Question: If we can generate a maximum of 4 Boolean functions using n Boolean variables, what is the minimum value of n?

Formula: Number of Boolean functions of $n$ variables is:

\[ 2^{2^n} \]

Condition: We are told the total functions must be ≤ 4:

\[ 2^{2^n} \leq 4 \]

✅ Try values of $n$:

• $n = 0$: $2^{2^0} = 2^1 = 2$ ✅
• $n = 1$: $2^{2^1} = 2^2 = 4$ ✅
• $n = 2$: $2^{2^2} = 2^4 = 16$ ❌

Minimum $n$ for which number of Boolean functions ≤ 4 is:

\[ \boxed{1} \]

✅ Final Answer: $\boxed{1}$

✅ Concept:

In modern processors, to improve performance, the CPU often overlaps the fetch, decode, and execute stages of multiple instructions. This overlapping is known as:

Instruction Pipelining

✅ Example:

• Instruction 1: Being executed
• Instruction 2: Being decoded
• Instruction 3: Being fetched

This overlapping allows multiple instructions to be processed simultaneously at different stages of the pipeline, improving throughput.

✅ Final Answer: $\boxed{\text{Instruction Pipelining}}$

✅ Concept:

The Accumulator is a special-purpose register used by the ALU to store intermediate results during arithmetic and logic operations.

✅ Explanation:

• It simplifies CPU design by reducing the number of memory accesses.
• Results of one operation are stored in the accumulator and used in the next.
• Widely used in simple or older CPU architectures.

✅ Final Answer: $\boxed{\text{Accumulator}}$

✅ Conversion (Using Powers of 2):

• 1 Terabyte (TB) = $2^{10}$ Gigabytes (GB)
• 1 Exabyte (EB) = $2^{10}$ Petabytes (PB)
• 1 Petabyte (PB) = $2^{10}$ Terabytes (TB)
• Therefore, 1 Exabyte (EB) = $2^{10} \times 2^{10} \times 2^{10} = 2^{30}$ Gigabytes (GB)

✅ Final Answer:

• 1 Terabyte (TB) = $\boxed{2^{10}}$ GB
• 1 Exabyte (EB) = $\boxed{2^{30}}$ GB

✅ Concept:

Cache memory is highly effective due to its ability to take advantage of the locality of reference. This refers to the tendency of programs to access a relatively small portion of memory repeatedly during execution.

✅ Explanation of Options:

• Memory Localization: Not the correct answer, as it's a general concept related to memory organization.
• Locality of Reference: The correct answer! This refers to the tendency of a program to repeatedly access the same memory locations, which cache memory leverages to speed up data access.
• Memory Size: Cache memory is small, and its size is actually one of the factors that makes it faster, but size alone does not determine its effectiveness.
• None of the Mentioned: This option is incorrect because "locality of reference" is the main reason cache memory is effective.

✅ Final Answer:

The most effective reason for cache memory is the $\boxed{\text{Locality of Reference}}$.

✅ Concept:

The fastest means of memory access for the CPU refers to the storage that allows the quickest retrieval of data. This is crucial for efficient processing and performance.

✅ Explanation of Options:

• Registers: Registers are the fastest form of memory access because they are directly in the CPU. However, they are very small and not used for regular memory storage.
• Cache: Cache memory is the most effective and frequently used type of memory, as it stores frequently accessed data for fast retrieval. It is still slower than registers but much faster than main memory.
• Main Memory: Main memory (RAM) is slower than both cache and registers, but it provides larger storage for currently active processes.
• Stack: The stack is used for storing function calls and local variables. While it is quick, it still doesn't surpass cache memory in terms of speed.

✅ Final Answer:

The fastest means of memory access for the CPU is $\boxed{\text{Registers}}$.

✅ Step 1: Convert Octal to Binary

Each octal digit is represented by 3 binary digits. Let's convert (2217)₈ to binary:

2₈ = 010, 2₈ = 010, 1₈ = 001, 7₈ = 111

Binary Representation: 010 010 001 111

✅ Step 2: Group Binary in 4-bit Sections

To convert binary to hexadecimal, group the binary digits in sets of 4, starting from the right:

010 010 001 111 becomes 0010 0100 0111

✅ Step 3: Convert Binary to Hexadecimal

Now convert each group of 4 bits into its hexadecimal equivalent:

• 0010 = 2
• 0100 = 4
• 0111 = 7

✅ Final Answer:

The hexadecimal equivalent of (2217)₈ is: (247)16

✅ Concept:

To fetch data from secondary memory, certain registers are used to handle the addresses and manage the data flow between secondary memory and the CPU.

✅ Explanation of Options:

• Program Counter (PC): The Program Counter holds the address of the next instruction to execute, not used for fetching data from secondary memory.
• Memory Address Register (MAR): This is the correct answer! The MAR holds the address in memory from which data will be fetched or written, including from secondary storage. It is responsible for managing the memory access.
• Memory Buffer Register (MBR): The MBR temporarily holds data being transferred to or from memory but does not directly fetch data from secondary memory.
• Accumulator: The Accumulator holds intermediate results of arithmetic or logical operations, not used for fetching data from secondary memory.

✅ Final Answer:

The register used to fetch data from secondary memory is the $\boxed{\text{Memory Address Register (MAR)}}$.

✅ Concept:

Binary multiplication is done similarly to decimal multiplication, where each bit is multiplied individually. Let's break down \( 00_2 \times 11_2 \) step by step.

✅ Step 1: Understanding Binary Multiplication

We multiply each digit in the first binary number by each digit in the second binary number. The multiplication follows the same rules as decimal multiplication but with only 0's and 1's.

✅ Step 2: Perform the Multiplication

Let's multiply the two binary numbers:

• First step: \( 0 \times 1 = 0 \)
• Second step: \( 0 \times 1 = 0 \)
• Third step: \( 0 \times 1 = 0 \)

✅ Step 3: Adding the Results

Since all the results are 0, the final multiplication result is also 0.

✅ Final Answer:

The result of \( 00_2 \times 11_2 \) is: 0

✅ Given:

The floating-point binary number is \( +1001.11_2 \).

We need to convert it into an 8-bit fraction and a 6-bit exponent format.

✅ Step 1: Normalize the Binary Number

We start by normalizing the binary number into scientific notation of the form:

\( 1.xxxx \times 2^n \)

Converting \( 1001.11_2 \) into scientific notation gives:

\( 1001.11_2 = 1.00111_2 \times 2^3 \)

The exponent is \( 3 \) (because the binary point is shifted 3 places to the left).

✅ Step 2: Convert the Exponent to Binary

The exponent is \( 3 \) in decimal. To represent this in binary using 6 bits, we get:

\( \text{Exponent} = 000100_2 \)

✅ Step 3: Convert the Fraction to 8 Bits

The fractional part of the normalized binary number is \( 00111 \). We need to extend it to 8 bits:

\( \text{Fraction} = 01001110_2 \)

✅ Final Answer:

The floating-point binary number \( +1001.11_2 \) in 8-bit fraction and 6-bit exponent format is:

Exponent: \( 000100_2 \), Fraction: \( 01001110_2 \)

✅ Given Values:

• Memory Access Time: 10 milliseconds
• Cache Hit Ratio: 15% = 0.15

✅ Formula for Effective Memory Access Time (EMAT):

\(\text{EMAT} = \text{Cache Hit Ratio} \times \text{Cache Access Time} + (1 - \text{Cache Hit Ratio}) \times (\text{Memory Access Time} + \text{Cache Access Time})\)

✅ Step 1: Plug Values into the Formula:

\(\text{EMAT} = 0.15 \times 10 + (1 - 0.15) \times (10 + 10)\)

\(\text{EMAT} = 0.15 \times 10 + 0.85 \times 20\)

\(\text{EMAT} = 1.5 + 17\)

\(\text{EMAT} = 18.5 \, \text{milliseconds}\)

✅ Final Answer:

EMAT = 18.5 milliseconds

✅ Given Information:

An 8-bit number in 2's complement form can represent values from \(-2^{n-1}\) to \(2^{n-1} - 1\).

✅ Step 1: Calculate the smallest integer:

Smallest value = \(-2^{8-1} = -2^7 = -128\)

✅ Final Answer:

The smallest integer that can be represented by an 8-bit number in 2's complement form is: -128

✅ Given Gates:

• I. NAND
• II. NOR

✅ NAND Gate:

The NAND gate is functionally complete. It can be used to construct any logic gate (AND, OR, NOT, etc.) by combining multiple NAND gates.

✅ NOR Gate:

The NOR gate is also functionally complete. It can be used to construct any logic gate (AND, OR, NOT, etc.) by combining NOR gates.

✅ Final Answer:

Both NAND and NOR gates are functionally complete.

✅ Given Information:

We are asked to find the total number of binary functions that can be defined using n Boolean variables.

✅ Step 1: Number of input combinations:

For n Boolean variables, the number of possible input combinations is 2ⁿ.

✅ Step 2: Number of possible outputs:

Each input combination can map to either 0 or 1, so there are 2 possible outputs for each of the 2ⁿ input combinations.

✅ Final Formula:

Total Binary Functions = 2⁽²ⁿ⁾

✅ Final Answer:

The total number of binary functions that can be defined using n Boolean variables is: 2⁽²ⁿ⁾

✅ Memory Unit Communication with CPU:

The memory unit which directly communicates with the CPU is known as:

✅ Primary Memory:

Primary memory, also called Main Memory, includes components like RAM (Random Access Memory) and Cache Memory.

✅ Why Primary Memory?

Primary memory is directly accessible by the CPU, allowing it to fetch data and instructions quickly for processing.

✅ Final Answer:

The memory unit that directly communicates with the CPU is: Primary Memory or Main Memory

 Maximum & Minimum in 16-bit 2's Complement

 Total Bits: 16

Format: 1 sign bit + 15 magnitude bits

• Maximum (positive): 0111 1111 1111 1111(2) = +32,767
• Minimum (negative): 1000 0000 0000 0000(2) = −32,768

✅ Final Answer:
Minimum = −32,768
Maximum = +32,767

Decimal to Binary Conversion

? Given: Convert (25.375)10 to binary.

? Step-by-step:

• Integer Part (25):
  Divide by 2 repeatedly:

  25 ÷ 2 = 12 remainder 1  
  12 ÷ 2 = 6  remainder 0  
  6  ÷ 2 = 3  remainder 0  
  3  ÷ 2 = 1  remainder 1  
  1  ÷ 2 = 0  remainder 1  
        

  → Binary: 11001
• Fractional Part (0.375):
  Multiply by 2 repeatedly:

  0.375 × 2 = 0.75   → 0  
  0.75 × 2  = 1.5    → 1  
  0.5 × 2   = 1.0    → 1  
        

  → Binary: .011

✅ Final Binary Answer: (25.375)10 = (11001.011)2

10-bit 2's Complement Representation of –35

Format: 10-bit signed integer using 2's complement representation.

Step-by-Step:

1. First, convert 35 to 10-bit binary: 0000100011
2. Find 1's complement: 1111011100
3. Add 1 (to get 2's complement): 1111011101

✅ Final Answer: 1111011101

 –35 in 10-bit 2's complement: 1111011101

Staircase Bulb Switch Logic

Question:
A staircase bulb is controlled by two switches — one at each floor. Each switch can independently turn the bulb ON or OFF, regardless of the other switch's position. What logic gate does this resemble?

✅ Correct Answer: Exclusive OR (XOR) Gate

Explanation:

In XOR logic, the output is ON (1) when inputs are different, and OFF (0) when inputs are the same. Similarly, the bulb glows when the switches are in different states and turns OFF when both are in the same state.

Logic Used: XOR Gate

10-bit Floating Point: +∞ Representation

Format: The 10-bit floating point is structured as follows:

• 1 bit for sign (S)
• 5 bits for exponent (E)
• 4 bits for fraction/mantissa (F)

IEEE Rule for +∞:

In IEEE floating-point format, +∞ is represented as:

• Sign bit (S): 0
• Exponent bits (E): all 1s → 11111
• Fraction bits (F): all 0s → 0000

✅ Final 10-bit Representation: 0111110000

Data Flow Management

Question:
What is the name of the storage device that compensates the difference in rates of flow of data from one device to another?

✅ Correct Answer: Buffer

Explanation:

A Buffer is a temporary storage area that helps in matching the speed of data transfer between a fast and a slow device, ensuring smooth data flow without loss.

Final Answer: Buffer

✅ Given Information:

The CPU generates 32-bit virtual addresses with a 4 KB page size. The TLB holds 128 page table entries and is 4-way set associative.

✅ Step 1: Virtual Address Breakdown:

We split the 32-bit virtual address into:

• Page Offset: 12 bits (because 4 KB = 2¹²)
• Page Number: 20 bits (32 - 12)

✅ Step 2: TLB Breakdown:

The TLB has 128 entries, and with 4-way set associativity, there are:

128 / 4 = 32 sets, so we need 5 bits for the index.

✅ Step 3: Tag Calculation:

The total number of bits for the virtual address is 32:

• Page Offset: 12 bits
• Index: 5 bits (because we have 32 sets)
• Tag: Remaining bits = 32 - 12 - 5 = 15 bits

✅ Final Answer:

The minimum size of the TLB tag is: 15 bits

✅ Number of Minterms in a n-variable Truth Table:

For n Boolean variables, the number of possible combinations of inputs is:

2ⁿ

Each combination corresponds to one minterm. Therefore, the total number of minterms in the truth table is also 2ⁿ.

✅ Final Answer:

The number of minterms in a n-variable truth table is: 2ⁿ

Register Responsible for Next Instruction

Question:
Which of the following registers is used to keep track of the address of the memory location where the next instruction is located?

✅ Correct Answer: Program Counter (PC)

Explanation:

The Program Counter (PC) holds the address of the next instruction to be fetched from memory and executed by the CPU. After fetching the current instruction, it automatically updates to point to the next one.

Final Answer: Program Counter

Time for Fetch & Execute

Definition:
The time required to fetch and execute one machine instruction is called a Machine Cycle / CPU Cycle.

Includes:

• Instruction fetch
• Instruction decode
• Operand fetch (if any)
• Execution
• Result store (if any)

✅ Final Answer: Machine Cycle / CPU Cycle.

Quick Solution

Each RAM chip: 32K × 1 = 32K bits = 4 KB

Total required: 256 KB

Number of chips needed: 256 ÷ 4 = 64

✅ Final Answer: 64 chips

Binary Division

Question: What is the quotient when 11010111 is divided by 101 in binary?

Step 1: Convert to decimal:

• 11010111 = 215
• 101 = 5

Step 2: Divide: 215 ÷ 5 = 43

Step 3: Convert 43 to binary = 101011

✅ Final Answer: 101011

Data Transfer Interface in Computer Systems

Question: Which interface transfers data between memory and I/O peripheral without CPU involvement?

✅ Correct Answer: Direct Memory Access (DMA)

Why DMA?

• The CPU only initiates the process.
• The DMA controller directly transfers data between I/O and memory.
• Frees the CPU for other tasks, improving efficiency.

❌ Incorrect Options (if any): Programmed I/O, Interrupt-driven I/O – both require CPU involvement.

Computer Architecture: Role of Cache Memory

Question: Cache memory functions as an intermediary between?

✅ Correct Answer: CPU and Main Memory (RAM)

Explanation:

• CPU is very fast, but RAM is slower in comparison.
• Cache memory holds frequently accessed data closer to the CPU.
• This reduces data access time and improves overall system performance.

Hierarchy: CPU ↔ Cache ↔ Main Memory (RAM) ↔ Secondary Storage

TLB Tag Size Calculation

Given:

• Virtual Address = 32 bits
• Page Size = 4 KB = \(2^{12}\) → Offset = 12 bits
• VPN = 32 − 12 = 20 bits
• TLB entries = 128, 4-way set associative → 32 sets

Set index bits: \( \log_2(32) = 5 \text{ bits} \)

TLB Tag = VPN − Set Index = 20 − 5 = 15 bits

✅ Final Answer: 15 bits

Boolean Simplification

Given Expression: P + QR

We ask: This is the simplified (reduced) form of which expression?

Try expanding: (P + Q)(P + R)

Using distributive law:

(P + Q)(P + R) = P(P + R) + Q(P + R) = P + PR + PQ + QR = P + PQ + PR + QR = P + QR (since P absorbs PQ and PR)

✅ Final Answer: (P + Q)(P + R)

Computer Architecture: Purpose of Cache Memory

Question: What is the primary purpose of cache memory in a computer system?

✅ Correct Answer: To increase the speed of data access by storing frequently used data closer to the CPU

Explanation:

• Cache memory is much faster than main memory (RAM).
• It holds frequently accessed data and instructions.
• Helps reduce CPU waiting time and improves performance.

Hierarchy: CPU → Cache → RAM → Storage

Detailed Register Allocation Analysis

Objective: Determine the minimum number of registers needed to execute the program without spilling.

Live Range Analysis:

• a: Line 1 → 4
• b: Line 6 → 7
• c: Line 3 → 6
• d: Line 8 → 9
• e: Line 7 → 8
• f: Line 5 → 9

Max live variables: 3 (after lines 6 and 7)

✅ Final Answer: 3 registers are required to execute the program without spilling.

Basic Computer Knowledge: Smallest Unit of Data

Question: Which of the following is the smallest unit of data in a computer?

✅ Correct Answer: Bit (Binary Digit)

Explanation:

• A bit is the most fundamental unit of data in computing.
• It represents a binary value: 0 or 1.
• All other units (Byte, Kilobyte, etc.) are multiples of bits.

Common Units of Data:

• 1 Bit = Smallest unit
• 1 Byte = 8 Bits
• 1 Kilobyte (KB) = 1024 Bytes
• 1 Megabyte (MB) = 1024 KB

? Note: Every piece of data in a computer—from text and images to video and sound—ultimately breaks down into bits.

2's Complement Addition (4-bit)

Given: 1101 and 0100 (in 2’s complement)

Step-by-step:

• 1101 = −3 (in decimal)
• 0100 = +4 (in decimal)
• Sum = −3 + 4 = +1
• +1 in 4-bit 2’s complement = 0001

✅ Final Answer: 0001

Memory Chip Calculation

Given:

• RAM chip size = 4K × 8-bit = 4096 bytes = 4 KB
• Required memory = 64 KB = 65536 bytes

Calculation:

Total chips needed = 65536 / 4096 = 16 chips

✅ Final Answer: 16 RAM chips are needed to build 64 KB of memory.

2's Complement Addition (8-bit)

Given:

• A = 11111111 → (−1)
• B = 11111111 → (−1)

Sum: −1 + (−1) = −2

Convert −2 to 8-bit 2's complement:

• +2 = 00000010
• Invert = 11111101
• Add 1 = 11111110

✅ Final Answer: 11111110

Custom Number System: Position of 10A

Given digits: 0, 1, A (base-3)

Convert 10A to base-10:

• 1 → 1
• 0 → 0
• A → 2
• 10A = 1×9 + 0×3 + 2 = 11

List of numbers in sequence:

1. 00
2. 01
3. 0A
4. 10
5. 11
6. 1A
7. A0
8. A1
9. AA
10. 100
11. 101
12. 10A

✅ Final Answer: 12

Instruction Formation in CPU Processing

Question: Which of the following is not a part of instruction formation?

Options:

• Opcode
• Register file
• Source operand
• Destination operand

✅ Correct Answer: Register File

Explanation:

• Opcode, Source operand, Destination operand — all are part of the instruction format.
• Register File — a hardware structure that stores registers, but it is not encoded into the instruction.

✅ IEEE 754 Double Precision Format:

The **exponent** field in the IEEE 754 double precision (Binary 64) format is 11 bits long and uses a **bias of 1023**.

✅ Step-by-Step Explanation:

• The exponent field has **11 bits**, which means it can represent values between **0 and 2047**.
• The exponent is stored in **biased form**: the stored value is the actual exponent + 1023.
• For **normalized numbers**, the exponent range is:
• Minimum exponent: \( 1 - 1023 = -1022 \)
• Maximum exponent: \( 2046 - 1023 = +1023 \)
• Special cases:
• **Exponent = 0**: Used for **denormalized numbers**.
• **Exponent = 2047**: Represents **infinity** or **NaN** (Not a Number).

✅ Final Answer:

The range of the exponent E in the IEEE 754 double precision (Binary 64) format is:

From -1022 to +1023 (for normalized numbers)

Binary to Decimal: 2's Complement Conversion

Given binary numbers:

• 11001 (5-bit)
• 1001 (4-bit)
• 111001 (6-bit)

Step-by-step (2's complement):

• Each starts with 1 → negative number
• Convert by inverting and adding 1
• All result in binary 0111 → decimal 7
• So final value = −7

✅ Final Answer: Each binary number corresponds to the decimal number −7.

✅ Finding the Memory Size:

The last memory location "FFFF" is a hexadecimal number.

Each hexadecimal digit represents 4 bits, so:

• "FFFF" is a 4-digit hexadecimal number, meaning it represents 16 bits or 2 bytes.
• "FFFF" in decimal is equivalent to 65535.
• Therefore, the total memory size is 65536 bytes or 64 KB.

✅ Final Answer:

The memory size is: 64 KB