Computer Networking: A Top-Down Approach (7th Edition)
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN: 9780133594140
Author: James Kurose, Keith Ross
Publisher: PEARSON
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"The main memory capacity is 256M bytes. A 2-way set associative cache contains 64kBytes and has a block size of 32Bytes. Assume the main memory is byte-addressable. Work out the address field structure and draw the diagram of the cache organization."

**Explanation:**

To work out the address field structure, follow these steps:

1. **Calculate Total Address Bits:**
   - Main memory is 256M bytes.
   - 256M bytes = \(2^{28}\) bytes since \(256 = 2^8\) and \(M = 2^{20}\).
   - Therefore, the address bus width needs to address \(2^{28}\) addresses, which requires 28 bits.

2. **Calculate Block Offset Bits:**
   - Block size is 32 bytes.
   - 32 bytes = \(2^5\) bytes.
   - Hence, the block offset requires 5 bits.

3. **Calculate Number of Sets:**
   - The cache is 64K bytes, with each block being 32 bytes.
   - 64K bytes = \(2^{16}\) bytes, and 32 bytes = \(2^5\) bytes.
   - Total number of blocks = \(2^{16}\) / \(2^5\) = \(2^{11}\).
   - It is a 2-way set associative cache, so:
     - Number of sets = Total blocks / Associativity = \(2^{11}\) / 2 = \(2^{10}\), which is 1024 sets.
   - The set index requires 10 bits.

4. **Calculate Tag Bits:**
   - Total address bits = Block offset bits + Set index bits + Tag bits
   - 28 = 5 + 10 + Tag bits
   - Tag bits = 28 - 15 = 13 bits

Therefore, the address field structure will be:
- Tag: 13 bits
- Set index: 10 bits
- Block offset: 5 bits

**Cache Organization Diagram:**

The cache diagram will have the following segments:

1. **Tag Array:**
   - Each entry in the tag array will be 13 bits wide.
   - There will be 1024 sets, each set will have 2 tags (since it is 2-way).

2. **Data Array:**
   - Each block in the data array will be
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Transcribed Image Text:"The main memory capacity is 256M bytes. A 2-way set associative cache contains 64kBytes and has a block size of 32Bytes. Assume the main memory is byte-addressable. Work out the address field structure and draw the diagram of the cache organization." **Explanation:** To work out the address field structure, follow these steps: 1. **Calculate Total Address Bits:** - Main memory is 256M bytes. - 256M bytes = \(2^{28}\) bytes since \(256 = 2^8\) and \(M = 2^{20}\). - Therefore, the address bus width needs to address \(2^{28}\) addresses, which requires 28 bits. 2. **Calculate Block Offset Bits:** - Block size is 32 bytes. - 32 bytes = \(2^5\) bytes. - Hence, the block offset requires 5 bits. 3. **Calculate Number of Sets:** - The cache is 64K bytes, with each block being 32 bytes. - 64K bytes = \(2^{16}\) bytes, and 32 bytes = \(2^5\) bytes. - Total number of blocks = \(2^{16}\) / \(2^5\) = \(2^{11}\). - It is a 2-way set associative cache, so: - Number of sets = Total blocks / Associativity = \(2^{11}\) / 2 = \(2^{10}\), which is 1024 sets. - The set index requires 10 bits. 4. **Calculate Tag Bits:** - Total address bits = Block offset bits + Set index bits + Tag bits - 28 = 5 + 10 + Tag bits - Tag bits = 28 - 15 = 13 bits Therefore, the address field structure will be: - Tag: 13 bits - Set index: 10 bits - Block offset: 5 bits **Cache Organization Diagram:** The cache diagram will have the following segments: 1. **Tag Array:** - Each entry in the tag array will be 13 bits wide. - There will be 1024 sets, each set will have 2 tags (since it is 2-way). 2. **Data Array:** - Each block in the data array will be
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