Question

Consider a memory system with 1M bytes of main memory and 16K bytes of cache memory. Assume that the processor generates a 20-bit memory address, and the cache block size is 16 bytes. If the cache uses direct mapping, how many bits will be required to store all the tag values? [Assume memory is byte addressable, 1K = \(2^{10}\), 1M = \(2^{20}\).]

• A. \( 6 \times 2^{10} \)
• B. \( 8 \times 2^{10} \)
• C. \( 2^{12} \)
• D. \( 2^{14} \)

Hint:

To calculate the number of tag bits, break the address into three parts: the tag, index, and block offset. Use the cache size and block size to determine the index and block offset bits.

Answer 1

The Correct Option is A

Given Data:
Main memory size = 1M bytes = \(2^{20}\) bytes
Cache memory size = 16K bytes = \(2^{14}\) bytes
Cache block size = 16 bytes = \(2^4\) bytes

Step 1: Compute Number of Cache Blocks
\[ \text{Number of cache blocks} = \frac{\text{Cache memory size}}{\text{Cache block size}} = \frac{2^{14}}{2^4} = 2^{10} \] So, the number of cache blocks is \( 2^{10} \).

Step 2: Compute Block Offset Bits
Since each block has 16 bytes, we need \( \log_2(16) = 4 \) bits to address bytes within a block.
\[ \text{Block offset bits} = 4 \]
Step 3: Compute Index Bits
The index bits represent the number of cache blocks: \[ \text{Index bits} = \log_2(2^{10}) = 10 \]
Step 4: Compute Tag Bits
The total number of address bits is 20. Using the formula: \[ \text{Address bits} = \text{Tag bits} + \text{Index bits} + \text{Block offset bits} \] \[ 20 = \text{Tag bits} + 10 + 4 \] Solving for tag bits: \[ \text{Tag bits} = 20 - 10 - 4 = 6 \]
Step 5: Compute Storage for Tag Values
Since there are \( 2^{10} \) cache blocks, and each block requires 6 tag bits, the total number of bits required to store all tag values is: \[ 6 \times 2^{10} \]
Final Answer: The number of bits required to store all the tag values is \( 6 \times 2^{10} \).