ch6 Synchronization Tools.ppt

Chapter 6: Synchronization Tools Operating System Concepts – 10th Edition Silberschatz, Galvin and Gagne ©2018  Chapter 6: Synchronization Tools Background The Critical-Section Problem...

Chapter 6: Synchronization Tools Operating System Concepts – 10th Edition Silberschatz, Galvin and Gagne ©2018  Chapter 6: Synchronization Tools Background The Critical-Section Problem Peterson’s Solution Hardware Support for Synchronization Mutex Locks Semaphores Liveness Evaluation Operating System Concepts – 10th Edition 6.2 Silberschatz, Galvin and Gagne ©2018  Objectives Describe the critical-section problem and illustrate a race condition Illustrate hardware solutions to the critical-section problem using compare-and-swap operations, and atomic variables Demonstrate how mutex locks, semaphores, monitors, and condition variables can be used to solve the critical section problem Evaluate tools that solve the critical-section problem in low-. Moderate-, and high-contention scenarios Operating System Concepts – 10th Edition 6.3 Silberschatz, Galvin and Gagne ©2018  Background Processes can execute concurrently  May be interrupted at any time, partially completing execution Concurrent access to shared data may result in data inconsistency Maintaining data consistency requires mechanisms to ensure the orderly execution of cooperating processes Illustration of the problem: Suppose that we wanted to provide a solution to the consumer-producer problem that fills all the buffers. We can do so by having an integer counter that keeps track of the number of full buffers. Initially, counter is set to 0. It is incremented by the producer after it produces a new buffer and is decremented by the consumer after it consumes a buffer. Operating System Concepts – 10th Edition 6.4 Silberschatz, Galvin and Gagne ©2018  Producer while (true) { while (counter == BUFFER_SIZE) ; buffer[in] = next_produced; in = (in + 1) % BUFFER_SIZE; counter++; } Operating System Concepts – 10th Edition 6.5 Silberschatz, Galvin and Gagne ©2018  Consumer while (true) { while (counter == 0) ; next_consumed = buffer[out]; out = (out + 1) % BUFFER_SIZE; counter--; } Operating System Concepts – 10th Edition 6.6 Silberschatz, Galvin and Gagne ©2018  Race Condition counter++ could be implemented as register1 = counter register1 = register1 + 1 counter = register1 counter-- could be implemented as register2 = counter register2 = register2 - 1 counter = register2 Consider this execution interleaving with “count = 5” initially: S0: producer execute register1 = counter {register1 = 5} S1: producer execute register1 = register1 + 1 {register1 = 6} S2: consumer execute register2 = counter {register2 = 5} S3: consumer execute register2 = register2 – 1 {register2 = 4} S4: producer execute counter = register1 {counter = 6 } S5: consumer execute counter = register2 {counter = 4} Operating System Concepts – 10th Edition 6.7 Silberschatz, Galvin and Gagne ©2018  Race Condition Processes P0 and P1 are creating child processs using the fork() system call Race condition on kernel variable next_available_pid which represents the next available process identifier (pid) Unless there is mutual exclusion, the same pid could be assigned to two different processes! Operating System Concepts – 10th Edition 6.8 Silberschatz, Galvin and Gagne ©2018  Critical Section Problem Consider system of n processes {p0, p1, … pn-1} Each process has critical section segment of code  Process may be changing common variables, updating table, writing file, etc  When one process in critical section, no other may be in its critical section Critical section problem is to design protocol to solve this Each process must ask permission to enter critical section in entry section, may follow critical section with exit section, then remainder section Operating System Concepts – 10th Edition 6.9 Silberschatz, Galvin and Gagne ©2018  Critical Section General structure of process Pi Operating System Concepts – 10th Edition 6.10 Silberschatz, Galvin and Gagne ©2018  Solution to Critical-Section Problem 1. Mutual Exclusion - If process Pi is executing in its critical section, then no other processes can be executing in their critical sections 2. Progress - If no process is executing in its critical section and there exist some processes that wish to enter their critical section, then the selection of the processes that will enter the critical section next cannot be postponed indefinitely 3. Bounded Waiting - A bound must exist on the number of times that other processes are allowed to enter their critical sections after a process has made a request to enter its critical section and before that request is granted  Assume that each process executes at a nonzero speed  No assumption concerning relative speed of the n processes Operating System Concepts – 10th Edition 6.11 Silberschatz, Galvin and Gagne ©2018  Solution to Critical-Section Problem What are three requirements of any solution to the critical section problem? 1. Mutual Exclusion - No two processes simultaneously in critical sections 1. Progress no process outside of a critical region may cause another process to block 3. Bounded Waiting No process must wait forever to enter its critical region  Assume that each process executes at a nonzero speed  No assumption concerning relative speed of the n processes Operating System Concepts – 10th Edition 6.12 Silberschatz, Galvin and Gagne ©2018  Critical-Section Handling in OS Two approaches depending on if kernel is preemptive or non- preemptive  Preemptive – allows preemption of process when running in kernel mode  Non-preemptive – runs until exits kernel mode, blocks, or voluntarily yields CPU Essentially free of race conditions in kernel mode Operating System Concepts – 10th Edition 6.13 Silberschatz, Galvin and Gagne ©2018  Peterson’s Solution Not guaranteed to work on modern architectures! (But good algorithmic description of solving the problem) Two process solution Assume that the load and store machine-language instructions are atomic; that is, cannot be interrupted The two processes share two variables:  int turn;  boolean flag The variable turn indicates whose turn it is to enter the critical section The flag array is used to indicate if a process is ready to enter the critical section. flag[i] = true implies that process Pi is ready! Operating System Concepts – 10th Edition 6.14 Silberschatz, Galvin and Gagne ©2018  Algorithm for Process Pi while (true){ flag[i] = true; turn = j; while (flag[j] && turn = = j) ; flag[i] = false; } Operating System Concepts – 10th Edition 6.15 Silberschatz, Galvin and Gagne ©2018  Peterson’s Solution (Cont.) Provable that the three CS requirement are met: 1. Mutual exclusion is preserved Pi enters CS only if: either flag[j] = false or turn = i 2. Progress requirement is satisfied 3. Bounded-waiting requirement is met Operating System Concepts – 10th Edition 6.16 Silberschatz, Galvin and Gagne ©2018  Peterson’s Solution Although useful for demonstrating an algorithm, Peterson’s Solution is not guaranteed to work on modern architectures. Understanding why it will not work is also useful for better understanding race conditions. To improve performance, processors and/or compilers may reorder operations that have no dependencies. For single-threaded this is ok as the result will always be the same. For multithreaded the reordering may produce inconsistent or unexpected results! Operating System Concepts – 10th Edition 6.17 Silberschatz, Galvin and Gagne ©2018  Peterson’s Solution Two threads share the data: boolean flag = false; int x = 0; Thread 1 performs while (!flag) ; print x Thread 2 performs x = 100; flag = true What is the expected output? Operating System Concepts – 10th Edition 6.18 Silberschatz, Galvin and Gagne ©2018  Peterson’s Solution 100 is the expected output. However, the operations for Thread 2 may be reordered: flag = true; x = 100; If this occurs, the output may be 0! The effects of instruction reordering in Peterson’s Solution This allows both processes to be in their critical section at the same time! Operating System Concepts – 10th Edition 6.19 Silberschatz, Galvin and Gagne ©2018  Synchronization Hardware Many systems provide hardware support for implementing the critical section code. Uniprocessors – could disable interrupts  Currently running code would execute without preemption  Generally too inefficient on multiprocessor systems  Operating systems using this not broadly scalable We will look at three forms of hardware support: 1. Memory barriers 2. Hardware instructions 3. Atomic variables Operating System Concepts – 10th Edition 6.20 Silberschatz, Galvin and Gagne ©2018  Hardware Instructions Special hardware instructions that allow us to either test-and-modify the content of a word, or two swap the contents of two words atomically (uninterruptibly.) Test-and-Set instruction Compare-and-Swap instruction Operating System Concepts – 10th Edition 6.21 Silberschatz, Galvin and Gagne ©2018  test_and_set Instruction Definition: boolean test_and_set (boolean *target) { boolean rv = *target; *target = true; return rv: } 1. Executed atomically 2. Returns the original value of passed parameter 3. Set the new value of passed parameter to true Operating System Concepts – 10th Edition 6.22 Silberschatz, Galvin and Gagne ©2018  Solution using test_and_set() Shared boolean variable lock, initialized to false Solution: do { while (test_and_set(&lock)) ; lock = false; } while (true); Operating System Concepts – 10th Edition 6.23 Silberschatz, Galvin and Gagne ©2018  compare_and_swap Instruction Definition: int compare _and_swap(int *value, int expected, int new_value) { int temp = *value; if (*value == expected) *value = new_value; return temp; } 1. Executed atomically 2. Returns the original value of passed parameter value 3. Set the variable value the value of the passed parameter new_value but only if *value == expected is true. That is, the swap takes place only under this condition. Operating System Concepts – 10th Edition 6.24 Silberschatz, Galvin and Gagne ©2018  Solution using compare_and_swap Shared integer lock initialized to 0; Solution: while (true){ while (compare_and_swap(&lock, 0, 1) != 0) ; lock = 0; } Operating System Concepts – 10th Edition 6.25 Silberschatz, Galvin and Gagne ©2018  Atomic Variables Typically, instructions such as compare-and-swap are used as building blocks for other synchronization tools. One tool is an atomic variable that provides atomic (uninterruptible) updates on basic data types such as integers and booleans. For example, the increment() operation on the atomic variable sequence ensures sequence is incremented without interruption: increment(&sequence); Operating System Concepts – 10th Edition 6.26 Silberschatz, Galvin and Gagne ©2018  Atomic Variables The increment() function can be implemented as follows: void increment(atomic_int *v) { int temp; do { temp = *v; } while (temp != (compare_and_swap(v,temp,temp+1)); } Operating System Concepts – 10th Edition 6.27 Silberschatz, Galvin and Gagne ©2018  Mutex Locks Previous solutions are complicated and generally inaccessible to application programmers OS designers build software tools to solve critical section problem Simplest is mutex lock Protect a critical section by first acquire() a lock then release() the lock  Boolean variable indicating if lock is available or not Calls to acquire() and release() must be atomic  Usually implemented via hardware atomic instructions such as compare-and-swap. But this solution requires busy waiting This lock therefore called a spinlock Operating System Concepts – 10th Edition 6.28 Silberschatz, Galvin and Gagne ©2018  Solution to Critical-section Problem Using Locks while (true) { acquire lock critical section release lock remainder section } Operating System Concepts – 10th Edition 6.29 Silberschatz, Galvin and Gagne ©2018  Mutex Lock Definitions acquire() { while (!available) ; available = false;; } release() { available = true; } These two functions must be implemented atomically. Both test-and-set and compare-and-swap can be used to implement these functions. Operating System Concepts – 10th Edition 6.30 Silberschatz, Galvin and Gagne ©2018  Semaphore Synchronization tool that provides more sophisticated ways (than Mutex locks) for process to synchronize their activities. Semaphore S – integer variable Can only be accessed via two indivisible (atomic) operations  wait() and signal()  (Originally called P() and V()) Definition of the wait() operation wait(S) { while (S value--; if (S->value < 0) { add this process to S->list; block(); } } signal(semaphore *S) { S->value++; if (S->value list; wakeup(P); } } Operating System Concepts – 10th Edition 6.35 Silberschatz, Galvin and Gagne ©2018  Problems with Semaphores Incorrect use of semaphore operations:  signal (mutex) …. wait (mutex)  wait (mutex) … wait (mutex)  Omitting of wait (mutex) and/or signal (mutex) These – and others – are examples of what can occur when sempahores and other synchronization tools are used incorrectly. Operating System Concepts – 10th Edition 6.36 Silberschatz, Galvin and Gagne ©2018  Liveness Processes may have to wait indefinitely while trying to acquire a synchronization tool such as a mutex lock or semaphore. Waiting indefinitely violates the progress and bounded-waiting criteria discussed at the beginning of this chapter. Liveness refers to a set of properties that a system must satisfy to ensure processes make progress. Indefinite waiting is an example of a liveness failure. Operating System Concepts – 10th Edition 6.37 Silberschatz, Galvin and Gagne ©2018  Liveness Deadlock – two or more processes are waiting indefinitely for an event that can be caused by only one of the waiting processes Let S and Q be two semaphores initialized to 1 P0 P1 wait(S); wait(Q); wait(Q); wait(S);...... signal(S); signal(Q); signal(Q); signal(S); Consider if P0 executes wait(S) and P1 wait(Q). When P0 executes wait(Q), it must wait until P1 executes signal(Q) However, P1 is waiting until P0 execute signal(S). Since these signal() operations will never be executed, P0 and P1 are deadlocked. Operating System Concepts – 10th Edition 6.38 Silberschatz, Galvin and Gagne ©2018  Liveness Other forms of deadlock: Starvation – indefinite blocking  A process may never be removed from the semaphore queue in which it is suspended Priority Inversion – Scheduling problem when lower-priority process holds a lock needed by higher-priority process Solved via priority-inheritance protocol Operating System Concepts – 10th Edition 6.39 Silberschatz, Galvin and Gagne ©2018  Priority Inheritance Protocol Consider the scenario with three processes P1, P2, and P3. P1 has the highest priority, P2 the next highest, and P3 the lowest. Assume a resouce P3 is assigned a resource R that P1 wants. Thus, P1 must wait for P3 to finish using the resource. However, P2 becomes runnable and preempts P3. What has happened is that P2 - a process with a lower priority than P1 - has indirectly prevented P3 from gaining access to the resource. To prevent this from occurring, a priority inheritance protocol is used. This simply allows the priority of the highest thread waiting to access a shared resource to be assigned to the thread currently using the resource. Thus, the current owner of the resource is assigned the priority of the highest priority thread wishing to acquire the resource. Operating System Concepts – 10th Edition 6.40 Silberschatz, Galvin and Gagne ©2018  Priority Inheritance Protocol Operating System Concepts – 10th Edition 6.41 Silberschatz, Galvin and Gagne ©2018  End of Chapter 6 Operating System Concepts – 10th Edition Silberschatz, Galvin and Gagne ©2018