ch6 - Chapter 6: Process Synchronization Module 6: Process...

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Chapter 6: Process Synchronization Chapter 6: Process Synchronization
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6.2 Silberschatz, Galvin and Gagne ©2005 Operating System Concepts – 7 th Edition, Feb 8, 2005 Module 6: Process Synchronization Module 6: Process Synchronization Background The Critical-Section Problem Peterson’s Solution Synchronization Hardware Semaphores Classic Problems of Synchronization Monitors Synchronization Examples Atomic Transactions
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6.3 Silberschatz, Galvin and Gagne ©2005 Operating System Concepts – 7 th Edition, Feb 8, 2005 Background Background Concurrent access to shared data may result in data inconsistency Maintaining data consistency requires mechanisms to ensure the orderly execution of cooperating processes 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 count that keeps track of the number of full buffers. Initially, count 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.
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6.4 Silberschatz, Galvin and Gagne ©2005 Operating System Concepts – 7 th Edition, Feb 8, 2005 Producer Producer while (true) { /* produce an item and put in nextProduced */ while (count == BUFFER_SIZE) ; // do nothing buffer [in] = nextProduced; in = (in + 1) % BUFFER_SIZE; count++; }
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6.5 Silberschatz, Galvin and Gagne ©2005 Operating System Concepts – 7 th Edition, Feb 8, 2005 Consumer Consumer while (true) { while (count == 0) ; // do nothing nextConsumed = buffer[out]; out = (out + 1) % BUFFER_SIZE; count--; /* consume the item in nextConsumed }
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6.6 Silberschatz, Galvin and Gagne ©2005 Operating System Concepts – 7 th Edition, Feb 8, 2005 Race Condition Race Condition count++ could be implemented as register1 = count register1 = register1 + 1 count = register1 count-- could be implemented as register2 = count register2 = register2 - 1 count = register2 Consider this execution interleaving with “count = 5” initially: S0: producer execute register1 = count {register1 = 5} S1: producer execute register1 = register1 + 1 {register1 = 6} S2: consumer execute register2 = count {register2 = 5} S3: consumer execute register2 = register2 - 1 {register2 = 4} S4: producer execute count = register1 {count = 6 } S5: consumer execute count = register2 {count = 4}
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6.7 Silberschatz, Galvin and Gagne ©2005 Operating System Concepts – 7 th Edition, Feb 8, 2005 Solution to Critical-Section Problem Solution to Critical-Section Problem 1. Mutual Exclusion - If process P i 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
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6.8
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ch6 - Chapter 6: Process Synchronization Module 6: Process...

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