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.
What is the primary goal of the Banker’s Algorithm ?
To allocate resources optimally
Deadlocks can be handled by prevention, avoidance, or detection and recovery; the Banker’s algorithm (Dijkstra) is the classic avoidance method.
It lets processes request resources freely but grants a request only if the resulting state is safe, so resources are handed out in the best possible way that never leads to deadlock — i.e. resources are allocated optimally under the safety constraint.
It does not stop processes from requesting resources; it decides when a request can be met.
Detecting a deadlock after it occurs and recovering from it is a different strategy (wait-for graph or detection algorithm); the Banker’s algorithm makes sure the deadlock never arises.
Maximising CPU utilisation is a goal of CPU scheduling, not of resource allocation.
To allocate resources optimally option is correct.
Similar Questions
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 ?
Which of the following is NOT a requirement for Banker’s algorithm to grant a resource request ?
What information is used to determine if a resource request can be granted ?
Important Questions from Deadlock Handling - Teaching
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
