Test by Bikram | Operating Systems | Test 2 | Question: 5

Question

Let us initialize counting semaphore $X$ to $5$. Assume that processes  $P_i$ where $i= 1$ to $15$ are coded as follows.

while (1)
{
P (x);
  {
    critical section
   }
V (x);
}

and suppose that $P_{16}$ is coded as follows:

while (1)
{
V (x);
    {
      critical section
    }
P (x);
}

The number of processes can be in the critical section at most at any point of time is ______

Comments

First of all P₁₆ enters and makes x=6  ( Currently 1 process is in CS), 

Same time, P₁ to P₆ - six processes can go to CS (because X=6);

So Total should be 7.

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@Bikram I liked this question. 

and its resemblance with gate 1997 question.

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What is the difference between the number of processes that can enter the critical section and the maximum number of processes that can be inside the critical section at the same time?

In this question, processes P1 to P15 can enter the critical section one after another as the semaphore becomes available. So, over time, many processes can enter.

But the question asks for the maximum number of processes that can stay inside the critical section simultaneously.

For example, when 5 processes are already inside, P16 executes V(X), allowing one more process to enter. Thus, we can have 7 processes simultaneously inside the critical section.

When one of those processes leaves and executes V(X), another waiting process can enter, but it replaces the process that left. Therefore, the number of processes entering over time can be greater than the maximum number present simultaneously.

Answer 1

At first P₁ to P₁₅ any 5 processes can enter Critical Section because $X = 5$. Each entry does a $P$ operation which decrements $X$ and after 5 $P$ operations $X = 0.$

Now, P₁₆ comes and go to its critical section as $V$ comes first and hence no wait. Now, due to $V$ operation ($X$ incremented) it also allows one more process to enter the critical section as its wait operation - $P$ can now succeed.

Total Number of process is in Critical Section = 1 ( P₁₆)  + 6 (P₁ to  P₁₅ ) = 7.

Comments

sir, how does while(-1) loop works ,

 I thought as it is while (-1) none of the process from (P1 - P15) will ever enter the while loop (and will end) , so only P16 will keep entering the loop and incrementing the X value.That way only P16 will be there in the critical section.

Then I ran a c program for while (-1)  , it ran normally like while(1) would run [ whereas when I ran for while(0) it skipped the while loop and ended the program] 

So if while(-1) is true ,

 First 5 processes will enter into Critical section and make X=0 , then P16 will enter and make X=1 , causing yet another Process to enter the CS and thus making total number of processes to 7.

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hello @harsh , i think you are correct.

please read my given answer , if u found any mistake just point it out and write your own answer ... i need to see again.

I think you are correct here.

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now that all the processes start with while(1) , the answer is 7.

Answer 2

In the above case when you run while(-1) , it runs normally like while(1) , that means 

while(-1) is true here ,

 First 5 processes will enter into Critical section and make X=0 , then P16 will enter and make X=1 , causing yet another Process to enter the CS and thus making total number of processes to 7.

PS: had the question been slightly changed to while(0) , then the answer would be 1 as 

none of the process from (P1 - P15) will ever enter the while loop (and will end) , so only P16 will keep entering the loop and incrementing the X value.That way only P16 will be there in the critical section.

Answer 3

Assume processes can enter as per Process IDs whenever possible.

 

Let $P_1$ to $P_5$ enter the critical section. $X=0$ now.

Let $P_{16}$ enter. $X=1$ now.

Let $P_6$ enyer. $X = 0$ again.

That's it.

 

7 processes can be in the critical section at maximu,=m.