GATE CSE 2008 | Question: 63

Question

The $P$ and $V$ operations on counting semaphores, where s is a counting semaphore, are defined as follows:

$P(s):$	$s=s-1;$ If $s < 0$ then wait;
$V(s):$	$s=s+1;$ If $s \leq0$ then wake up process waiting on s;

Assume that $P_b$ and $V_b$ the wait and signal operations on binary semaphores are provided. Two binary semaphores $x_b$ and $y_b$ are used to implement the semaphore operations $P(s)$ and $V(s)$ as follows:

$P(s):$	$\quad P_b(x_b);$ $\quad s = s-1;$ $\quad \text{if } (s<0)$ $\quad \{$ $\qquad V_b(x_b);$ $\qquad  P_b(y_b); $ $\quad \}$ $\quad \text{ else } V_b(x_b); $
$ V(s):$	$\quad P_b(x_b);$ $\quad s= s+1;$ $\quad \text{if } (s\leq0) V_b(y_b) ;$ $\quad V_b(x_b);$

The initial values of $x_b$ and $y_b$ are respectively

• A. $0$ and $0$
• B. $0$ and $1$
• C. $1$ and $0$
• D. $1$ and $1$

Comments

@Chhotu @Vishal Rana @Akash Kanase @kenzou @Arjun @srestha @Bikram @Ayush Upadhyaya @Arvnd

Can someone comment on the order of these instructions -

1. if (s<0)
   {
   $V_{b}(X_{b})$
   $P_{b}(Y_{b})$
   } //Why can't I execute $P_{b}(Y_{b})$; before $V_{b}(X_{b})$; in P(S).
2. if (s≤0)$V_{b}(Y_{b})$;
   $V_{b}(X_{b})$; //Why can't I execute $V_{b}(X_{b})$; before $V_{b}(Y_{b})$; in V(S).

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@ayushsomani

1.

If you write Pb(Yb) before Vb(Xb) that means you are calling Down on Yb which may cause this process to go to sleep before releasing the lock on Xb and in case, it happens then no other process will be able to enter P(s) or V(s). That's why we first release the lock on Xb and then call Pb so that other processes do not face any problem while accessing the P(s) or V(s) even if this process goes to sleep.

2.

for this case, let's take an example:

if the value of S was -1 that means there is one process in the blocked queue and now when we execute S=S+1; its value will become 0 and we need to wake some sleeping process up so that it can go in the ready queue. But now, if we release the lock on Xb before calling wakeUp(executing Vb(Yb)) then In case if this process preempts right here and some other process access the P(s) then that process will make S value -1 by executing S=S-1; and it will also get blocked by IF statement in P(s), and now there are two processes in blocked queue but S=-1 says that there is only one process in blocked queue which should not happen.

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I have a doubt here :-

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Lets 4 process are there P0 to P3 and S=2.

P0 and P1 successfully call P(S) and S become 0.

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P2, P3 call P(S), S becomes -2.  Because of s<0 goes into if condition calls Vb(Xb) and got preempted.

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Now P0 after executing CS, calls V(S). S becomes  S = -2+1=-1. Vb(Yb) gets called. Binary semaphore queue is empty so none to wake up. Yb value becomes Yb= 0+1=1.

Next P1 after executing CS calls V(S). S becomes S= -1+1=0. Vb(Yb) gets called. Yb value still remains 1. Because binary semaphore.

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Now P0 again starts exeution from where it got preempted. And calls Pb(Yb). Yb values becomes Yb= 1-1=0. P0 goes to execute CS.

Same way P1 starts executing from where it got preempted and call Pb(Yb). Now Yb is 0 so it waits in binary semaphore queue until P0 completes executing CS and makes Yb 1 again.

But according to our counting semaphore 2 processes can be there in the CS. But in this last scenario it is allowing only 1.

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Is there any wrong concept in this doubt?

Can anyone please help out?

Answer 1

Answer is (C) .

Reasoning :-

First let me explain what is counting semaphore & How it works. Counting semaphore gives count, i.e. no of processes that can be in Critical section at same time. Here value of $S$ denotes that count. So suppose $S = 3$, we need to be able to have $3$ processes in Critical section at max. Also when counting semaphore $S$ has negative value we need to have Absolute value of $S$ as no of processes waiting for critical section.

(A) & (B) are out of option, because $Xb$ must be $1$, otherwise our counting semaphore will get blocked without doing anything. Now consider options (C) & (D).

Option (D) :-

$Yb = 1, Xb = 1$

Assume that initial value of $S = 2$. (At max $2$ processes must be in Critical Section.)

We have $4$ processes, $P1, P2, P3 \& P4.$

$P1$ enters critical section , It calls $P(s) , S = S - 1 = 1.$ As $S > 1$, we do not call $Pb(Yb)$.

$P2$ enters critical section , It calls $P(s) , S = S - 1 = 0.$ As $S >0$  we do not call $Pb(Yb).$

Now $P3$ comes, it should be blocked but when it calls $P(s) , S = S - 1 = 0-1 = -1$ As $S < 0$  ,Now we do call $Pb(Yb)$. Still $P3$ enters into critical section & We do not get blocked as $Yb$'s Initial value was $1$.

This violates property of counting semaphore. $S$ is now $-1$, & No process is waiting. Also we are allowing $1$ more process than what counting semaphore permits.

If $Yb$ would have been $0, P3$ would have been blocked here & So Answer is (C).

$Pb(yb);$

Comments

@akash

I just want to understand what happens when P4 enters critical section?

As when P3 entered critical section value of s was -1 and P3 was blocked by binary semaphore Yb which is the expected behavior. But when P4 enters the critical section value of s becomes -2 and now even P4 is blocked by the same semaphore Yb. 

So when V(s) is called expected behavior is either of P3 or P4 gets unblocked but not both. But as V(s) will make Yb to 1. Both P3 and P4 gets unblocked which should not be the case.

I am able to understand until P3 enters critical section but according to me this solution fails when P4 enters. Please correct me if there is some gap in my understanding of the question.

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You are contradicting your own point first you have taken s=3 then you are performing operations for s=2, may be your answer is correct but that isn't correct logic.

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@prayas 

Important Note: There is no busy waiting in semaphores as the processes which get blocked are put in waiting queue and hence no spinlocks as well but context switch overhead is still present.

Yes, Improper use of Semaphores with wait queues can cause deadlock.

Ref: http://academic.udayton.edu/

Answer 2

Answer is (c)

Xb must be 1 because both P(S) and V(S) operations perform Pb(Xb) first. So if Xb= 0 then all the processes performing these operations will be blocked.

Yb must be 0. otherwise two processes can be in critical section at the same time, when s=1 and Yb=1 . So Yb must not be 1.

Comments

@Arjun sir , i have a doubt

suppose S=3, we need to be able to have 3 processes in Critical section at max. Also when counting semaphore S has negative value we need to have Absolute value of S as no of processes waiting for critical section. Suppose P0,P1,P3 want to enter in critical section in order then they will call wait(); one after one and wait();is being called 3 times and S value will be 0 at this time all  3 will be there in critical section Now, if two more process P4,P5 will come and call the wait(); one afer one  then s value will be -2 and they will be block and added to list of waiting process now, suppose if P0 comes out of critical section then it will call signal(); So S will be increment to -1 and it will see that S<=0 and it will call wakeup so, the process will be unblocked in same order(P3,thenP4)

MYDOUBT>now, if P3 wants to enter in critical section then it will call wait(); and S value will be -2 but S-value<0 then how will P3 will enter in critical section it will be again added to list of processs ,and if again P4comes out of critical section then again such thing will happen,

please tell me where i am making  mistake ?

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Sir, I think in this case more than one process can be inside critical section because here we are implementing Counting Semaphore.

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Which portion of program is considered critical section here

Answer 3

In the given above question, we need to analyse that here counting semaphore is implemented with the help of binary semaphore.

Now let us compare both $P(s)$ code, if we see we can find that in second one they are using a binary semaphore for $s$ which implies that at a time only one process will access counting semaphore variable. 

So options a and b can be eliminated as if $x_b=0,$ then no process can access $s$.

so $x_b$ is 1 .

Now the answer can be either c or d. 

Now for $y_b$:

In the description of P(s), it is mentioned that if $s < 0,$ process must wait and so in our implementation $y_b$ must be 0 as then only $P_b(y_b)$ does wait (definition of P operation on binary semaphore). So, initial value of $y_b$ must be 0. 

So, option C.

Here $x_b$ is used to ensure that only one process access S at a time and $y_b$ ensures a wait operation if $s$ becomes negative. ($s$ being a counting semaphore it does allow multiple processes to access the critical section based on the initial value. If initial value of $s$ is 1, it becomes a binary semaphore)

 

Comments

mostly correct, but binary semaphore value going negative? S < 0, then processes must wait as per question. Now, for waiting yb must be 0.

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Oh yes . binary value can take a value betwee 0 and 1 only . i got it . thanks :)

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:D

Answer 4

option A & option B: fails coz if $x_b = 0$ initially then it's a state of deadlock, nothing happens.

option D:
question demans the functionality that if s<0 then Wait
So, if $y_b = 1$ initially then on execution of P(s) in case when $s=0$ initially; It will successfully first decrement $x_b$ then $s$ and then signal $x_b$ then decrement $y_b$ and exit. NO waiting by P(s) happened even though s<0. This failed to implement the functionality.

option C:
suggests that keep $y_b = 0$ initially; So, if P(s) is executed given that $s=0$ initially then it will be suspended on performing $P_b(y_b)$ operation, it will continue to wait until $s$ is incremented, increment of $s$ will still be possible and also functionality of resuming a sleeping process will happen. both Functionality implemented successfully here.

answer = option C

Comments

Which part is considered critical section here?