# Consider a cache memory of 16 words. Each block consists of 4 words. Size of the main memory is 256 bytes.

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Consider a cache memory of 16 words. Each block consists of 4 words. Size of the main memory is 256 bytes. Draw associative mapping and calculate TAG, and WORD size.

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verified

Main Memory Size = 256 bytes = $$2^8 bytes$$

Cache Memory Size = 16 Words = $$2^4 bytes$$

Block Size = 4 Words = $$2^2 bytes$$

To Find -

TAG Field Size = ?

WORD Field Size = ?

Associative Mapping Structure = ?

Solution -

Physical Address = 8 bits obtained from the size of Main Memory ($$2^8$$)

WORD Field Size = 2 bits obtained from the size of Cache Block Size ($$2^2$$)

TAG Field Size = Number of bits in the Physical Address - Number of bits in the WORD Field

= 8 - 2 = 6 bits

Associative Mapping Structure looks like as follows:

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