task is to model men and woman as threads which execute the functions male() and female() to use the bathroom. Find a solution for the synchronization problem using semaphores and write it down in pseudo code. It is sufficient if the solution is deadlock free. Make sure that the usage of your semaphores and the logic behind the solution is well documented. A starvation free solution is a bit more complex and not required to obtain the points.

3 Below is the implementation of the function male(). The function female() is similar. semaphore empty = 1; /* controls access to the bathroom */ semaphore male_mutex = 1; /* mutex for male_counter */ semaphore male_multiplex = 3; /* limits # of men in the bathroom */ int male_counter; /* # of men in bathroom or waiting */ semaphore female_mutex = 1; /* mutex for female_counter */ semaphore female_multiplex = 3; /* limits # of women in the bathroom */ int female_counter; /* # of women in bathroom or waiting */ male() { down(&male_mutex); male_counter++; if (male_counter == 1) { down(&empty); /* make this a male bathroom or wait */ up(&male_mutex); down(&male_multiplex); /* limit # of people in the bathroom */ use_bathroom(); up(&male_multiplex); /* let the next one in */ down(&male_mutex) male_counter--; if (male_counter == 0) { up(&empty); /* may become a female bathroom now */ up(&male_mutex); Problem F.4: deadlock avoidance (banker s algorithm) (4+10+6=20 points) The Banker s algorithm can be used to avoid deadlocks by granting resource requests only if the system remains in a safe state, which can not lead to deadlocks. Consider a system with n = 4 processes (P 1, P 2, P 3, P 4 ). The current state of the system is as follows: Max = Need = Avail = (2, 2, 3, 3) a) Determine the current allocation matrix and the total number of instances of the 4 resource types. b) Is the current state safe? Justify your answer by executing the Banker s algorithm. c) Process P 1 requests 2 additional instances of resource two. Can this request be granted? a) Currently allocated resources: Alloc = Max Need =

4 The total number of resource instances is given by the vector (4, 10, 4, 7). b) Check whether the state is safe: Avail = (2, 2, 3, 3) Need = Process P 4 can get the claimed resources, terminate and release the assigned resources. This Avail = (3, 2, 3, 4) Need = Process P 3 can get the claimed resources, terminate and release the assigned resources. This Avail = (3, 4, 4, 4) Need = Process P 2 can get the claimed resources, terminate and release the assigned resources. This Avail = (3, 9, 4, 7) Need = Process P 1 can finally get the claimed resources, terminate and release the assigned resources. Since there is a sequence in which all processes can get their maximum claims, the state is safe and the resource request can be granted. c) Check whether the state after granting the resource request would be safe: Avail = (2, 0, 3, 3) Need = Process P 4 can get the claimed resources, terminate and release the assigned resources. This Avail = (3, 0, 4, 4) Need = Process P 3 can get the claimed resources, terminate and release the assigned resources. This Avail = (3, 2, 4, 4) Need = The maximum resource requests of the remaining processes cannot be satisfied anymore. The followup state is unsafe and hence the resource request must be denied.

5 Problem F.5: page replacement algorithms (10+10=20 points) Consider an operating system which uses paging to implement virtual memory. The physical memory system can hold 3 frames. A multi-threaded application generates the following reference string: w = 1, 5, 2, 5, 3, 5, 1, 2, 3, 4, 4, 5, 3, 5, 2 Assume that the memory frames are initially unused. a) Determine the entries in the page table under the Least Frequently Used (LFU) page replacement strategy. LFU replaces the page which has been used least frequently since it was loaded the last time. If multiple pages qualify, the FIFO strategy is used among these pages to break the tie. How many page faults to occur? b) Determine the entries in the page table under Belady s optimal page replacement strategy. If multiple pages have the same forward distance, the page with the smallest number is selected. How many page faults do occur? a) LFU replacement strategy (9 page faults) * * * * * * * * * b) Belady s optimal replacement strategy (7 page faults) * * * * * * *

6 Problem F.6: inter-process communication ( =10 points) Indicate which of the following statements are correct or incorrect by marking the appropriate boxes. For every correctly marked box, you will earn two points. For every incorrectly marked box, you will loose one point. Statements which are not marked or which are marked as and will be ignored. The minimum number of points you can achieve is zero. A pipe is a uni-directional communication channel between a parent and a child process. A machine instruction accessing an invalid memory address leads to hardware interrupt from the memory management unit which will be catched by the operating system kernel and cause a signal to be delivered to the process which executed the machine instruction. Unblocked signals can interrupt the normal control flow at arbitrary points in times. Datagram sockets do not need to call the bind() function since the remote address is passed to the sendto() function and retrieved from the recvfrom() function. The accept() socket function returns a new socket upon successful completion. A pipe is a uni-directional communication channel between a parent and a child process. Unblocked signals can interrupt the normal control flow at arbitrary points in times. A machine instruction accessing an invalid virtual memory address leads to hardware interrupt which will be catched by the operating system kernel and cause a signal to be delivered to the process which executed the machine instruction. Datagram sockets do not need to call the bind() function since the remote address is passed to the sendto() function and retrieved from the recvfrom() function. The accept() socket function returns a new socket upon successful completion.

Operating Systems and Networks Course: International University Bremen Date: Dr. Jürgen Schönwälder Deadline:

Operating Systems and Networks Course: 320202 International University Bremen Date: 2004-03-26 Dr. Jürgen Schönwälder Deadline: 2004-03-26 Midterm Examination Problem M.1: processes (10 points) Indicate