GATE CSE 2017 Set 2 | Question: 33

Question

A system shares $9$ tape drives. The current allocation and maximum requirement of tape drives for that processes are shown below:

$$\begin{array}{ccc}  \textbf{Process} & \textbf{Current Allocation} & \textbf{Maximum Requirement} \\ \text{P1} & 3 & 7 \\ \text{P2} & 1 & 6 \\ \text{P3} & 3 & 5 \\  \end{array}$$

Which of the following best describes current state of the system?

• A. Safe, Deadlocked
• B. Safe, Not Deadlocked
• C. Not Safe, Deadlocked
• D. Not Safe, Not Deadlocked

Comments

it is not possible to be deadlocked if the state is safe so clearly option A is wrong.
and B is the correct answer
order -p3--p2--p1 or p3--p1--p2

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There can never be a deadlock in safe state ..This clearly eliminate Option A 

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Deepak Poonia Sir plz kindly see this issue...image is not loading fully

Answer 1

Process	Current Allocation	Max Requirement	Need
P1	3	7	4
P2	1	6	5
P3	3	5	2

Since,Total number of available tapes =9 and currently allocated tapes (3+1+3)=7 .Available no of tapes =2.

Request for p3 can be fulfilled as Need of tapes < Available tapes .Now available number of tapes=(3+2). Further,

request of  p1 or p2 can be fulfilled as need (4 or 5) <= available no of tapes. Hence ,system is in safe condition

and not deadlock. Option B is correct.

Answer 2

Answer should be D. because if P₁ asks for 2 more then there is deadlock. 

Comments

then dont give them :)

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@Arjun Sir, I laughed for 5 mins straight looking at the comment!! :D

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:) thats the exact answer rt? I mean thats what OS should do..

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Yes absolutely!!

Answer 3

Option A is straightforward to eliminate , the reason being there can never be a situation in case of safe state leading to deadlock, that’s why we study Banker’s algorithm

Coming to options C and D, when the system is in an unsafe state, we can’t really comment using the banker’s algorithm, so it is also eliminated.

 

So,only option B is remaining,so it has be right :)