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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.

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.

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:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
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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
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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