ISRO CS 2025 Q.69: solution and answer

ISRO 2025 · CSQ.69 · 1 mark · MCQ

Consider a system with 3 processes P0 to P2 and 3 resource types A, B, and C. Resource type A has 10 instances, resource type B has 5 instances and resource type C has 7 instances. The following snapshot of the system has been taken at a given time.

Process	Allocation (A B C)	Max (A B C)	Available (A B C)
P0	0 1 0	7 5 3	5 4 5
P1	2 0 0	3 2 2
P2	3 0 2	9 0 2

Which of the following is true?

(A)
The system is not in safe state
(B)
The system is blocked
(C)
The system is in safe state
(D)
The request cannot be granted to all the processes

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Answer and solution

ANSWER

The system is in a safe state, with safe sequence $⟨ P 1, P 0, P 2 ⟩$ . Option (c) is true.

Official answer: CSource: ISRO's final answer key of 15 May 2026

Banker's algorithm: $Need = Max - Allocation$ . So $P 0 = (7, 5, 3) - (0, 1, 0) = (7, 4, 3)$ , $P 1 = (3, 2, 2) - (2, 0, 0) = (1, 2, 2)$ , $P 2 = (9, 0, 2) - (3, 0, 2) = (6, 0, 0)$ .
Check the snapshot is consistent: total allocated is $(0 + 2 + 3, 1 + 0 + 0, 0 + 0 + 2) = (5, 1, 2)$ . Adding Available $(5, 4, 5)$ gives $(10, 5, 7)$ , which equals the stated instances of A, B and C.
Start with $Work = (5, 4, 5)$ . $P 0$ needs $(7, 4, 3)$ , and $7 > 5$ in A, so $P 0$ cannot run yet.
$P 1$ needs $(1, 2, 2) \leq (5, 4, 5)$ , so $P 1$ can run to completion. It releases its allocation $(2, 0, 0)$ , giving $Work = (7, 4, 5)$ .
Now $P 0$ needs $(7, 4, 3) \leq (7, 4, 5)$ , so $P 0$ can finish. It releases $(0, 1, 0)$ , giving $Work = (7, 5, 5)$ .
$P 2$ needs $(6, 0, 0) \leq (7, 5, 5)$ , so $P 2$ can finish. It releases $(3, 0, 2)$ , giving $Work = (10, 5, 7)$ , the full system total.
Every process can finish in the order $⟨ P 1, P 0, P 2 ⟩$ (the order $⟨ P 1, P 2, P 0 ⟩$ also works). A safe sequence exists, so the system is in a safe state: option (c).
Option (a) is false because a safe sequence exists. Option (b) is false because a safe state cannot lead to deadlock, so the processes are not blocked. Option (d) is false because no request is being made, and every process's remaining need can be met in the safe order.

KEY IDEA

Compute Need = Max - Allocation, then repeatedly pick a process whose Need fits within Work, let it finish, and add its allocation back to Work. If all processes can finish, the state is safe.

WATCH OUT FOR

Comparing Max (not Need) with Available, or stopping when the first process in order, P0, cannot run. A process that cannot run now may run later, after other processes release resources, so keep scanning the remaining processes.

From the ISRO CS 2025 paper. Every question of this paper · The full paper (PDF, ISRO) · All 2 ISRO Deadlock questions, with notes