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 Max Available 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\(R_{1}\) \(R_{2}\) \(R_{3}\) \(P_{0}\) 0 1 0 \(P_{1}\) 2 0 0 \(P_{2}\) 3 0 2 \(P_{3}\) 2 1 1 \(P_{4}\) 0 2 2 \(R_{1}\) \(R_{2}\) \(R_{3}\) 7 5 3 3 2 2 9 0 2 2 2 2 4 3 3 \(R_{1}\) \(R_{2}\) \(R_{3}\) 3 3 2
Safe state
Apply the Banker's algorithm. Need = Max - Allocation gives \(P_{0}(7,4,3), P_{1}(1,2,2), P_{2}(6,0,0), P_{3}(0,1,1), P_{4}(4,3,1)\).
The request \((1,0,2)\) is within Need and Available, so after granting it: Available \(= (2,3,0)\), \(P_{1}\) Allocation \(= (3,0,2)\), Need \(= (0,2,0)\).
\(P_{1}\) can finish: Available \(= (2,3,0) + (3,0,2) = (5,3,2)\).
\(P_{3}\) finishes: \((5,3,2) + (2,1,1) = (7,4,3)\); then \(P_{4}\): \((7,4,3) + (0,2,2) = (7,4,5)\).
\(P_{0}\) finishes: \((7,4,5) + (0,1,0) = (7,5,5)\); then \(P_{2}\): \((7,5,5) + (3,0,2) = (10,5,7)\).
Check: the final Available \((10,5,7)\) equals the total instances, and the sequence \(\langle P_{1}, P_{3}, P_{4}, P_{0}, P_{2} \rangle\) completes every process.
Hence, Safe state.
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 __________.
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 is the primary goal of the Banker’s Algorithm ?
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 __________.
Which data structure does the Banker’s Algorithm use to maintain the state of available, maximum and allocated resources ?
