Final answer:
A 128-entry direct-mapped cache with an 8-byte block size and 16-bit address will have 6 tag bits, 7 index bits, and 3 offset bits. The design incorporates a comparator for tag comparison, a decoder for index selection, multiplexers, and control logic.
Step-by-step explanation:
To draw a 128-entry direct-mapped cache with 8-byte block size assuming a 16-bit address, we need to consider how the address is split into tag, index, and offset fields. A 16-bit address means there are 216 possible addresses. With an 8-byte block size, the offset would be 3 bits since 23 = 8. There are 128 entries in the cache, which requires 7 bits for the index because 27 = 128. The remaining bits are used for the tag, which in this case would be 16 - 3 - 7 = 6 bits.
A direct-mapped cache's design involves different combinational circuits, including a comparator, multiplexers, and decoders. The index bits are used to select a specific cache line using a decoder. The tag stored in the cache line is then compared to the tag derived from the address using a comparator. If they match and the valid bit is set, then a cache hit has occurred, and the data is sent to the processor. Otherwise, a cache miss occurs, and the requested data must be fetched from the main memory and then placed into the cache.