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Mrs.G.Chandraprabha,M.Sc.,M.Phil.,
Assistant Professor,
Department of Information Technology,
V.V.Vanniaperumal College for Women,
Virudhunagar.
2
 Bounded-Buffer Problem
 Readers and Writers Problem
 Dining-Philosophers Problem
CS420:
Operating
Systems
3
 The buffer pool has a maximum size, this problem is often called
the Bounded buffer problem.
 This problem is generalised in terms of the Producer Consumer problem,
where a finite buffer pool is used to exchange messages between
producer and consumer processes.
 Solution to this problem is, creating two counting semaphores "full" and
"empty" to keep track of the current number of full and empty buffers
respectively.
 In this Producers mainly produces a product and consumers consume the
product, but both can use of one of the containers each time.
 The main complexity of this problem is that we must have to maintain the
count for both empty and full containers that are available.
CS420:
Operating
Systems
4
• In a generalized Bounded Buffer problem, N buffers, each can
hold one data item
• Utilize three semaphores to control access to buffer between the
producer and the consumer
- Semaphore mutex locks access to critical region of code
where buffer is modified
• Initialized to the value 1
- Semaphore full keeps track of how many items are actually in the
buffer
• Initialized to the value 0
- Semaphore empty keeps track of how many available slots there
are in the buffer
• Initialized to the value N
CS420:
Operating
Systems
5
an item
do {
....
// produce
....
// inited to N
wait (empty);
wait (mutex);
....
buffer
// add
....
signal
signal
item to the
(mutex);
(full);
} while (TRUE);
inited to 0
do {
....
wait (full); // full
wait (mutex);
....
// remove item from buffer
....
signal (mutex);
signal (empty);
....
// consume the item
....
} while (TRUE);
Producer Process Consumer Process
CS420:
Operating
Systems
6
 The dining philosopher's problem involves
the allocation of limited resources to a
group of processes in a deadlock-free and
starvation-free manner.
 There are five philosophers sitting around
a table, in which there are five
chopsticks/forks kept beside them and a
bowl of rice in the centre, When a
philosopher wants to eat, he uses two
chopsticks - one from their left and one
from their right. When a philosopher
wants to think, he keeps down both
chopsticks at their original place.
CS420:
Operating
Systems
8
• Possible Solution??
- Instruct each philosopher to behave as follows:
 Think until the left chopstick is available; when it is pick it up
 Think until the right chopstick is available; when it is pick it up
 Eat some rice
 Put the left chopstick
 down Put the right
 chopstick down Go
 back to thinking
CS420:
Operating
Systems
9
5] );
% 5] );
do {
wait ( chopstick[i] );
wait ( chopStick[ (i + 1) %
...
// eat
...
signal ( chopstick[i] );
signal ( chopstick[ (i + 1)
...
// think
...
} while (TRUE);

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Synchronization problems

  • 1. Mrs.G.Chandraprabha,M.Sc.,M.Phil., Assistant Professor, Department of Information Technology, V.V.Vanniaperumal College for Women, Virudhunagar.
  • 2. 2  Bounded-Buffer Problem  Readers and Writers Problem  Dining-Philosophers Problem
  • 3. CS420: Operating Systems 3  The buffer pool has a maximum size, this problem is often called the Bounded buffer problem.  This problem is generalised in terms of the Producer Consumer problem, where a finite buffer pool is used to exchange messages between producer and consumer processes.  Solution to this problem is, creating two counting semaphores "full" and "empty" to keep track of the current number of full and empty buffers respectively.  In this Producers mainly produces a product and consumers consume the product, but both can use of one of the containers each time.  The main complexity of this problem is that we must have to maintain the count for both empty and full containers that are available.
  • 4. CS420: Operating Systems 4 • In a generalized Bounded Buffer problem, N buffers, each can hold one data item • Utilize three semaphores to control access to buffer between the producer and the consumer - Semaphore mutex locks access to critical region of code where buffer is modified • Initialized to the value 1 - Semaphore full keeps track of how many items are actually in the buffer • Initialized to the value 0 - Semaphore empty keeps track of how many available slots there are in the buffer • Initialized to the value N
  • 5. CS420: Operating Systems 5 an item do { .... // produce .... // inited to N wait (empty); wait (mutex); .... buffer // add .... signal signal item to the (mutex); (full); } while (TRUE); inited to 0 do { .... wait (full); // full wait (mutex); .... // remove item from buffer .... signal (mutex); signal (empty); .... // consume the item .... } while (TRUE); Producer Process Consumer Process
  • 6. CS420: Operating Systems 6  The dining philosopher's problem involves the allocation of limited resources to a group of processes in a deadlock-free and starvation-free manner.  There are five philosophers sitting around a table, in which there are five chopsticks/forks kept beside them and a bowl of rice in the centre, When a philosopher wants to eat, he uses two chopsticks - one from their left and one from their right. When a philosopher wants to think, he keeps down both chopsticks at their original place.
  • 7. CS420: Operating Systems 8 • Possible Solution?? - Instruct each philosopher to behave as follows:  Think until the left chopstick is available; when it is pick it up  Think until the right chopstick is available; when it is pick it up  Eat some rice  Put the left chopstick  down Put the right  chopstick down Go  back to thinking
  • 8. CS420: Operating Systems 9 5] ); % 5] ); do { wait ( chopstick[i] ); wait ( chopStick[ (i + 1) % ... // eat ... signal ( chopstick[i] ); signal ( chopstick[ (i + 1) ... // think ... } while (TRUE);