Process Synchronization
Race Condition, Critical Section
Inter Process Communication
• 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.
Producer and Consumer
while (true) { while (true) {
/* produce an item in next
produced */ while (counter == 0)
; /* do nothing */
while (counter == BUFFER_SIZE); next_consumed = buffer[out];
/* do nothing */ out = (out + 1) %
BUFFER_SIZE;
buffer[in] = next_produced;
counter--;
in = (in + 1) % BUFFER_SIZE;
/* consume the item in next
counter++;
consumed */
}
}
Race Condition
• counter++ could be implemented as A situation where several processes
access and manipulate the same data
register1 = counter concurrently and the outcome of the
register1 = register1 + 1
counter = register1 execution depends on the particular
order in which the access takes place,
• counter-- could be implemented as is called a race condition
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}
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
Critical Section
• General structure of process Pi
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
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
Peterson’s Solution
• 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[2]
• 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!
Algorithm for Process Pi
do {
flag[i] = true;
turn = j;
while (flag[j] && turn = = j);
critical section
flag[i] = false;
remainder section
} while (true);
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
