ISRO CSE 2018 | Question: 21

Question

Consider a system having $m$ resources of the same type. These resources are shared by $3$ processes $A, B, C,$ which have peak time demands of $3, 4, 6$ respectively. The minimum value of $m$ that ensures that deadlock will never occur is:

1. $11$
2. $12$
3. $13$
4. $14$

Comments

option A is correct m=11.

---

11 is answer

Answer 1

The sufficient condition for ensuring that deadlock doesn't occur:

                                  S_i <n+m

where S_i = Total request resource of i^th process

         n = Total number of process

        m = Total number of resource

if above condition is false then DL may occur, if the condition is true then-No DL.

for the above problem, total demand is: 3+4+6=13

13<3+m

10<m, the minimum value of m that ensure deadlock is never occurring is 11.

Answer 2

option a

Comments

This is the right and easiest technique.

---

beautiful

Answer 3

Option A is correct.

Maximum requirement of A, B and C are 3, 4 and 6 instances of a resource, respectively. Minimum instances of the resource required without creating a deadlock:

(3-1)+(4-1)+(6-1) + 1 = 11. The last one added is shared among all three.

Answer 4

Option A, minimum value of m=11 that ensures deadlock will never occur.

So we have three process A, B and C and the peak demands is 3, 4 and 6 respectively, if we allocate A=2 , B=3 and C=5 so with 10 resources there is deadlock but if we allocate one more so deadlock will never occur.

Answer 5

Answer is (C) 13.

If anyone's peak demand satisfy we don't get deadlock.

Comments

they ask for the minimum, therefore the answer is 11, not 13.