MODULE 3 process synchronizationnnn.pptx

MODULE 3
Process Synchronization

Outline
• Critical-Section Problem
• Peterson’s Solution
• Synchronization Hardware
• Mutex Locks
• Semaphores
• Classic Problems of Synchronization.

Race Condition
• Processes P0 and P1 are creating child processes using the fork()
system call
• Race condition on kernel variable next_available_pid which
represents the next available process identifier (pid)
• Unless there is a mechanism to prevent P0 and P1 from accessing the
variable next_available_pid the same pid could be assigned to two
different processes!

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

Critical-Section Problem (Cont.)
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 process 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
Requirements for solution to critical-section problem

Interrupt-based Solution
• Entry section: disable interrupts
• Exit section: enable interrupts
• Will this solve the problem?
• What if the critical section is code that runs for an hour?
• Can some processes starve – never enter their critical section.
• What if there are two CPUs?

Software Solution 1
• Two process solution
• Assume that the load and store machine-language
instructions are atomic; that is, cannot be interrupted
• The two processes share one variable:
• int turn;
• The variable turn indicates whose turn it is to enter the
critical section
• initially, the value of turn is set to i

Algorithm for Process Pi
while (true){
while (turn = = j);
/* critical section */
turn = j;
/* remainder section */
}

Correctness of the Software Solution
• Mutual exclusion is preserved
Pi enters critical section only if:
turn = i
and turn cannot be both 0 and 1 at the same time
• What about the Progress requirement?
• What about the Bounded-waiting requirement?

Peterson’s Solution
• 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
while (true){
flag[i] = true;
turn = j;
while (flag[j] && turn = = j)
;
flag[i] = false;
}

Correctness of Peterson’s Solution
• 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

Peterson’s Solution and Modern Architecture
• Although useful for demonstrating an algorithm, Peterson’s
Solution is not guaranteed to work on modern architectures.
• To improve performance, processors and/or compilers may reorder
operations that have no dependencies
• Understanding why it will not work is useful for better
understanding race conditions.
• For single-threaded this is ok as the result will always be the
same.
• For multithreaded the reordering may produce inconsistent or
unexpected results!

Modern Architecture Example
• 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?
100

Modern Architecture Example (Cont.)
• However, since the variables flag and x are
independent of each other, the instructions:
flag = true;
x = 100;
for Thread 2 may be reordered
• If this occurs, the output may be 0!

Peterson’s Solution Revisited
• The effects of instruction reordering in Peterson’s Solution
• This allows both processes to be in their critical section at the same time!
• To ensure that Peterson’s solution will work correctly on modern computer
architecture we must use Memory Barrier.

Memory Barrier
• Memory model are the memory guarantees a computer
architecture makes to application programs.
• Memory models may be either:
• Strongly ordered – where a memory modification of one processor is
immediately visible to all other processors.
• Weakly ordered – where a memory modification of one processor
may not be immediately visible to all other processors.
• A memory barrier is an instruction that forces any change in
memory to be propagated (made visible) to all other
processors.

Memory Barrier Instructions
• When a memory barrier instruction is performed, the system
ensures that all loads and stores are completed before any
subsequent load or store operations are performed.
• Therefore, even if instructions were reordered, the memory
barrier ensures that the store operations are completed in
memory and visible to other processors before future load or
store operations are performed.

Memory Barrier Example
• Returning to the example of slides 6.17 - 6.18
• We could add a memory barrier to the following
instructions to ensure Thread 1 outputs 100:
• Thread 1 now performs
while (!flag)
memory_barrier();
print x
• Thread 2 now performs
x = 100;
memory_barrier();
flag = true
• For Thread 1 we are guaranteed that that the value of
flag is loaded before the value of x.
• For Thread 2 we ensure that the assignment to x occurs
before the assignment flag.

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. Hardware instructions
2. Atomic variables

Hardware Instructions
• Special hardware instructions that allow us to either
test-and-modify the content of a word, or to swap the
contents of two words atomically (uninterruptedly.)
• Test-and-Set instruction
• Compare-and-Swap instruction

The test_and_set Instruction
• Definition
boolean test_and_set (boolean
*target)
{
boolean rv = *target;
*target = true;
return rv:
}
• Properties
• Executed atomically
• Returns the original value of passed parameter
• Set the new value of passed parameter to true

Solution Using test_and_set()
• Shared boolean variable lock, initialized to false
• Solution:
do {
while (test_and_set(&lock))
; /* do nothing */
lock = false;
} while (true);
• Does it solve the critical-section problem?

The 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;
}
• Properties
• Executed atomically
• Returns the original value of passed parameter value
• 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.

Solution using compare_and_swap
• Shared integer lock initialized to 0;
• Solution:
while (true){
while (compare_and_swap(&lock, 0, 1) != 0)
; /* do nothing */
lock = 0;
}
• Does it solve the critical-section problem?

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
• Boolean variable indicating if lock is available or not
• Protect a critical section by
• First acquire() a lock
• Then release() the lock
• 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

Solution to CS Problem Using Mutex Locks
while (true) {
acquire lock
critical section
release lock
remainder section
}

Semaphore
• Synchronization tool that provides more sophisticated ways (than
Mutex locks) for processes 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 <= 0)
; // busy wait
S--;
}
• Definition of the signal() operation
signal(S) {
S++;
}

Semaphore (Cont.)
• Counting semaphore – integer value can range over
an unrestricted domain
• Binary semaphore – integer value can range only
between 0 and 1
• Same as a mutex lock
• Can implement a counting semaphore S as a binary
semaphore
• With semaphores we can solve various
synchronization problems

Semaphore Usage Example
• Solution to the CS Problem
• Create a semaphore “mutex” initialized to 1
wait(mutex);
CS
signal(mutex);
• Consider P1 and P2 that with two statements S1 and S2
and the requirement that S1 to happen before S2
• Create a semaphore “synch” initialized to 0
P1:
S1;
signal(synch);
P2:
wait(synch);
S2;

Semaphore Implementation
• Must guarantee that no two processes can execute the wait()
and signal() on the same semaphore at the same time
• Thus, the implementation becomes the critical section problem
where the wait and signal code are placed in the critical
section
• Could now have busy waiting in critical section implementation
• But implementation code is short
• Little busy waiting if critical section rarely occupied
• Note that applications may spend lots of time in critical sections
and therefore this is not a good solution

Semaphore Implementation with no Busy waiting
• With each semaphore there is an associated waiting queue
• Each entry in a waiting queue has two data items:
• Value (of type integer)
• Pointer to next record in the list
• Two operations:
• block – place the process invoking the operation on the appropriate
waiting queue
• wakeup – remove one of processes in the waiting queue and place it
in the ready queue

Implementation with no Busy waiting (Cont.)
• Waiting queue
typedef struct {
int value;
struct process *list;
} semaphore;

Implementation with no Busy waiting (Cont.)
wait(semaphore *S) {
S->value--;
if (S->value < 0) {
add this process to S->list;
block();
}
}
signal(semaphore *S) {
S->value++;
if (S->value <= 0) {
remove a process P from S->list;
wakeup(P);
}
}

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
semaphores and other synchronization tools are used
incorrectly.

Classical Problems of Synchronization
• Classical problems used to test newly-proposed
synchronization schemes
• Bounded-Buffer Problem
• Readers and Writers Problem
• Dining-Philosophers Problem

Bounded-Buffer Problem
• n buffers, each can hold one item
• Semaphore mutex initialized to the value 1
• Semaphore full initialized to the value 0
• Semaphore empty initialized to the value n

Bounded Buffer Problem (Cont.)
• The structure of the producer process
while (true) {
...
/* produce an item in next_produced */
...
wait(empty);
wait(mutex);
...
/* add next produced to the buffer */
...
signal(mutex);
signal(full);
}

Bounded Buffer Problem (Cont.)
• The structure of the consumer process
while (true) {
wait(full);
wait(mutex);
...
/* remove an item from buffer to next_consumed
*/
...
signal(mutex);
signal(empty);
...
/* consume the item in next consumed */
...
}

Readers-Writers Problem
• A data set is shared among a number of concurrent processes
• Readers – only read the data set; they do not perform any updates
• Writers – can both read and write
• Problem – allow multiple readers to read at the same time
• Only one single writer can access the shared data at the same time
• Several variations of how readers and writers are considered – all
involve some form of priorities

Readers-Writers Problem (Cont.)
• Shared Data
• Data set
• Semaphore rw_mutex initialized to 1
• Semaphore mutex initialized to 1
• Integer read_count initialized to 0

• The structure of a writer process
while (true) {
wait(rw_mutex);
...
/* writing is performed */
...
signal(rw_mutex);
}

• The structure of a reader process
while (true){
wait(mutex);
read_count++;
if (read_count == 1) /* first reader */
wait(rw_mutex);
signal(mutex);
...
/* reading is performed */
...
wait(mutex);
read_count--;
if (read_count == 0) /* last reader */
signal(rw_mutex);
signal(mutex);
}

Readers-Writers Problem Variations
• The solution in previous slide can result in a situation
where a writer process never writes. It is referred to as
the “First reader-writer” problem.
• The “Second reader-writer” problem is a variation the
first reader-writer problem that state:
• Once a writer is ready to write, no “newly arrived reader” is
allowed to read.
• Both the first and second may result in starvation.
leading to even more variations
• Problem is solved on some systems by kernel providing
reader-writer locks

Dining-Philosophers Problem
• N philosophers’ sit at a round table with a bowel of rice in the
middle.
• They spend their lives alternating thinking and eating.
• They do not interact with their neighbors.
• Occasionally try to pick up 2 chopsticks (one at a time) to eat from
bowl
• Need both to eat, then release both when done
• In the case of 5 philosophers, the shared data
• Bowl of rice (data set)
• Semaphore chopstick [5] initialized to 1

Dining-Philosophers Problem Algorithm
• Semaphore Solution
• The structure of Philosopher i :
while (true){
wait (chopstick[i] );
wait (chopStick[ (i + 1) % 5] );
/* eat for awhile */
signal (chopstick[i] );
signal (chopstick[ (i + 1) % 5] );
/* think for awhile */
}
• What is the problem with this algorithm?

Monitor Solution to Dining Philosophers
monitor DiningPhilosophers
{
enum {THINKING; HUNGRY, EATING} state [5];
condition self [5];
void pickup (int i) {
state[i] = HUNGRY;
test(i);
if (state[i] != EATING) self[i].wait;
}
void putdown (int i) {
state[i] = THINKING;
// test left and right neighbors
test((i + 4) % 5);
test((i + 1) % 5);
}

Solution to Dining Philosophers (Cont.)
void test (int i) {
if ((state[(i + 4) % 5] != EATING) &&
(state[i] == HUNGRY) &&
(state[(i + 1) % 5] != EATING) ) {
state[i] = EATING ;
self[i].signal () ;
}
}
initialization_code() {
for (int i = 0; i < 5; i++)
state[i] = THINKING;
}
}

• Each philosopher “i” invokes the operations pickup() and
putdown() in the following sequence:
DiningPhilosophers.pickup(i);
/** EAT **/
DiningPhilosophers.putdown(i);
• No deadlock, but starvation is possible
Solution to Dining Philosophers (Cont.)

MODULE 3 process synchronizationnnn.pptx

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