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)
Understanding Deadlock Avoidance and Safe States
This problem deals with resource allocation in an operating system to prevent a situation called deadlock. A deadlock occurs when two or more processes are blocked forever, waiting for resources held by each other. One way to avoid deadlock is to ensure the system is always in a "safe state".
A system is in a safe state if there exists a sequence of processes that can complete execution without causing a deadlock. This sequence is called a safe sequence. We can use the Banker's Algorithm principle to find such a sequence.
Initial System State Analysis
The system has a total of 12 magnetic tape drives. We have three processes, p0, p1, and p2, with their maximum resource needs and currently allocated resources (Current Needs) at time t0 given as:
| Process | Maximum Needs | Current Allocation (Current Needs) |
|---|---|---|
| p0 | 10 | 5 |
| p1 | 4 | 2 |
| p2 | 9 | 2 |
First, let's calculate the total number of magnetic tape drives currently allocated:
Total Allocated = Current Allocation of p0 + Current Allocation of p1 + Current Allocation of p2
Total Allocated = $5 + 2 + 2 = 9$ magnetic tape drives.
The number of available magnetic tape drives is the total number of drives minus the total allocated drives:
Available Resources = Total Resources - Total Allocated
Available Resources = $12 - 9 = 3$ magnetic tape drives.
Next, we need to calculate the 'Need' for each process, which is the additional number of magnetic tape drives each process might request to reach its maximum need.
Need = Maximum Needs - Current Allocation
- p0 Need = $10 - 5 = 5$
- p1 Need = $4 - 2 = 2$
- p2 Need = $9 - 2 = 7$
Let's summarize the Needs and Available resources:
| Process | Need |
|---|---|
| p0 | 5 |
| p1 | 2 |
| p2 | 7 |
Initial Available Resources = 3.
Checking for Safe Sequences
A sequence of processes $\langle p_i, p_j, p_k, \dots \rangle$ is a safe sequence if, for each process $p_x$ in the sequence, the current Available resources are greater than or equal to $p_x$'s Need. If $p_x$ can be satisfied, we assume it finishes and releases its currently allocated resources, increasing the Available pool for the next process in the sequence.
Evaluating Option 1: (p0, p1, p2)
- Check p0: p0's Need is 5. Current Available is 3. Since $5 > 3$, p0 cannot be satisfied with the current available resources. This sequence is not safe.
Evaluating Option 2: (p1, p0, p2)
Let's check if processes can complete in this order:
- Check p1: p1's Need is 2. Current Available is 3. Since $2 \le 3$, p1 can be satisfied.
Assume p1 finishes. It releases its Current Allocation (2 drives).
New Available = Current Available + p1 Current Allocation = $3 + 2 = 5$. - Check p0: p0's Need is 5. Current Available (after p1 finishes) is 5. Since $5 \le 5$, p0 can be satisfied.
Assume p0 finishes. It releases its Current Allocation (5 drives).
New Available = Current Available + p0 Current Allocation = $5 + 5 = 10$. - Check p2: p2's Need is 7. Current Available (after p0 finishes) is 10. Since $7 \le 10$, p2 can be satisfied.
Assume p2 finishes. It releases its Current Allocation (2 drives).
New Available = Current Available + p2 Current Allocation = $10 + 2 = 12$.
Since all processes can complete in the sequence (p1, p0, p2), this is a safe sequence.
Evaluating Option 3: (p2, p1, p0)
- Check p2: p2's Need is 7. Current Available is 3. Since $7 > 3$, p2 cannot be satisfied with the current available resources. This sequence is not safe.
Evaluating Option 4: (p0, p2, p1)
- Check p0: p0's Need is 5. Current Available is 3. Since $5 > 3$, p0 cannot be satisfied with the current available resources. This sequence is not safe.
Conclusion: Identifying the Safe Sequence
Based on our evaluation, only the sequence (p1, p0, p2) allows all processes to complete without risking a deadlock, given the initial allocation and maximum needs. Therefore, (p1, p0, p2) is a safe sequence.
Revision Table: Deadlock Concepts
| Concept | Description |
|---|---|
| Deadlock | A situation where processes are blocked indefinitely, waiting for resources held by other blocked processes. |
| Safe State | A state in which the system can allocate resources to each process (up to its maximum need) in some order and still avoid a deadlock. |
| Safe Sequence | An order of processes in a safe state such that each process can complete its execution if granted its resources. |
| Banker's Algorithm | An algorithm used in deadlock avoidance that checks if granting a resource request leads to a safe state. |
Additional Information: Banker's Algorithm Details
The Banker's Algorithm is a classic deadlock avoidance algorithm. It's named this way because it's similar to how a banker would manage loans to clients. The algorithm checks if allowing a process to request resources keeps the system in a safe state.
For the Banker's Algorithm to work, processes must declare their maximum resource needs beforehand. The algorithm requires information about:
- Total resources in the system.
- Resources currently allocated to each process.
- Maximum resources needed by each process.
- Resources currently available.
By keeping track of this information, the algorithm can determine if a resource request is 'safe' to grant, preventing the system from entering an unsafe state that could lead to deadlock.
Important Questions from Deadlock Handling - Teaching
A computer has six tapes drives with n processes competing for them. Each process may need two drives. What is the maximum value of n for the system to be deadlock free?
Suppose a system has 12 instances of some resources with n processes competing for that resource. Each process may require 4 instances of the resource. The maximum value of n for which the system never enters into deadlock is
Consider a system with five processes P 0, through P 4, and three resource types A, B and C. Resource type A has seven instances, resource type B has two instances and resource type C has six instances suppose at time T 0we Have the following allocation.
Process
Allocation
Request
Available
A
B
C
A
B
C
A
B
C
P 0
0
1
0
0
0
0
0
0
0
P 1
2
0
0
2
0
2
P 2
3
0
3
0
0
0
P 3
2
1
1
1
0
0
P 4
0
2
2
0
0
2
If we implement Deadlock detection algorithm we claim that system is ______.
