Question

a) Suppose we have a 64 KB, direct-mapped cache with 8-word blocks. Determine how many
bits are required for the tag, index, and offset fields for a 32-bit memory address.
b) If instead, we use a 64 KB, 4-way set-associative cache with 8-word blocks, how many bits
will be required for the tag, index, and offset fields for a 32-bit address?
c) What type of cache is shown in problem 2? How many bits are required for this cache’s tag,
index, and offset fields for a 32-bit memory address?

Answer 1

Answer:- a) The block size is 8-word = 8 x 4 byte = 32-byte.
Thus the number of bits in offset field is = log₂(32) = 5.

Number of index bits = log₂(64 KB / 32) = log₂(2¹⁶ / 2⁵) = 11 bits.

Number of tag bits = 32 - (11 + 5) = 16.

Answer:- b) The block size is 8-word = 8 x 4 byte = 32-byte.
Thus the number of bits in offset field is = log₂(32) = 5.

Number of index bits = log₂(64 KB / 32*4) = log₂(2¹⁶ / 2⁷) = 9 bits. Divided by 4 as 4-blocks per set.

Number of tag bits = 32 - (9 + 5) = 18.

c) Problem 2 is not given here. Please add. Thank you.