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+28 votes

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?

  1. $\forall i,\: s_i, < m$

  2. $\forall i, \:s_i <n$

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

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

in Operating System by Veteran (52.3k points)
edited by | 4.1k views

Pigeonhole Principle :)

how pigeonhole principal ??
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.

4 Answers

+87 votes
Best answer
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: $C$
by Veteran (61k points)
edited by
is the answer is a?
Answer is option c
why (max requirement -1) ? please elaborate
because if every process holds one resource less than it really requires, and one extra resource is present, any of the process can take that resource, finish itself and release its resources which can further satisfy the need of other processes.
meaning of reserved and released one at a time
Can someone tell me when to take <= and >? I'm confused as to which variable needs to be taken as constant.
Didn't get your question?

Is it safe to say that option A is a NECESSARY condition but not a SUFFICIENT condition to avoid deadlock? @Arjun sir @Bikram sir @Digvijay Pandey sir



Dining Philosopher's problem -> m = 5 forks, n = 5 philosophers, s$_{i}$ = 2 forks

+9 votes

There are N processes and they share M resources.

Sum of the requirement of all processes will be " Sum of all Si" 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.

by Junior (973 points)
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 ?
Thqq could u please tell me good resource for file systems concept that i can answer any question from gate
+4 votes

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 

by Boss (10.2k points)

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 

+3 votes
If we allocate one resources less to each process

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

s1 + s2 + s3 -n  <  m < m+n
by Junior (647 points)
edited by
Why we should allocate on resource less than the required resources?

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