Question

An 8-way set associative cache of size 64 KB (1 KB = 1024 bytes) is used in a system with a 32-bit address. The address is sub-divided into TAG, INDEX, and BLOCK OFFSET. Find the number of bits in the TAG.

Hint:

In associative caches: Tag bits = (Address bits) $-$ (Index bits + Block offset bits). Index bits depend on number of sets, while block offset bits depend on block size.

Answer 1

Step 1: Cache parameters

Cache size = 64 KB = 2¹⁶ bytes
Address size = 32 bits
Associativity = 8-way

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Step 2: Use power-of-two representation

Assume block size = 64 bytes = 2⁶ bytes

Total cache blocks:
2¹⁶ / 2⁶ = 2¹⁰ blocks

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Step 3: Determine number of sets directly

Each set contains 8 blocks = 2³ blocks

Number of sets:
2¹⁰ / 2³ = 2⁷ sets

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Step 4: Bit allocation from address

Block offset bits = log₂(64) = 6

Set index bits = log₂(2⁷) = 7

Tag bits:
32 − (6 + 7) = 19

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Final Answer:

Number of tag bits = 19