Banker's Algorithm in Operating Systems

Banker's Algorithm in Operating Systems

The Banker's Algorithm is a resource allocation algorithm designed to avoid deadlock, proposed by Edsger Dijkstra. It determines whether to allocate resources by simulating resource allocation and checking if the system remains in a safe state. Below, I will explain its core concepts, algorithm steps, and a specific example in detail.

1. Core Concepts

  • Available Resources (Available): The current quantity of each type of resource available in the system.
  • Maximum Demand (Max): The maximum quantity of each type of resource required by each process.
  • Allocated Resources (Allocation): The quantity of each type of resource currently allocated to each process.
  • Need (Need): The quantity of each type of resource still needed by each process (Need = Max - Allocation).
  • Safe State: A state where there exists a sequence of process execution (a safe sequence) such that the system can allocate resources according to this sequence and complete all processes without causing deadlock.

2. Steps of the Banker's Algorithm

Step 1: Initialization

Define resource vectors and process matrices. For example:

  • Resource types: A, B, C (e.g., printers, memory blocks, etc.).
  • Number of processes: P0, P1, P2, P3, P4.
  • Available resource vector Available = [3, 3, 2] (indicating 3, 3, and 2 units remaining for resource types A, B, and C, respectively).

Step 2: Checking Resource Requests

When a process requests resources, the algorithm follows this order of judgment:

  1. Is the request valid?: The requested amount must not exceed the process's declared maximum demand (Need).
  2. Are resources sufficient?: The requested amount must not exceed the currently available resources (Available).
  3. Tentative allocation: Assume the resources are allocated and update the state:
    • Available = Available - Request (reduce available resources).
    • Allocation = Allocation + Request (increase allocated resources).
    • Need = Need - Request (reduce needed resources).
  4. Perform safety check: Determine whether the system remains in a safe state after the tentative allocation.

Step 3: Safety Check Algorithm

  1. Initialize two vectors:
    • Work = Available (a copy of the current available resources).
    • Finish = [false, false, ...] (marks whether a process has finished).
  2. Loop to find a process that satisfies the following conditions:
    • Finish[i] = false (the process has not finished).
    • Need[i] ≤ Work (the resources needed by the process are ≤ the current available resources).
  3. If such a process is found:
    • Work = Work + Allocation[i] (release the resources occupied by the process).
    • Finish[i] = true (mark the process as finished).
    • Repeat Step 2 until all processes are finished (safe) or no process meets the conditions (unsafe).

Step 4: Decision

  • If the safety check passes (a safe sequence exists), allocate the resources.
  • If it fails, deny the request, and the process must wait.

3. Specific Example

Assume the system has 3 resource types (A, B, C) and 5 processes (P0-P4), with the initial state as follows:

Process Allocation (A,B,C) Max (A,B,C) Available (A,B,C)
P0 0,1,0 7,5,3 3,3,2
P1 2,0,0 3,2,2
P2 3,0,2 9,0,2
P3 2,1,1 2,2,2
P4 0,0,2 4,3,3

Calculate the Need matrix (Max - Allocation):

  • P0: (7,5,3) - (0,1,0) = (7,4,3)
  • P1: (3,2,2) - (2,0,0) = (1,2,2)
  • P2: (9,0,2) - (3,0,2) = (6,0,0)
  • P3: (2,2,2) - (2,1,1) = (0,1,1)
  • P4: (4,3,3) - (0,0,2) = (4,3,1)

Safety check process:

  1. Work = Available = (3,3,2), Finish = [false, false, false, false, false].
  2. Find processes where Need ≤ Work:
    • P1: Need=(1,2,2) ≤ Work=(3,3,2) → satisfied. Update Work=(3,3,2)+(2,0,0)=(5,3,2), Finish[1]=true.
    • P3: Need=(0,1,1) ≤ Work=(5,3,2) → satisfied. Update Work=(5,3,2)+(2,1,1)=(7,4,3), Finish[3]=true.
    • P4: Need=(4,3,1) ≤ Work=(7,4,3) → satisfied. Update Work=(7,4,3)+(0,0,2)=(7,4,5), Finish[4]=true.
    • P0: Need=(7,4,3) ≤ Work=(7,4,5) → satisfied. Update Work=(7,4,5)+(0,1,0)=(7,5,5), Finish[0]=true.
    • P2: Need=(6,0,0) ≤ Work=(7,5,5) → satisfied. Update Work=(7,5,5)+(3,0,2)=(10,5,7), Finish[2]=true.
  3. All Finish values are true, and the safe sequence is [P1, P3, P4, P0, P2] (not unique).

4. Key Points Summary

  • Advantages: Avoids deadlock and ensures system safety.
  • Disadvantages: Requires prior knowledge of maximum resource demands, and the computational overhead is relatively high (suitable for scenarios with few resource types).
  • Practical Applications: Primarily used in teaching or theoretical analysis; deadlock detection and recovery strategies are more commonly employed in real-world systems.