GATE CSE 2015 Set 2 | Question: 23

Question

A system has $6$ identical resources and $N$ processes competing for them. Each process can request at most $2$ requests. Which one of the following values of $N$ could lead to a deadlock?

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Comments

Did gate gave bonus marks for this question?

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Important comments to understand this question 

https://gateoverflow.in/8114/gate-cse-2015-set-2-question-23?show=413769#a413769

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They made wrong question. Even with 4 process it can never go in deadlock. All philosophy and whatnot but the question is wrong. Answer should be 6.

Answer 1

supos there are n process p1,p2,p3....pn share m identical resource which can be release/allocate one at the time.

the max resource required for process pi is si, where si>0

the sufficient condition for ensuring that DL doesn't occure:  Σsi(for i=1 to n) < (m+n)

where m= number of resource

      n= number of process

      si= total req(request resource for ith process)

Comments

And what is answer coming ?

2N<N+6

N should be min 3 ti satisfy deadlock,rt?

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@Hira Thakur and @srestha

 shouldn’t 3 i.e. option C satisfy the answer as we can make resource allocation graph for this    ?

Answer 2

Whenever this type of question came, related to resource and deadlock or process and deadlock the simple way to solve is

Allocate each process a resource equal to their maximum need - 1, This type of allocation gives you maximum no resources that lead to deadlock but if you just add 1 to this number you will get minimum no of resources causing no deadlock.

for example 

let us say the total no of resources is R and we have to N process and each can take almost M resource the what is the value of N? 

N * (M-1) + 1 <= R 

for this question, the maximum no of processes is 5 without deadlock, the minimum no of the process causing deadlock is 6 but it is not given in option so go with 4

Answer 3

Option D will be right option for it.

Suppose we have three process namely P1,P2 and P3

P1=2unit

P2=2Unit

P3=2unit

but here u will get no deadlock will be happen here.But if we will add one more process like below

P1=1unit

P2=1Unit

P3=1unit

P4=1Unit

it will be lead to deadlock always.

Comments

how you have 2 units of resource left...which can satisfy any two of them..and hence system is safe

Answer 4

I believe given options are wrong because..

a) 1 Process , 6 Resources No Deadlock.

b) 2 Process , 6 Resources No Deadlock.

C) 3 Process , 6 Resources No Deadlock.         Every process can take 2 resources

D. 4 Process , 6 Resources No Deadlock.  Assign each process one resource still 2R will remaining causing no DL.

The Number of process that would lead to DL would 6. Minimum of Process that would cause no dead lock would be 5.

Please correct me If I am wrong.!

Comments

it is a duplicate question

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how to check? if a question was answered previously on this Site before answering ?

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type the question in google custom searching which is available in this site ( even in this page )... then references will come from this site only... but if some body post the images, then this search can't be efficient... that's why some people are suggesting type the question instead of posting the screen shots