Starvation is possible or not?

Question

Comments

picture not clear ?? wt is the value of semaphore X is it 0???

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x =0

Answer 1

option d

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Explanation:

Semaphore X=0;

Process 1                                                Process 2

Repeat                                                     Repeat

    1. V(X)                                                   1. P(X)

    2. Compute;                                            2. Compute;

    3. P(X)                                                   3. V(X) 

Forever                                                    Forever  

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The process 1 is performing up() operation in step 1, whereas process 2 is performing down() operation.

Semaphore is initialized to 0;

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1. Both the process keeps on rotating because of forever loop.
2. Definitely Process 2 can not enter as Down() operation at step 1; 
3. Process 1 will only enter and it will perform up() making X=1;
4. As soon as X becomes 1 which is favourable for Process 2 to enter (remember it is still in continous rotaion will find X=1), it will enter or resume.
5. Now Process 1 will Down() the semaphore and Process 2 will perform Up(); 

Comments

Let P1 perform up() and get pre-empted. Now P2 is scheduled. It completes the execution till up() once and then again performs down() because of loop. This time, when P1 is scheduled, it will find semaphore = 0 and hence can't perform down().

What if this cycle keeps on repeating and P1 never gets a chance to perform down() operation?

Is it considered starvation?

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Above scenario is right, but how will you assure that by the time P2 comes back, P1 hasn't got any chance to resume from it's suspended state. Although progress and mutual exclusion is satisfied here but Bounded waiting is not.

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We can't ensure the sequence of operation but there is a possibility. Given options talk about the possibilities only

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If there is no bounded waiting, P1 can be  blocked forever for whatever reason P2 was unable to release resources,resulting in starvation.