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Consider a system with 5 processes P0 through P4 and three resource types A, B and C. The following snapshot of the system has been taken 


 Allocation Max Available

A B C A B C 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


  1. Find the need matrix.
  2. Is the system safe? Explain.

Banker’s Algorithm: Need Matrix and Safe State Check (BTech CSE, Operating Systems)

Given

Processes: P0–P4; Resource types: A, B, C

Available (A, B, C) = (3, 3, 2)

Allocation and Max (from the snapshot):

Process   Allocation (A B C)   Max (A B C)
P0        (0 1 0)              (7 5 3)
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)

1) Need Matrix (Need = Max − Allocation)

Process   Need (A B C)
P0        (7 4 3)    = (7 5 3) − (0 1 0)
P1        (1 2 2)    = (3 2 2) − (2 0 0)
P2        (6 0 0)    = (9 0 2) − (3 0 2)
P3        (0 1 1)    = (2 2 2) − (2 1 1)
P4        (4 3 1)    = (4 3 3) − (0 0 2)

2) Safety Check using Banker’s Algorithm

Initial Work = Available = (3, 3, 2)

  1. Find a process whose Need ≤ Work:
    • P1 Need (1,2,2) ≤ (3,3,2) → safe to run
    After P1 finishes, Work = Work + Allocation[P1] = (3,3,2) + (2,0,0) = (5,3,2)
  2. With Work = (5,3,2), choose:
    • P3 Need (0,1,1) ≤ (5,3,2) → run P3
    After P3, Work = (5,3,2) + (2,1,1) = (7,4,3)
  3. With Work = (7,4,3), choose:
    • P0 Need (7,4,3) ≤ (7,4,3) → run P0
    After P0, Work = (7,4,3) + (0,1,0) = (7,5,3)
  4. With Work = (7,5,3), choose:
    • P2 Need (6,0,0) ≤ (7,5,3) → run P2
    After P2, Work = (7,5,3) + (3,0,2) = (10,5,5)
  5. With Work = (10,5,5), choose:
    • P4 Need (4,3,1) ≤ (10,5,5) → run P4
    After P4, Work = (10,5,5) + (0,0,2) = (10,5,7)

Conclusion

  • Need Matrix: 
    P0 (7,4,3), P1 (1,2,2), P2 (6,0,0), P3 (0,1,1), P4 (4,3,1)
  • Safe Sequence found: 
    P1 → P3 → P0 → P2 → P4
  • Therefore, the system is in a safe state because every process can finish in the above order without causing deadlock.