GATE CSE 1997 | Question: 6.7

Question

An operating system contains $3$ user processes each requiring $2$ units of resource $R$. The minimum number of units of $R$ such that no deadlocks will ever arise is

• A. $3$
• B. $5$
• C. $4$
• D. $6$

Comments

$3$ process each requiring $2$ units of resource $R$.

$\therefore 3(2-1)+1\leq R=>4\leq R$.  

The min no of units of $R$ such that deadlock never arises = $4$

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Answer 1

If we have $X$ number of resources where $X$ is sum of $r_i-1$ where $r_i$ is the resource requirement of process $i$, we might have a deadlock. But if we have one more resource, then as per Pigeonhole principle, one of the process must complete and this can eventually lead to all processes completing and thus no deadlock. 

Here, $n = 3$ and $r_i = 2$ for all $i$. So, in order to avoid deadlock minimum no. of resources required

$=\displaystyle{\sum_{i=1}^{3}} (2-1) + 1 = 3 + 1 = 4.$

PS: Note the minimum word, any higher number will also cause no deadlock. 

Correct Answer: $C$

Comments

good

Answer 2

Answer: C

(a) 1-1-1 allocation of resources can cause a deadlock.

(c) 2-1-1 is the general resource allocation which will never cause a deadlock as the process getting 2 resources will release its resources after its task is over.

Answer 3

For a system to be deadlock free ,

Sum of max need of processes < No. of processes + No. of resources

Given , 3 user processes each requiring 2 units of resource , so max need of processes = 3 *2 = 6 .

Number of process = 3 , Number of resources = R

so 6 < 3 + R

=> 3 < R

so R must be 4 to be Not in deadlock( minimum value ) . Any value more than 4 is maximum value ..

Comments

@Bikram But there still could be deadlock with 4 processes suppose p1 is holding resource r1 and required r2 to complete. 

p2 is holding r2 and needs r1 to complete and finally, p3 is having r3 and r4. So it will complete its execution. But isn’t p1 and p2 will be in Deadlock?

 

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@Rusty_01 Maybe i'm late, but to answer your doubt, the Question says "instances of same resource R", so we are not talking about different resources here.
Let me know if there's anything wrong with what I've pointed out here.

Answer 4

Minimum resourses by which deadlock will not happened will be 4

Explanation::

Max resourses by which deadlock happen is like

P1=1

P2=1

P3=1

So  Max resourses by which deadlock happen is =3 so minimum number of resourses by which deadlock did not happen=4

Answer 5

Let use the formulae,

n*k - (n - 1)  or   (Total Number Process * Total Number of Resource ) - (Total Process -1 )

 

then,

= 3*2 - (3 - 2)

= 4

 

For Example : There are 8 process and each require 2 resources to complete the task. So the Minimum number of resource which guarantee the dead lock free is 10

=8*2 -(8-2)

=10

Answer 6

Step(1)Create circular wait of processes where each process in the set is waiting for other to complete. How this is done is, give each process exactly 1 less the maximum number of resource it demands.

$P_1$	$P_2$	$P_3$
1	1	1

Now every process in the set is holding and waiting and the circular wait is created. Deadlock!!

Step (2). Now give one extra resource, so that any one process is able to complete, hence circular wait is broken.!

so 3+1=4 resources required to never fall into a deadlock

and Maximum 3 resources can be there for the deadlock to happen.