Deadlocks Practice Questions

Statement A: Banker's algorithm is a method for deadlock avoidance, not prevention. Avoidance involves ensuring the system remains in a safe state before granting a resource request, while prevention involves breaking one of the four necessary deadlock conditions from the outset. This statement incorrectly classifies the algorithm, so it is FALSE.

Statement B: Breaking the 'Hold and Wait' condition is a valid deadlock prevention strategy. You can enforce this by either making a process request all its resources at once, or forcing it to release all held resources before requesting a new one. This statement is TRUE.

Statement C: In a Resource Allocation Graph (RAG), an assignment edge represents a resource allocated to a process. The edge is directed from the resource to the process ($R \rightarrow P$). A request edge is directed from a process to a resource ($P \rightarrow R$). The statement incorrectly describes the direction of the assignment edge, so it is FALSE.

Statement D: A safe state guarantees that there is at least one sequence (a safe sequence) in which all processes can complete their execution without leading to a deadlock. This is the fundamental definition of a safe state, used by algorithms like Banker's. This statement is TRUE.

To ensure a system is deadlock-free for $n=5$ processes, each requiring a maximum of $m=2$ instances of resource $R$, the total resources $k$ must satisfy the condition $k > \sum_{i=1}^{n} (m_i - 1)$.
In the worst-case scenario for deadlock, each process holds $1$ instance ($2-1=1$) and waits for the second, occupying $5 \times 1 = 5$ instances total.
If $k=5$, every process holds one resource and waits for another, resulting in a deadlock.
To be deadlock-free, we need at least one extra instance to allow at least one process to complete its task and release its resources: $k = 5 + 1 = 6$.
Thus, the minimum value of $k$ is $6$.
[CORRECT_OPTION: 6]

To identify deadlocks, we can draw a Resource Allocation Graph (RAG) and look for cycles involving assigned and claimed resources.
Assignment edges (solid lines): $R_1 \to P_1$, $R_2 \to P_2$, $R_3 \to P_3$, $R_4 \to P_4$.
Claim edges (dashed lines): $P_1 \to R_2$, $P_2 \to R_3$, $P_2 \to R_4$, $P_3 \to R_1$, $P_4 \to R_2$.

Let's draw the graph and identify cycles:

1. Cycle 1: $P_1 \to R_2 \to P_2 \to R_3 \to P_3 \to R_1 \to P_1$. This cycle indicates a deadlock.
2. Cycle 2: $P_2 \to R_4 \to P_4 \to R_2 \to P_2$. This cycle indicates another deadlock.

Now, let's analyze the options:
(A) Aborting $P_1$: If $P_1$ is aborted, $R_1 \to P_1$ is removed and $P_1 \to R_2$ is removed.
Cycle 1: $P_1 \to R_2 \to P_2 \to R_3 \to P_3 \to R_1 \to P_1$ is broken.
Cycle 2: $P_2 \to R_4 \to P_4 \to R_2 \to P_2$ still exists. So, the system is not deadlock-free. (A) is FALSE.

(B) Aborting $P_3$: If $P_3$ is aborted, $R_3 \to P_3$ is removed and $P_3 \to R_1$ is removed.
Cycle 1: $P_1 \to R_2 \to P_2 \to R_3 \to P_3 \to R_1 \to P_1$ is broken.
Cycle 2: $P_2 \to R_4 \to P_4 \to R_2 \to P_2$ still exists. So, the system is not deadlock-free. (B) is FALSE.

(C) Aborting $P_2$: If $P_2$ is aborted, $R_2 \to P_2$ is removed, and $P_2 \to R_3$ and $P_2 \to R_4$ are removed.
Cycle 1: $P_1 \to R_2 \to P_2 \to R_3 \to P_3 \to R_1 \to P_1$ is broken as $P_2$ is removed.
Cycle 2: $P_2 \to R_4 \to P_4 \to R_2 \to P_2$ is broken as $P_2$ is removed.
All cycles are broken. So, the system becomes deadlock-free. (C) is TRUE.

(D) Aborting $P_1$ and $P_4$: If both $P_1$ and $P_4$ are aborted.
$R_1 \to P_1$ and $P_1 \to R_2$ are removed.
$R_4 \to P_4$ and $P_4 \to R_2$ are removed.
Cycle 1: $P_1 \to R_2 \to P_2 \to R_3 \to P_3 \to R_1 \to P_1$ is broken by removing $P_1$.
Cycle 2: $P_2 \to R_4 \to P_4 \to R_2 \to P_2$ is broken by removing $P_4$.
All cycles are broken. So, the system becomes deadlock-free. (D) is TRUE.

Let's analyze each statement regarding deadlocks:
(A) "Circular wait is a necessary condition for the formation of deadlock." This statement is TRUE. The four necessary conditions for deadlock are mutual exclusion, hold and wait, no preemption, and circular wait. All four must be present for a deadlock to occur.
(B) "In a system where each resource has more than one instance, a cycle in its wait-for graph indicates the presence of a deadlock." This statement is FALSE. A wait-for graph can only detect deadlocks directly if there is only one instance of each resource. If multiple instances of resources exist, a cycle in the wait-for graph indicates a potential deadlock, but not necessarily an actual deadlock. For multiple instances, a resource allocation graph is needed to accurately detect deadlocks.
(C) "If the current allocation of resources to processes leads the system to unsafe state, then deadlock will necessarily occur." This statement is FALSE. An unsafe state is a state where the system might lead to a deadlock because there is no sequence of resource allocations that allows all processes to complete. However, a deadlock is not guaranteed to occur from an unsafe state; it merely indicates that the system could enter a deadlock if processes request resources in a particular order.
(D) "In the resource-allocation graph of a system, if every edge is an assignment edge, then the system is not in deadlock state." This statement is TRUE. An assignment edge (or resource instance edge) means a resource has been allocated to a process. A request edge means a process is requesting a resource. If every edge is an assignment edge, it implies that no process is currently waiting for any resource. If no process is waiting, then there can be no circular wait, and thus no deadlock.

The statements that are TRUE are (A) and (D).
Therefore, the correct option should be (a, d).

Raymond's algorithm organizes processes into a logical tree structure where the unique token acts as the permission to enter the $CS$. The algorithm eliminates deadlock because the tree structure and token-based request propagation strictly prevent circular wait conditions. However, token movement depends on local request queues at each node; since there is no global sequence to order every concurrent request across the entire tree, request propagation can be delayed by network conditions or contention. In such distributed asynchronous environments, a specific process may potentially be bypassed, meaning that while deadlock is structurally avoided, starvation is not fundamentally guaranteed to be impossible.

A Resource Allocation Graph (RAG) is a directed graph used to model the relationship between processes and resources in an operating system. It comprises nodes representing processes ($P$) and resource instances ($R$), with edges denoting allocation ($R \rightarrow P$) or request ($P \rightarrow R$). A deadlock occurs if there is a circular wait, which corresponds to a cycle within the RAG. Detecting these cycles algorithmically allows the system to determine deadlock occurrence, making the RAG the primary aid for this purpose.

Processes $p$ and $q$ have a maximum additional need of 0 ($Y_p=Y_q=0$), so they can finish execution. Upon completion, they will release their currently held resources, $X_p$ and $X_q$. The total number of available resources in the system will then become $X_p + X_q$. For no other process $k$ to be able to complete, its maximum required additional resources, $Y_k$, must be greater than the available resources. This must hold for all other processes. Therefore, the condition is $X_p + X_q < \text{Min}\{Y_k | 1 \le k \le n, k \ne p, k \ne q\}$.

To determine if the system is in a safe state, we use the Banker's algorithm. First, we calculate the Need matrix as Max - Allocation.
Need matrix: $P_0$: (3,3,0), $P_1$: (1,0,2), $P_2$: (0,3,0), $P_3$: (3,4,1).
The available resources are (3,3,0).
The safety algorithm checks if there is a sequence of processes that can complete. A safe sequence exists, for example, <$P_0, P_2, P_1, P_3$>.

1. $P_0$'s need (3,3,0) $\le$ Available (3,3,0). $P_0$ can execute. New Available = (3,3,0) + (1,0,1) = (4,3,1).
2. $P_2$'s need (0,3,0) $\le$ Available (4,3,1). $P_2$ can execute. New Available = (4,3,1) + (1,0,3) = (5,3,4).
   This process continues for $P_1$ and $P_3$. Since a safe sequence can be found, the system is in a safe state.

To always avoid deadlock, a system must be in a safe state. A safe state is guaranteed if the sum of maximum resource needs of all processes is less than the sum of the total number of processes and total resources.
Let n be the number of processes (3), m be the number of resources (4), and K be the maximum need of each process.

The condition for avoiding deadlock is:
\sum_{i=1}^{n} \text{max_need}_i < n + m
$3 \times K < 3 + 4$
$3K < 7$
$K < \frac{7}{3} \approx 2.333$

Since K must be an integer, the largest value of K that satisfies this condition is 2.

To ensure deadlock never occurs, we must ensure that the system can always satisfy the remaining demand of at least one process. A deadlock state can potentially occur if each process $i$ holds $n_i - 1$ resources. The maximum number of resources that can be held without guaranteeing completion of any process is the sum of these values:
$\sum (n_i - 1) = (3 - 1) + (4 - 1) + (6 - 1) = 2 + 3 + 5 = 10$
If the total number of resources $m$ is greater than this sum, at least one process will always be able to acquire its remaining needed resources and complete, thus preventing deadlock. Therefore, the minimum value is $m = 10 + 1 = 11$.

This problem can be solved using the Banker's algorithm for deadlock avoidance.
First, calculate the available resources: Total drives (9) - Allocated drives (3+1+3=7) = 2 available drives.
Next, calculate the remaining need for each process (Need = Max - Allocation):

• P1 needs 7 - 3 = 4
• P2 needs 6 - 1 = 5
• P3 needs 5 - 3 = 2

With 2 available drives, only P3's need can be met. P3 runs to completion and releases its 3 allocated drives.
Available resources become 2 + 3 = 5.
Now, P1's need of 4 can be met. P1 runs and releases its 3 drives.
Available resources become 5 + 3 = 8.
Finally, P2's need of 5 can be met.
Since a safe sequence <P3, P1, P2> exists, the system is in a safe state and is not deadlocked.

Dijkstra's Banker's algorithm is a resource allocation strategy designed to ensure a system never enters an unsafe state. By simulating the allocation of resources, it checks if granting a process's request allows for a sequence of execution where all processes can finish. If a request leads to an unsafe state, the system denies it to prevent the possibility of a deadlock occurring. Unlike deadlock recovery, which fixes a system after a deadlock, this algorithm proactively manages resource distribution to avoid circular wait conditions. Consequently, it is a definitive method for deadlock avoidance.

A deadlock requires a circular wait condition, which involves multiple threads and multiple resources (locks).

• With only one thread (x=1), a deadlock cannot occur as there is no other thread to create a circular dependency with.
• With only one lock (y=1), a deadlock is also not possible. One thread will acquire the lock, and any other threads will simply wait for it to be released. No circular wait can be formed.
  The minimum condition for a deadlock is with two threads (x=2) and two locks (y=2). A deadlock occurs if Thread 1 acquires Lock 1 and waits for Lock 2, while Thread 2 acquires Lock 2 and waits for Lock 1.

Given $P = 3$ processes, each requiring a maximum of $M = 2$ resources, with a total of $R = 4$ resources available.
In the worst-case scenario, each process is allocated $M - 1 = 1$ resource.
The total resources held by these processes is $\sum_{i=1}^{3} (2 - 1) = 3$.
The remaining free resources are $R - 3 = 4 - 3 = 1$.
Since there is at least $1$ free resource, any process can request and acquire its final unit to complete its task, subsequently releasing its held resources.
Because completion is always guaranteed in this state, deadlock can never occur.

Let $R$ be a single resource instance. For a deadlock to occur, the circular wait condition must hold, which requires a cycle in the resource allocation graph involving at least two resources. With only one instance of $R$, processes can only be in a simple waiting queue; they cannot form a circular dependency cycle because only one process can hold $R$ at a time. Since a cycle is a necessary condition for deadlock, no number of competing processes can induce a deadlock on a single resource instance. Thus, none of the provided options A, B, or C are correct.

Deadlock can be prevented by breaking one of the four necessary conditions: mutual exclusion, hold and wait, no preemption, or circular wait.

• Policy I prevents the "hold and wait" condition by forcing a process to acquire all resources at once or release any it holds.
• Policies II, III, and IV all prevent the "circular wait" condition. By enforcing a strict ordering (either increasing or decreasing) on resource requests, a cycle of processes waiting on each other cannot form.
  Since all four policies effectively break at least one of the necessary conditions, any of them can be used to prevent deadlock.

For a deadlock to occur, a circular wait condition must be established where every process in a set is waiting for a resource held by another process in the set. With a maximum request of 2 resources per process, a deadlock can only happen if every process holds 1 resource and is waiting for a second one.

Let N be the number of processes and R be the number of resources (R=6). If each of the N processes holds 1 resource, N resources are allocated. For a deadlock, there must be no free resources to grant any request, meaning all 6 resources are allocated. This happens when N=6. If N < 6, there will always be at least one free resource, which can be allocated to a waiting process, breaking the potential deadlock.

To determine which requests can be granted, we apply the Banker's algorithm by checking for a safe state after each request.

Initially, Available = (3, 2, 2). We calculate the Need matrix: Need = Max - Allocation.
For P0: Need = (8, 4, 3) - (0, 0, 1) = (8, 4, 2)
For P1: Need = (6, 2, 0) - (3, 2, 0) = (3, 0, 0)
For P2: Need = (3, 3, 3) - (2, 1, 1) = (1, 2, 2)

Evaluate REQ1: P0 requests (0, 0, 2)

1. Check if REQ1 <= Available: (0, 0, 2) <= (3, 2, 2). This is true.
2. Assume REQ1 is granted:
   Available = (3, 2, 2) - (0, 0, 2) = (3, 2, 0)
Allocation[P0] = (0, 0, 1) + (0, 0, 2) = (0, 0, 3)
Need[P0] = (8, 4, 2) - (0, 0, 2) = (8, 4, 0)
3. Check for a safe sequence:
   • Only P1 can be executed since Need[P1] = (3, 0, 0) <= Available = (3, 2, 0).
   • After P1 finishes, Available = (3, 2, 0) + (3, 2, 0) = (6, 4, 0).
   • Now, Need[P0] = (8, 4, 0) and Need[P2] = (1, 2, 2). Neither P0 nor P2 can be executed because their needs exceed the current Available = (6, 4, 0).
   • Thus, no safe sequence exists, so REQ1 cannot be permitted.

Evaluate REQ2: P1 requests (2, 0, 0) (Revert to initial state for this independent request)

1. Check if REQ2 <= Available: (2, 0, 0) <= (3, 2, 2). This is true.
2. Assume REQ2 is granted:
   Available = (3, 2, 2) - (2, 0, 0) = (1, 2, 2)
Allocation[P1] = (3, 2, 0) + (2, 0, 0) = (5, 2, 0)
Need[P1] = (3, 0, 0) - (2, 0, 0) = (1, 0, 0)
3. Check for a safe sequence:
   • P1 can be executed: Need[P1] = (1, 0, 0) <= Available = (1, 2, 2).
   • After P1 finishes, Available = (1, 2, 2) + (5, 2, 0) = (6, 4, 2).
   • P2 can be executed: Need[P2] = (1, 2, 2) <= Available = (6, 4, 2).
   • After P2 finishes, Available = (6, 4, 2) + (2, 1, 1) = (8, 5, 3).
   • P0 can be executed: Need[P0] = (8, 4, 2) <= Available = (8, 5, 3).
   • A safe sequence (P1, P2, P0) exists.
   • Thus, REQ2 can be permitted.

To ensure deadlock never occurs, the system must avoid the worst-case state where every process $P_i$ is allocated $D_i - 1$ resources, leaving them all waiting for one additional resource. The total number of allocated resources in this state is calculated as $\sum_{i=1}^{3} (D_i - 1) = (3 - 1) + (4 - 1) + (5 - 1) = 2 + 3 + 4 = 9$. At this point, 9 resources are occupied, and any process's next request cannot be fulfilled, leading to potential deadlock. To guarantee that at least one process can complete its execution, we must provide at least one additional resource. Thus, the minimum number of resources required is $9 + 1 = 10$.

This is a classic resource allocation problem, often solved using the Banker's Algorithm. To prevent deadlocks, the system must always be in a safe state.

Given:

• Number of processes (programs) = 3
• Each program requires 3 tape units for its operation.

To avoid deadlock, the system needs to ensure that even if all programs hold their current allocations and then each requests one more unit, there is still one program that can complete its execution.

Let $N$ be the number of programs ($N=3$).
Let $M$ be the maximum demand for tape units by each program ($M=3$).

The minimum number of tape units to ensure a deadlock-free system is given by the formula:
$\text{Minimum Tape Units} = N \times (M - 1) + 1$
Substituting the given values:
$\text{Minimum Tape Units} = 3 \times (3 - 1) + 1$
$\text{Minimum Tape Units} = 3 \times 2 + 1$
$\text{Minimum Tape Units} = 6 + 1 = 7$
Therefore, the system must have at least 7 tape units to prevent deadlocks.

[CORRECT_OPTION: 7]