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23 votes
23 votes

Consider a system having $m$ resources of the same type. These resources are shared by $3$ processes $A, B$, and $C$ which have peak demands of $3, 4$, and $6$ respectively.  For what value of $m$ deadlock will not occur?

  1. $7$
  2. $9$
  3. $10$
  4. $13$
  5. $15$
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5 Answers

Best answer
46 votes
46 votes
$\mathbf{13 \text{ and } 15.}$

Consider the worst scenario: all processes require one more instance of the resource. So, $P1$ would have got $2, P2 - 3$ and $P3 - 5$. Now, if one more resource is available at least one of the processes could be finished and all resources allotted to it will be free which will lead to other processes also getting freed. So, $2 + 3 + 5 = 10$ would be the maximum value of $m$ so that a deadlock can occur.
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16 votes
16 votes
Worst case

3-1 =2

4-1=3

6-1=5

10 then may be deadlock but we don't want deadlock so we required minimum 11 resources

So ans is any number $\geq$ 11
10 votes
10 votes

We have m resourses now suppose we have three process like below

P1,P2 and P3

P1=2

P2=3

P3=5

Max resourses by which deadlock happen=10

So minimum resourses by which deadlock didn't happen =10+1>=11 therefore answer 13 according to option.

9 votes
9 votes
ans is 13 and 15
 
Formula-
Total of max needs < (no. of resource instances + no. of processes)
(3+4+6)< m+3.
 
 
m>10. hence m should be eleven or above.
Only 13 and 15  qualify the criteria.

 

Answer:

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