In general, a cache is named according to the amount of data it contains (i.e., a 4 KiB cache can hold 4 KiB of data); however, caches also require SRAM to store metadata such as tags and valid bits. You design this cache memory and will examine how a cache's configuration affects the total amount of SRAM needed to implement it as well as the performance of the cache. For all parts, assume that the that addresses and words are 64 bits. (A) Calculate the total number of bits required to implement a 32 KiB cache with two-word blocks. (B) Calculate the total number of bits required to implement a 96 KiB cache with 16-word blocks. How much bigger is this cache than the 32 KiB cache described in (A) above? (Notice that, by changing the block size, we increased the amount of data without doubling the total size of the cache.) (C) Explain why this 96 KiB cache, despite its larger data size, might provide slower performance than the first cache.

consider someone is using direct-mapped cache memory, each external memory address maps to what?     a specific cache memory location     one of many different cache memory locations     a unique cache memory location, not shared with other external addresses     a randomly-selected cache memory location

What are the obstacles to developing a cache replacement mechanism that is compatible with any given address sequence?

Direct Mapping Example: CPU is searching an Instruction stored at RAM address 1110011010 in cache and doesn't find it. What happens then? Tag 0000000000 ?? 0000000001 ?? Block:0, (j=0) 0000000010 Instruction-1 Line1 = 1 Tag Tag 0000000011 Instruction-2 0000000100 Instruction-3 Instruction-4 Block:1, G=1) Solution: 0000000101 Line 2 (= 2) Tag 0000000110 Instruction-5 0000000111 Instruction-6 0000001100 Instruction-7 0000001101 Instruction-8 Block:2 0000001110 Data-1 (j=2) 0000001111 Data-2 Line.3 (3) 0000011100 Data-3 0000011101 Data-4 Tag 0000011110 Data-5 0000011111 Data-6 Cache Size = 64B Size of a line = 4B Total number of Lines in Cache: m 16 RAM Size = 1KB %3D 1111111100 %3D Size of a Block= 4B Total number of Blocks, M = 256 j = 0, 1, 2, ..255 (M-1) Block 255. G =255) 1111111101 1111111110 i = 0, 1, 2, ... 15 (m-1) 1111111111