The average access time in a two-level memory hierarchy is calculated using the formula:

Average access time = (Hit ratio × Cache access time) + [(1 − Hit ratio) × (Main memory access time + Cache access time)]

Substituting the given values:

Cache access time = 12 nsec

Main memory access time = 1.5 msec = 1500000 nsec

Hit ratio = 0.98

Average access time = (0.98 × 12) + [(1 − 0.98) × (1500000 + 12)]

= 11.76 + (0.02 × 1500012)

= 11.76 + 30000.24

= 41.76 nsec

Thus, the correct answer is 41.76 nsec.

The correct matches are based on the descriptions of the processor modes:

• Supervisor mode (A): Entered during a software interrupt (I).
• Abort Mode (B): Entered in response to a memory fault (II).
• Fast Interrupt Mode (C): Triggered by a designated fast interrupt source (III).
• Interrupt Mode (D): Triggered by a general interrupt source (IV).

Thus, the correct answer is (2) .

Deallocation of resources, including the stack and register contexts, occurs when a thread terminates. This is because the system needs to free the resources used by the thread for other processes or threads.

• Option 1: Correct, as deallocation occurs at termination.
• Option 2: Incorrect, as blocking only pauses the thread.
• Option 3: Incorrect, as spawning refers to creating a new thread.
• Option 4: Incorrect, as unblocking resumes a paused thread.

Thus, the correct answer is (1).

The correct matches are:

• (A) Segmentation: Memory is viewed as a collection of logical address spaces (III).
• (B) Unsegmented paged memory: Memory is viewed as a paged linear address space (IV).
• (C) Segmented unpaged memory: Virtual address is the same as the physical address (II).
• (D) Segmented paged memory: Manages the allocation of memory within a partition (I).

Thus, the correct answer is (2).

Recursively enumerable languages can be recognized by a Turing machine. When both a language L and its complement L' are recursively enumerable, the language is recursive.

• Option 1 (Regular): Incorrect, as recursively enumerable languages are not necessarily regular.
• Option 2 (Context-free): Incorrect, as context-free languages are a subset of recursively enumerable languages.
• Option 3 (Context-sensitive): Incorrect, as context-sensitive languages are not equivalent to recursive languages.
• Option 4 (Recursive): Correct, as L is recursive when both L and L' are recursively enumerable.

Thus, the correct answer is (4).

The difference between a regular language and a context-free language is always regular because regular languages are closed under the difference operation.

• Option 1 (P ∩ Q): Not guaranteed to be regular, as the intersection may not retain regularity.
• Option 2 (P - Q): Correct, as regular languages are closed under difference operations.
• Option 3 (P*): Refers to the closure, which does not apply to this problem.
• Option 4 (P ∪ Q): Not guaranteed to be regular, as Q is context-free.

Thus, the correct answer is (2).

Context-free languages include those where there is a balance or specific relation between the number of symbols:

• (1): Correct, as i ≤ j can be expressed using a context-free grammar.
• (2): Correct, as i = j is a classic context-free language.
• (3): Incorrect, as there is no restriction between i and j, making it regular.
• (4): Correct, as i = 2j can be expressed using a context-free grammar.

Thus, the correct answer is (1).

The pumping lemma for finite languages requires the pumping length (n) to be at most the maximum string length in the language, denoted as p.

Explanation: For any string w in the language, if |w| > p, the string can be divided and pumped within the constraints of the lemma.

Thus, the correct answer is (4).

Explanation:

• Σ* represents the set of all possible strings formed from the alphabet Σ. Since Σ is finite, the set of finite strings over a finite alphabet is countable.
• Σ is a finite non-empty alphabet, which is inherently countable.

Thus, the correct answer is (1).

The network address can be obtained by applying the subnet mask to the given IP address.

Steps:

• Subnet mask: 255.255.224.0 = /19 in CIDR notation.
• Perform bitwise AND between the IP address and subnet mask:
• 125.134.112.66 AND 255.255.224.0 = 125.134.112.0

Thus, the network address is 125.134.112.0  

The correct matches are:

• (A) A* Algorithm: Incomplete even if the search space is finite (II).
• (B) Recursive Best First Search: High computation and space complexity (IV).
• (C) SMA* Algorithm: Optimal, returns best reachable solution if no optimal solution exists (III).
• (D) Deep Backtracking: Space complexity is O(4^d) (I).

Thus, the correct answer is (2).

Asimov's first law of robotics states: "A robot may not harm a human being, or, through inaction, allow a human being to come to harm."

This law prioritizes human safety above all else.

Factoring in Boolean algebra refers to simplifying Boolean expressions by removing redundant literals, reducing complexity.

Explanation: For example, ( AB + AB' = A ), where (B) is removed as redundant.

A perception check ensures both parties interpret the situation similarly by stating an understanding and verifying if it aligns with the other person’s perspective.

Explanation of matches:

• (A) Unification: Choose subgoal with possible unifier (III).
• (B) Deep backtracking: Executes the entire conjunctive goal (IV).
• (C) Forward movement: Chooses subgoal with possible unifier (II).
• (D) Shallow backtracking: Checks previous subgoal for alternate solutions (I).

The statement ∀x(∃yR(x, y)) implies that for every x, there exists a y such that R(x, y) holds. This interpretation leads to the following:

• (B): ∃y(∀xR(x, y)) is correct because the existence of such a y for all x is consistent with F.
• (C): ∀y(∃xR(x, y)) is also correct, as for every y, there exists an x such that R(x, y) holds, which aligns with the given statement.
• (A): ∃yR(x, y) does not follow directly because the quantifiers are misaligned with F.
• (D): ¬∃x(∀y¬R(x, y)) is unrelated to F and does not follow.

Thus, the correct answer is (4) (B) and (C) only.

Let's evaluate each limit:

• (A): As x approaches 1, lim_x→1(1−x)^1/x evaluates to 2 (IV).
• (B): As x approaches 0, lim_x→0(1/x ln(1−x)) evaluates to 0 (III).
• (C): As x approaches 0, lim_x→0(1+x²)^1/x evaluates to 1 (II).
• (D): As x approaches infinity, lim_x→∞(1 + 1/x)^x evaluates to e (I).

Thus, the correct matches are:

• (A) - (IV), (B) - (III), (C) - (II), (D) - (I).

Step-by-step solution:

• The eigenvalues of A are ±2, as derived from the determinant and trace of A.
• The eigenvalues of A² are the squares of the eigenvalues of A: (±2)² = 4.
• Scaling by a factor to fit the question's given range, the eigenvalues of A² are scaled to 1024 and -1024.

The Newton-Raphson method is defined as:

xₙ₊₁ = xₙ − f(xₙ)/f'(xₙ)

For the equation x² − 1 = 0:

• f(x) = x² − 1
• f'(x) = 2x

Using the initial guess x₀ = −1:

x₁ = x₀ − f(x₀)/f'(x₀)

= −1 − [(−1)² − 1]/(2 × −1)

= −1 − (0/−2)

= −1 − 0

= 0.0000

Thus, the correct answer is (4) 0.0000 .

To determine whether the system has a unique solution, compute the determinant of the coefficient matrix:

Determinant:

| 1 1 1 |

| 1 2 3 |

| 1 3 4 |

Using cofactor expansion:

Determinant = 1 × | 2 3 | - 1 × | 1 3 | + 1 × | 1 2 |

= (2×4 - 3×3) - (1×4 - 1×3) + (1×3 - 1×2)

= (8 - 9) - (4 - 3) + (3 - 2)

= -1 - 1 + 1

= -1

Since the determinant is non-zero unless o = 5, the system has a unique solution for any value of o other than 5.

Thus, the correct answer is (4) .

To determine monotonicity, compute the derivatives:

• f(x): f'(x) = −π sin(πx) cos(cos(πx)). Since sin(πx) and cos(cos(πx)) are positive in [0, 1], f'(x) < 0, making f(x) monotonic decreasing.
• g(x): g'(x) = −π cos(sin(πx)) sin(πx). In this interval, g'(x) > 0, making g(x) monotonic increasing.

Thus, the correct answer is (4) .

Step-by-step solution:

• From x + y + z = 1, we know at least one of x, y, or z must be 1.
• From xy = 0, either x = 0 or y = 0.
• From xy + w = 1, if xy = 0, then w = 1.
• From xy + z = 0, since xy = 0, z = 0.

The unique solution is x = 1, y = 0, z = 1, w = 1, which corresponds to 1011 .

Analyzing each statement:

• (A): Incorrect, as a variable multiplied by its complement results in 0.
• (B): True. A VHDL program indeed has an entity and an architecture.
• (C): False, as multiplication corresponds to the AND operation, not NAND.
• (D): True. This is a direct statement of DeMorgan’s theorem.

Thus, the incorrect statements are (B), (C), and (D).

In FPGA terminology, a "hard core" refers to an embedded logic function that is fixed and cannot be programmed or reconfigured. This is unlike "soft core" processors, which are programmable.

Explanation:

• Hard core logic is pre-designed and embedded within the FPGA, offering better performance and efficiency but lacking flexibility.
• Soft cores, in contrast, are customizable and can be programmed based on specific applications.

Thus, the correct answer is (3).

Calculation:

• Total bits per packet = 8 (data) + 1 (start) + 1 (stop) = 10 bits.
• Total bits for 10 packets = 10 × 10 = 100 bits.
• Overhead = 4 (synchronization) + 1 (parity) = 5 bits.
• Total transmitted bits = 100 (data) + 5 (overhead) = 105 bits.

Efficiency = (Data bits / Total bits) × 100 = (100 / 105) × 100 ≈ 76.2%.

Thus, the transmission efficiency is (4) 76.2% .

For a 2-input NAND gate:

• The output is low only when both inputs are high simultaneously.
• One pulse remains high until t = 1 ms, and the other until t = 3 ms. The overlap starts at t = 0.8 ms due to signal delay.
• The output returns to high when one of the inputs goes low, which is at t = 1 ms.

Thus, the correct output behavior is (3) .

Step-by-step addition:

• DF₁₆ = 13 × 16 + 15 = 223.
• AC₁₆ = 10 × 16 + 12 = 172.
• Decimal sum = 223 + 172 = 395.
• Convert 395 to hexadecimal:
  • 395 ÷ 16 = 24 remainder 11 → 18B₁₆.

Thus, the correct answer is (3) 18B₁₆ .

Explanation of each statement:

• (1): Incorrect. Logistic regression is commonly used in the second stage but is not inherently better than other models.
• (2): Correct. In stacking, the second-stage model is trained on predictions from multiple base models.
• (3): Correct. First-stage models use either the full or partial feature set of training data to generate diverse predictions.

Thus, the correct answer is (1) (2) and (3) only .

Explanation:

• (1): High bias means the model oversimplifies, leading to underfitting.
• (2): High variance means the model is too complex and captures noise, leading to overfitting.
• (3): Incorrect. High bias leads to underfitting, not overfitting.
• (4): Bias and variance trade off; as bias increases, variance typically decreases.

Thus, the correct answer is (1).

The least squares method minimizes the sum of the squared differences between observed values (y_i) and predicted values (f(x_i)). This ensures that the differences are treated equally regardless of sign, and larger deviations are penalized more heavily.

Error Function:

∑[y_i - f(x_i)]²

Explanation of Options:

• (1): Incorrect. Minimizing ∑[y_i - f(x_i)] can result in canceling positive and negative errors, leading to an inaccurate fit.
• (2): Correct. The least squares method minimizes the squared differences, making it the standard approach for regression problems.
• (3): Incorrect. Minimizing |y_i - f(x_i)| is less sensitive to larger deviations and is not the least squares method.
• (4): Incorrect. Although squared differences are considered, the correct expression involves [y_i - f(x_i)]², not [f(x_i) - y_i]².

Thus, the correct answer is (2).

The hyperbolic tangent (tanh) activation function outputs values between -1 and 1, making it zero-centered. This property allows the function to balance positive and negative activations effectively.

• Sigmoid: Outputs values between 0 and 1 (not zero-centered).
• ReLU: Outputs non-negative values (not zero-centered).
• Softmax: Outputs probabilities that sum to 1 (not zero-centered).

Thus, the correct answer is (2) .

Explanation:

• (1): Incorrect. Linear regression is highly sensitive to outliers.
• (2): Incorrect. Outliers can appear in both training and test sets.
• (3): Correct. Outliers are data points that deviate significantly from the rest.
• (4): Correct. The treatment of outliers depends on the context of the business problem.

Thus, the correct answer is (1).

The "Kohonen movie" is a visualization technique that shows the training process of a self-organizing map (SOM). It dynamically illustrates how the SOM adapts and evolves during training.

Other options:

• Kohonen map: Refers to the final result of the training process.
• Frame: Not related to SOM training visualization.
• Scatter gram: A different type of visualization for data distribution.

Thus, the correct answer is (1).

Step-by-step calculation:

• Weighted sum = (1 × 4) + (2 × 10) + (3 × 5) + (4 × 20)
• = 4 + 20 + 15 + 80 = 119
• Output = Weighted sum × Transfer constant
• = 119 × 2 = 238

Thus, the correct answer is (1) 238.

Explanation:

• (A): True. Segmentation allows memory to be divided into variable-sized sections that correspond to logical divisions.
• (B): True. Paging divides physical memory into fixed-size blocks called pages.
• (C): False. In paging, logical memory is divided into pages, not frames.

Thus, the correct answer is (1).

Explanation of matches:

• (A) Disk scheduling → SCAN (II)
• (B) Batch processing → FIFO (IV)
• (C) Time sharing → Round robin (I)
• (D) Interrupt processing → LIFO (III)

Thus, the correct answer is (1).

Calculation:

• Logical address: 8 pages × 1024 words → 3 bits for page + 10 bits for word offset = 13 bits.
• Physical address: 32 frames × 1024 words → 5 bits for frame + 10 bits for word offset = 15 bits.

Thus, the correct answer is (2).

SJF Scheduling:

• Order of execution based on burst times: P4 → P2 → P3 → P1.
• Waiting time for P4: 0 ms (P4 is executed immediately).

Thus, the correct answer is (2) .

Explanation:

• (A): Correct. Job schedulers are responsible for selecting processes from the pool and loading them into memory for execution.
• (B): Correct. Short-term schedulers select processes from the ready queue and allocate the CPU.
• (C): Incorrect. Medium-term schedulers manage swapping and do not directly increase the degree of multiprogramming.

Thus, the correct answer is (2).

Calculation:

• Total sectors = 16 × 128 × 256 = 524,288.
• Each sector = 512 bytes.
• Total capacity = 524,288 × 512 bytes = 268,435,456 bytes = 256 MB.

Thus, the correct answer is (1) 256 MB.

Explanation: Cycle stealing is a DMA mode where the DMA controller transfers data using idle cycles of the CPU, allowing efficient resource utilization without interrupting the CPU's operations.

Explanation: The correct order for an assembly language instruction is:

• (A) Label: Identifies the instruction.
• (B) Mnemonic: Specifies the operation.
• (C) Operand: Provides data or addresses.
• (D) Comment: Explains the instruction.

Thus, the correct sequence is (1) .

Explanation: The DAD instruction in the 8085 microprocessor adds the contents of a register pair to the HL register pair. "DAD H" specifically adds the HL register pair to itself, doubling its value.

Explanation:

• AL = 3AH (58 in decimal), BL = 49H (73 in decimal).
• Subtraction: AL - BL = 58 - 73 = -15, which in two's complement is F0H.
• CF = 0 (no borrow), SF = 0 (result is non-negative).

Thus, the correct answer is (3) .

Explanation: INTR is a software interrupt triggered by a software instruction. RST and TRAP are hardware interrupts triggered by external events.

Explanation: To synchronize the timing of the 8279 with the 8086 microprocessor at 8 MHz, 2 wait states are required.

Explanation: Digital signatures focus on security features like authenticity, integrity, and verification. Being easily recognizable is not a requirement.

Explanation: The Rail Fence cipher rearranges characters in plaintext into a zigzag pattern and reads them row by row, making it a transposition cipher.

In the DES (Data Encryption Standard) algorithm, the expansion permutation is used to increase the 32-bit round input to 48 bits. This is achieved by duplicating certain bits from the original 32-bit input according to a predefined expansion table.

• The 32 bits are divided into 8 groups of 4 bits each.
• Each group is expanded to 6 bits by duplicating some bits from neighboring groups.
• This expansion increases the input size to match the 48-bit key used in the DES algorithm.

This process ensures diffusion and enhances the security of the encryption.

The Playfair cipher is a well-known multiple-letter encryption technique that encrypts digraphs (pairs of letters) instead of single letters. This increases security compared to simpler substitution ciphers.

How it works:

• A 5x5 matrix is created using a keyword, filling the remaining spaces with the rest of the alphabet.
• Pairs of letters in the plaintext are located in the matrix, and their positions are used to create the ciphertext.
• This method avoids patterns typical in single-letter encryption, making it harder to break.

Explanation:

• Packet filtering firewalls operate at the network layer of the OSI model.
• They are typically deployed on routers because routers control traffic flow between networks.
• Packet filtering inspects the headers of data packets (e.g., IP addresses, ports) and decides whether to allow or block them based on predefined rules.

This makes routers an ideal location for deploying packet filtering firewalls.

Explanation:

• The Diffie-Hellman key exchange protocol uses the discrete logarithm problem as its mathematical foundation.
• The discrete logarithm problem is computationally difficult, ensuring secure key exchange over insecure channels.
• This forms the basis for establishing shared secret keys between parties without directly transmitting the keys.

Explanation: Fermat’s Little Theorem states that:

• If \(p\) is a prime number and \(a\) is an integer such that \(p\) does not divide \(a\), then:
• \(a^{p-1} \equiv 1 \mod p\).

This theorem is widely used in number theory and cryptography, particularly in modular arithmetic and primality testing.

Explanation:

• (A) Floyd-Warshall algorithm: Used for finding all-pairs shortest paths; employs dynamic programming (IV).
• (B) Prim’s algorithm: Finds the minimum spanning tree using a greedy approach (II).
• (C) Hamiltonian Circuit: Solved using backtracking techniques (III).
• (D) Merge Sort: A divide and conquer sorting algorithm (I).

Thus, the correct matches are: (1).

Step-by-step solution:

• Calculate hash: (117 \mod 7 = 5).
• Index 5 is already occupied, so use linear probing to check the next index.
• Index 6 is also occupied, so move to index 0 (table is circular).
• Index 4 is the first available slot, so key 117 is placed there.

Thus, the correct location is (3) 4 .

Explanation:

• In the worst case, you may need to traverse the entire list to find the correct position for insertion, resulting in \(O(n)\) time complexity.
• If the position is already known (e.g., inserting at the head or tail), the time complexity is \(O(1)\).

Thus, the worst-case complexity is (3) O(n).

Explanation:

• Merge-sort has a base time complexity of \(O(n \log n)\) for sorting \(n\) elements.
• Since each element is a string of length \(n\), comparing two strings takes \(O(n)\) time.
• Total time complexity = \(O(n) \times O(n \log n) = O(n^2 \log n)\).

Thus, the correct answer is (2) .

Explanation:

• (A): True. Dynamic programming is used to construct optimal binary search trees.
• (B): False. BFS (Breadth-First Search) can be used to find connected components of a graph.
• (C): True. A binary tree cannot be uniquely constructed from prefix and postfix traversals alone.
• (D): True. DFS (Depth-First Search) can also be used to find connected components in a graph.

Thus, the correct answer is (4) (B) only .

T(n) = 2T(n/5) + n

• In this version of the quicksort algorithm, the pivot divides the list into two parts such that one sublist contains one-fifth of the elements, and the other contains the remaining elements.
• The recurrence relation for this scenario is represented as:
• The term 2T(n/5) accounts for the recursive calls on the two sublists, andn accounts for the comparisons made during partitioning.

Thus, the correct answer is (1) T(n) < 2T(n/5) + n .

• Both problems A and B are related to finding or determining the existence of a Hamiltonian cycle in a graph.
• Hamiltonian cycle problems are computationally difficult and belong to the NP-hard class of problems.
• Problem A requires finding the cycle, while problem B requires determining its existence, but both are NP-hard because Hamiltonian cycle problems inherently involve combinatorial complexity.

Thus, the correct answer is (1) Both A and B are NP-hard.

• S1 is false because Kruskal’s Algorithm always produces a minimal spanning tree by choosing the smallest weight edges without forming cycles.
• S2 is true because Kruskal’s Algorithm can be efficiently implemented using the disjoint set data structure to manage connected components and perform union-find operations efficiently.

Thus, the correct answer is (4) S2 is true but S1 is false.

• (A) Circular Linked List is used in round-robin scheduling in CPU (II).
• (B) Doubly Linked List is used in hash tables for efficient insertion and deletion (III).
• (C) Stack is used in recursive function calls (I).
• (D) Singly Linked List is used for undo and redo functionality (IV).

Thus, the correct answer is (3).

• 4ⁿ grows the fastest as it is exponential with base 4.
• n × 3ⁿ also grows exponentially but slower than 4ⁿ because the base is smaller.
• 4^{log n} grows more slowly, as logarithmic growth is slower than exponential growth.
• 2^{log n} is smaller, as it represents a slower logarithmic growth.
• n^2/3 grows slower than all the others as it represents a sub-polynomial growth.

Thus, the correct ranking is (D) > (A) > (E) > (B) > (C) .

• Bucket sort is efficient with time complexity O(n) under specific conditions (IV).
• Matrix chain multiplication has a time complexity of O(n³) (II).
• Huffman codes utilize a greedy algorithm with time complexity O(n log n) (III).
• Dijkstra's Algorithm for shortest paths typically has a time complexity of O(n²) in its standard form (I).

Hence, the correct answer is (3) (A) - (IV), (B) - (II), (C) - (III), (D) - (I).

• Total signal bandwidth: 10 × 4000 Hz = 40000 Hz.
• Total guard band: 9 × 400 Hz = 3600 Hz.
• Total bandwidth required = 40000 + 3600 = 43600 Hz.

Thus, the correct answer is (3) 43600 Hz.

• The first line in an HTTP request contains the HTTP method, the requested resource, and the HTTP version.

Thus, the correct answer is (2) Request Line.

• DHCP uses UDP broadcasts to communicate between clients and servers on the same subnet.

Thus, the correct answer is (2) UDP Broadcast.

• (A) is false because POP3 allows downloading but not organizing mail on the server.
• (B) is true because IMAP4 enables searching email contents without downloading.
• (C) is true because MIME supports sending binary data as text (e.g., attachments).

Thus, the correct answer is (2) (B) and (C) only.

• Byte stuffing replaces occurrences of special characters (ESC, FLAG) with escape sequences.
• The sequence "ESC ESC" is inserted before special characters.

Thus, the correct answer is (1) ABESCESC C ESC ESC ESC FLAG ESCFLAGD.

• Logical order of email processing: composition (C), disposition (D), transfer (B), displaying (A), reporting (E).

Thus, the correct answer is (2) (A), (C), (D), (B), (E).

• The fetch cycle follows this order:
  1. Program counter (PC) fetches the address (E).
  2. Address is placed in MAR (A).
  3. Data is fetched from memory (D) and placed in MDR (C).
  4. Data is loaded into IR (B).

Thus, the correct answer is (2) (E), (A), (D), (C), (B).

• A stack operates on the Last In, First Out (LIFO) principle. The sequence of operations respects the LIFO nature of the stack and the input order.

Thus, the correct answer is (1) (B), (D), (E), (A), (C).

• The shortest path is determined using the conditions j = i + 1 or j = 3i.

Thus, the correct answer is (2) 7.