Question

1. (10 pts) Explain how a 32-bit byte memory address should be divided into Tag/Index/Offset fields for each of the cache configurations below. Note: 1KB -210 bytes. You must explain how many bits to assign to each field and the ordering of the three fields. You get at most 50% of the credit if you give the length of each field without an explanation. 1) A fully associative cache with cache block size - 2 words and cache size - 512KB. 2) A direct-mapped cache with cache block size - 2 words and cache size 512KB. 3) A four-way set associative cache with cache block size 2 words and cache size- 512KB

Text:

Explain how a 32-bit byte memory address should be divided into Tag/Index/Offset fields for each of the cache configurations below. Note: 1KB = 210 bytes. You must explain how many bits to assign to each field and the ordering of the three fields. You get at most 50% of the credit if you give the length of each field without an explanation.

1) A fully associative cache with cache block size = 2 words and cache size = 512KB.

2) A direct-mapped cache with cache block size = 2 words and cache size = 512KB.

3) A four-way set associative cache with cache block size = 2 words and cache size = 512KB.

Answer 1

Main memory size = 32 bit word size = 4 byte Fully Associative cache block size = 2 words = 2 times 4 = 8 bytes. Cache size = 512 KB = 2^9 times 2^10 = 2^19 bytes Block offset = log_2 8 = 3 bit Index bit = 0 bit(in fully Associative) Tag = 32 - (index + offset) = (32 - 3) = 29 bit tag = 29 bit Index = 0 offset = 3 bit 32 bit Direct mapped cache Cache size = 512 K B = 2^9 times 2^10 = 2^19 Byte block size = 2 words = (2 times 4) = 8 bytes block offset = log 2 8 = 3bit Number of blocks in cache = cache size/Block size = 2^19/2^3 = 2^16 block Index bit = log_2 (No of blocks) = log_2 2^16 = 16 bit Tag bit = 32 - (16 + 3) Rightarrow 32 - 19 = 13 bit 4 - way set Associative cache size = 512 K B = 2^19 bytes Associativity = 4 block size = 2 words = 8 bytes block offset = log_2 8 = 3 bit Number of Block = cache size/Block size = 2^19/2^3 = 2^16 blocks Number of sets = Number of blocks/Associativity Rightarrow 2^16/4 = 2^14 set