Question
Write and explain the Banker’s algorithm. Consider the following snapshot of a system
| Allocation | Max | Available | ||||||||||
| A | B | C | D | A | B | C | D | A | B | C | D | |
| PO | 0 | 0 | 1 | 2 | 0 | 0 | 1 | 2 | 1 | 5 | 2 | 0 |
| P1 | 1 | 0 | 0 | 0 | 1 | 7 | 5 | 0 | ||||
| P2 | 1 | 3 | 5 | 4 | 2 | 3 | 5 | 6 | ||||
| P3 | 0 | 6 | 3 | 2 | 0 | 6 | 5 | 2 | ||||
| P4 | 0 | 0 | 1 | 4 | 0 | 6 | 5 | 6 | ||||
Answer the following questions using Banker's algorithm:
i. What is the content of the matrix need?
ii. Is the system in a safe state?
iii.If a request from P1 arrives for (0, 4, 2, 0), can the request be granted immediately?
Answer :
Word Count : 934
### i. What is the content of the matrix `Need`? The Need matrix is calculated by subtracting the Allocation matrix from the Max matrix for each process and each resource. Mathematically: \[ \text{Need} = \text{Max} - \text{Allocation} \] For each process, we calculate the `Need` for each resource (A, B, C, D): \[ \text{Need}[i][j] = \text{Max}[i][j] - \text{Allocation}[i][j] \] Here is the calculation for each process: - P0: - Need for A = Max(A) - Allocation(A) = 0 - 0 = 0 - Need for B = Max(B) - Allocation(B) = 0 - 0 = 0 - Need for C = Max(C) - Allocation(C) = 1 - 1 = 0 - Need for D = Max(D) - Allocation(D) = 2 - 2 = 0 - P1: - Need for A = Max(A) - Allocation(A) = 1 - 1 = 0 - Need for B = Max(B) - Allocation(B) = 7 - 0 = 7 - Need for C = Max(C) - Allocation(C) = 5 - 0 = 5 - Need for D = Max(D) - Allocation(D) = 0 - 0 = 0 - P2: - Need __________ ____ ____ __________ _________ __________ __________ ____ _____ _______.
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### i. What is the content of the matrix `Need`? The Need matrix is calculated by subtracting the Allocation matrix from the Max matrix for each process and each resource. Mathematically: \[ \text{Need} = \text{Max} - \text{Allocation} \] For each process, we calculate the `Need` for each resource (A, B, C, D): \[ \text{Need}[i][j] = \text{Max}[i][j] - \text{Allocation}[i][j] \] Here is the calculation for each process: - P0: - Need for A = Max(A) - Allocation(A) = 0 - 0 = 0 - Need for B = Max(B) - Allocation(B) = 0 - 0 = 0 - Need for C = Max(C) - Allocation(C) = 1 - 1 = 0 - Need for D = Max(D) - Allocation(D) = 2 - 2 = 0 - P1: - Need for A = Max(A) - Allocation(A) = 1 - 1 = 0 - Need for B = Max(B) - Allocation(B) = 7 - 0 = 7 - Need for C = Max(C) - Allocation(C) = 5 - 0 = 5 - Need for D = Max(D) - Allocation(D) = 0 - 0 = 0 - P2: - Need __________ ____ ____ __________ _________ __________ __________ ____ _____ _______.
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