Question 40

Determine the address bit breakdown:

• Main memory = 16 MB = $2^{24}$ bytes $\implies$ 24-bit physical address.
• Block size = 256 bytes = $2^8$ bytes $\implies$ Block offset = 8 bits (bits 0 to 7).
• Total cache size = 64 KB = $2^{16}$ bytes.
• Number of lines in cache = $\frac{64 \text{ KB}}{256 \text{ B}} = 256$ lines.
• 4-way set associative $\implies$ Number of sets = $\frac{256}{4} = 64 = 2^6$ sets.
• Set index field = 6 bits (bits 8 to 13).
• Tag field = $24 - (8 + 6) = 10$ bits (bits 14 to 23).

Extract Set Index (bits 8 to 13) for each address:

• $\text{A1} = \text{0x42C8A4}$:
  • Byte 1 (bits 8..15) is 0xC8 = 1100 1000
  • Lower 6 bits (bits 8..13) are 00 1000 = $8_{10}$
• $\text{A2} = \text{0x546888}$:
  • Byte 1 (bits 8..15) is 0x68 = 0110 1000
  • Lower 6 bits (bits 8..13) are 10 1000 = $40_{10}$
• $\text{A3} = \text{0x6A289C}$:
  • Byte 1 (bits 8..15) is 0x28 = 0010 1000
  • Lower 6 bits (bits 8..13) are 10 1000 = $40_{10}$
• $\text{A4} = \text{0x5E4880}$:
  • Byte 1 (bits 8..15) is 0x48 = 0100 1000
  • Lower 6 bits (bits 8..13) are 00 1000 = $8_{10}$

Comparing the sets:

• A1 and A4 map to set 8.
• A2 and A3 map to set 40.
• Therefore, 'A2 and A3 are mapped to the same cache set' is TRUE.