A system has 5 processes [P0 – P4] and 3 resource types (A, B, C). The current…
2025
A system has 5 processes [P0 – P4] and 3 resource types (A, B, C). The current allocation and maximum need matrices are :
Allocation = [[0, 1, 0], [2, 0, 0], [3, 0, 2], [2, 1, 1], [0, 0, 2]]
Max = [[7, 5, 3], [3, 2, 2], [9, 0, 2], [2, 2, 2], [4, 3, 3]]
Available = [3, 3, 2]
Which of the following sequence is not a safe sequence according to the Banker’s algorithm ?
- A.
P1 → P3 → P4 → P0 → P2
- B.
P3 → P1 → P4 → P0 → P2
- C.
P4 → P3 → P1 → P0 → P2
- D.
P1 → P4 → P3 → P2 → P0
Attempted by 23 students.
Show answer & explanation
Correct answer: C
First, compute the Need matrix by subtracting Allocation from Max for each process.
P4 requires [4, 3, 1], but Available resources are only [3, 3, 2]. P4 needs more Resource A than available.
Since P4 cannot execute first, any sequence starting with it is unsafe. Option C matches this condition.