✅ 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:

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

✅ 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

Solution:

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

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

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

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

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.

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.