Question

A direct mapped memory cache is implemented on a 16-bit architecture. the chache can hold 8 blocks of memory, each of which is 32 bytes long. how many bits are necessary in order to represent the tag of each of these blocks? explain how the cache works and how powers of two are used to parse a byte address and find the correct data stored in the cache.

Answer 1

For a 16-bit architecture with a cache that holds 8 blocks of 32 bytes each, 8 bits are needed for the cache block tags. This is calculated by subtracting the number of bits needed for the block and byte offsets from the total address length.

Step-by-step explanation:

To calculate the number of bits needed for the tag in a direct mapped memory cache, let's first understand how the cache works. In a 16-bit address architecture, each memory address is represented by 16 bits. Given the cache can hold 8 blocks with each block being 32 bytes, we can use powers of two to determine the layout of the addressing:

• A 32-byte block size is 25 bytes since 25 = 32.
• With 8 blocks, we need 3 bits for the block offset because 23 = 8.
• Each block contains 32 bytes, so we need 5 bits for the byte offset because 25 = 32.

Now, subtracting the block offset (3 bits) and the byte offset (5 bits) from the total address length (16 bits), we find that 16 - 3 - 5 = 8 bits are left for the tag. Therefore, 8 bits are necessary to represent the tag of each block in the cache.