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

incorrect options plzz correct it..

---

Confused regarding Deadlock..
Given total number of drives=  9
         P1 max. requirement = 7
         P2 max. requirement  = 6
         P3 max. requirement = 5
Now For three processes to be not in deadlock with max. requirements of 7 6 and 5,
No. of resources required= (6+5+4) + 1 = 16
But given No. of resources = 9
So deadlock will happen.

---

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

---

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

Answer 1

$$\small\begin{array}{|c|c|c|c|} \hline \textbf{Process} & \textbf{Current Allocation} & \textbf{Max Requirement}& \textbf{Need} \\\hline \text{P1} & 3 & 7 & 4 \\\hline \text{P2} & 1 & 6 & 5 \\\hline \text{P3} & 3 & 5  & 2\\\hline  \end{array}$$Given there are total $9$ tape drives,

So, according to the above table we can see we have currently allocated ($7$ tape drive), so currently Available tape drives $= 2$

So, $P3$ can use it and after using it will release it $3$ resources New Available $=5$

then $P1$ can use it and will release it $3$ resources so New Available $= 8$

and lastly $P2$ so, all the process are in SAFE STATE and there will be NO DEADLOCK

Safe Sequence will be $\bf P3\to P2\to P1$ or $\bf P3\to P1 \to P2.$

Answer will be (B) only.

Comments

@Arjun Sir, @Bikram Sir, what if one instance of tape is given to P1, and another instance is given to P2, and now deadlock will occur right? Please someone reply me.

---

@VS @rio

This answer will surely clear you'r doubts👍
https://cs.stackexchange.com/a/45149

---

In safe state,deadlock is not possible.(eliminates option A)

But,in unsafe state,we cannot say anything about it;deadlock may or may not occur(eliminates options C and D)

Thus,we can answer it without even solving that answer will be B.

Correct me if I'm wrong.

Answer 2

b) safe state , not deadlocked

we can execute in the order p3,p1 ,p2

 or p3,p2, p1

Answer 3

hope it might help.........

Answer 4

Require P1 =7-3 =4

           P2 =6-1 =5  

           P3 =5-3 =2

Available 9-(3+1+3) =2

P3->P2->P1

OR

P3->P1->P2

Hence Safe and NO Deadlocked

Ans B