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
(p 1, p 0, p 2)
Step 1 — available and remaining resources:
Available drives = 12 − (5 + 2 + 2) = 3. Remaining needs of p₀, p₁ and p₂ are 5, 2 and 7 respectively.
Step 2 — safety test: p₁ can finish with 3 available and release its 2 allocated drives, raising availability to 5. Then p₀ can finish and release 5, raising availability to 10; finally p₂ can finish.
Thus (p₁, p₀, p₂) is a safe sequence.
Important Questions from Deadlock Handling - Teaching
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 ?
