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24 votes
24 votes

A computer system has $6$ tape devices, with n processes competing for them. Each process may need $3$ tape drives. The maximum value of n for which the system is guaranteed to be deadlock-free is:

  1. $2$
  2. $3$
  3. $4$
  4. $1$
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4 Answers

Best answer
12 votes
12 votes
Allocate max-$1$ resources to all processes and add one more resource to any process (Pigeon hole principle) so that this particular process can be completed (resources can be freed) and there is no deadlock.

Max resources required is $3$.

$\therefore (3-1)*n+1=6$

$n=\left \lfloor \frac{5}{2} \right \rfloor=2$

Correct Answer: $A$
18 votes
18 votes

Answer: (A).

For $n=3$, $2-2-2$ combination of resources leads to deadlock.

For $n=2$, $3 - 3$ is the maximum need and that can always be satisfied.

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2 votes
2 votes
Don’t use formulas until and unless it’s necessary!

This can easily be solved with the help of intuition. Ultimately, we want that at least one process should get executed completely, so 1st process consumes 3 resources. We are left with 3 more, now if we take 2 more processes then what might happen is each of 3 processes consume 2-2 resources each because of which deadlock occurs. So, at max only 1 more process we can select and it will be consuming the remaining 3 resources. So, total of 2 processes which give n =2, option A. I hope it helped. :)
1 votes
1 votes

We need to only that option in which system will always be deadlock free..

Case 1 if we have 4 processes.  indeed we do have possibility of deadlock with 4 processes.

p1 p2 p3 p4
1 1 1 1
1 1    


Case 2 if we have 3 processes. Again with 3 processes we can have deadlock

p1 p2 p3
1 1 1
1 1 1


Case 3 if we have 2 processes. With 2 processes we can never have deadlock in any configuration..

p1 p2
1 1
1 1
1 1


We have to discard all the cases where we can have deadlock..2 is the only option in which system will be deadlock free.

Answer:

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