Modified by Dr M A Berbar Chapter 6: Process Synchronization Background The Critical-Section Problem Synchronization Hardware Semaphores Classical Problems.

Modified by Dr M A Berbar Chapter 6: Process Synchronization Background The Critical-Section Problem Synchronization Hardware Semaphores Classical Problems of Synchronization Critical Regions Monitors Synchronization in Solaris 2 & Windows 2000

Modified by Dr M A Berbar Background Concurrent access to shared data may result in data inconsistency. Maintaining data consistency requires mechanisms to ensure the orderly execution of cooperating processes. Shared-memory solution to bounded-buffer problem allows at most n – 1 items in buffer at the same time. A solution, where all N buffers are used is not simple.  Suppose that we modify the producer-consumer code by adding a variable counter, 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.

Modified by Dr M A Berbar Background: process synchronization Concurrent access to shared data may result in data inconsistency. Maintaining data consistency requires mechanisms to ensure the orderly execution of cooperating processes. Shared-memory solution to bounded-buffer problem (Chapter 3) allows at most n – 1 locations in buffer at the same time. A solution, where all n locations are used is not simple. Problem: Solution:

Modified by Dr M A Berbar Bounded-Buffer – Producer Process from ch3 item nextProduced; while (1) { while (((in + 1) % BUFFER_SIZE) == out) ; /* do nothing; buffer is full */ buffer[in] = nextProduced; in = (in + 1) % BUFFER_SIZE; } 0 11 0 producer consumer 0 11 out in out Empty buffer Full buffer

Modified by Dr M A Berbar Bounded-Buffer Suppose that we modify the producer-consumer code by adding a variable counter, initialized to 0 and incremented each time a new item is added to the buffer Shared data: #define BUFFER_SIZE 10 typedef struct {... } item; item buffer[BUFFER_SIZE]; int in = 0; int out = 0; int counter = 0; /* # of items in buffer */ New variable

Modified by Dr M A Berbar Bounded-Buffer Producer process item nextProduced; while (1) { while (counter == BUFFER_SIZE) ; /* do nothing */ buffer[in] = nextProduced; in = (in + 1) % BUFFER_SIZE; // -1 < in < BUFFER_SIZE counter++; }

Modified by Dr M A Berbar Bounded-Buffer Consumer process item nextConsumed; while (1) { while (counter == 0) ; /* do nothing because nothing in the buffer */ nextConsumed = buffer[out]; out = (out + 1) % BUFFER_SIZE; counter--; }

Modified by Dr M A Berbar Bounded-Buffer Producer - Consumer Producer process Producer process item nextProduced; while (1) { while (counter == BUFFER_SIZE) ; /* do nothing */ buffer[in] = nextProduced; in = (in + 1) % BUFFER_SIZE; counter++; } Consumer process Consumer process item nextConsumed; while (1) { while (counter == 0) ; /* do nothing */ nextConsumed = buffer[out]; out = (out + 1) % BUFFER_SIZE; counter--; } producer consumer 0 11 out in

Modified by Dr M A Berbar Bounded Buffer The statements counter++; counter--; must be performed atomically. Atomic operation means an operation that completes in its execution without interruption. Race Condition

Modified by Dr M A Berbar Bounded Buffer The statement “count++” may be implemented in machine language as: register1 = counter register1 = register1 + 1 counter = register1 The statement “count--” may be implemented as: register2 = counter register2 = register2 – 1 counter = register2 Race Condition

Modified by Dr M A Berbar Bounded Buffer If both the producer and consumer attempt to update the buffer concurrently, the assembly language statements may get interleaved. Interleaving depends upon how the producer and consumer processes are scheduled. Race Condition

Modified by Dr M A Berbar Bounded Buffer Assume counter is initially 5. One interleaving of statements is: To: producer: execute register1 = counter (register1 = 5) T1: producer: execute register1 = register1 + 1 (register1 = 6) T2: consumer: execute register2 = counter (register2 = 5) T3: consumer: execute register2 = register2 – 1 (register2 = 4) T4: producer: execute counter = register1 (counter = 6) T5: consumer: execute counter = register2 (counter = 4) The value of count may be either 4 or 6, where the correct result should be 5. Race Condition

Modified by Dr M A Berbar Race Condition Race condition: The situation where several processes access – and manipulate shared data concurrently. The final value of the shared data depends upon which process finishes last. To prevent race conditions, concurrent processes must be synchronized.

Modified by Dr M A Berbar The Critical-Section Problem n processes all competing to use some shared data Each process has a code segment, called critical section, in which the shared data is accessed. Problem – ensure that when one process is executing in its critical section, no other process is allowed to execute in its critical section.

Modified by Dr M A Berbar Initial Attempts to Solve Problem Only 2 processes, P 0 and P 1 General structure of process P i (other process P j ) do { entry section critical section exit section reminder section } while (1); Processes may share some common variables to synchronize their actions. Requesting permission to enter its critical section

Modified by Dr M A Berbar 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.

Modified by Dr M A Berbar Algorithm 1 Shared variables:  int turn; initially turn = 0  turn == i  P i can enter its critical section Process P i do { while (turn != i) /* do nothing */ ; critical section /* turn == i */ turn = j ; /* give permission for process j*/ /* and prevent Pi to enter CS even Pj is not ready to goto its CS */ reminder section } while (1);

Modified by Dr M A Berbar Algorithm 1 Satisfies mutual exclusion, but not progress … Satisfies mutual exclusion as this solution ensures that only one process at a time can be in its critical section. Not satisfies progress For example: if turn == 0 and P1 is ready to enter its critical section again, P1 cannot do so, even though P0 may be in its remainder section. The problem with algorithm 1 is that it does not retain sufficient information about the state of each process; it remembers only which process is allowed to enter its critical section.

Modified by Dr M A Berbar Algorithm 2 Shared variables  boolean flag[2]; initially flag [0] = flag [1] = false.  flag [i] = true  P i ready to enter its critical section Process P i do { flag[i] := true; while (flag[j]) /* do nothing */ ; critical section flag [i] = false; remainder section } while (1); Satisfies mutual exclusion, but not progress requirement.

Modified by Dr M A Berbar Algorithm2 Satisfies mutual exclusion, but not progress requirement. For example: To : Po sets flag[0] = true T1 : P1 sets flag[1] = true Now Po and P1 are looping for ever in their respective while statements. This algorithm is crucially dependent on exact timing of the two processes.

Modified by Dr M A Berbar Peterson's solution Combined shared variables of algorithms 1 and 2.  boolean flag[2];  initially flag [0] = flag [1] = false.  Int turn; Process P i do { flag [i] := true; turn = j; while (flag [j] and turn == j) /* do nothing */ ; critical section /* The condition of while is false flag [i] = false; remainder section } while (1); Meets all three requirements; solves the critical-section problem for two processes.

To prove property 1: Pi enters its critical section only if either flag[j] == false or turn == i “ the condition of while is false” What about the problem of To : Po sets flag[0] = true T1 : P1 sets flag[1] = true Variable turn can be either 0 or 1, but cannot be both. Property 2 is satisfied because Po and P1 are not looping for ever in their respective while statements, and it does not prevent the last executed process to enter again its CS as long as the other process is not ready to go to its CS Property 3 is satisfied because if process Pi finished its CS and Pj has made a request to enter its critical section and that request is granted, Pi not allowed to go again to its CS.

Modified by Dr M A Berbar Multiple Process Solution Bakery Algorithm Before entering its critical section, process receives a number. Holder of the smallest number enters the critical section. If processes P i and P j receive the same number, if i < j, then P i is served first; else P j is served first. The numbering scheme always generates numbers in increasing order of enumeration; i.e., 1,2,3,3,3,3,4,5... Critical section for n processes

Modified by Dr M A Berbar Bakery Algorithm Notation <  lexicographical order (ticket #, process id #)  (a,b) < c,d) if a < c or if a == c and b < d  max (a 0,…, a n-1 ) is a number, k, such that k  a i for i - 0, …, n – 1 Shared data boolean choosing[n]; int number[n]; Data structures are initialized to false and 0 respectively

Modified by Dr M A Berbar Bakery Algorithm do { choosing[i] = true; /* the process i is ready to enter its critical section /* then get a number for the process number[i] = max(number[0], number[1], …, number [n – 1])+1; /* note : number[] is not sorted choosing[i] = false; for (j = 0; j < n; j++) { while (choosing[j]) /* do nothing */ ; while ((number[j] != 0) && ( (number[j], j) < (number[i],i) ) ) ; } /* arrived here means there is no process have number less than number[i] critical section number[i] = 0; remainder section } while (1); Continues looping until P j leaves its critical section go to check if there is any process want to go its critical section and have number less than P i There is a process number j is going to go its critical section

Modified by Dr M A Berbar What is missing? “Real” solutions are based on stronger prerequisites than just variable sharing. Besides, we don’t want to guess what is “atomic” when programming. So, we want: Hardware support—special instructions. OS kernel provided synchronization primitives. At the lowest level (hardware), there are two elementary mechanisms: – interrupt masking – atomic read-modify-write

Modified by Dr M A Berbar A simple minded alternative Let’s start by using hardware instructions to mask interrupts. If we don’t let the CPU interrupt (i.e., take away the control from) the current process, the solution for N processes would be as simple as below: Process i... while( TRUE ) { disableInterrupts(); enableInterrupts();... } Unfortunately, there is only one system-wide critical section active at a time. Why?

Modified by Dr M A Berbar Disadvantages of masking the interrupts Masking the interrupts is not feasible in a multiprocessor environment. Disabling interrupts on a multiprocessors can be time-consuming, as the message is passed to all the processors. This message passing delays entry into each critical section, and system efficiency decreases.

Modified by Dr M A Berbar Hardware support Many CPUs today provide hardware instructions to read, modify, and store a word atomically. Most common instructions with this capability are: – TAS - test-and-set (Motorola 68K) – CAS - compare-and-swap (IBM 370 and Motorola 68K) – XCHG - exchange (x86) The basic idea is to be able to read out the contents of a variable (i.e., memory location), and set it to something else all in one execution cycle. Hence not interruptable. The use of these special instructions makes life easier!

Solution to Critical-section Problem Using Locks do { acquire lock critical section release lock remainder section } while (TRUE);

Modified by Dr M A Berbar Synchronization Hardware Test and modify the content of a word atomically. // Test And Set function TASet boolean TestAndSet (boolean *target) { boolean rv = *target; *target = TRUE; return rv: } Return the current value of global lock variable and set the global lock variable to true atomically.

Solution using TestAndSet Shared boolean variable lock., initialized to false. Solution: do { while ( TestAndSet (&lock )) ; // do nothing // critical section lock = FALSE; // remainder section } while (TRUE); Only one global guard variable is associated with each critical section (i.e., there can be many critical sections).

Swap Instruction Definition: void Swap (boolean *a, boolean *b) { boolean temp = *a; *a = *b; *b = temp: } atomically.

Solution using Swap Shared Boolean variable lock initialized to FALSE; Each process has a local Boolean variable key Solution: do { key = TRUE; while ( key == TRUE) Swap (&lock, &key ); // critical section lock = FALSE; // remainder section } while (TRUE);

Modified by Dr M A Berbar Mutual Exclusion with Swap Shared data (initialized to false): boolean lock; boolean waiting[n]; Process P i do { key = true; while (key == true) Swap(lock,key); critical section lock = false; remainder section } Continue looping until setting lock to False Atomic operation These algorithms do not satisfy the bounded-waiting requirement

Figure Bounded-waiting mutual exclusion with TASet Boolean waiting[n]; // initialized to false Boolean lock; // initialized to false to prove that mutual exclusion is met do { waiting[i] = true; key = true; while (waiting[i] && key) key = TASet(lock); waiting[i] = false; critical section j = (i+1) % n; while ((j != i) && !waiting[j]) j = (j+1) % n; if (j == i) lock = false; else waiting [j] = false; remainder section } while (1); Return the current value of global lock variable and set the global lock variable to true Pi can enter its critical section only if either waiting [i] == false or key == false. The value of key can become false only if the TASet is executed. The lock global variable initialized false, so the first process to execute the TASet will set its key to false; and set lock to True so all others must wait. maintaining the mutual-exclusion requirement. If waiting[j] == true or j ==i finished looping During running critical section other processes trying to enter their critical section It designates the first process in this ordering that is in the entry section (waiting [j] == true) as the next one to enter the critical section. Any process waiting to enter its critical section will thus do so within n -1 turns. scans the array waiting in the cyclic ordering (i + 1, i + 2,..., n - 1,0,..., i -1). To prove the progress requirement is met, we note that the arguments presented for mutual exclusion also apply here, since a process exiting the critical section either sets lock to false, or sets waiting [j] to false. Both allow a process that is waiting to enter its critical section to proceed.

Bounded-waiting Mutual Exclusion with TestandSet() do { waiting[i] = TRUE; key = TRUE; while (waiting[i] && key) key = TestAndSet(&lock); waiting[i] = FALSE; // critical section j = (i + 1) % n; while ((j != i) && !waiting[j]) j = (j + 1) % n; if (j == i) lock = FALSE; else waiting[j] = FALSE; // remainder section } while (TRUE);

Modified by Dr M A Berbar Bounded-waiting mutual exclusion with TASet The first process to execute the TASet will find key = false; and lock  True all others must wait (mutual exclusion). The variable waiting [i] can become false only if another process leaves its critical section (progress); only one waiting [i] is set to false, maintaining the mutual-exclusion requirement. To prove the progress requirement is met, we note that the arguments presented for mutual exclusion also apply here, since a process exiting the critical section either sets lock to false, or sets waiting [j] to false. Both allow a process that is waiting to enter its critical section to proceed. To prove the bounded-waiting requirement is met, we note that, when a process leaves its critical section, it scans the array waiting in the cyclic ordering (i + 1, i + 2,..., n - 1,0,..., i -1). It designates the first process in this ordering that is in the entry section (waiting [j] == true) as the next one to enter the critical section. Any process waiting to enter its critical section will thus do so within n -1 turns. Unfortunately for hardware designers, implementing atomic TASet instructions on multiprocessors is not a trivial task. Such implementations are discussed in books on computer architecture.

Modified by Dr M A Berbar Semaphores Synchronization tool that does not require busy waiting. Semaphore S – integer variable can only be accessed via two indivisible (atomic) operations Two standard operations modify S: wait() and signal() Less complicated wait (S){ while S  0 //do no-op; S--;} signal (S){ S++;} wait (S); 0 1 0 signal (S); 0 1

Modified by Dr M A Berbar Critical Section of n Processes Mutual exclusion implementation with semaphores Shared data: semaphore mutex; //initially mutex = 1 Also known as mutex locks Process Pi: do { wait(mutex); critical section signal(mutex); remainder section } while (1); under the classical definition of semaphores with busy waiting the semaphore value is never negative wait (S); 0 1 0 signal (S); 0 1

Semaphore Implementation Must guarantee that no two processes can execute wait () and signal () on the same semaphore at the same time Thus, 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.

Modified by Dr M A Berbar The main disadvantage of previous algorithms The main disadvantage of the mutual-exclusion solutions of previous algorithms, and of the semaphore definition given here, is that they all require busy waiting. entry code While a process is in its critical section, any other process that tries to enter its critical section must loop continuously in the entry code. This continual looping is clearly a problem in a real multiprogramming system, where a single CPU is shared among many processes. Busy waiting wastes CPU cycles 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).!

Modified by Dr M A Berbar When the Spinlocks will be useful ?!! Spinlocks are useful in multiprocessor systems. The advantage of a spinlock is that no context switch is required when a process must wait on a lock, and a context switch may take considerable time. Thus, when locks are expected to be held for short times, spinlocks are useful Thus, when locks are expected to be held for short times, spinlocks are useful.

Modified by Dr M A Berbar To overcome the need for busy waiting When a process executes the wait operation and finds that the semaphore value is not positive, it must wait. However, rather than busy waiting, the process can block itself…how ? The block operation places a process into a waiting queue associated with the semaphore, and the state of the process is switched to the waiting state. Then, control is transferred to the CPU scheduler, which selects another process to execute. A process that is blocked, waiting on a semaphore S, should be restarted when some other process executes a signal operation at the end of its critical section. The process is restarted by a wakeup operation, which changes the process from the waiting state to the ready state. The process is then placed in the ready queue. (The CPU may or may not be switched from the running process to the newly ready process, depending on the CPU-scheduling algorithm.)

Modified by Dr M A Berbar Busy-wait Problem Definition : Processes waste CPU cycles while waiting to enter their critical sections Solution: wait block 1. Modify wait operation into the block operation. The process can block itself rather than busy-waiting. 2. Place the process into a wait queue associated with the critical section signalwakeup 3. Modify signal operation into the wakeup operation. 4. Change the state of the process from wait to block.

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. (suspends the process that invokes it.)  wakeup – remove one of processes in the waiting queue and place it in the ready queue. wakeup(P) resumes the execution of a blocked process P.

Modified by Dr M A Berbar Implementation Each semaphore has an integer value and a list of processes. When a process must wait on a semaphore, it is added to the list of processes. A signal operation removes one process from the list of waiting processes and awakens that process.

Semaphore Implementation with no Busy waiting (Cont.) Implementation of wait: wait(semaphore *S) { S->value--; if (S->value < 0) { add this process to S->list; block(); } Implementation of signal: signal(semaphore *S) { S->value++; if (S->value <= 0) { remove a process P from S->list; wakeup(P); } Note 1 Semaphore waiting list this implementation may have negative semaphore values. If the semaphore value is negative, its magnitude is the number of processes waiting on that semaphore.

Modified by Dr M A Berbar bounded waiting Each semaphore contains an integer value and a pointer to a list of PCBs. One way to add and remove processes from the list, that ensures bounded waiting would be to use a FIFO queue, where the semaphore contains both head and tail pointers to the queue Note 1 The critical aspect of semaphores is that wait and signal operations are executed atomically. This situation can be solved in either of two ways. In a uni-processor environment, we can simply inhibit interrupts during the time the wait and signal operations are executing, instructions from different processes cannot be interleaved, until interrupts are re-enabled and the scheduler can regain control. software solutions critical sections consist of the wait and signal procedures In a multiprocessor environment, inhibiting interrupts does not work. Instructions from different processes (running on different processors) may be interleaved in some arbitrary way. If the hardware does not provide any special instructions, we can employ any of the correct software solutions for the critical-section problem, where the critical sections consist of the wait and signal procedures

Modified by Dr M A Berbar

Semaphore as a General Synchronization Tool Execute B in P j only after A executed in P i Use semaphore flag initialized to 0 Code: P i P j   Await(flag) signal(flag)B

Modified by Dr M A Berbar We have removed busy waiting from the entry to the critical sections of application programs. Furthermore, we have limited busy waiting to only the critical sections of the wait and signal operations, and these sections are short (if properly coded, they should be no more than about 10 instructions). busy waiting occurs rarely, and then for only a short time. In some application programs their critical sections may be long (minutes or even hours)

Modified by Dr M A Berbar 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. Let we consider a system consisting of two processes, Po and P1, each accessing two semaphores, S and Q, set to the value1 P 0 P 1 wait(S);wait(Q);  wait(Q);wait(S);  signal(S);signal(Q); signal(Q)signal(S); Suppose that Po executes wait (S), and then P1 executes wait (Q). When Po executes wait(Q), it must wait until P1 executes signal(Q). Similarly, when P1 executes wait(S), it must wait until Po executes signal(S). Since these signal operations cannot be executed, Po and P1 are deadlocked.

Modified by Dr M A Berbar Starvation – indefinite blocking. A process may never be removed from the semaphore queue in which it is suspended. Another problem related to deadlocks is indefinite blocking or starvation, a situation where processes wait indefinitely within the semaphore. Priority Inversion - Scheduling problem when lower- priority process holds a lock needed by higher-priority process

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

Bounded-Buffer Problem  Shared data The mutex semaphore provides mutual execlusion for accesses to the buffer pool and is initialzed to the value 1.  The empty and full semaphores count the number of empty and full buffers. Initially: full = 0, empty = n, mutex = 1

Producer/consumer Processes

do { … produce an item in nextp … wait(empty); wait(mutex); … add nextp to buffer … signal(mutex); signal(full); } while (1); do { wait(full) wait(mutex); … remove an item from buffer to nextc … signal(mutex); signal(empty); … consume the item in nextc … } while (1); Producer Consumer Initially: full = 0, empty = n, mutex = 1 wait(semaphore *S) { S->value--; if (S->value < 0) { add this process to S->list; block(); } signal(semaphore *S) { S->value++; if (S->value list; wakeup(P); }

Producer/consumer Processes do { … produce an item in nextp … wait(empty); wait(mutex); … add nextp to buffer … signal(mutex); signal(full); } while (1); do { wait(full) wait(mutex); … remove an item from buffer to nextc … signal(mutex); signal(empty); … consume the item in nextc … } while (1); Producer Consumer Initially: full = 0, empty = n, mutex = 1 empty --; If empty +ve there are empty places; If empty < 0 then the buffer if full and go to wait queue of empty semaphore mutex --; If mutex ==0 no problem; If mutex < 0 then the buffer is used by the consumer then go to wait queue of mutex semaphore increment full because an item has been addedfull --; If full is +ve then there are items else wait because there is no items in the buffer mutex --; If mutex == 0 no problem; If mutex < 0 then the buffer is used by the producer and go to wait queue of mutex semaphore wait(semaphore *S) { S->value--; if (S->value < 0) { add this process to S->list; block(); } signal(semaphore *S) { S->value++; if (S->value list; wakeup(P); }

Readers-Writers Problem

 The first readers-writers problem, requires that no reader will be kept waiting unless a writer has already obtained permission to use the shared object. All readers waiting at the end of writer execution should be given priority over the next writer.  The second readers-writers problem requires that, if a writer is waiting to access the object, no new readers may start reading.  A solution to either problem may result in starvation. In the first case, writers may starve; in the second case, readers may starve.

Readers-Writers Problem  Shared data semaphore mutex, wrt; Int readCount; Initially mutex = 1, wrt = 1, readCount = 0

Readers-Writers Problem wait(mutex); readCount++; if (readCount == 1) wait(wrt); signal(mutex); … reading is performed … wait(mutex); readCount--; if (readCount == 0) signal(wrt); signal(mutex): wait(wrt); … writing is performed … signal(wrt); mutex --; If mutex ==0 no problem; If mutex < 0 then there is a reader updating the readCount so go to wait queue of mutex semaphore Initially mutex = 1, wrt = 1, readCount = 0 The mutex semaphore is used to ensure mutual exclusion when the variable readcount is updated. The readCount variable keeps track of how many processes are currently reading the object. If first or last reader then wrt --; If wrt ==0 no writer working; If wrt < 0 then the buffer is used by the writer and go to wait queue of wrt semaphore. if a writer is in the critical section and n readers are waiting, then one reader is queued on wrt, and n - 1 readers are queued on mutex. mutex --; If mutex ==0 no problem; If mutex < 0 then there is a reader updating the readCount so go to wait queue of mutex semaphore wrt --; If wrt ==0 no problem; If wrt < 0 then the buffer is used by another writer and go to wait queue of wrt semaphore. The readCount ==1 when the first reader tries to enters the critical section The readCount ==0 when the last reader tries to exits the critical section release a reader release a writer

The semaphore wrt functions  The semaphore wrt functions as a mutual- exclusion semaphore for the writers.  It is also used by the first or last reader that enters or exits the critical section.  It is not used by readers who enter or exit while other readers are in their critical sections.

Dining - Philosophers Problem  Shared data semaphore chopstick[5]; Initially all values are 1 It is a simple representation of the need to allocate several resources among several processes in a deadlock and starvation free manner One simple solution is to represent each chopstick by a semaphore. A philosopher tries to grab the chopstick by executing a wait operation on that semaphore; She release her chopstick by executing the signal operation on the appropriate semaphore.

Dining-Philosophers Problem  Shared data  Bowl of rice (data set)  Semaphore chopstick [5] initialized to 1

Dining-Philosophers Problem  Philosopher i : do { wait(chopstick[i]) wait(chopstick[(i+1) % 5]) … eat … signal(chopstick[i]); signal(chopstick[(i+1) % 5]); … think … } while (1); The contents of chopstick[i] decremented and tested is there is another philosophers is using the chopstick[i] ?

Problems with Semaphores  Incorrect use of semaphore operations:  signal (mutex) …. wait (mutex)  wait (mutex) … wait (mutex)  Omitting of wait (mutex) or signal (mutex) (or both)

Modified by Dr M A Berbar Chapter 6: Process Synchronization Background The Critical-Section Problem Synchronization Hardware Semaphores Classical Problems.

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Modified by Dr M A Berbar Chapter 6: Process Synchronization Background The Critical-Section Problem Synchronization Hardware Semaphores Classical Problems.

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Presentation on theme: "Modified by Dr M A Berbar Chapter 6: Process Synchronization Background The Critical-Section Problem Synchronization Hardware Semaphores Classical Problems."— Presentation transcript:

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