MadeEasy Test Series: Operating System - Resource Allocation

Question

Consider the following resources allocation graph with four processes $P_{1}, P_{2}, P_{3}, P_{4}$ and three resources $R_{1}, R_{2}, R_{3}$. System has two instances of resources $R_{1}$, two instances of resources $R_{2}$ and two instances of resources $R_{3}$.

Which of the following statement is correct for the above system?

• A. No safe sequence exist.
  
• B. More than one safe sequence exists.
  
• C. Exactly one safe sequence exists.
  
• D. More than two safe sequence exists.

Comments

There will be more than one safe sequence (4 to be precise), try with banker's algorithm..

Answer 1

After P3 more than one option possible as P2,P1,P4 all can be satisfied so many combinations possible hence the answer. Hope it helps!

Answer 2

D) more than 2 safe sequence.

First P3 shall complete and then other 3 processes in any order.

safe sequences:

(P3 P1 P2 P4), (P3 P1 P4 P2)

(P3 P2 P4 P1), (P3 P2 P1 P4)

(P3 P4 P2 P1), (P3 P4 P1, P2)

Answer 3

(D)More than 2 safe sequences exist.

Explanation: After creating the Request and allocated table from the request allocated graph given above we can see R1, R2 and R3 are left with only 0,1 and 0 resources respectively. With this, we only can satisfy the need for P3. After P3 executes it releases 1 instance of R2 and one instance of R3 making the available resources of R1, R2, and R3 as 0,2 and 1 respectively. With this, we can satisfy the requirement of all the processes so we can schedule any one of them so there will be more than two safe sequences possible.

Answer 4

Yess more than 1 safe sequence possible.