GATE Computer Science Engineering(CSE) 2027 Test: IPC, Synchronization

Each P operation will decrease the semaphore value by 1 and V operation increases it by 1. If x is 18, then 7 Poperations will make semaphore value 0. If this is followed by 7 V operations the value comes back to 7. So, after 18 P and 18 V operations, the value of the semaphore will be 7. The remaining 2 P operations result in the semaphore values 5.

‘Aging’ is a process that increases the priority of jobs corresponding to the time process is waiting for the CPU. So, the process with longer waiting time turns to have more priority over the process that is just arriving. This is done to ensure there is no STARVATION and termination of processes in a finite time.

Let the mutex be initialized to 1. Any one of the 9 processes P_i, i = 1 , 2 , 3 , ... 9 can get into the critical section after executing P (mutex) which decrements the mutex value to 0. At this time P10 can enter into the critical section as it uses V (mutex) instead of P(mutex) to get into the critical section. As a result of this, mutex will be incremented by 1. Now any one of the 9 processes P_i, i — 1, 2 , 3 , ... 9 (excepting the one that is already inside the critical section) can get into the critical section after decrementing the mutex to 0. None of the remaining processes can get into the critical section.
If the mutex is initialized to 0, only 2 processes can get into the critical section. So the largest number of processes is 3.

Given set of processes depicts a standard producer-consumer relationship. Increasing the number of buffers will help in providing more slots to P₁ to fill a buffer and more slots to P₂ to empty a buffer. Hence, increase the rate at which requests are satisfied.
This will not affect the likelihood of deadlock:

e = N=> empty = N [The complete array is empty initially]
f= 0 = > full = 1 [No item is there in the array] b = 1 => Only one process can execute at a time

• The purpose of semaphore 'f' is to ensure that dequeue is not executed on an empty queue. Process 2 will block itself on first statement i.e. p(t).
• The purpose of semaphore ‘e’ is to ensure that not more than N elements being present in queue or if N element are already there in queue, process 1 must block itself.
• The purpose of semaphore ‘b’ is to ensure mutual exclusion for queue operations.

• Since binary semaphore used, so at a time either reader or writer can active inside critical section.
• Readsers can use binary semaphore without interversion of writers and can enter into critical section, so statement is true.

If after checking the condition (critical - flag == FALSE) P₁ preempts without making critical flag TRUE and in turn gives option for P₂ to check the condition and since critical - flag is FALSE till this point. P₂ will also enter the if block and make the critical - flag as TRUE and start assessing critical -region ().
If P₁ again comes back, it will start from time 2 directly and rewrite critical - flag as TRUE again and will start assessing critical - region (). So, it is possible for P₁ and P₂ to access critical - region concurrently.
Since, mutual exclusion is mandatory condition for deadlock, deadlock will never occur in this case. However, consistency can’t be ensured in such cases.

Since there is a strict alteration between producer and consumer. So values generated and stored in 'x' by producer will always be consumed before the producer can generate a new value.

If R₁ accesses the critical section:
1. No mutual exclusion needed for R₂ and R₃.
2, Mutual exclusion must be there for W₁and W₂.

Number of P(C) wait process = 6 i.e. decrement

Number of V(C) signal process = 2 i.e. increment

Therefore value of ‘C’ at the end of program execution
= Initial value - Number of P(C) + Number of V(C)
= 10 - 6 + 2 = 6

Number of wait process = 20 i.e. decrement Number of signal process = 15 i.e. increment Therefore value of semaphore at the end of program execution
= Initial value - Number of P(C) + Number of V(C)
= 7 - 20 + 15 = 2