GATE Overflow Test Series | Mock GATE | Test 1 | Question: 20

Question

Consider processes $P_i, i = 1,\dots, 20$ coded as follows:

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

Here, mutex is a binary semaphore initialized to $0.$ The code for $P_{21}$ is identical except it uses $\text{P (mutex)}$ in place of $\text{V (mutex)}$ and vice versa. What is the largest number of processes that can be inside the critical section at any moment ___________

Comments

@Arjun  @Sachin Mittal 1     where i am wrong.

 for process i=1 to 20

1. repeat
2. V(mutex)
3. {Critical section}
4. P(mutex)
5. forever

and for process 21  

1. repeat
2. P(mutex)
3. {Critical section}
4. V(mutex)
5. forever

here binary semaphore initilized mutex   =   0 

p1 arrive and mutex=1; enter in cs and prempt now p21 arrive mutex become 0 and p21 enter in to cs  again p2 arrive and mutex will be 1  { coz mutex is binary semaphone so no other value exist}   overall process at a time in cs will be 3

where i am wrong

if counting semaphone then answer can be 21 at a time.

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Similar question: https://gateoverflow.in/2264/gate-cse-1997-question-6-8

Answer 1

Since the first statement for the $20$ processes is $V(mutex)$ or signal, all of them can enter the critical section. If it happens that the $21^{st}$ process comes in between this (before all the $20$ processes does $P$ operation), it is also allowed to enter the critical section. So, at max $21$ processes can be in critical section.

Comments

Thanks @palashbehra5 for replying back and adding the code snippet. I am very confused in synchronization topic and always get a wrong answer :(. I was trying to understand the solution and draw a conclusion so that mistake won’t be repeated.

Can you please provide any link or PDF to read this topic, if you have.

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http://www.facweb.iitkgp.ac.in/~isg/OS/SLIDES/ch5-Process_Synchronization.pdf

or just: https://duckduckgo.com/?q=process+synchronization+pdf&t=newext&atb=v272-1&ia=web

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Thanks @palashbehra5, I will check this out.

Answer 2

Since Code not given as for loop etc of the processes from i 1 to 20 ….its only code for every process from 1 to 20, so we may make every other process enter CS whose code is reverse of above then we can have total of maximum 40 processes finally together all in the CS. else if in loop structure its given then it must be only 21 maximum.