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GATE | GATE CS 2013 | Question 65

  • Difficulty Level : Hard
  • Last Updated : 28 Jun, 2021

A shared variable x, initialized to zero, is operated on by four concurrent processes W, X, Y, Z as follows. Each of the processes W and X reads x from memory, increments by one, stores it to memory, and then terminates. Each of the processes Y and Z reads x from memory, decrements by two, stores it to memory, and then terminates. Each process before reading x invokes the P operation (i.e., wait) on a counting semaphore S and invokes the V operation (i.e., signal) on the semaphore S after storing x to memory. Semaphore S is initialized to two. What is the maximum possible value of x after all processes complete execution?

(A) -2
(B) -1
(C) 1
(D) 2

Answer: (D)

Background Explanation:

A critical section in which the process may be changing common variables, updating table, writing a file and perform another function. The important problem is that if one process is executing in its critical section, no other process is to be allowed to execute in its critical section. Each process must request permission to enter its critical section. A semaphore is a tool for synchronization and it is used to remove the critical section problem which is that no two processes can run simultaneously together so to remove this two signal operations are used named as wait and signal which is used to remove the mutual exclusion of the critical section. as an unsigned one of the most important synchronization primitives, because you can build many other Decrementing the semaphore is called acquiring or locking it, incrementing is called releasing or unlocking.

Solution :

Since initial value of semaphore is 2, two processes can enter critical section at a time- this is bad and we can see why.
Say, X and Y be the processes.X increments x by 1 and Z decrements x by 2. Now, Z stores back and after this X stores back. So, final value of x is 1 and not -1 and two Signal operations make the semaphore value 2 again. So, now W and Z can also execute like this and the value of x can be 2 which is the maximum possible in any order of execution of the processes. (If the semaphore is initialized to 1, processed would execute correctly and we get the final
value of x as -2.)

Option (D) is the correct answer.

Another Solution:

Processes can run in many ways, below is one of the cases in which x attains max value Semaphore S is initialized to 2
Process W executes S=1, x=1 but it doesn’t update the x variable.
Then process Y executes S=0, it decrements x, now x= -2 and signal semaphore S=1
Now process Z executes s=0, x=-4, signal semaphore S=1
Now process W updates x=1, S=2 Then process X executes X=2
So correct option is D

Another Solution:

S is a counting semaphore initialized to 2 i.e., Two process can go inside a critical section protected by S. W, X read the variable, increment by 1 and write it back. Y, Z can read the variable, decrement by 2 and write it back.
Whenever Y or Z runs the count gets decreased by 2. So, to have the maximum sum, we should copy the variable into one of the processes which increases the count, and at the same time the decrementing processed should run parallel, so that whatever they write back into memory can be overridden by incrementing process. So, in effect decrement would never happen.

Related Links: for explanation

This solution is contributed by Nitika Bansal

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