Question

A 32-bit word computer system employs a 1 Gbyte main memory and a 2 Mbyte cache. Determine the number of bits in each field of the Memory Address Register (MAR) as seen by the cache in the following organizations (show your calculations) for the following problems:

1. Fully Associative Mapping with Line Size of 8 words.

2. Direct Mapping with Line Size of 8 words.

3. 2-Way Associative Mapping with Line Size of 4 words.

Answer 1

To calculate the MAR fields for the given cache organizations, we must determine the number of lines or sets of cache and subtract the size required for the offset field from the total address length. The Fully Associative cache has 27 bits for the tag. Direct Mapping needs 16 bits for line, 3 bits for offset, and 13 bits for tag. 2-Way Associative Mapping has 15 bits for set, 2 bits for offset, and 15 bits for the tag.

Step-by-step explanation:

To address the question about a 32-bit word computer system with a 1 Gbyte main memory and a 2 Mbyte cache, we must calculate the number of bits in each field of the Memory Address Register (MAR) for different cache organizations. The unit for one million bytes is a megabyte (MB), and one gigabyte (GB) consists of 1024 MB. Now, we need to calculate the number of bits for MAR based on the cache organization and line size.

• Fully Associative Mapping with Line Size of 8 words:

Since the cache is fully associative and has a line size of 8 words (each word being 32 bits), the cache will have (2 MB / (8 words * 4 bytes/word)) = 65,536 lines. As we are considering the cache as seen by the cache, no bits are required for the line or set field. Therefore, the tag field will have (32 - log2(8 words * 4 bytes/word)) = 32 - log2(32) = 32 - 5 = 27 bits.

• Direct Mapping with Line Size of 8 words:

For direct mapping, the line field needs to address all possible lines in the cache. Since there are (2 MB / (8 words * 4 bytes/word)) = 65,536 lines, the line field will have log2(65,536) = 16 bits. With 8 words per line, the offset field will have log2(8) = 3 bits. Subtracting these from the 32-bit address gives 32 - 16 - 3 = 13 bits for the tag field.

• 2-Way Associative Mapping with Line Size of 4 words:

For 2-way associative mapping, the number of sets will be half the number of lines in direct mapping. With 4 words per line, we'll have (2 MB / (4 words * 4 bytes/word)) / 2 (ways) = 32,768 sets, thus needing log2(32,768) = 15 bits for the set field. Each line is log2(4) = 2 bits. Therefore, the tag field requires 32 - 15 - 2 = 15 bits.