Aspire Faculty ID #11727 · Topic: CUET 2024 · Just now
CUET 2024

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.

Solution

Banker's Algorithm Solution

Step 1: Need Matrix

The Need Matrix is calculated as:

Need = Max - Allocation

ProcessR1R2R3
P1452
P2101
P3600
P4010

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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