algorithm,process,operating-system,synchronization,race-condition , Bound wait to solve race condition


Bound wait to solve race condition

Question:

Tag: algorithm,process,operating-system,synchronization,race-condition

I am trying to Give a race condition example , then write an algorithm to impose synchronization and write an algorithm that implement the Bounded wait solution?! I tried the case of when two admins A and B in the school receive 2 students to register them if they hit the save in the same time then the 2 students will have the same ID

Then i used the semaphore to solve it as following :-

Start
Initialization
Do
{
Wait(semaphore);
Submitting the order to generate the ID; \\ critical section
Signal(semaphore);
}while (true);

I did not know that is it correct and satisfy the bound wait?!!!


Answer:

Bounded Waiting is defined as :-

There exists a bound, or limit, 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.

Returning to your problem, it is also an example of bounded waiting,but only between two processes. It doesn't make realise properly that several of the processes are contending for execution of their critical-section. A better example for bounded waiting.would be :-

When n admins A[1],A[2],..,A[n] in the school receive students to register them, if they hit the save in the same time then the students will have the same ID. So, at a time a single admin is allowed to execute its critical section code(i.e., to register the student).

Then, returning to the solution using the Semaphores, you could do the following :-

The n processes share a semaphore,mutex, initialised to 1. Each process A[i] is organised as :-

do{
wait(MUTEX);
// critical section
signal(MUTEX);
// remainder section
} while(TRUE);

This is also one of the way, but, unfortunately, it doesn't clearly give any idea about bounded waiting. Here, you can introduce some additional constraint satisfaction by bringing some improvements in the wait() and signal() function. I would guide you the other way as mentioned below.

You can better achieve this using Pieterson's solution:-

 do{
  flag[i]=TRUE;
  turn = j;
  while(flag[j] && turn == j);
   // critical section
  flag[i]=FALSE;
   // remainder section
 } while(TRUE);

This provides a better solution to bounded waiting. A process A[i] can be prevented from entering their critical section only if it is stuck in the while loop with the condition flag[j]==true and turn == j; -- this loop is the only possibility. Similarly, every process will have the respective codes arranged in the terms of other process'e execution schedule.

Each will enter the critical section at a certain time. And, hence, all of the processes will be finally scheduled and execute their critical section, unless any error occurs like deadlock(that is the other facet).

Hence, a process can wait for maximum n-1 rounds, and will finally be provided the chance to execute its critical section --- thereby satisfying bounded waiting. This is a good solution for the case of avoiding race-conditions ----- thereby providing unique Registration ID to each student.


CONTENTS TAKEN FROM the book Operating System Concepts(Galvin,Silberschatz and Gagne)


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