Read the below passage and answer the questions . The Banker’s Algorithm is a critical deadlock avoidance method in operating systems, designed to facilitate resource allocation without causing deadlock. It operates by maintaining information about the maximum resources. Each process may require, the current allocated resources and the available resources in the system. The algorithm checks each resource request to determine, if granting it would leave the system in a safe state, meaning that there is always a sequence in which all processes can complete their execution without getting stuck due to resource unavailability. Each process must specify its maximum demand for each resource type before it starts execution. When a process requests additional resources, the algorithm checks if granting the request will keep the system in a safe state. If so, the resources are allocated otherwise the process must wait until its request can be safely fulfilled.
Which of the following is NOT a requirement for Banker’s algorithm to grant a resource request ?
The process must be the only one requesting resources.
The resource-request part of the Banker’s algorithm checks three conditions before granting a request from process i.
Step 1 — Requesti ≤ Needi: the request must not exceed the maximum the process declared it can claim; otherwise it is an error.
Step 2 — Requesti ≤ Available: the requested resources must be available now; otherwise the process waits.
Step 3 — after a tentative allocation the safety algorithm must find a safe sequence, i.e. the system must be in a safe state after granting the request.
Nothing requires that the process be the only one requesting resources; many processes request at the same time and each request is evaluated on its own against the current state, so this is the condition that is not required.
The process must be the only one requesting resources. option is correct.
An operating system has 13 tape drives. There are three processes P1, P2 & P3. Maximum requirement of P1 is 11 tape drives, P2 is 5 tape drives and P3 is 8 tape drives. Currently, P1 is allocated 6 tape drives, P2 is allocated 3 tape drives and P3 is allocated 2 tape drives. Which of the following sequences represent a safe state ?
Consider a system with twelve magnetic tape drives and three processes \(P_{1}\), \(P_{2}\) and \(P_{3}\). Process \(P_{1}\) requires maximum ten tape drives, process \(P_{2}\) may need as many as four tape drives and \(P_{3}\) may need upto nine tape drives. Suppose that at time \(t_{1}\), process \(P_{1}\) is holding five tape drives, process \(P_{2}\) is holding two tape drives and process \(P_{3}\) is holding three tape drives. At time \(t_{1}\), system is in :
A system has four processes and five allocatable resources. The current allocation and maximum needs are as follows :
| Allocated | Maximum | Available | |
| Process A | 1 0 2 1 1 | 1 1 2 1 3 | 0 0 \(x\) 1 1 |
| Process B | 2 0 1 1 0 | 2 2 2 1 0 | |
| Process C | 1 1 0 1 0 | 2 1 3 1 0 | |
| Process D | 1 1 1 1 0 | 1 1 2 2 1 |
The smallest value of \(x\) for which the above system in safe state is __________.
Consider a system with five processes \(P_{0}\) through \(P_{4}\) and three resource types \(R_{1}\), \(R_{2}\) and \(R_{3}\). Resource type \(R_{1}\) has 10 instances, \(R_{2}\) has 5 instances and \(R_{3}\) has 7 instances. Suppose that at time \(T_{0}\), the following snapshot of the system has been taken :
Allocation
| \(R_{1}\) | \(R_{2}\) | \(R_{3}\) | |
| \(P_{0}\) | 0 | 1 | 0 |
| \(P_{1}\) | 2 | 0 | 0 |
| \(P_{2}\) | 3 | 0 | 2 |
| \(P_{3}\) | 2 | 1 | 1 |
| \(P_{4}\) | 0 | 2 | 2 |
Max
| \(R_{1}\) | \(R_{2}\) | \(R_{3}\) |
| 7 | 5 | 3 |
| 3 | 2 | 2 |
| 9 | 0 | 2 |
| 2 | 2 | 2 |
| 4 | 3 | 3 |
Available
| \(R_{1}\) | \(R_{2}\) | \(R_{3}\) |
| 3 | 3 | 2 |
Assume that now the process \(P_{1}\) requests one additional instance of type \(R_{1}\) and two instances of resource type \(R_{3}\). The state resulting after this allocation will be
Which data structure does the Banker’s Algorithm use to maintain the state of available, maximum and allocated resources ?
What is the significance of the Banker’s algorithm in terms of resource management ?
What is the primary goal of the Banker’s Algorithm ?
What information is used to determine if a resource request can be granted ?
Suppose a system has 12 magnetic tape drives and at time t 0, three processes are allotted tape drives out of their as given below:
Maximum Needs | Current Needs | |
p 0 | 10 | 5 |
p 1 | 4 | 2 |
p 2 | 9 | 2 |
At time t 0, the system is in safe state. Which of the following is safe sequence so that deadlock is avoided?
NOTE:
Current Need is currently allocated resource to a process
An operating system has 13 tape drives. There are three processes P1, P2 & P3. Maximum requirement of P1 is 11 tape drives, P2 is 5 tape drives and P3 is 8 tape drives. Currently, P1 is allocated 6 tape drives, P2 is allocated 3 tape drives and P3 is allocated 2 tape drives. Which of the following sequences represent a safe state ?
Consider a system with twelve magnetic tape drives and three processes \(P_{1}\), \(P_{2}\) and \(P_{3}\). Process \(P_{1}\) requires maximum ten tape drives, process \(P_{2}\) may need as many as four tape drives and \(P_{3}\) may need upto nine tape drives. Suppose that at time \(t_{1}\), process \(P_{1}\) is holding five tape drives, process \(P_{2}\) is holding two tape drives and process \(P_{3}\) is holding three tape drives. At time \(t_{1}\), system is in :
A system has four processes and five allocatable resources. The current allocation and maximum needs are as follows :
| Allocated | Maximum | Available | |
| Process A | 1 0 2 1 1 | 1 1 2 1 3 | 0 0 \(x\) 1 1 |
| Process B | 2 0 1 1 0 | 2 2 2 1 0 | |
| Process C | 1 1 0 1 0 | 2 1 3 1 0 | |
| Process D | 1 1 1 1 0 | 1 1 2 2 1 |
The smallest value of \(x\) for which the above system in safe state is __________.
Consider a system with five processes \(P_{0}\) through \(P_{4}\) and three resource types \(R_{1}\), \(R_{2}\) and \(R_{3}\). Resource type \(R_{1}\) has 10 instances, \(R_{2}\) has 5 instances and \(R_{3}\) has 7 instances. Suppose that at time \(T_{0}\), the following snapshot of the system has been taken :
Allocation
| \(R_{1}\) | \(R_{2}\) | \(R_{3}\) | |
| \(P_{0}\) | 0 | 1 | 0 |
| \(P_{1}\) | 2 | 0 | 0 |
| \(P_{2}\) | 3 | 0 | 2 |
| \(P_{3}\) | 2 | 1 | 1 |
| \(P_{4}\) | 0 | 2 | 2 |
Max
| \(R_{1}\) | \(R_{2}\) | \(R_{3}\) |
| 7 | 5 | 3 |
| 3 | 2 | 2 |
| 9 | 0 | 2 |
| 2 | 2 | 2 |
| 4 | 3 | 3 |
Available
| \(R_{1}\) | \(R_{2}\) | \(R_{3}\) |
| 3 | 3 | 2 |
Assume that now the process \(P_{1}\) requests one additional instance of type \(R_{1}\) and two instances of resource type \(R_{3}\). The state resulting after this allocation will be