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25 votes
25 votes

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

  1. $3$
  2. $5$
  3. $4$
  4. $6$
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6 Answers

Best answer
18 votes
18 votes

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: $C$

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18 votes
18 votes
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.
9 votes
9 votes
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 ..
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6 votes
6 votes

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:

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