GATE Overflow Test Series | Operating Systems | Test 1 | Question: 27

Question

Consider the following snapshot of a system running $n$ processes.

Process $i$ is holding $x_i$ instances of a resource $R$,$ 1\leq i\leq n$. Currently, all instances of $R$ are occupied. Further, for all $i$, process $i$ has placed a request for an additional $y_i$ instances while holding the $x_i$ instances it already has. There are exactly two processes $p$ and $q$ and such that $y_p=y_q=0$. Which one of the following can serve as a sufficient condition to guarantee that the system would not be going to a deadlock?

• A. $x_{p}+x_{q}> \max_{k\neq p,q}y_{k}$
• B. $\max(x_{p},x_{q})>1$
• C. $x_{p}+x_{q}\geq \min_{k\neq p,q}y_{k}$
• D. None of the above

Comments

Great question !!

Answer 1

The question asks for "sufficient" condition for the system not to reach a deadlock.

This means if this condition is satisfied there should be no deadlock. For this the system must currently be in a safe state and also proceed through a sequence of safe states.

Neither the question statement guarantees a safe state nor the given options guarantees a safe sequence (option A ensures that the if the current state is safe, then the next state is also safe but it does not guarantee that current state is safe).

So correct answer is D. None of the above.

Comments

@Himanshu555 Option $A$ guarantees my system is in a safe state. Why? Because if you can completely allocate all the resources to the maximum resource needing process $P_{max}$, you can always finish up that process and reallocate the freed up resources to other process which have less resource requirements than $P_{max}$. This is not true for option $C$. Try to find out why not. 

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I think,because according to option c  if true  we can  fulfill  minimum yi request of some process ,then it is not guaranteed that the resources free up  from it fulfill need of other process,hence we can not  guaranteed to give safe sequence with it,so not a safe state guaranteed.

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Perfect