Phrases Previous Year Questions (PYQs) – 2021 – Page 1 of 12

✅ 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: $\boxed{{\displaystyle \text{Instruction Pipelining}}}$

✅ 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: $\boxed{{\displaystyle \text{Accumulator}}}$

✅ 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) = $\boxed{{\displaystyle 2^{10}}}$ GB
• 1 Exabyte (EB) = $\boxed{{\displaystyle 2^{30}}}$ 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 $\boxed{{\displaystyle \text{Locality of Reference}}}$.

✅ 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 $\boxed{{\displaystyle \text{Registers}}}$.

✅ 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 $\boxed{{\displaystyle \text{Memory Address Register (MAR)}}}$.

✅ 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

$\text{For n}=10000={10}^4$ 

$$Time=\frac{\text{No of tasks}}{\text{Speed of computer}}$$ 

$$Time=\frac{2n^2}{{10}^6}$$ 

$$=2\times {10}^2$$ 

Programmed I/O: In program-controlled I/O, the processor program controls the complete data transfer. So only when an I/O transfer instruction is executed, the transfer could take place. It is required to check that device is ready/not for the data transfer in most cases.  Usually, the transfer is to & from a CPU register & peripheral. Here, CPU constantly monitors the peripheral. Here, until the I/O unit indicates that it is ready for transfer, the CPU wait & stays in a loop. It is time-consuming as it keeps the CPU busy needlessly.