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6.3 Peterson’s Solution The two processes share two variables: Int turn; Boolean flag[2] The variable turn indicates whose turn it is to enter the critical.

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Presentation on theme: "6.3 Peterson’s Solution The two processes share two variables: Int turn; Boolean flag[2] The variable turn indicates whose turn it is to enter the critical."— Presentation transcript:

1 6.3 Peterson’s Solution 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!

2 Algorithm for Process Pi while (true) { flag[i] = TRUE; turn = j; while ( flag[j] && turn == j) ; CRITICAL SECTION flag[i] = FALSE; REMAINDER SECTION }

3 Algorithm for Process Pi do { acquire lock critical section release lock remainder section }

4 6.5 Semaphore It’s a hardware based solution Semaphore S –integer variable Two standard operations modify S: wait() and signal()

5 6.5 Semaphore Can only be accessed via two indivisible (atomic) operations wait (S) { while S <= 0 ; // no-op S--; } signal (S) { S++; }

6 6.5 Semaphore Binary semaphore –integer value can range only between 0 and 1; can be simpler to implement Counting semaphore –integer value can range over an unrestricted domain

7 6.5 Semaphore Provides mutual exclusion Semaphore S; // initialized to 1 do { wait (S); //Critical Section signal (S); //Remainder Section } while (true)

8 6.5 Semaphore Must guarantee that no two processes can execute wait ()and signal ()on the same semaphore at the same time Atomic = non-interruptable

9 6.5 Semaphore The main disadvantage of the semaphore is that it requires busy waiting, which wastes CPU cycle that some other process might be able to use productively This type of semaphore is also called a spinlock because the process “spins” while waiting for the lock

10 6.5 Semaphore To overcome the busy waiting problem, we create two more 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.

11 Diagram of Process State

12 Semaphore Implementation with no Busy waiting

13 Deadlock and Starvation Deadlock –two or more processes are waiting indefinitely for an event that can be caused by only one of the waiting processes Starvation–indefinite blocking. A process may never be removed from the semaphore queue in which it is suspended.

14 Deadlock example

15 6.6 Classical Problems of Synchronization Bounded-Buffer Problem Readers and Writers Problem Dining-Philosophers Problem

16 6.6 Classical Problems of Synchronization 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.

17 Bounded Buffer Problem

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19 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. First readers-writers problem: requires that no reader will be kept waiting unless a writer has already obtained permission to use the shared object

20 Readers-Writers Problem Shared Data Data set Semaphore mutexinitialized to 1. Semaphore wrtinitialized to 1. Integer readcountinitialized to 0.

21 Readers Readers-Writers Problem

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23 Dining- Philosophers Problem

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