GATE IT 2005 | Question: 41

Question

Given below is a program which when executed spawns two concurrent processes :
semaphore $X : = 0 ;$
/* Process now forks into concurrent processes $P1$ & $P2$ */

$\begin{array}{|l|l|}\hline \text{$P1$}  &  \text{$P2$}  \\\hline  \text{repeat forever } & \text{repeat forever} \\ \text{$V (X) ;$ } & \text{$ P(X) ;$} \\  \text{Compute;  } & \text{Compute;}\\  \text{$P(X) ;$  } & \text{$V(X) ;$} \\\hline \end{array}$

Consider the following statements about processes $P1$ and $P2:$

1. It is possible for process $P1$ to starve.
2. It is possible for process $P2$ to starve.

Which of the following holds?

• A. Both (I) and (II) are true.
• B. (I) is true but (II) is false.
• C. (II) is true but (I) is false
• D. Both (I) and (II) are false

Comments

will you please explain your line

" Since P2 can't execute signal on P1 's X. "

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https://gateoverflow.in/3788/gate2005-it-41?show=279141#a279141

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yes, bounded waiting will not satisfied by this code.

Answer 1

Option A is correct , Both the cases I have mentioned below.

Answer 2

Correct me If I am wrong..

If P1 does not execute then P2 has to starve for infinity time. If P1 enters Compute stage making x=1, then P2 can enter into Compute stage making x=0. Now P1 can not come out of the Program because it stuck in while loop due to x=0. If P2 takes infinity time to compute then P1 has to starve in the while loop infinity time. Hence both could starve.

Answer 3

They ask for any possibility not for a single case so
Both processes can starve here
p1 can starve if p1 is in c.s. and then p2 repeatedly executing.
p2 can starve if p1 repeatedly executing

So option A is correct