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The following program consists of $3$ concurrent processes and $3$ binary semaphores. The semaphores are initialized as $S0=1, S1=0$ and $S2=0.$

 Process $P0$ Process $P1$ Process $P2$ while (true) { wait (S0); print '0'; release (S1); release (S2); } wait (S1); release (S0); wait (S2); release (S0);

How many times will process $P0$ print '$0$'?

1. At least twice
2. Exactly twice
3. Exactly thrice
4. Exactly once
edited | 5.4k views
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if s is a binary semaphore and the initial value of s is 1 then can 'signal' be performed on s ?  what will happen if signal(s) is performed?
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because of the word concurrent we can execute both the process together, but my doubt is if s=1 then can we release(s) again? because it is a binary semaphore & already its value is 1. please someone explain
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when p0 execute second time  it will  free s1,and s2 so agian there will  chance to run p0 again n  this process continue forever and infinite zero will be printed ?
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If there hadn't been the while loop in P0 then answer would have been Exactly once?

There are only $2$ possible execution sequences here:

1. $P_0\to P_1\to P_0\to P_2\to P_0 \qquad P_0$ is executed thrice here and will print $0$ thrice.
2. $P_0\to P_1\to P_2\to P_0\qquad P_0$ is executed twice only.

Hence, answer is "at least twice".

edited
+4
in binay semaphore can we do up operations (signal) any number of times?
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the process P0 will print 0 at least twice and at most thrice...
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I understood the first explanation but can after p0, p1 and p2 be performed regularly?

I mean to say that if p1 releases the lock and make semaphore S0 value 1 then can now again p2 also release the lock ? or It needs to wait until S0 again decremented?

If I am right then in your second example also 0 will be printed three times.
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is it necessary for p0 to go again if it is not willing to go then how a is the answer
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The magic here lies in the fact realizing how a signal operation on binary semaphore can be "lost".
+7

This is the best answer but there should be one point which need to be made focused and that is P1,P2 are not in loop, therefore, they will be executed only once while p0 is in infinite loop

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But it cannot print zero 4 times then how is atleast twice possible?
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@kamal

At least 2 and at most 3 times
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If P1 and P2 executes first and then P0??

First P0 will enter the while loop as S0 is 1. Now, it releases both S1 and S2 and one of them must execute next. Let that be P1. Now, P0 will be waiting for P1 to finish. But in the mean time P2 can also start execution. So, there is a chance that before P0 enters the second iteration both P1 and P2 would have done release (S0) which would make S1 1 only (as it is a binary semaphore). So, P0 can do only 1 more iteration printing '0' two times.
If P2 does release (S0) only after P0 starts its second iteration, then P0 would do three iterations printing '0' three times.

If the semaphore had 3 values possible (an integer semaphore and not a binary one), exactly three '0's would have been printed.

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atleast 1 is never possible?
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Can't understand "If the semaphore had 3 values possible, exactly three '0's would have been printed" How we can comment "exactly 3 0's " ?

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@Gate-Keeda At least 1 is possible (because it holds for exactly 3 also). But less than 2 is not possible here.
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@shree Here semaphore can have value either 0 or 1 only (because it is a binary semaphore). If we allow, 0, 1 and 2 as the values then both P1 and P2 will both release S0 (thus incrementing it twice), and thus P0 will execute totally 3 times. Thus exactly 3 0's will be printed.
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Why it is mandatory to again execute P0? what if I just execute each process once?
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In P0 there is an infinite while loop
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sorry to bother you much. But I'm not getting why atleast once is not possible? Because it isn't in the options?
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It will print 0 at least 2 times (in any circumstance). Now, even if "at least one" is there in option, at least 2 is the better answer.
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sorry i didn't get it.why exaxtly 3 0's ?.it will keep on executing wait and release and a infinte number of 0's can be printed.
+5
Infinite times 0 could not printed bcoz process P1 and P2 are not under while loop means they are executed only once. when process P0 execute second time it release s1 and s2 but process P1 and P2 not executed second time and program terminated.
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Is it neccesary to execute process 1 and 2?What if p1 and p2 don't want to execute
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because of the word concurrent we can execute both the process together, but my doubt is if s=1 then can we release(s) again? because it is a binary semaphore & already its value is 1. please someone explain.
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Can anyone please tell why there is no preemption here? ....means I understood both the cases and both are without consideration of preemption....please someone help...

Thanks
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@Arjun sir plz make that bracket in bold trust me half of the doubts will be vanished.

(as it is a binary semaphore)

Option A is True.
Initially P0 will execute because only S0=1. It will print single 0.

Now when S1 and S2 are releases by P0 then any one of them can be executed.

Let us suppose P1 executes and releases S0(Now value of S0 is 1).

Now there are two possibilities either P0 or P2 can execute.

Let us take P2 executes and releases S0, so at the end P0 execute and print 0 (means two 0's) but if P0 executes before P2 then total of 3 0's will print(one at the time of P0 and then P2 which releases S0 so P0 executes again).
So the perfect answer is at least two 0's.

http://stackoverflow.com/questions/12069305/how-many-times-will-process-p0-print-0

reshown
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because of the word concurrent we can execute both the process together, but my doubt is if s=1 then can we release(s) again? because it is a binary semaphore & already its value is 1. please someone explain.
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Is atleast one possible? if not, why?
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You are saying that the least possibility is 1 rt?

First, we need to also have a look at the options and answer accordingly.

We have to try the maximum times we can execute P0.

Here we can execute P0 atleast twice. i.e. >=2.
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yes, I was asking if "atleast one" was in option, it would have been correct or not?
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yes. It would have been more correct than atleast twice.
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Semaphores S0, S1, S2 have their own block queue or is it common for all?
+3
Answer should be either 2 or 3. So, at least 2. 1 is never possible.
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every semaphore has its own separate queue...
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@Arjun Sir, what if P2 and P3 do not execute?
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@MINIPanda,

I think you are asking for P1 and P2 . If they will not execute then P0 will be executed only once and it will wait for release(S0).

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Oh yes it will be P1 and P2.

And if that is the case then had "at least once" been in the options, it would have been more appropriate right?
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@MINIPanda,

as you are saying P1 and P2 don't execute, then after the first iteration of P0 who will perform release(S0) for next iteration of it :)

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@MINIPanda,

don't get confused with "atleast".In the options "atleast" is there only bcoz P0 can be executed either twice or thrice so merging both choices we are getting "atleast twice"
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Ok..I understand what you are trying to say. I was just computing all the possible cases that may arise. If you consider mine ( where P0 can only iterate once) along with the other two cases (where P0 iterates twice and thrice) then I thought it would indicate "atleast once" as the answer.
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Yes, if that case would be included there, then it will be "atleast once".
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because of the word concurrent we can execute both the process together, but my doubt is if s=1 then can we release(s) again? because it is a binary semaphore & already its value is 1. please someone explain.
a is most appropriate ans here
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because of the word concurrent we can execute both the process together, but my doubt is if s=1 then can we release(s) again? because it is a binary semaphore & already its value is 1. please someone explain
0
Hi, here atleast 1 cannot be the answer because, in any order we try to execute the 3 processes, say if P0 executes first then after its execution, S1=S2=1, and S0=0, so if P0 executes again since its in while I can execute again, it gets blocked and waiting for someone to releaseS0.

then whenever P1 or P2 get a chance they will not be blocked since S1=S2=1, so they will release S0 and make S0=1,

Now there r 2 cases, either P1 and P2 occur one after the other then finally P0 comes in this case S1=1 only so only once P0 can execute and it will print second 0.

Or P1 then P0 then again P2 then P0, in this way P0 gets two more times to run so total 3 zeros. So anyhow atleast 2 two times its printed.

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