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Study Stuff for MCA Examinations - NIMCET
Study Stuff for MCA Examinations - NIMCET
Previous Year Question (PYQs)
The current allocation and Maximum requirement of different types of resources for four processes are given below:
Consider the following four statements.
(A) P2 → P4 → P1 → P3 is a safe sequence
(B) P4 → P2 → P1 → P3 is a safe sequence
(C) P4 → P2 → P3 → P1 is a safe sequence
(D) P1 → P4 → P2 → P3 is a safe sequence
Identify correct statements from the given options.
(A) P2 → P4 → P1 → P3 is a safe sequence
(B) P4 → P2 → P1 → P3 is a safe sequence
(C) P4 → P2 → P3 → P1 is a safe sequence
(D) P1 → P4 → P2 → P3 is a safe sequence
Identify correct statements from the given options.
Solution
Banker's Algorithm Solution
Step 1: Need Matrix
The Need Matrix is calculated as:
Need = Max - Allocation
| Process | R1 | R2 | R3 |
|---|---|---|---|
| P1 | 4 | 5 | 2 |
| P2 | 1 | 0 | 1 |
| P3 | 6 | 0 | 0 |
| P4 | 0 | 1 | 0 |
Step 2: Initial Available Resources
The Available resources at the start are:
- R1 = 4
- R2 = 4
- R3 = 5
Step 3: Evaluate Each Sequence
Sequence (A): P2 → P4 → P1 → P3
- P2: Need = [1, 0, 1], Available = [4, 4, 5]. Allocate resources. New Available = [7, 5, 7].
- P4: Need = [0, 1, 0], Available = [7, 5, 7]. Allocate resources. New Available = [10, 7, 9].
- P1: Need = [4, 5, 2], Available = [10, 7, 9]. Allocate resources. New Available = [14, 8, 11].
- P3: Need = [6, 0, 0], Available = [14, 8, 11]. Allocate resources. New Available = [17, 12, 14].
Result: Sequence (A) is valid.
Sequence (B): P2 → P1 → P3 → P4
- P2: Need = [1, 0, 1], Available = [4, 4, 5]. Allocate resources. New Available = [7, 5, 7].
- P1: Need = [4, 5, 2], Available = [7, 5, 7]. Allocate resources. New Available = [11, 6, 9].
- P3: Need = [6, 0, 0], Available = [11, 6, 9]. Allocate resources. New Available = [14, 10, 12].
- P4: Need = [0, 1, 0], Available = [14, 10, 12]. Allocate resources. New Available = [17, 12, 14].
Result: Sequence (B) is valid.
Sequence (C): P4 → P2 → P3 → P1
- P4: Need = [0, 1, 0], Available = [4, 4, 5]. Allocate resources. New Available = [7, 6, 7].
- P2: Need = [1, 0, 1], Available = [7, 6, 7]. Allocate resources. New Available = [10, 7, 9].
- P3: Need = [6, 0, 0], Available = [10, 7, 9]. Allocate resources. New Available = [14, 8, 11].
- P1: Need = [4, 5, 2], Available = [14, 8, 11]. Allocate resources. New Available = [17, 12, 14].
Result: Sequence (C) is valid.
Sequence (D): P1 → P4 → P2 → P3
- P1: Need = [4, 5, 2], Available = [4, 4, 5]. Cannot proceed as Need > Available.
Result: Sequence (D) is invalid.
Final Answer:
Safe Sequences: (A), (B), (C)
Invalid Sequence: (D)
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