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Question 31 10 pts Suppose we have a byte-addressable computer using 4-way set associative mapping with 24-bit main memory ad

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Answer #1

1.)

  • Given number of bits in main memory address = 24 bits.
  • Block size = 32 bits = 2^5.
  • So, block offset size = 5 bits.
  • Given, number of cache block = 64.
  • So, total number of sets in cache = number of block/set size = 64/4 = 16 sets = 2^4 sets
  • set size = 4 bits
  • tag field size = total address size - (block offset size + set field size) = 24 - (5 + 4) = 15 bits.
  • Answer:

block offset field size = 5 bits

set field size = 4 bits

tag field = 15 bits

2.)

  • Magnetic tape is the oldest and most cost effective storage device of all mass storage devices.
  • Answer: Magnetic tape.
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