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1- A 64-bit computer system employs a 16Gbyte main memory and a 32 Kilo word cache. Determine the number of bits in each fiel

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Answer #1

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-

  • Set size = 4
  • Cache memory size = 16 Kwords =215words
  • Block size = Frame size = Line size = 1word = 20

Thus, Number of bits in block offset = 0 bits

  • Main memory size = 16GB = 234 BYTES

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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