# GATE1993-7.9, UGCNET-Dec2012-III: 41

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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. $15$

recategorized
0
plz ans
–1
i think 15 but not sure
0

correct answer should be 11 (minimum )

$\mathbf{13 \text{ and } 15.}$

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.

edited by
0
Can't it be 7?? Because if p1 and p2 both have sufficient number of resources, so they can run to end and after finishing they will release the resources occupied by them. Now, P3 can take 6 resources and can run.
6
No. Because we dont have a control to distribute the resources (not doing banker's algorithm).
7
sir ignoring the option correct answer should be $11$ (minimum )?
2
yes.
0
So by your explanation answer will be 11 --minimum value of m that ensures that deadlock will never occur.
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

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.

0
Well, Why not $15$?
0
Not in options
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.

0
sir, this formula is giving the right answer. but, may i know the source of this formula for further study?
1 vote

# R >= Sum(p×(n-1))+1

Here R= total no. Of resources

n= Max demand of resources for each process p

m >= ([1×(3-1)]+[1×(4-1)]+[1×(6-1)]) +1

m>=(2+3+5)+1

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