# GATE | GATE CS 2019 | Question 32

• Last Updated : 15 Feb, 2019

Consider three concurrent processes P1, P2 and P3 as shown below, which access a shared variable D that has been initialized to 100.

The process are executed on a uniprocessor system running a time-shared operating system. If the minimum and maximum possible values of D after the three processes have completed execution are X and Y respectively, then the value of Y–X is __________.

Note: This was Numerical Type question.
(A) 80
(B) 130
(C) 50
(D) None of these

Explanation: Minimum value (X) of D will possible when,

2. P1 executes D=D+20, D=120.
3. P3 executes D=D+10, D=130.
4. Now, P2 has D=100, executes, D = D-50 = 100-50 = 50. P2 writes D=50 final value.

So, minimum value (X) of D is 50.

Maximum value (Y) of D will possible when,

2. P2 reads D=100, executes, D = D-50 = 100-50 = 50.
3. Now, P1 executes, D = D+20 = 100+20 = 120.
4. And now, P3 reads D=120, executes D=D+10, D=130. P3 writes D=130 final value.

So, maximum value (Y) of D is 130.

Therefore,

```= Y - X
= 130 - 50
= 80 ```

So, option (A) is correct.

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