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 -- this loop is the only possibility. Similarly,

`flag[j]==true and turn == j;`

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* -----

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