Question

Problem 6. Suppose we have a computer with 32 megabytes of main memory, 256 bytes of cache, and a block size of 16 bytes. For each configuration below, determine the memory address format, indicating the number of bits needed for each appropriate field (i.e. tag, block, set, offset). Show any relevant calculations. Direct cache mapping and memory is byte-addressable a) Direct cache mapping and memory is word-addressable with a word size of 16 bits b) c) 2-way set associative cache mapping and memory is byte-addressable. d) 4-way set associative cache mapping and memory is word-addressable with a word size of 32 bits. Fully associative cache mapping and memory is byte-addressable. e) Fully associative cache mapping and memory is word addressable with a word size of 64 bits f)

Answer 1

b) Direct cache mapping f memory word size = 16 bits = 2B Q. Main Memory: 32MB-2B Cache = 256B = 288 Block size = 16 B = 24B a) Direct Cache Memory & memory is byte addressable. Total no. of blocks inside - 28:24. - cache 24 .: Block offset = 4bit Index Offset 4 bit .: Tag bits = 25-(4+4)= 17 bits. b) Direct cache mapping & memory is word- addressable with Block size: 16 B = words. .: Block Offset - 3 bit. Cache size = 28,27 word .: Total no. of block inside , q* = 4 bit. Main Memory = 225-22+ word . Tag bits : 24-(4+3) 224-7= 17 bits c) 2-way set associative. Total no. of set inside, Totalno.of blocks: 24:23 .: Index offset - 3 bits. & Tag bits: 25- (3+ 4) = 18 bits. Cache 2

4-way set associative & memory is word addressable with word size 32 bit. Word size: 4B. - 13lock size 16 B = 4 word. Cache size - 256B : 256 = 64 word. Main memory = 2258 .223 word. Total no. of block inside cache = 64:2:24 Total no. of set inside cache a 24 -22 .: Index offset: 2 bit Block offeset = 2 bit .: Jag bits: 28- (4) - 19 bits . © fully associative with byte addressable memory Main Memory = 288. Block Offset: 4 bit Index offset - Obit ) Tag bits: 25-4 = 19. Fully associative with word addressable of word size 64 bit Word size 64 bit .: Block size = 16 B = Qword. Main Memory= 225 22 word .: Block offsetI bit Index offset = obit Tag bits = P2-C1)=21. = 8B
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