Modified gate1997

Question

My question is little different.

Each Process Pi, i = 1....9 is coded as follows

repeat
    P(mutex)
    {Critical section}
    V(mutex)
forever

The code for P10 is identical except it uses V(mutex) in place of P(mutex). What is the largest number of processes that can be inside the critical section at any moment?

If instead of mutex a semaphore would have been used, then was it possible that p1, p2 , p3 or any other process instead of process p10 could have opted out while all are in the critical section.?

My doubt started from the following note which I read from a source:

"Mutex can be released only by thread that had acquired it, while you can signal semaphore from any other thread (or process), so semaphores are more suitable for some synchronization problems like producer-consumer."

Comments

A mutex is essentially the same thing as a binary semaphore and sometimes uses the same basic implementation. The differences between them are in how they are used. While a binary semaphore may be used as a mutex, a mutex is a more specific use-case, in that only the thread that locked the mutex is supposed to unlock it. This constraint makes it possible to implement some additional features in mutexes:

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"The thread that locked the mutex is supposed to unlock it." This is the line due to which my doubt arises for this question.

Okay I will write down and draw all these on a piece of paper and will post .

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You can write down all the doubts here.

Answer 1

Answer is D 
Ifinitial value is 1//execute P1 or P10 first  
Ifinitial value is 0, P_10 can execute and make the value 1. 
sincethe both code (ie p1 to p9 and p10)can be executed any number of times and code for p10 is 
 

repeat
{
    v(mutex)
    C.S.
    V(mutex)
}
forever 

now let me say P1 is in c.s  
then p10 comes executes the CS(up on mutex) 
now P2 comes (down on mutex) 
now P10 moves out of CS (again binary semaphore will be 1 ) 
now P3 comes (down on mutex) 
now P10 come (up on mutex) 
now P4 comes (down on mutex)  
so if we take p10 in out of CS recursively all 10 process can be in CS at same time 
using Binary semaphore only

Comments

Have you not read my question fully.

The original question has solution already. I had some doubt.  Isn't this the case with mutex that the only process can give signal to mutex which has applied wait on the mutex ?