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Given below is a program which when executed spawns two concurrent processes :
semaphore $X : = 0 ;$
/* Process now forks into concurrent processes $P1$ & $P2$ */

$\begin{array}{|l|l|}\hline \text{$P1$}  &  \text{$P2$}  \\\hline  \text{repeat forever } & \text{repeat forever} \\ \text{$V (X) ;$ } & \text{$ P(X) ;$} \\  \text{Compute;  } & \text{Compute;}\\  \text{$P(X) ;$  } & \text{$V(X) ;$} \\\hline \end{array}$

Consider the following statements about processes $P1$ and $P2:$

  1. It is possible for process $P1$ to starve.
  2. It is possible for process $P2$ to starve.

Which of the following holds?

  1. Both (I) and (II) are true.
  2. (I) is true but (II) is false.
  3. (II) is true but (I) is false
  4. Both (I) and (II) are false
in Operating System edited by
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4 Comments

will you please explain your line

" Since P2 can't execute signal on P1 's X. "

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yes, bounded waiting will not satisfied by this code.
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10 Answers

76 votes
76 votes
Best answer

Check : What is Starvation?
 

Here $P_2$ can go in infinite waiting while process $P_1$ executes infinitely long.

Also, it can be the case that the Process $P_1$ starves for $\infty$ long time on the semaphore S, after it has successfully executed its critical section once, while $P_2$ executes infinitely long.

Both $P_1$ and $P_2$ can starve for $\infty$ long period of time.

Answer is option A.

edited by
74 votes
74 votes
A is the answer.

Case 1:Here P1 is performing signal operation so it is first one to start...Now it may happen that after executing CS it makes X=0 again tries to enter CS my making X=1 so it is possible that P2 can starve(just a possibility)

Case 2:If P1 executes signal operation and its execution is suspended temporarily, P2 executes wait and enter CS  and then execute signal operation making X=1 now P2 can enter critical section by executing wait operation..this may happen for infinite amount of time..so in this P1 may starve
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4 Comments

Check this line of selected ans.

Here P2 can go in infinite waiting while process P1 executes infinitely long.

 This itself means a chance of deadlock.

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@ mam, Deadlocks means both the processes shouldn't be able to access critical section. But starvation means a process can't access because of another process.. Am I wrong?
 

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yes, right for some cases
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73 votes
73 votes

Answer :

Option A (If we consider the Classical Busy Waiting Definition of Semaphores (Such Semaphores are called Spinlocks, Given in Galvin Book))

or

Option D (If we consider the Blocking Definition of Semaphores(Queue implementation as some Students call it)).

NOTE that Galvin provides both the definitions and calls both the definitions (Busy Waiting definition and Blocking Queue Definition) as Semaphores. Hence, the confusion arises which definition to use. But....

Many other authors(William Stallings or Tanenbaum) use the term semaphore only for blocking solutions(Queue implementation as some Students call it) and would call Busy waiting solutions spin locks.  

Let's see everything in detail. 


Semaphores can be of one of the Two types : Binary Semaphores or Counting Semaphores(General Semaphore)

Furthermore, There are Two definitions of Semaphores, One is with busy waiting (Called Classical Definition) and One with Blocking solutions. Some authors use the term semaphore only for blocking solutions and would call Busy waiting solutions spin locks.

So, We have Four possible scenarios here regarding Semaphores. 

1. Binary semaphores with Busy Waiting

2. Binary semaphores with Blocking Solution

3. Counting semaphores with Busy Waiting

4. Counting semaphores with Blocking Solution

Let's analyze the given Question using each of the above definitions. 


                                                1. Considering Binary semaphores with Busy Waiting :

Definition of "Binary Semaphores with Busy Waiting" :

In classical definition of Binary Semaphore, Binary semaphore is just a simple Boolean Variable.

A binary semaphore $S$ takes on two possible values ``1(Open/Available)' and ``0(Closed/Unavailable)''

  • Two usual Atomic semaphore operations are supported $P,V$
  • $P(S)$ is
       P(S) {
             while (S==0) {}  // Loop as long as S is 0.
             S = 0 ;   // This is NOT the body of the while, It means Set S with value 0 if S was 1 earlier.
    }
  • V(S) is simply S<-- 1

i.e. $V(S)  \{$

          $S = 1;$

           $\}$ 

Analyzing the given Program using this Definition :

In this case i.e. When Considering Binary Semaphore with Busy Waiting, It is possible for both processes to Starve. 

It is possible for $P2$ to starve :

Let me show you a possibility of $P2$ getting starved for CS.

Run $P2$ first, and since initially value of semaphore $X$ is $0,$ so, By the above definition of $P$ operation, Process $P2$ will be stuck in the loop. Now, When $P1's$ turn will come, It will execute $V(X)$ and the $X$ will become $1.$ Now, $P1$ will enter into CS. And Let it execute $P(X)$ and now $X$ will become 0. Now at this moment, preempt $P1$ and run $P2.$ Since value of semaphore $X$ is still 0, It will continue looping in the while loop. And turn will go to $P1$ again, which will then execute $V(X)$ and make $X= 1.$ Now the same scenario may repeat as above. i.e. Preempt $P1$ after executing $P(X)$ and run $P2.$ and so on.. So, $P2$ will starve for Critical Section(CS) this way.

It is possible for $P1$ to starve :

We can make a similar situation for $P1$ as well. 

Let me show you a possibility of $P1$ getting starved for CS.

Run $P1$ first, and since initially value of semaphore $X$ is $0,$ so, By the above definition of $V$ operation, $X$ will become 1. Now, $P1$ will enter into CS. And Now preempt it and run $P2.$ Let it execute $P(X)$ and now $X$ will become 0. Now $P2$ will enter into CS. Now preempt $P2$ and let $P1$ run.  Since the value of semaphore $X$ is $0,$ so, By the above definition of $P$ operation, Process $P1$ will be stuck in the loop. Now, When $P2's$ turn will come, It will execute $V(X)$ and the $X$ will become $1.$ Now, let $P2$  execute $P(X)$ and now $X$ will become 0. And $P2$ will enter into CS. Now at this moment, preempt $P2$ and run $P1.$ Since value of semaphore $X$ is still 0, It will continue looping in the while loop. And turn will go to $P2$ again, which will then execute $V(X)$ and make $X= 1.$ Now the same scenario may repeat as above. i.e. Preempt $P2$ after executing $P(X)$ and run $P1.$ and so on.. So, $P1$ will starve for Critical Section(CS) this way.

 

So, When Binary Semaphores with Busy Waiting Considered, Answer will be Option $A.$


                                                  2. Considering Binary Semaphores with Blocking Solution.

 Definition of "Binary Semaphores with Blocking Solution" :

The main disadvantage of the semaphore definition with busy waiting given above is that it requires Busy Waiting. 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. 

To overcome the need for busy waiting, we can modify the definition of the wait() and signal() semaphore operations. When a process executes the wait () operation and finds that the semaphore value is not positive, it must wait. However, rather than engaging in busy waiting, the process can block itself. 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. 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.) To implement semaphores under this definition, we define a semaphore as a "C' structure :

Each semaphore has a Boolean value and a list of processes list. 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.

The wait(), and Signal() semaphore operations can now be defined as following (Image source : William Stallings Book) :

The block() operation suspends the process that invokes it. The wakeup(P) operation resumes the execution of a blocked process P. These two operations are provided by the operating system as basic system calls.

Analyzing the given program using this definition :

In this case i.e. When Considering Binary Semaphore with Blocking Solution, It is NOT possible for any process to Starve.

The Idea/Hint is that If a process blocks itself on Semaphore $X$ then the other process will wake it up. And Hence, as soon as a process shows interest in accessing CS by executing its Entry code, It will get into CS within a bound of 1. So, Here, Bounded Waiting is Also Satisfied. So, If you analyze the given program using Blocking solution definition of Binary semaphore little bit more, you'll find that Starvation, Deadlock etc are Not possible and BW is Satisfied. 

Note that Bounded Waiting is Not related to Starvation/Deadlock in any way. 

 

So, When Binary Semaphores with Blocking Solution is Considered, Answer will be Option $D.$


                                                      3 . Considering Counting Semaphore with Busy Waiting :

Definition of "Counting Semaphores with Busy Waiting" :

In this scenario, A semaphore S is an integer variable that, apart from initialization, is accessed only through two standard atomic operations: wait () and signal ().

The definition of wait () is as follows: 

$P(S)$ is

  •    P(S) {
             while (S<=0) {}  // Loop as long as S is 0.
             S-- ;   // This is NOT the body of the while, It means decrment S if S was > 0 earlier.
    }
  • V(S) is simply S++

i.e. $V(S)  \{$

          $S++;$

           $\}$ 

Analyzing the given Program using this Definition :

In this case i.e. When Considering Counting Semaphore with Busy Waiting, It is possible for both processes to Starve. 

It is possible for $P2$ to starve :

Let me show you a possibility of $P2$ getting starved for CS.

Run $P2$ first, and since initially value of semaphore $X$ is $0,$ so, By the above definition of $P$ operation, Process $P2$ will be stuck in the loop. Now, When $P1's$ turn will come, It will execute $V(X)$ and the $X$ will become $1.$ Now, $P1$ will enter into CS. And Let it execute $P(X)$ and now $X$ will become 0. Now at this moment, preempt $P1$ and run $P2.$ Since value of semaphore $X$ is still 0, It will continue looping in the while loop. And turn will go to $P1$ again, which will then execute $V(X)$ and make $X= 1.$ Now the same scenario may repeat as above. i.e. Preempt $P1$ after executing $P(X)$ and run $P2.$ and so on.. So, $P2$ will starve for Critical Section(CS) this way.

It is possible for $P1$ to starve :

We can make a similar situation for $P1$ as well. 

Let me show you a possibility of $P1$ getting starved for CS.

Run $P1$ first, and since initially value of semaphore $X$ is $0,$ so, By the above definition of $V$ operation, $X$ will become 1. Now, $P1$ will enter into CS. And Now preempt it and run $P2.$ Let it execute $P(X)$ and now $X$ will become 0. Now $P2$ will enter into CS. Now preempt $P2$ and let $P1$ run.  Since the value of semaphore $X$ is $0,$ so, By the above definition of $P$ operation, Process $P1$ will be stuck in the loop. Now, When $P2's$ turn will come, It will execute $V(X)$ and the $X$ will become $1.$ Now, let $P2$  execute $P(X)$ and now $X$ will become 0. And $P2$ will enter into CS. Now at this moment, preempt $P2$ and run $P1.$ Since value of semaphore $X$ is still 0, It will continue looping in the while loop. And turn will go to $P2$ again, which will then execute $V(X)$ and make $X= 1.$ Now the same scenario may repeat as above. i.e. Preempt $P2$ after executing $P(X)$ and run $P1.$ and so on.. So, $P1$ will starve for Critical Section(CS) this way.

 

So, When Counting Semaphores with Busy Waiting Considered, Answer will be Option $A.$


                                                        4. Considering Counting Semaphore with Blocking Solutions :

 Definition of "Counting Semaphores with Blocking Solution" :

 

Analyzing the given program using this definition :

In this case i.e. When Considering Counting Semaphore with Blocking Solution, It is NOT possible for any process to Starve.

The Idea/Hint is that If a process blocks itself on Semaphore $X$ then the other process will wake it up. And Hence, as soon as a process shows interest in accessing CS by executing its Entry code, It will get into CS within a bound of 1. So, Here, Bounded Waiting is Also Satisfied. So, If you analyze the given program using Blocking solution definition of Counting semaphore little bit more, you'll find that Starvation, Deadlock etc are Not possible and BW is Satisfied. 

Note that Bounded Waiting is Not related to Starvation/Deadlock in any way. 

 

So, When Counting Semaphores with Blocking Solution is Considered, Answer will be Option $D.$



P.S. :

Note 1 : By default, Semaphore means Counting Semaphore.

Note 2 : The above codes are not real, i.e., it is not an implementation of P/V. It is, instead, a definition of the effect P/V is to have.

Note 3 : Answer could be any of A, D. Both are correct. I do not know which was the official answer of GATE that year Or was it $A\,\,or\,\,D.$

Note 4 : Though both options $A$ or $D$ are correct depending on which definition of Semaphore is considered, But in my opinion, better answer should be Option $D$ (i.e. Blocking Queue definition considered) Because most authors like William Stallings or Tanenbaum consider the Blocking queue definition only for Semaphores. The busy waiting definition is called Spinlock according to most authors (even Galvin also calls the Busy waiting definition of Semaphores as Spinlocks)

Note 5 : Some students analyze the program for Starvation or Bounded waiting by arguing that "A process could run continuously and thus other process will starve.." .. Well, this is Not a correct analysis. Refer the definitions of Bounded waiting or Starvation in Galvin and you'll find that the concept that such students are missing is "" A Process is $"willing"$ to enter the CS""

 

Resources :

https://www2.cs.uic.edu/~jbell/CourseNotes/OperatingSystems/5_Synchronization.html

https://cs.nyu.edu/courses/spring02/V22.0202-002/lecture-06.html

Operating System Concepts, written by Peter B. Galvin, Greg Gagne and Abraham Silberschatz.

Modern Operating Systems by AS Tanenbaum 

Operating Systems: Internals and Design Principles  Book by William Stallings (Best Resource for Semaphores in my opinion)

edited by

4 Comments

Thank you so much sir!

What a detailed explanation as always !
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@Arjun Sir plz make this answer as the best answer.

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Awesome explanation sir, gives a clear understanding.
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9 votes
9 votes

Ans A)

P1 starts , and go infinitely . So P2 starves here.

P2 cannot start process any time.right?

Now, P1 starts and makes semaphore value X=1 and executes CS once. Now, P2 takes this semaphore and executes infinitely. So, P1 starves here

Now, P1 starts and makes semaphore value X=1 and executes CS once. Now, P2 takes that semaphore and it also executes it's critical section then Pagain takes this semaphore and looks which one executes next V(s) of P1 or P(s) of P1 or P(x) for P2 .

So, more than one process can be in CS here

So, Mutual Exclusion violated , But Progress and Bounded Waiting satisfied here

4 Comments

 srestha  

ME satisfies here

reason:when p1 enter CS we cant preempt  in between Thus P2 will never comes into between 

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 set2018 as soon as V(x); P1 enters CS and at the same moment in P2 P(X); happens so both are in CS
explain how ME is satisfied?

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edited by

From ur answer:

1."If a process enter CS at least once then No starvation"   is False. ???    i think correct one will be "Execute cs completely" means no starvation am i right?

2.There may be case when though Process is in CS and pre-emted then after that it is not able to enter that might lead to Starvation" ??

3. "Starvation for Pi if and only if Pi never get chance to execute CS by trying all possible means"

This is False???

please verify.

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Answer:

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