Suppose we have a byte-addressable computer with a cache that holds 16 blocks of 4 bytes each. Assuming address has 8 bits, to which cache set would the hexadecimal address OxBE map if the computer mapping? that each memory uses 2-way set associative

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**Cache Memory Mapping Problem**

**Problem Statement:**

Suppose we have a byte-addressable computer with a cache that holds 16 blocks of 4 bytes each. Assuming that each memory address has 8 bits, to which cache set would the hexadecimal address 0xBE map if the computer uses 2-way set associative mapping?

**Detailed Explanation:**

1. **Memory Address Bits:**
   - The memory address size is 8 bits.

2. **Cache Organization:**
   - The cache holds 16 blocks.
   - Each block holds 4 bytes.
   - Cache is organized as 2-way set associative.

3. **Block Calculation:**
   - Number of sets in the cache = (Number of cache blocks) / (Associativity).
   - For a 2-way set associative cache with 16 blocks: Number of sets = 16 / 2 = 8 sets.
   - We need 3 bits to address the 8 sets (since 2^3 = 8).

4. **Address Breakdown:**
   - For a set-associative cache, the address is broken down into three main parts: Tag, Index (Set), and Block Offset.
   - Block Offset: With each block holding 4 bytes (2^2), 2 bits are needed for the block offset.
   - Index: With 8 sets, 3 bits are needed for the set index.
   - Tag: The remaining bits after accounting for the block offset and index will represent the tag.

   Therefore, for an 8-bit address:
   - Block Offset (2 bits).
   - Index (3 bits).
   - Tag (remaining bits, i.e., 3 bits).

5. **Hexadecimal to Binary Conversion:**
   - Hexadecimal address 0xBE needs to be converted to binary.
   - Hexadecimal B = 1011, E = 1110, hence 0xBE = 10111110 in binary.

6. **Binary Address Breakdown:**
   - Binary: 10111110
   - Tag = First 3 bits: 101
   - Index = Next 3 bits: 111
   - Block Offset = Last 2 bits: 10

   The index part (111 in binary) gives the set number in the cache.

7. **Set Calculation:**
   - Converting binary 111 to decimal gives 7.

**Conclusion:**
Transcribed Image Text:**Cache Memory Mapping Problem** **Problem Statement:** Suppose we have a byte-addressable computer with a cache that holds 16 blocks of 4 bytes each. Assuming that each memory address has 8 bits, to which cache set would the hexadecimal address 0xBE map if the computer uses 2-way set associative mapping? **Detailed Explanation:** 1. **Memory Address Bits:** - The memory address size is 8 bits. 2. **Cache Organization:** - The cache holds 16 blocks. - Each block holds 4 bytes. - Cache is organized as 2-way set associative. 3. **Block Calculation:** - Number of sets in the cache = (Number of cache blocks) / (Associativity). - For a 2-way set associative cache with 16 blocks: Number of sets = 16 / 2 = 8 sets. - We need 3 bits to address the 8 sets (since 2^3 = 8). 4. **Address Breakdown:** - For a set-associative cache, the address is broken down into three main parts: Tag, Index (Set), and Block Offset. - Block Offset: With each block holding 4 bytes (2^2), 2 bits are needed for the block offset. - Index: With 8 sets, 3 bits are needed for the set index. - Tag: The remaining bits after accounting for the block offset and index will represent the tag. Therefore, for an 8-bit address: - Block Offset (2 bits). - Index (3 bits). - Tag (remaining bits, i.e., 3 bits). 5. **Hexadecimal to Binary Conversion:** - Hexadecimal address 0xBE needs to be converted to binary. - Hexadecimal B = 1011, E = 1110, hence 0xBE = 10111110 in binary. 6. **Binary Address Breakdown:** - Binary: 10111110 - Tag = First 3 bits: 101 - Index = Next 3 bits: 111 - Block Offset = Last 2 bits: 10 The index part (111 in binary) gives the set number in the cache. 7. **Set Calculation:** - Converting binary 111 to decimal gives 7. **Conclusion:**
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