📝 Single Instance of Each Resource Type Deadlock Detection (43 MCQs)

📖 From Operating System • 7. Deadlocks • 43 questions available

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What is Single Instance of Each Resource Type Deadlock Detection?

Definition:
For single-instance resources, deadlock detection reduces to finding cycles in the Resource Allocation Graph using depth-first search or similar traversal.

Example:
DFS on graph $G$ detects back edge $(u, v)$ indicating cycle $v \to ... \to u \to v$ , confirming deadlock among involved processes.

Reason:
Cycle detection in single-instance RAGs is efficient ( $O(n + m)$ ) and provides exact deadlock identification without complex matrix computations.

Easy

Medium

Hard

📝 All Single Instance of Each Resource Type Deadlock Detection MCQs

Q1. For which type of resource system is the Wait-For Graph algorithm specifically designed?

A.Systems with multiple instances of each resource.

B.Systems with only one instance of each resource type. ✅

C.Systems with no resource sharing.

D.Systems with dynamic resource creation.

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The text explicitly states that if all resources have only a single instance, we can define a deadlock detection algorithm using a wait-for graph. This simplification relies on the one-to-one mapping between resources and holders.

Q2. What is a Wait-For Graph derived from?

A.The Process Control Block.

B.The Resource-Allocation Graph. ✅

C.The Memory Map.

D.The File System Table.

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: A Wait-For Graph is obtained directly from the Resource-Allocation Graph (RAG) by removing resource nodes and collapsing the edges that passed through them.

Q3. What happens to resource nodes when creating a Wait-For Graph?

A.They are highlighted.

B.They are duplicated.

C.They are removed. ✅

D.They are converted into process nodes.

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: The defining step in creating a WFG from a RAG is the removal of all resource nodes. Only process nodes remain in the WFG.

Q4. In a Wait-For Graph, what do the vertices represent?

A.Resources

B.Processes ✅

C.Both Resources and Processes

D.Memory Addresses

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Since resource nodes are removed, the only remaining vertices in a Wait-For Graph represent the processes in the system.

Q5. What does an edge from Pi to Pj in a Wait-For Graph signify?

A.Pi is holding a resource needed by Pj.

B.Pi is waiting for Pj to release a resource. ✅

C.Pj is waiting for Pi to release a resource.

D.Pi and Pj are competing for the same free resource.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: An edge $P_i \to P_j$ implies that process $P_i$ is blocked, waiting for a resource that is currently held by process $P_j$ . $P_i$ cannot proceed until $P_j$ releases that resource.

Q6. How is an edge Pi → Pj formed in the Wait-For Graph?

A.If Pi requests any resource.

B.If Pj releases any resource.

C.If the RAG contains edges Pi → Rq and Rq → Pj for some resource Rq. ✅

D.If Pi and Pj are in the same priority queue.

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The edge formation is a result of 'collapsing' the resource node. If $P_i$ is waiting for resource $R_q$ ( $P_i \to R_q$ ) and $P_j$ holds $R_q$ ( $R_q \to P_j$ ), then $P_i$ is effectively waiting for $P_j$ , creating the direct edge $P_i \to P_j$ in the WFG.

Q7. What is the necessary and sufficient condition for deadlock in a single-instance system using a WFG?

A.The graph is disconnected.

B.The graph contains a cycle. ✅

C.The graph has no edges.

D.The graph is complete.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: For single-instance resources, a deadlock exists if and only if there is a cycle in the Wait-For Graph. The cycle represents a circular chain of processes waiting for each other.

Q8. If a Wait-For Graph has no cycles, what can be concluded?

A.A deadlock has occurred.

B.The system is in an unsafe state.

C.No deadlock currently exists. ✅

D.All resources are free.

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The absence of cycles in a WFG for single-instance resources guarantees that there is no circular wait. Therefore, no deadlock currently exists in the system.

Q9. What is the time complexity of the cycle detection algorithm for a WFG?

A.O(n)

B.O(n log n)

C.O(n²) ✅

D.O(2^n)

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The text states that an algorithm to detect a cycle in the graph requires an order of $n^2$ operations, where $n$ is the number of vertices (processes) in the graph.

Q10. What does 'n' represent in the O(n²) complexity formula for cycle detection?

A.The number of resources.

B.The number of vertices (processes) in the graph. ✅

C.The number of edges.

D.The total memory size.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: In the context of the Wait-For Graph, $n$ refers to the number of vertices. Since resource nodes are removed, the vertices correspond to the processes in the system.

Q11. Why is the Wait-For Graph simpler than the Resource-Allocation Graph for detection?

A.It has more nodes.

B.It removes resource nodes, reducing graph size. ✅

C.It uses colors to indicate status.

D.It only shows allocated resources.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: By eliminating resource nodes, the WFG reduces the total number of nodes and edges. This simplification makes cycle detection algorithms more efficient and easier to implement for single-instance systems.

Q12. If Process P1 waits for Resource R1, and Process P2 holds Resource R1, what edge exists in the WFG?

A.P1 → P2 ✅

B.P2 → P1

C.P1 → R1

D.R1 → P2

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Following the collapse rule: $P_1 \to R_1$ (request) and $R_1 \to P_2$ (assignment) in RAG becomes $P_1 \to P_2$ in WFG. This indicates P1 is waiting for P2.

Q13. Can a Wait-For Graph contain resource nodes?

A.Yes, always.

B.Yes, if requested.

C.No, never. ✅

D.Only for critical resources.

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: By definition, a Wait-For Graph is constructed by removing all resource nodes from the Resource-Allocation Graph. It contains only process nodes.

Q14. What action must the system take to use the WFG for detection?

A.Periodically invoke a cycle-search algorithm. ✅

B.Continuously allocate new resources.

C.Terminate all waiting processes.

D.Ignore the graph structure.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The system needs to maintain the WFG and periodically run an algorithm to search for cycles. This periodic check determines if a deadlock has formed since the last check.

Q15. If a cycle exists in the WFG, what does it imply about the processes involved?

A.They are all finished.

B.They are all waiting for each other in a circular fashion. ✅

C.They are running efficiently.

D.They are not using any resources.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: A cycle $P_1 \to P_2 \to ... \to P_n \to P_1$ means $P_1$ waits for $P_2$ , $P_2$ waits for $P_3$ , ..., and $P_n$ waits for $P_1$ . None can proceed because each is waiting for the next to release a resource.

Q16. Is the presence of a cycle in a WFG sufficient to prove deadlock in single-instance systems?

A.Yes ✅

B.No

C.Only if there are 3 or more processes

D.Only if resources are scarce

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: For single-instance resources, a cycle is both necessary and sufficient for deadlock. Unlike multiple-instance systems, there are no alternative instances to break the wait, so the cycle guarantees a deadlock.

Q17. What information is lost when converting RAG to WFG?

A.Which specific resource is being waited for. ✅

B.Which process is waiting.

C.Which process is holding.

D.The existence of a wait.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The WFG abstracts away the specific resource identity. An edge $P_i \to P_j$ tells you $P_i$ waits for $P_j$ , but not *which* resource causes the wait. The RAG retains this detail.

Q18. If P1 → R1 and R1 → P2, and P2 → R2 and R2 → P1, what exists in the WFG?

A.Two separate edges P1→P2 and P2→P1

B.A cycle P1→P2→P1 ✅

C.No edges

D.A single edge P1→P1

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: $P_1 \to R_1 \to P_2$ becomes $P_1 \to P_2$ . $P_2 \to R_2 \to P_1$ becomes $P_2 \to P_1$ . Together, they form a cycle $P_1 \to P_2 \to P_1$ , indicating a deadlock between P1 and P2.

Q19. Why is the algorithm called 'Wait-For' Graph?

A.Because it waits for user input.

B.Because edges represent waiting relationships between processes. ✅

C.Because it waits for resources to become free.

D.Because it is slower than other graphs.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The name reflects the semantic meaning of the edges: an edge from A to B means A is *waiting for* B to release a resource.

Q20. What is the primary advantage of using WFG over RAG for single-instance detection?

A.It provides more detailed resource info.

B.It simplifies the graph structure for faster cycle detection. ✅

C.It allows multiple instances.

D.It prevents deadlocks.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Removing resource nodes reduces the graph's complexity (fewer nodes/edges). This makes the $O(n^2)$ cycle detection faster and simpler to implement compared to searching for cycles in the larger, bipartite RAG.

Q21. If a system has 10 processes, what is the maximum number of vertices in its WFG?

A.5

B.10 ✅

C.20

D.Depends on resources

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The WFG contains only process nodes. If there are 10 processes, there are exactly 10 vertices in the WFG, regardless of the number of resources.

Q22. Can a self-loop (Pi → Pi) exist in a valid WFG?

A.Yes, if Pi requests a resource it already holds.

B.No, because a process cannot wait for itself. ✅

C.Yes, if there is only one resource.

D.No, because resources are single instance.

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: In standard modeling, a process does not wait for itself. It either holds the resource or requests it from another holder. A self-loop would imply a logical error or a specific resource type not covered by simple single-instance models (like reentrant locks handled differently). In the context of this basic algorithm, edges are between distinct processes.

Q23. What triggers the need to update the Wait-For Graph?

A.Every CPU cycle.

B.Every resource request or release. ✅

C.Every minute.

D.Only when a deadlock is detected.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The WFG must reflect the current state. Therefore, it must be updated whenever a process requests a resource (adding potential edges) or releases a resource (removing potential edges).

Q24. If P1 waits for R1, R1 is free. Is there an edge from P1 in the WFG?

A.Yes

B.No ✅

C.Only if P1 is old

D.Only if P1 is new

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: An edge in WFG exists only if $P_i$ waits for a resource held by $P_j$ . If R1 is free, P1 is not waiting for any *process*. It might be granted the resource immediately. No process-to-process wait exists, so no edge originates from P1 for this resource.

Q25. What is the relationship between RAG edges and WFG edges?

A.One-to-one

B.Many-to-one ✅

C.One-to-many

D.No relationship

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Multiple RAG paths involving different resources could theoretically contribute to WFG structure, but specifically, a pair of RAG edges ( $P_i \to R_q, R_q \to P_j$ ) collapses into a single WFG edge ( $P_i \to P_j$ ). Thus, it simplifies many components into one edge.

Q26. If the WFG is empty (no edges), what is the state?

A.Deadlocked

B.All processes are waiting

C.No process is waiting for another process ✅

D.System crash

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: No edges mean no process is waiting for any other process to release a resource. All processes are either running or waiting for free resources (which will be granted).

Q27. Does the WFG algorithm work for multiple-instance resources?

A.Yes, perfectly.

B.No, a cycle is not sufficient for deadlock. ✅

C.Yes, if modified.

D.No, it is too slow.

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: For multiple instances, a cycle in a WFG (if constructed) is necessary but not sufficient. A process might wait for a holder, but if another instance of that resource is available or becomes available from a third party, the wait can be resolved without the holder releasing. Thus, simple cycle detection fails.

Q28. What is the 'collapsing' of edges?

A.Deleting all edges.

B.Merging request and assignment edges into a single process-to-process edge. ✅

C.Combining two processes.

D.Removing cycles.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Collapsing refers to the transformation where the intermediate resource node is removed, and the incoming request edge and outgoing assignment edge are joined to form a direct edge between the requesting and holding processes.

Q29. If P1 → P2 and P2 → P3 and P3 → P1 exist in WFG, how many processes are deadlocked?

A.1

B.2

C.3 ✅

D.0

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: All three processes are part of the cycle. P1 waits for P2, P2 for P3, P3 for P1. All three are blocked indefinitely, so 3 processes are deadlocked.

Q30. What is the main cost of maintaining a WFG?

A.Hardware cost

B.Run-time overhead to update the graph on every request/release ✅

C.Network cost

D.Storage cost for resources

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The system must constantly update the graph structure as processes request and release resources. This computational effort is the run-time overhead associated with the detection scheme.

Q31. If a cycle is detected, what is the next step in the overall scheme?

A.Prevention

B.Avoidance

C.Recovery ✅

D.Ignorance

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Detection identifies the deadlock. The next phase in the 'Detection and Recovery' scheme is Recovery, which takes actions to break the cycle.

Q32. Why is O(n²) considered acceptable for this algorithm?

A.Because n is usually very large.

B.Because n (number of processes) is typically small enough for quadratic time to be fast. ✅

C.Because it is the fastest possible.

D.Because it uses parallel processing.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: In most systems, the number of concurrent processes (n) is not massive (e.g., hundreds or low thousands). An $O(n^2)$ algorithm is computationally feasible and fast enough for periodic execution or event-triggered checks.

Q33. Can a process be part of a cycle and still be running?

A.Yes

B.No ✅

C.Only if it has high priority

D.Only if it has low priority

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: If a process is part of a cycle in the WFG, it means it is waiting for another process in the cycle. Waiting implies it is blocked, not running. Therefore, all processes in a deadlock cycle are blocked.

Q34. What distinguishes the WFG from the RAG in terms of node types?

A.WFG has only process nodes; RAG has process and resource nodes. ✅

B.WFG has only resource nodes; RAG has process nodes.

C.Both have the same nodes.

D.WFG has virtual nodes.

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: This is the structural difference: RAG is bipartite (Processes and Resources), while WFG is unipartite (Processes only).

Q35. If P1 waits for R1, and P2 waits for R1, and R1 is held by P3. What are the WFG edges?

A.P1 → P3 and P2 → P3 ✅

B.P1 → P2 and P2 → P3

C.P3 → P1 and P3 → P2

D.P1 → P3 only

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: $P_1 \to R_1$ and $R_1 \to P_3$ creates $P_1 \to P_3$ . $P_2 \to R_1$ and $R_1 \to P_3$ creates $P_2 \to P_3$ . Both P1 and P2 are waiting for P3.

Q36. Is the WFG directed or undirected?

A.Undirected

B.Directed ✅

C.Both

D.Neither

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The edges represent a specific direction of waiting (Pi waits for Pj). Therefore, the graph is directed. $P_i \to P_j$ is not the same as $P_j \to P_i$ .

Q37. What happens to a WFG edge if the holding process releases the resource?

A.It remains.

B.It is reversed.

C.It is removed. ✅

D.It becomes dashed.

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: If Pj releases the resource, Pi is no longer waiting for Pj (it might get the resource or wait for someone else). The specific dependency $P_i \to P_j$ is broken, so the edge is removed from the WFG.

Q38. Can the WFG detect deadlocks in real-time?

A.Yes, if checked continuously. ✅

B.No, it is purely theoretical.

C.Only on Sundays.

D.Only for single processes.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: While typically done periodically, the algorithm *can* be invoked continuously or on every state change to provide real-time detection, though this increases overhead significantly.

Q39. What is the primary limitation of the WFG method?

A.It is too complex.

B.It only works for single-instance resources. ✅

C.It requires too much memory.

D.It is too slow.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The simple cycle-detection in WFG is valid *only* for single-instance resources. It cannot correctly identify deadlocks in multiple-instance systems without significant modification.

Q40. If a system has 5 processes and 5 resources (single instance), what is the max number of edges in WFG?

A.5

B.10

C.20 ✅

D.25

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: In a directed graph with n vertices, the maximum number of edges is $n(n-1)$ . For 5 processes, $5 \times 4 = 20$ edges. (Assuming no self-loops).

Q41. What is the benefit of 'periodic' invocation of the detection algorithm?

A.It saves memory.

B.It balances detection latency with computational overhead. ✅

C.It prevents deadlocks.

D.It simplifies the graph.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Running detection continuously is expensive. Running it rarely risks long deadlock durations. Periodic invocation strikes a balance, checking often enough to resolve deadlocks reasonably quickly without consuming excessive CPU time.

Q42. If P1 → P2 exists, and P2 finishes and releases all resources, what happens to the edge?

A.It stays.

B.It is removed. ✅

C.It changes to P2 → P1.

D.It becomes null.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: If P2 finishes, it releases resources. P1 is no longer waiting for P2 (P1 will likely get the resource or move to wait for someone else). The dependency on P2 is gone, so the edge is removed.

Q43. Which statement is true about WFG cycles?

A.They indicate potential future deadlocks.

B.They indicate current deadlocks in single-instance systems. ✅

C.They indicate safe states.

D.They indicate resource abundance.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: In single-instance systems, a cycle in the WFG is a definitive sign that a deadlock *currently* exists, not just that it might happen.

🔗 Related Topics (MCQs)

📝 Deadlocks Introduction in Operating System MCQs 📝 Deadlocks System Model in Operating System MCQs 📝 Deadlock Characterization in Operating System MCQs 📝 Necessary Conditions for Deadlock Characterization MCQs 📝 Methods for Handling Deadlocks in Operating System MCQs

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