GATE CSE 2005 | Question: 71

Question

Suppose $n$ processes, $P_1, \dots P_n$ share $m$ identical resource units, which can be reserved and released one at a time. The maximum resource requirement of process $P_i$ is $s_i$, where $s_i > 0$. Which one of the following is a sufficient condition for ensuring that deadlock does not occur?

• A. $\forall i,\: s_i, < m$

• B. $\forall i, \:s_i <n$

• C. $\displaystyle{\sum_{i=1}^n} \: s_i  < (m+n)$

• D. $\displaystyle{\sum_{i=1}^n} \: s_i  < (m \times n)$

Comments

Pigeonhole Principle :)

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how pigeonhole principal ??

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I see "Pigeonhole Principle" with a smiley in a lot of questions. There are a lot of questions and there will be someone with just a comment "Pigeonhole Principle :)".

I want to ask those people, what is with this comment, okay it's related with pigeonhole principle but why just tell us the name. why don't explain the complete relation with pigeonhole principle. that would be really nice.

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not pal its ple

principle

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$Pigeonhole$ $principle$ is a way how you allocate resources to your processes

thus C is right answer

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Aakash Pigeonhole Principle :)

Answer 1

To ensure deadlock never happens allocate resources to each process in following manner:
Worst Case Allocation (maximum resources in use without any completion) will be $(\text{max requirement} -1)$ allocations for each process. i.e., $s_i-1$ for each $i$
Now, if $\displaystyle{\sum_{i=1}^{n}{(s_i - 1)}} \leq m$ dead lock can occur if $m$ resources are split equally among the $n$ processes and all of them will be requiring one more resource instance for completion.

Now, if we add just one more resource, one of the process can complete, and that will release the resources and this will eventually result in the completion of all the processes and deadlock can be avoided. i.e., to avoid deadlock

$\displaystyle{\sum_{i=1}^{n}{(s_i - 1)}} + 1 \leq m$

$\implies \displaystyle{\sum_{i=1}^{n}{s_i }} - n + 1 \leq m$

$\implies \displaystyle{\sum_{i=1}^{n}{s_i }}  < (m + n).$

Correct Answer: $C$

Comments

@Pascua option A can still have deadlock consider a case n=4, m=5 

P1-->s1=2

 P2-->s2=2

 P3-->s3=1

 P4-->s4=4

So, ∀i,si,<m holds and now if every process holds 1 less than its max need, we get m resources split among n processes fully and all of them requiring 1 more instance of resource for completion but here, no extra resource is available so deadlock.

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In @Digvijay Pandey answer “Now, if we add just one more resource, one of the process can complete, and that will release the resources and this will eventually result in the completion of all the processes and deadlock can be avoided”, should we should add 1 on the RHS so that resources become m+1, why has he added 1 on the LHS… Please explain

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OPTION A is not true .Here m is total number of resources available.So if option a is satisfied ,it does not ensure deadlock free.But if option a is not satisfied,surely there is a possibility of deadlock.Since if all m resources are allocated to Pi and it still needs resources, then all processes along with Pi are waiting for resources and thus blocked.

Just think of this question as the one in which you find min number of processes that ensures deadlock free execution for given number of resources and solve.

SO ,Option C is correct answer.

Answer 2

There are N processes and they share M resources.

Sum of the requirement of all processes will be " Sum of all S_i" ie ∑i=1nSi∑i=1nSi . Now If you dont want the deadlock to occur,then you should share M resources such that each process will get 1 less than their maximum requirement and then add 1 resource to any one of the processes. This way you can avoid deadlock.

So the expression should be like this : ∑i=1nSi∑i=1nSi- N+ 1= M meaning we are summing all requirements, then subtracting N,that is 1 resource from each of the processes' requirement, so that all Processes will be waiting. then adding 1, that is allocating 1 resource to any 1 of the processes so that it can complete its execution and leave its resources so that other process can use. This should be equal to M.

Now if you move N to RHS then expression will be like ∑i=1nSi∑i=1nSi-1 = M+N.

If clearly, now if we ignore 1, then the expression wil be reduced to ∑i=1nSi∑i=1nSi < M+N. Right.

Comments

Thanks for this nice explanation :)

And just to confirm, in this line

"So the expression should be like this : ∑i=1nSi∑i=1nSi- N+ 1= M" --> shouldn't LHS<=RHS instead of LHS=RHS ?

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Thqq could u please tell me good resource for file systems concept ..so that i can answer any question from gate

Answer 3

If we allocate one resources less to each process

(s1-1) + (s2-1) + (s3-1) + .............. (sn-1)  <  m

s1 + s2 + s3 +......sn -n  <  m

s1+s2....sn < m+n

Comments

Why we should allocate on resource less than the required resources?

Answer 4

maximum number of resourses by which deadlock never happen in our system

m> ∑ Si -n   ;;;; where i>=1 and i<=n 

So minimum number of resourses by which we can conclude that deadlock never happen

m> ∑ Si -n +1  ;;;; where i>=1 and i<=n 

m+n >  ∑ Si +1  ;;;; where i>=1 and i<=n 

Comments

maximum number of resourses by which deadlock never happen in our system

m> ∑ Si -n   ;;;; where i>=1 and i<=n 

can you please explain the above line