Deadlock Avoidance

Banker's Algorithm

A deadlock avoidance algorithm simulating resource allocation safety in multi-process systems.

What is Banker's Algorithm?

The Banker's Algorithm is a classic resource allocation and deadlock avoidance algorithm. Designed by Edsger Dijkstra, it tests for safety by simulating the allocation of predetermined maximum possible amounts of all resources. It then performs an 's-state' safety check to ensure that the system can never fall into an unsafe state where deadlock is guaranteed.

Key Characteristics:

  • Prevents deadlock by dynamically checking system states.
  • Maintains matrices for Allocation, Max Needs, and Available resources.
  • Computes processes' outstanding Needs dynamically.
  • Guarantees that resource allocation only occurs if a safe sequence exists.

Complexity Analysis

Avg Time Complexity: O(R * P^2) where R is resources count and P is processes count
Space Complexity: O(R * P) matrix storage allocations
Best Case: O(R * P) (resources immediately satisfy processes in direct order)
Worst Case: O(R * P^2) (exhaustive search checking all potential permutations)

How it Works Step-by-Step

  1. Calculate Need Matrix: Compute Need[i][j] = Max[i][j] - Allocation[i][j] for all i, j.
  2. Initialize Status: Set Work = Available, and Finish[i] = False for all processes.
  3. Find Candidate Process: Find process i such that Finish[i] == False and Need[i] <= Work.
  4. Allocate & Reclaim: If found, set Work = Work + Allocation[i], set Finish[i] = True, and go back to step 3.
  5. State Evaluation: If all processes are finished (Finish[i] == True for all i), the system is in a Safe State. Otherwise, it is Unsafe.

Code Implementation

Worked Trace Example

With 3 processes (P0, P1, P2) and 1 resource type (Available = 3):
- Allocations: P0=1, P1=2, P2=0. Max requirements: P0=2, P1=4, P2=3.
- Needs: P0=1, P1=2, P2=3.
1. Check P0: Need (1) <= Work (3)? Yes. Work becomes 3 + 1 = 4. Finish[P0] = True.
2. Check P1: Need (2) <= Work (4)? Yes. Work becomes 4 + 2 = 6. Finish[P1] = True.
3. Check P2: Need (3) <= Work (6)? Yes. Work becomes 6 + 0 = 6. Finish[P2] = True.
Result: Safe sequence P0 -> P1 -> P2.

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