Question

Make an assumption that your cache is either:

• Fully associative
• Direct mapped
• Two-way set-associative
• Four-way set-associative

determine:

the size of the Tag and Word for Associative cache;

OR

the size of the Tag, Line, and Word for Direct-Mapped Cache ;

Or

the size of Tag, Set, and Word for K-Way Set-Associative Cache.

You may make any assumptions necessary including the number of Words in each block (recommend 2 or 4 or 8)

Answer 1

Answer:

I am assuming K-Way Set-Associative Cache.

A cache in the primary storage hierarchy contains cache lines that are grouped into sets. If each set contains k lines then we say that the cache is k-way associative.

A data request has an address specifying the location of the requested data. Each cache-line sized chunk of data from the lower level can only be placed into one set. The set that it can be placed into depends on its address.

This mapping between addresses and sets must have an easy, fast implementation. The fastest implementation involves using just a portion of the address to select the set. When this is done, a request address is broken up into three parts:

• An offset part identifies a particular location within a cache line.
• A set part identifies the set that contains the requested data.
• A tag part must be saved in each cache line along with its data to distinguish different addresses that could be placed in the set.

Example:

A computer uses 32-bit byte addressing. The computer uses a 2-way associative cache with a capacity of 32KB. Each cache block contains 16 bytes. Calculate the number of bits in the TAG, SET, and OFFSET fields of a main memory address.

Answer

Since there are 16 bytes in a cache block, the OFFSET field must contain 4 bits (2⁴ = 16). To determine the number of bits in the SET field, we need to determine the number of sets. Each set contains 2 cache blocks (2-way associative) so a set contains 32 bytes. There are 32KB bytes in the entire cache, so there are 32KB/32B = 1K sets. Thus the set field contains 10 bits (2¹⁰ = 1K).

Finally, the TAG field contains the remaining 18 bits (32 - 4 - 10). Thus a main memory address is decomposed as shown below:

Tag = 18

Set = 10

Offset = 4

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