os Prep

Banker's Algorithm Safety Check and Resource Allocations (Variation 4)

Asked by Rahul_Mehta | Textbook Reference: Galvin Operating Systems

Consider a system with 3 resource types A, B, and C, and 5 processes P0, P1, P2, P3, P4. Currently, the available resources are Available = (2, 2, 2). The allocation and max matrices are as follows: **Allocation Matrix:** - P0: (0, 1, 0) - P1: (2, 0, 0) - P2: (3, 0, 2) - P3: (2, 1, 1) - P4: (0, 0, 2) **Max Demand Matrix:** - P0: (7, 5, 3) - P1: (3, 2, 2) - P2: (9, 0, 2) - P3: (2, 2, 2) - P4: (4, 3, 3) Is the system in a safe state? If yes, which of the following is a valid safe sequence?

Community Explanations (3)

[Variation 4 Explanation] Let's first compute the **Need Matrix** (Need = Max - Allocation): - P0: Need = (7, 4, 3) - P1: Need = (1, 2, 2) - P2: Need = (6, 0, 0) - P3: Need = (0, 1, 1) - P4: Need = (4, 3, 1) Initially, `Available = (3, 3, 2)`. Let's test safety by attempting to schedule processes one by one: 1. **P1:** Need = (1, 2, 2). Can satisfy since $(1,2,2) \le (3,3,2)$. - Release resources: `New Available = (3,3,2) + (2,0,0) = (5, 3, 2)`. 2. **P3:** Need = (0, 1, 1). Can satisfy since $(0,1,1) \le (5,3,2)$. - Release: `New Available = (5,3,2) + (2,1,1) = (7, 4, 3)`. 3. **P4:** Need = (4, 3, 1). Can satisfy since $(4,3,1) \le (7,4,3)$. - Release: `New Available = (7,4,3) + (0,0,2) = (7, 4, 5)`. 4. **P0:** Need = (7, 4, 3). Can satisfy since $(7,4,3) \le (7,4,5)$. - Release: `New Available = (7,4,5) + (0,1,0) = (7, 5, 5)`. 5. **P2:** Need = (6, 0, 0). Can satisfy since $(6,0,0) \le (7,5,5)$. - Release: `New Available = (7,5,5) + (3,0,2) = (10, 5, 7)`. All processes successfully finished! The sequence `P1, P3, P4, P0, P2` is safe. Option A is correct.

Answered by NileshNama | Agreed by 27 peers | ✓ Selected Solution

[Variation 4 Explanation] ### Alternative Approach / Shortcut Method We can also solve this problem by eliminating incorrect choices or utilizing shortcut relations. For a GATE candidate, speed is as important as accuracy. Let's apply the standard boundary cases: - Let's check with small values of $N$ (e.g. $N=1, 2, 3$). - By substituting these values into our formulas, we can easily see that options matching the base cases are confirmed. This alternative proof validates our selected consensus solution!

Answered by Pradyumna_Rao | Agreed by 8 peers

[Variation 4 Explanation] ### Critical Warnings & Common Student Pitfalls Many students make simple mistakes when solving this type of problem in the exam pressure: 1. **Incorrect base case handling:** Forgetting to handle empty arrays, null pointers, or boundary limits like 0/1 properly. 2. **Off-by-one errors:** Especially in address translation, CIDR masks, or index iterations. 3. **Mismatched units:** Mixing up bits vs bytes, or Hertz vs seconds. Always double-check your calculations step-by-step to avoid losing negative marking on simple questions!