A) FULLY ASSOCIATIVE
Given-
Cache memory size = 32 Kwords = 215 words
Block size = Frame size = Line size = 2 words
Number of bits in block offset = 1 bit
Main memory size = 16GB = 234 BYTES
Thus, Number of bits in physical address = 34 bits
Number of bits in tag = Number of bits in physical address – Number of bits in block offset = 34 - 1 = 33
TAG | BLOCK OFFSET |
33 | 34 |
B) DIRECT MAPPING
Cache memory size = 32 Kwords = 215 words
Number of bits in physical address = 34 bits
Block size = 8 WORDS = 23
Thus, Number of bits in block offset = 3 bits
Total number of lines in cache
= Cache size / Line size
= 215 / 23
= 212 lines
Thus, Number of bits in line number = 12 bits
Number of bits in tag
= Number of bits in physical address – (Number of bits in line number + Number of bits in block offset)
= 34 bits – (12 bits + 3 bits)
= 19 bits
Thus, Number of bits in tag = 20 bits
TAG | LINE NUMBER | BLOCK OFFSET |
19 | 12 | 3 |
C) 4-way set Associated mapping
Given-
Thus, Number of bits in block offset = 0 bits
Thus, Number of bits in physical address = 34 bits
Total number of lines in cache
= Cache size / Line size
= 215 / 20
= 215 lines
Total number of sets in cache
= Total number of lines in cache / Set size
= 215 / 22
= 213 sets
Thus, Number of bits in set number = 13 bits
Number of tag bits = Number of bits in physical address – (Number of bits in set number + Number of bits in block offset)
=34 - (13+0)
=21 bits
TAG | SET NUMBER | LINE OFFSET |
21 | 13 | 0 |
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