GATE CSE 1993 | Question: 7.9, UGCNET-Dec2012-III: 41

Question

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?

• A. $7$
• B. $9$
• C. $10$
• D. $13$
• E. $15$

Comments

Give the maximum possible resources to each of the processes such that deadlock will not occur if you add one more resource to any of the available processes. So, peak demand is 3,4,6. We will get 2,3,5, now adding a single resource to any of the processes will surely be deadlock-free. So, our result would be 2 +3+5+1 = 11. 11 resources would be sufficient. So, options D and E both are correct.

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let's subtract one from each peak demand and add them and add the total obtained to 1

(3-1)+(4-1)+(6-1) + 1<= m

2 + 3 + 5 + 1 <= m

m >= 11

Therefore, m = 13 or 15

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In worst case,

$2+3+5=10$ will lead to deadlock situation.

if $m \geq 11$ then deadlock will not occur.

Ans should be $(D),(E)$. But minimum value will be $11$.

Answer 1

Condition for deadlock free

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

m>=11 : for deadlock free

So, answer should be 13,15